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Students can Download Chapter 13 Hydrocarbons Questions and Answers, Plus One Chemistry Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Chemistry Chapter Wise Questions and Answers Chapter 13 Hydrocarbons

Plus One Chemistry Hydrocarbons One Mark Questions and Answers

Question 1.
Which of the following cannot be prepared by Wurtz reaction?
a) CH₄
b) C₂H₆
c) C₃H₈
d) C₄H₈
Answer:
a) CH₄

Question 2.
The cyclic polymerization of propane produces __________ .
Answer:
1, 3, 5-trimethylbenzene

Question 3.
The reaction

a) Hydration
b) Dehydration
c) Dehydrogenation
d) Dehalogenation
Answer:
b) Dehydration

Question 4.
Say TRUE or FALSE.
Calcium carbide on hydrolysis gives ethylene.
Answer:
False

Question 5.
3-Hexyne reacts with Na/liquid NH₃ to produce
a) cis-3-Hexene
b) trans-3-Hexene
c) 3-Hexylamine
d) mm2-Hexylamine
Answer:
b) trans-3-Hexene

Question 6.
Fill in the blanks after finding the correct relationship
CH₃ – O – CH₃ : Ether, CH₃-CH₂-OH: …………….
Answer:
Alcohol

Question 7.
Choose the correct answer from the brackets given below:
1) General formula of alkene (C_nH_2n, C_nH_4n-2)
2) The 2 different forms of elemental carbon (Bitumen, Diamond, Charcoal, Coke, Led, Graphite)
3) Find the odd man out. Give reason (C₂H₄, C₃H₆, C₄H₈, C₂H₆)
Answer:
1) C_nH_2n
2) Diamond, graphite
3) C₂H₆. It is an alkane. All others are alkenes.

Question 8.
The structural formula of a compound and the name given by a student to it is given below:

Answer:
The name given by the student is wrong. The correct name of the compound is 3-Ethyl -2-methyl nonane.

Question 9.
Which is not the isomer of CH₃-CH₂-O-CH₂-CH₃? (1)
a) CH₃-O-CH₂-CH₂-CH₃
b) CH₃-CH₂-CH₂-CH₂-OH
c) CH₃-CH₂-CO-CH₃
d) CH₃-CH₂(OH)-CH-CH₃
Answer:
c) CH₃CH₂COCH₃

Question 10.
Gammexane has the formula
Answer:
C₆H₆Cl₆

Question 11.
Bayer’s reagent is __________ .
Answer:
dil. alkaline KMnO₄

Question 12.
The electrophile attacking benzene during nitration is __________ .
Answer:
NO₂⁺

Question 13.
The compound that is least readily nitrated is __________ .
a) phenol
b) Toluene
c) Ethylbenzene
d) Benzoic acid
e) Xylene
Answer:
d) Benzoic acid

Question 14.
The hydrocarbon formed when Beryllium carbide is treated with water is __________ .
Answer:
Methane

Plus One Chemistry Hydrocarbons Two Mark Questions and Answers

Question 1.
The lUPAC names of 2 compounds with their structural formulae is given below. Make out the errors and correct them.
A. 2, 2-Dimethyl-3-hexyne

B. 1-Butyne
CH₃-CH₂-CH=CH₃
Answer:
A is wrong. The strucutral formula of 2,2-Dimethyl -3- hexyne is

Question 2.
Is there an organic compound named 2-Ethylpentane? Why? If no, write the correct answer.
Answer:
No. When an ethyl group comes with the second carbon atom, the longest chain will have 6 carbon atoms. Hence its correct name will be 3-Methyl hexane.

Question 3.
Write down 2 similarities and 2 difference of the hydrocarbons given below:

Answer:

Similarities	Dissimilarities
Both are unsaturated compounds	Both have different functional group
Both hydrocarbons have same word root	One hydrocarbon is alkene and the other is alkyne

Question 4.
An equation for combined Chemical reaction of acetylene is given below:

Find the products X’ and ‘Y’?
Answer:

Question 5.
In a Chemistry class, teacher asked students to write the geometrical form of dicarboxylic acid with the formula C₄H₄O₄. For this question, one student wrote

Both of them argued fortheir answers. Hearing this argument teacher told that both of them are right and she also explained the reason for it. Can you write the answer given by the teacher?
Answer:
Two geometrical isomers are possible for dicarboxylic acid. They are cis and transform. In the cis form, similar groups are present on the same side of the double bond whereas in transform, identical groups are present on different side of the double bond.

Question 6.
Match the following:

Inductive effect	CH3-CH = CH2
Electrometric effect	C6H5-NO2
Hyper conjugation	CH3-CH2 – Br
Resonance effect	CH3-CH2(+)

Answer:

Inductive effect	CH3 – CH2 (+)
Electrometric effect	CH3 – CH2 – Br
Hyper conjugation	CH3-CH = CH2
Resonance effect	C6H5-NO2

Question 7.
Addition of HBr to propene yields 2-bromopropane, what happens if benzoyl peroxide is added to the above reaction.
Answer:
When propene is allowed to react with HBr in the presence of peroxide, 2 bromo propane is obtained as the minor product and this is called peroxide effect. Anti markonikov’s rule.

Plus One Chemistry Hydrocarbons Three Mark Questions and Answers

Question 1.
Reactions
CH₃-C ≡ C-CH₃ + H₂
Reaction: 2
X + HCl → Y
a) What type of reaction is reaction 1 ?
b) Find X and Y.
c) Which are the different products obtained from the reaction of X with oxygen?
Answer:
a) Addition reaction

Question 2.

Here are some functional groups. You are supposed to form 3 structures of hydrocarbons using 3 different functional groups and try to find their names.
Answer:

Question 3
CH₃ – CH₂ – CH₂ – CH₂ – OH, CH₃ – CH₂ – CH – CH₃
i)For which isomerism can the example above be considered?
ii) Define it.
Answer:
i)Position isomerism
ii) Position isomerism arises as a result of the difference in the position of double bond, triple bond or functional group.

Question 4.
Explain the following with necessary chemicals equations.
i) Wurtz reaction
ii) Kolbe’s reaction
iii) Ozonolysisof alkenes
Answer:
i) Wurtz reaction – when alkyl halide is allowed to react with metallic sodium in presence of dry ether an alkane is obtained.

ii) Kolbes reaction – When a solution of sodium acetate is electrolized, ethane is obtained.

iii) Ozonolysis ofalkene: When an alkene is allowed to react with ozone, an ozonide is obtained. This on hydrolysis gives Aldehyde or ketone. The whole process is called ozonolysis.

Question 5.
b) Complete the reaction:

Answer:

Question 6
1. Explain the reaction between sodium metal and bromoethane in dry ether. (3)
2. Draw Sawhorse and Newman’s projections of the different conformers of ethane.
Answer:
1. 2CH₃CH₂Br + 2Na → CH₃-CH₂-CH₂-CH₃ + 2NaBr
2.

Question 7.
Analyse the given reactions, give the major products. Justify your answer.
a) HBris added to 1-Butene two products are obtained.
b) Action of excess chlorine with benzene in dark.
c) Addition of chlorine to benzene in uv light.
Answer:

c) When Benzene is allowed to react with chlorine in presence of sunlight. Benzene hexa chloride (BHC) is obtained.

Question 8.
Predict the product in the following reactions and identify the rules:-

Answer:
a) When propene is allowed to react with HBr in the presence of peroxide, 2-Bromo propane is obtained as the minor product. (Peroxide effect, Kharasch effect or Anti Markownikoffs rule of addition).

Question 9.
Write any two necessary condition for a compound to be aromatic. Convert Acetylene to benzene.
Answer:
Cyclic, Planar and should contain (4n+2)π electrons.

Plus One Chemistry Hydrocarbons Four Mark Questions and Answers

Question 1.
Match the following:

Answer:

Question 2.
Fill in the blanks:

Answer:

Question 3.
Given below are the structures of some hydrocarbons. Pick out the correct IUPAC names for them from the box.

Answer:
1) 4 – Ethyl -2, 3-dimethyl heptane
2) 3, 4-Dimethyl hex – 3-ene
3) 4, 5, 5- triethyl 3 methyl 2 heptyne
4) 3ethyl-2, 5, 5, trimethylheptane

Question 4.
a) I am an unsaturated hydrocarbon.
b) My wordroot is pent.
c) My suffix is ene. The double bond lies between 2nd and 3rd carbon atoms.
d) l have a branch of methyl group on my second carbon atom. Who am I?
Answer:

Question 5.
From the given table find out the isomer pairs and which type of isomerism they have?

Answer:
Isomer pairs
a – ii – functional group isomerism
b – i – chain isomerism
c-iv-position isomerism
d – iii – metamerism Question 6

Qn 6.
Given below is the structural formula of a compound written by a student.

i) Draw all the possible conformers of the compound?
ii) Arrange them in the order of stability?
Answer:
i)

ii) Eclipsed

Question 7.
A cross word puzzle.
Down
1. Two or more compounds having the same
molecular formula but different physical or chemical properties are known as isomers and the phenomenon is known as
2. Hydrocarbons having the general formula C_nH_2n.
3. IUPAC name of C₆H₅CH₃.
4. ………….. is added to the word root to show whether the hydrocarbon is saturated or unsaturated.
5. Hydrocarbons contain carbon-carbon triple bond.

Cross
6. From where does the inorganic compounds mainly originate from?
7. Name the alkene with the moelcular formula C₁₀H₂₀
8. Prefix of the functional group carboxylic acid.
9. Compounds with two rings.
10. Compounds having same molecular formula but different arrangements of carbon atoms on either side of the same functional group are called ……..

Answer:

Question 8.
1. Name the product obtained when HBr is added to propene. Why?
2. Acetylene is more acidic than ethylene or ethane. Why?
Answer:
1. When propene is treated with HBr, both 2-bromopnopane and 1-bromopropane are formed as products. The major product is2-Bromo-propane. This is due to the rule known as Markonikoffs rule of addition. It states that when a hydrogen halide is added to an unsymmetrical alkene, the halogen atom will goes to the doubly bonded carbon containing lesser number of hydrogen atoms.

2. In Acetylene, the hybridisation of carbon is sp and hence 50% s-character is present.

Question 9.
a) How will you prepare butane in the laboratory using ethyl bromide (CH₃CH₂Br) as one of the raw materials. Write relevant equation.

Identify the product ‘X’. Statethe law that explains the formation of X.
Answer:

The law of Anti-Markownikoff’s rule of addition explains the formation of 1 -Bromopropane.

Question 10.
1. Addition of HBrto propene yields 2-Bromopropane, while in the presence of Benzoyl peroxide. The same reaction yields 1-Bromopropane. Give reason. Justify your answer.
2. Three compounds are given. Benzene, m- dinitrobenzene and toluene. Identify the compound which will undergo nitration most easily and why?
Answer:
1. When propene is allowed to react with HBr in the presence of “peroxide” 2-Bromopropane is obtained as the minor product. (Peroxide effect or Kharach effect or Anti-Markownikoff’s role of addition) Peroxide effect is applicable only in the case of HBr.

2. Toluene. This is because the -CH₃ group, being an activating group activates the benzene ring towards electrophilic substitution in toluene.

Question 11.
a) Write IUPAC names of the products obtained by addition reactions of HBr to hex-1-ene:
i) In the absence of peroxide.
ii) In the presence of peroxide.
b) Howwill you convert:
i) Benzene to toluene
ii) Benzene to nitrobenzene
Answer:

Question 12.
The equations for two chemical reactions are given below:
i) CH ≡ CH + HCl → A → B
ii) OH₄ + O₂ → C + D
Which are the products A, B, C, and D?
Answer:

Question 13.
The 2 conformations shown below belongs to the compound cyclohexane. Infinite number of conformations are possible for cyclo hexane. But these 2 conformations given below has a peculiarity. Try to find it. Also, define the terms conformers and the phenomenon conformation?

Answer:
Infinite number of conformations are possible for cyclohexane. Out of it chair conformation is the most stable one and the boat conformation is the least stable form.

The different arrangement of a compound which arises as a result of rotation about carbon single bond are called conformers and the phenomenon is called conformation.

Plus One Chemistry Hydrocarbons NCERT Questions and Answers

Question 1.
How do you account for formation of ethane during chlorination of methane? (3)
Answer:
Chlorination of methane takes place through a free radical chain mechanism as given below:

From the above mechanism, it is evident that during chain propagation step, CH3free radicals areproduced. In the chain termination step, the two free CH₃ radicals may combine together to form ethane (CH₃-CH₃) molecule.

Question 2.
For the following compounds, write structural formulas and IUPAC names for all possible isomers having the number of double or triple bond as indicated: (4)
a) C₄H₈ (one double bond)
b) C₃H₈ (one triple bond)
Answer:
a) Isomers of C₄H₈ having one double bond are:

Question 3.
What are the necessary conditions for any compound to show aromaticaly? (3)
Answer:
The conditions for a compound to show aromaticity are:
i) The molecule must be cyclic
ii) It must have a conjugated system of (4n+2) π- electrones
iii) The molecule must be planar so that delocalization of π-electrones can take place.

Question 4
Explain why the following system are not aromatic?

Answer:

It dose not contain all the π- electrons in the ring. Therefore, it is not an aromatic compound.

It contains only four electrons, therefore, the system is not aromatic because it dose not contain (4n + 2) π- electrones.

Cyclootatetraene is a conjugated system having 8 π-electrons.
Therefore, the molecule is not contain (4n+2) π-electrons.

Question 5.
In the alkane, H₂CCH₂C(CH₃)₂ CH₂CH(CH₃)₂, identify 1°, 2°, 3° carbon atoms and give the total number of atoms bonded to each one of these.
Answer:

1° Carban atoms = 5
Hydrogen atoms attached to 1° carbon atoms = 15 2° Carbon atoms = 2
Hydrogen atoms attached to 2° carbon atoms = 4 3° Carbon atom = 1
Hydrogen atoms attached to 3° carbon atom = 1

Question 6.
What effect does branching of an alkane chain has on its melting point?
Answer:
As the branching increases melting point increases.

Question 7.
Why does benzene undergo electrophilic subsitution easily and nucleophilic substitutions with difficulty? (2)
Answer:
Benzene molecule has two n cloud rings, one above and the other below the plane of atoms. Therefore, it is likely to be attached by electrophiles which subsequently brings about substitution.

The nucleophiles would be repelled by the π-electron rings and hence benzene reacts with nucleophiles with difficulty.

Students can Download Chapter 5 States of Matter Notes, Plus One Chemistry Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Chemistry Notes Chapter 5 States of Matter

Introduction
The observable characteristics of chemical systems represent bulk properties of matter. Chemical properties do not depend the physical state of matter but chemical reactions do.

Inter Molecular Forces
Intermolecular forces are the forces of attraction and repulsion between interacting particles (atoms and molecules). This term does not include the electrostatic forces that exist between the two oppositely charged ions and the forces that hold atoms of a molecule together i.e., covalent bonds.

Dispersion Forces Or London Forces
A nonpolar atom or molecule has a positive centre surrounded by a symmetrical negative electron cloud. The displacement of electron cloud creates an instantaneous dipole temporarily. This instantaneous dipole distorts the electron distribution of other atoms or molecules which are close to it and induces dipole in them also. In this way, a large number of nonpolar molecules become temporarily polar and they mutually attracted by weak attractive forces. These forces are very weak and are known to operate in all types of molecules.

Dipole-Dipole Forces
These type of interactions occur in polar molecules having permanent dipoles such as HCl, HBr, H₂S, etc. Such molecules possess partial charges of opposite sign at their ends. The positive end of one molecule attracts the negative end of the other molecule and vice versa. A simple example is the of H-Cl in which chlorine being more electronegative acquires slight negative charge whereas hydrogen becomes slightly positively charged. The dipole-dipole inter-action then takes place in H-Cl as follows.

Dipole-Induced Dipole Forces
These type of interactions are found in a mixture, containing polar and nonpolar molecules. When a nonpolar molecule is brought neara polar molecule, the positive end of the polar molecule attracts the electron cloud of the nonpolar molecule. Thus a polarity is induced in the nonpolar molecule. Then there will be attractive interacting between the polar molecule and the induced dipole of the nonpolar molecule.

Hydrogen Bond
Hydrogen bond can be defined as the attractive force which binds hydrogen atom of one molecule with the electronegative atom (F, O or N) of another molecule. When hydrogen is bonded to strongly electronegative element ‘X’, the electron pair shared between the two atoms moves far away from hydrogen atom. As a result, the hydrogen atom becomes, highly electropositive with respect to the other atom ‘X’. Since there is displacement of electrons towards X, the hydrogen acquires fractional positive charge (δ⁺) while ‘X’ attain fractional negative charge (δ^–). This results in the formation of a polar molecule having electrostatic force of attraction which can be represented as: H^δ+ – X^δ-

The magnitude of H-bonding depends on the physical state of the compound. It is maximum in the solid state and minimum in the gaseous state. Thus, the hydrogen bonds have strong influence on the structure and properties of the compounds.

Types of Hydrogen Bonds
There are two types of hydrogen bonds

1. Intermolecular hydrogen bond
2. Intramolecular hydrogen bond

1. Intermolecular hydrogen bond:
It is formed between two different molecules of the same or different compounds. For example, H-bond in case of HF molecule, alcohol or water molecules, etc.

2. Intramolecular hydrogen bond:
It is formed when hydrogen atom is in between the two highly electronegative (F, O, N) atoms present within the same molecule. For example, in o-Nitrophenol the hydrogen is in between the two oxygen atoms as shown below:

Thermal Energy
Thermal energy is the energy of a body arising from motion of its atoms or molecules. It is directly proportional to the temperature of the substance. The movement of particles using thermal energy is called thermal motion.

Intermolecular Forces Vs Thermal Interactions
Intermolecular forces tend to keep the molecules together but thermal energy of the molecules tends to keep them apart. Three states of matter are the result of balance between intermolecular forces and the thermal energy of the molecules.

The Gaseous State

• Gases are highly compressible.
• Gases exert pressure equally in all directions.
• Gases have much lower density than the solids and liquids.
• The volume and the shape of gases are not fixed. These assume volume and shape of the container.
• Gases mix evenly and completely in all proportions without any mechanical aid.

The Gas Laws

Boyle’s Law (Pressure – Volume Relationship)
On the basis of his experiments, Robert Boyle reached to the conclusion that at constant temperature, the pressure of a fixed amount of gas
varies inversely with its volume. This is known as Boyle’s law. It can be written as p ∝ \(\frac{1}{V}\)
where temperature(T) and number of moles(n)are constant.

If a fixed amount of gas at constant temperature T occupying volume V₁ at pressure p₁ undergoes expansion, so that volume becomes V₂ and pressure becomes p₂, then according to Boyle’s law :
p₁V₁ = p₂V₂ = constant
\(\Rightarrow \frac{p_{1}}{p_{2}}=\frac{V_{2}}{V_{1}}\)

Here T is constant and the graph is called isotherm

Charles’ Law (Temperature – Volume Relationship)
Charles’ law, which states that pressure remaining constant, the volume of a fixed mass of a gas is directly proportional to its absolute temperature. According to this law

Here we use new temperature scale called the kelvin temperature scale or Absolute temperature scale, t °C in Celsius scaleis equal to (273.15+t) kelvin in kelvin scale.

Each line of the volume vs temperature graph is called isobar. The lowest hypothetical or imaginary temperature at which gases are supposed to occupy zero volume is called Absolute zero.
We can see that the volume of the gas at – 273.15 °C will be zero.

Gay Lussac’s Law (Pressure-Temperature Relationship)
The relationship between pressure and temperature was given by Joseph Gay Lussac and is known as Gay Lussac’s law. It states that at constant volume, pressure of a fixed amount of a gas varies directly with the temperature. Mathematically,
P ∝ T
⇒ \(\frac{p}{T}\) = constant
This relationship can be derived from Boyle’s law and Charles’ Law.Each line of Pressure vs temperature (kelvin) graph at constant molar volume is called isochore.

Avogadro Law (Volume -Amount Relationship)
In 1811 Italian scientist Amedeo Avogadro tried to combine conclusions of Dalton’s atomic theory and Gay Lussac’s law of combining volumes which is now known as Avogadro law. It states that equal volumes of all gases under the same conditions of temperature and pressure contain equal number of molecules.

Mathematically we can write v α n where n is the number of moles.

The number of molecules in one mole of a gas has been determined to be 6.022 *1023and is known as Avogadro constant. A gas that follows Boyle’s law, Charles’ law and Avogadro law strictly is called an ideal gas.

Ideal Gas Equation
The combination of Boyle’s law, Charles’ law, and Avagadro’s law leads to an equation which gives the combined effect of change of temperature and pressure on the volume of a gas.
According to Boyle’s law, V α \(\frac{1}{P}\) ——- (i) (at constant T and n)
According to Charles’ Law, V α T ——- (ii) (at constant P and n)
According to Avogardro’s Law, V α n ——- (iii) (at constant T and P

Where R is a constant known as the universal gas constant. The equation is known as ideal gas equation.

Density and Molar Mass of a Gaseous Substance
Ideal gas equation can be rearranged as follows:

we get \(\frac{d}{M}=\frac{p}{R T}\)
(where d is the density)
On rearranging equation we get the relationship for calculating molar mass of a gas.
\(M=\frac{d R T}{p}\)

Dalton’s Law of Partial Pressures
The law was formulated by John Dalton in 1801. It states that the total pressure exerted by the mix-ture of non-reactive gases is equal to the sum of the partial pressures of individual gases.
P_Total = P₁ + P₂ + P₃ + ………. (at constant T, V)

where p_total is the total pressure exerted by the mixture of gases and p₁, p₂, p₃ etc. are partial pressures of gases.

Partial pressure in terms of mole fraction

Kinetic Molecular Theory Of Gases
Maxwell, Boltzmann, and others put forward a theoretical model of the gas. The theory is known as
Kinetic molecular theory of gases or microscopic
model of gases.
Postulates of kinetic molecular theory.

1.  All gases are made up of a large number of extremely small particles called molecules.
2. The molecules are separated from one another by large distances so that the actual volume of the molecules is negligible as compared to the total volume of gas.
3. The molecules are in a state of continuous rapid motion in all directions. During their motion, they keep on colliding with one another and also with the walls of the container.
4. Molecular collisions are perfectly elastic i.e. there is no net loss or gain of energy in their collisions. However, there may be redistribution of energy during such collisions.
5. There are no attractive forces between the molecules. They move completely independent of each other.
6. The pressure exerted by the gas is due to the bombardment of its molecules on the walls of the container.
7. At any instant, different molecules possess different velocities and hence different energies. However, the average kinetic energy of the molecules is directly proportional to its absolute temperature.

Behaviour Of Real Gases:
Deviation From Ideal Gas Behaviour
There are two types of curves are seen in the graph. In the curves for dihydrogen and helium, as the pressure increases the value of pV also increases. The second type of plot is seen in the case of other gases like carbon monoxide and methane. In these plots first, there is a negative deviation from ideal behaviour, the pV value decreases with increase in pressure and reaches to a minimum value characteristic of a gas. After that pV value starts increasing. The curve then crosses the line for ideal gas and after that shows positive deviation continuously. It is thus, found that real gases do not follow ideal gas equation perfectly under all conditions.

We find that two assumptions of the kinetic theory do not hold good. These are

1. There is no force of attraction between the molecules of a gas.
2. Volume of the molecules of a gas is negligibly small in comparison to the space occupied by the gas.

If assumption (a) is correct, the gas will never liquify. This means that forces of repulsion are powerful enough and prevent squashing of molecules in tiny volume. If assumption (b) is correct, the pressure vs volume graph of experimental data (real gas) and that theoritically calculated from Boyles law (ideal gas) should coincide.

The volume occupied by the molecules also becomes significant because instead of moving in volume V, these are now restricted to volume (V-nb) where nb is approximately the total volume occupied by the molecules themselves. Here, b is a constant. Having taken into account the corrections for pressure and volume, we can rewrite equation as This equation is known as van der Waals’ equation.

Value of ‘a’ is measure of magnitude of intermolecular attractive forces within the gas and is independent of temperature and pressure.
Real gases show ideal behaviour when conditions of temperature and pressure are such that the intermolecular forces are practically negligible. The real gases show ideal behaviour when pressure approaches zero.

The deviation from ideal behaviour can be measured in terms of compressibility factor Z, which is the ratio of product pV and nRT. Mathematically
\(z=\frac{p V}{n R T}\)

For ideal gas Z = 1 at all temperatures and pressures because pV = nRT.
At high pressure, all the gases have Z > 1. These are more difficult to compress. At intermediate pressures, most gases have Z < 1. The temperature at which a real gas obeys ideal gas law over an appreciable. range of pressure is called Boyle temperature or Boyle point.

Liquefaction Of Gases
The highest temperature at which liquefaction of the gas first occurs is called Critical temperature (T_c). Volume of one mole of the gas at critical temperature is called critical volume (V_c) and pressure at this temperature is called critical pressure (P_c).
The critical temperature, pressure, and volume are called critical constants.

Liquid State
Intermolecular forces are stronger in liquid state than in gaseous state.

Vapour Pressure
The pressure exerted by the vapour on the walls of the container is known as vapour pressure.

Surface Tension
Liquids tend to minimize their surface area. The molecules on the surface experience a net downward force and have more energy than the molecules in the bulk, which do not experience any net force. This characteristic property of liquids is known as surface tention. Liquids tend to have minimum number of molecules at their surface due to surface tention.

If surface of the liquid is increased by pulling a molecule from the bulk, attractive forces will have to be overcome. This will require expenditure of energy. The energy required to increase the surface area of the liquid by one unit is defined as surface energy.

Viscosity
It is a common observation that certain liquids flow faster than others. For example, liquid like water, ether, etc. flow rapidly while liquids like glycerine, castor oil, honey, etc. flow slowly. These differences in flow rates result from a property known as viscosity. Every liquid has some internal resistance to flow. This internal resistance to flow possessed by a liquid is called its viscosity. Liquids which flow slowly have high internal resistance and are said to have high viscosity. On the other hand, liquids which flow rapidly have low internal resistance and are said to have low viscosity.

Viscosity is also related to intermolecular forces in liquids. If the intermolecular forces are large, the vis-cosity will be high. Viscosity of a liquid decreases with rise in temperature. This is because at higher temperature the attractive forces between molecules are overcome by the increased kinetic energies of the molecules.

Ncert Supplementary Syllabus

Kinetic Energy And Molecular Speeds
Molecules of gases remain in continuous motion. While moving they collide with each other and with the walls of the container. This results in change of their speed and redistribution of energy. So the speed and energy of all the molecules of the gas at any instant are not the same. Thus, we can obtain only average value of speed of molecules. If there are n number of molecules in a sample
and their individual speeds are u₁, u₂, ……… u_n, then average speed of molecules u_av can be calculated as follows:
\(u_{a v}=\frac{u_{1}+u_{2}+\ldots u_{n}}{n}\)

Maxwell and Boltzmann have shown that actual distribution of molecular speeds depends on temperature and molecular mass of a gas. Maxwell derived a formula for calculating the number of molecules possessing a particular speed. Fig. A(1) shows schematic plot of number of molecules vs. molecular speed at two different temperatures T₁ and T₂ (T₂ is higher than T₁). The distribution of speeds shown in the plot is called Maxwell-Boltzmann distribution of speeds.

The graph shows that number of molecules possessing very high and very low speed is very small. The maximum in the curve represents speed possessed by maximum number of molecules. This speed is called most probable speed, u_mp. This is very close to the average speed of the molecules. On increasing the temperature most probable speed increases. Also, speed distribution curve broadens at higher temperature. Broadening of the curve shows that number of molecules moving at higher speed increases. Speed distribution also depends upon mass of molecules. At the same temperature, gas molecules with heavier mass have slower speed than lighter gas molecules. For example, at the same temperature, lighter nitrogen molecules move faster than heavier chlorine molecules. Hence, at any given temperature, nitrogen molecules have higher value of most probable speed than the chlorine molecules. Though at a particular temperature the individual speed of molecules keeps changing, the distribution of speeds remains same.

The kinetic energy of a particle is given by the expression:
Kinetic Energy = \(\frac{1}{2}\) mu²
Therefore, if we want to know average translational kinetic energy, \(\frac{1}{2} m \overline{u^{2}}\) , for the movement of a gas particle in a straight line, we require the value of mean of square of speeds, \(\overline{u^{2}}\), of all molecules. This is represented as follows:
\(u^{2}=\frac{u_{1}^{2}+u_{2}^{2}+\ldots . u_{n}^{2}}{n}\)

The mean square speed is the direct measure of the average kinetic energy of gas molecules. If we take the square root of the mean of the square of speeds then we get a value of speed which is different from most probable speed and average speed. This speed is called root mean square speed and is given by the expression as follows:
\(u_{m s}=\sqrt{u_{2}}\)

Root mean square speed, average speed and the most probable speed have following relationship:
u_rms u_av u_mp

The ratio between the three speeds is given below:
u_mp : u_av : u_rms :: 1 : 1.128 : 1.224

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Kerala Plus One Chemistry Notes Chapter 8 Redox Reactions

Introduction
The reaction which involve both oxidation and reduction reactions is called Redox reaction.

Classical Idea Of Redox Reactions Oxidation And Reduction Reactions
“Oxidation” is defined as the addition of oxygen/electronegative element to a substance or removal of hydrogen/electropositive element from a substance. Examples of oxidation:

1. Addition of oxygen 2Mg + O2 → 2MgO
2. Removal of hydrogen 2H₂S + O₂ → 2S + 2H₂O
3. Addition of electronegative element Mg + Cl₂ → MgCl₂

The term reduction been broadened these days to include removal of oxygen/electronegative element from a substance or addition of hydrogen /electropositive element.

1. Removal of electronegative element FeCl₃ + H₂ → 2FeCl₂ + 2HCl
2. Removal of Oxygen (2H₂O → 2Hg + O₂)
3. Addition of Hydrogen (H₂ + Cl₂ → 2HCl)

Redox Reactions In Terms Of Electron Transfer Reactions
According to electronic concept, the processes which involves loss of electrons are called oxidation reactions. Similarly, processes which involve gain of electrons are called reduction reactions.
The atom which reduced, act as oxidising agent and the atom which oxidised act as reducing agent.For example;
2Na(s) + Cl₂(g) → 2Na⁺Cl^–(s) or 2NaCl(s)
Here Na is oxidised and Cl is redused.

Competitive Electron Transfer Reaction
Place a strip of metallic zinc in an aqueous solution of copper nitrate. You may notice that the strip becomes coated with reddish metallic copper and the blue colour of the solution disappears. Formation of Zn²⁺ ions among the products can easily be judged when the blue colour of the solution due to Cu2+ has disappeared. The reaction is,
Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s) zinc is oxidised, releasing electrons, something must be reduced, accepting the electrons lost by zinc. Copper ion is reduced by gaining electrons from the zinc.

Oxidation Number
Oxidation number of an element may be defined as the charge which an atom of the element has or appears so have when present in the combined state in a compound.

1. Electrons shared between two like atoms are divided equally between the sharing atoms.
2. Electrons shared between two unlike atoms are counted with the more electronegative atom. Atoms can assume positive, zero or negative values of oxidation numbers depending on their state of combination. Oxidation number can be a fraction in some cases.

The rules for calculation of oxidation number are:
1. In elements, in the free or the uncombined state, each atom bears an oxidation number of zero. Evidently each atom in H2 has the oxidation number zero.

2. For ions composed of only one atom, the oxidation number is equal to the charge on the ion. Thus Na+ ion has an oxidation number of +1, Mg²⁺ion, +2, Fe³⁺ ion, +3, Cl^– ion, -1, O^2- ion, -2; and so on. In their compounds all alkali metals have oxidation number of +1, and all alkaline earth metals have an oxidation number of +2. Aluminium is regarded to have an oxidation number of +3 in all its compounds.

3. The oxidation number of oxygen in most compounds is-2. However, we come across two kinds of exceptions here.in peroxides (e.g., H₂O₂, Na₂O₂), each oxygen atom is assigned an oxidation number of—1, in superoxides (e.g., KO₂, RbO₂) each oxygen atom is assigned an oxidation number of -(½). The second exception appears rarely, i.e. when oxygen is bonded to fluorine. In such compounds e.g., oxygen difluoride (OF₂) and dioxygen difluoride (O₂F₂), the oxygen is assigned an oxidation number of +2 and +1, respectively. The number assigned to oxygen will depend upon the bonding state of oxygen but this number would now be a positive figure only.

4. The oxidation number of hydrogen is +1, except when it is bonded to metals in binary compounds (that is compounds containing two elements). For example, in LiH, NaH, and CaH₂, its oxidation number is —1.

5. In all its compounds, fluorine has an oxidation number of-1. Other halogens (Cl, Br, and I) also have an oxidation number of-1, when they occur as halide ions in their compounds. Chlorine, bromine and iodine when combined with oxygen, for example in oxoacids and oxoanions, have positive oxidation numbers.

6. The algebraic sum of the oxidation number of all the atoms in a compound must be zero. In polyatomic ion, the algebraic sum of all the oxidation numbers of atoms of the ion must equal the charge on the ion. Thus, the sum of oxidation number of three oxygen atoms and one carbon atom in the carbonate ion, (CO₃)^2- must equal -2. A term that is often used interchangeably with the oxidation number is the oxidation state. Oxidation state of a metal is a compound is sometimes represented by Stock notation. According to this, the oxidation number is written as Roman numeral in parenthesis after the symbol of the metal in the molecular formula. e.g.,Fe(ll)0, Sn(IV), Cl₄,Mn(IV)O₂.

Problem
Using Stock notation, represent the following compounds HAUCl₄, Ti₂O, FeO, Fe₂O₃, Cul, CuO, MnO and MnO₂.

Solution
By applying various rules of calculating the oxidation number of the desired element in a compound, the oxidation number of each metallic element in its compound is as follows:
HAuCl₄ → Au has 3
Tl₂O → Tl has 1
FeO → Fe has 2
Fe₂O₃ → Fe has 3
Cul → Cu has 1
CuO → Cu has 2
MnO → Mn has 2
MnO₂ → Mn has 4

Therefore, these compounds may be represented as
HAU(III)Cl₄, Tl₂(I)O, Fe(II)O, Fe₂(III)O₃, Cu(I)l, Cu(II)O, Mn(II)O, Mn(IV)O₂.

In terms of oxidation number, oxidation may be defined as a chemical change in which there occurs an increase in the oxidation number of an atom or atoms. Reduction may be defined as a chemical change in which there occurs a decrease in the oxidation number of an atom or atoms. Thus, a redox reaction may be defined as a reaction in which the oxidation number of atoms undergoes a change.

Types Of Redox Reactions
1. Combination Reactions:
A combination reaction may be denoted in the manner
A + B → C

2. Decomposition Reaction:
Decomposition reactions are the opposite of combination reactions.
For example, 2H₂O → 2H₂ + O₂

3. Displacement Reaction:
In a displacement reaction, an ion (or an atom) in a compound is replaced by an ion (or an atom) of another element. It may be denoted as:
X +YZ → XZ + Y
Displacement reactions fit into two categories:
metal displacement and non-metal displacement.

a) Metal displacement:
A metal in a compound can be displaced by another metal in the uncombined state.
CuSO₄(aq) + Zn(s) → Cu(s) + ZnSO₄(aq)

b) Non-metal displacement:
The non-metal displacement redox reactions include hydrogen displacement and a rarely occurring reaction involving oxygen displacement.
2Na(s) + 2H₂O(I) → 2NaOH(aq) + H₂(g)

The power of these elements as oxidising agents decreases as we move down from fluorine to iodine in group 17 of the periodic table.

Note:
fluorine is the strongest oxidising agent; there is no way to convert F^– ions to F₂ by chemical means. The only way to achieve F₂ from F^– is to oxidise electrolytically,

4. Disproportionation Reactions:
In a disproportionation reaction an element in one oxidation state is simultaneously oxidised and reduced.

Balancing Of Redox Reactions
There are two ways to balance a redox equation.
They are oxidation number method and Half Reaction Method.

a) Oxidation Number Method
The various steps involved in this method are:

1. Write the skeletal equation and assign oxidation numbers to each element. Identify the elements undergoing change in oxidation number.
2. Find out the increase or decrease of oxidation number per atom. Multiply the increase or decrease of oxidation number with number of atoms undergoing the change.
3. Multiply the formulae of the oxidising agent and the reducing agent by suitable integers so as to equalize the total increase or decrease in oxidation number as determined in the above step.
4. Balance the equation with respect to all atoms other the term reduction has than oxygen and hydrogen.
5. Balance oxygen atoms by adding equal number of H₂O molecules to the side deficient in oxygen atoms.
6. For reaction taking place in acidic medium, add H⁺ ions to the side of deficient in hydrogen atoms.
7. For reaction taking place in basic medium, add H₂O molecules to the side deficient in hydrogen atoms and simultaneously add equal number of OH ions on the other side of the equation.

Problem
Permanganate ion reacts with bromide ion in basic medium to give manganese dioxide and bromate ion. Write the balanced ionic equation forthe reaction.
Solution:
The skeletal ionic equation is:
MnO₄^–(aq) + Br^–(aq) → MnO₂(s) + BrO₃^–(aq)

Assign oxidation numbers for Mn and Br

this indicates that permanganate ion is the oxidant and bromide ion is the reductant.

Calculate the increase and decrease of oxidation number, and make the increase equal to the decrease.

As the reaction occurs in the basic medium, and the ionic charges are not equal on both sides, add 2 OH^– ions on the right to make ionic charges equal.
2MnO₄^–(aq) + Br^–(aq) → 2MnO₂(s) + BrO₃^–(aq) + 2OH^–(aq)

Finally, count the hydrogen atoms and add appropri- ‘ ate number of water molecules (i.e. one H20 molecule) on the left side to achieve balanced redox change.
2MnO₄^–(aq) + Br^–(aq) → 2MnO₂(s) + Br0₃^–(aq) + 2OH^–(aq)

b) Half Reaction Method
This method involves identifying the oxidation and reduction reactions in the given skeletal equation and then splitting the reaction accordingly as two half reactions. Each half reaction is then balanced systematically in various steps as outlined below.

Step 1.
Write the skeletal equation and identify the oxidant and reductant.

Step 2.
Write the half reactions for oxidation and reduction separately.

Step 3.
Balance the half reaction with respect to atoms that undergo change in oxidation number. Add electron to whichever side is necessary, to make up for difference in ON.

Step 4.
Balance O-atoms by adding proper number of H₂O molecules to the side deficient in oxygen atoms.

Step 5.
For ionic equations in acid medium, add sufficient H⁺ ions to the side deficient in hydrogen. If the reaction occurs in basic medium, add sufficient H₂O molecules to the side deficient in H atoms to balance H atoms and equal number of hydroxyl ions on the opposite side.

Step 6.
Equalise the number of electrons lost or gained by multiplying the half reaction with suitable integer and add the half reactions to get the final balanced equation.

Problem
Permanganate (VII) ion, MnO₄^– in basic solution oxidises iodide ion, l^– to produce molecular iodine (l₂) and manganese (IV) oxide (MnO₂). Write a balanced ionic equation to represent this redox reaction.
Solution:

Redox Reactions As The Basis For Titrations
In redox systems, the titration method can be adopted to determine the strength of a reductant/ oxidant using a redox sensitive indicator. The usage of indicators in redox titration is illustrated below:
1. In one situation, the reagent itself is intensely coloured, e.g., permanganate ion, MnO₄^–. Here MnO₄^– – acts as the self indicator. The visible endpoint, in this case, is achieved after the last of the reductant (Fe²⁺ or C₂O₄^2-) is oxidised and the first lasting tinge of pink colour appears at MnO₄^– concentration as low as 10^-6 mol dm^-3 (10^-6 mol L^-1), This ensures a minimal ‘overshoot’ in colour beyond the equivalence point, the point where the reductant and the oxidant are equal in terms of their mole stoichiometry.

2. If there is no dramatic auto-colour change (as with Mn04 – titration), there are indicators which are oxidised immediately after the last bit of the reactant is consumed, producing a dramatic colour change. The best example is afforded by Cr2072-, which is not a self-indicator, but oxidises the indicator substance diphenylamine just after the equivalence point to produce an intense blue colour, thus signalling the endpoint.

Redox Reactions And Electrode Pro-Cesses
When zinc rod is dipped in copper sulphate solution, zinc gets oxidised to Zn²⁺ while Cu²⁺ ions are reduced to Cu due to direct transfer of electrons. However, if a zinc rod dipped in ZnSO₄ solution taken in a breaker is connected externally by a conducting wire to a copper rod placed in CuSO₄ solution in another beaker, electrons are transferred indirectly from Zn to Cu. Now, each beaker contains both the oxidised and reduced form of the same substance ‘ called a redox coupe. In this experiment the redox couples developed are Zn²⁺/Zn and Cu²⁺/Cu When the solutions in the two beakers (called electrodes) are joined by a salt bridge (a U-tube containing a solution of KCl, solidified in presence of agar-agar), electrons flow from Zn to Cu while current flows in the reverse direction. The salt bridge provides electrical continuity between the solutions without allowing them to mix with each other. The flow of current is due to a potential difference between Cu and Zn electrodes (or half cells). This experimental set up gives an electrochemical cell or galvanic cell.

The potential of an electrode is a measure of its ability to lose (oxidation) or gain (reduction) electrons. When the concentrations of solutions in the half cells are unity and the temperature is 298 K, the potential of each electrode is known as Standard Electrode Potential (E°). By convention, E° of hydrogen electrode is zero volts and the potential of other electrodes will be a measure of the relative tendency of the active species to be in oxidised/reduced form. A negative E°shows that the redox couple is a stronger reducing agent than H⁺/H₂ couple.

A positive E° shows that the redox couple is a weaker reducing agent than H⁺/H₂ couples. The values of standard reduction potentials of various electrodes are given in the increasing order in an electrochemical series (electromotive series)

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Kerala Plus One Chemistry Notes Chapter 4 Chemical Bonding and Molecular Structure

Introduction
Matter is made up of different type of elements. The attractive force which holds the constituents together is called a chemical bond.

Kossel-Lewis Approach To Chemical Bonding
The bond between constituents are formed by the sharing of a pair of electrons or their transfer. G.N. Lewis introduced simple notations to represent these outer shell electrons in an atom. These notations are called Lewis symbols.

Significance of Lewis Symbols :
The number of dots around the symbol represents the number of valence electrons. This number of valence electrons helps to calculate the common or group valence of the element. The group valence of the elements is generally either equal to the number of dots in Lewis symbols or8 minus the number of dots or valence electrons.

Kossel, in relation to chemical bonding, drew attention to the following facts:
The bond formed, as a result of the electrostatic attraction between the positiveand negative ions was termed as the electrovalent bond. The electrovalence is thus equal to the number of unit charge (s) on the ion.
In terms of Lewis structures

1. According to electronic theory of chemical bond¬ing, atoms can combine either by transfer of valence electrons from one atom to another (gaining or losing) or by sharing of valence electrons in order to have an octet in their valence shells. This is known as octet rule.

2. Covalent Bond, Langmuir in 1919 refined the Lewis postulations by abandoning the idea of the stationary cubical arrangement of the octet, and by introducing the term covalent bond.

By Lewis – Langmuir theory the formation of chlorine molecule is as follows :

In water molecule covalent bond is as follows:

when two atoms share one electron pair they are said to be joined by a single covalent bond. If two atoms share two pairs of electrons, the covalent bond between them is called a double bond. And when combining atoms share three electron pairs as in the case of N₂ molecule a triple bond a triple bond is formed.

• The total number of electrons required for writing the structures are obtained by adding the valence electrons of the combining atoms.
• For anions, each negative charge would mean addition of one electron. For cations, each positive charge would result in subtraction of one electron from the total number of valence electrons.
• The least electronegative atom occupies the central position in the molecule/ion.

Formal charge
Formal charge (F.C.) on an atom in a Lewis structure = total number of valence electrons in the free atom— total number of non bonding (lone pairjelectrons—(1/2) total number of bonding(shared)electrons.

Let us consider the ozone molecule (O₃).
The Lewis structure of O₃ may be drawn as:

The atoms have been marked as 1,2 and 3. The formal charge on:

• The central O atom marked 1 = 6 – 2- \(\frac{1}{2}\)(6) = +1
• The end O atom marked 2 = 6 – 4 – \(\frac{1}{2}\)(4) = o
• The end O atom marked 3 = 6 – 6 – \(\frac{1}{2}\)(2) = -1

Hence, we represent O₃ along with the formal changes as follows:

Limitations of the Octet Rule
There are three types of exceptions to the octet rule. The incomplete octet of the central atom In some compounds, the number of electrons surrounding the central atom is less than eight. This is especially the case with elements having less than four valence electrons.
Some compounds are BCl₃, AlCl₃ and BF₃.

Odd-electron molecules
In molecules with an odd number of electrons like nitric oxide, NO and nitrogen dioxide, NO₂, the octet rule is not satisfied for all the atoms.

The expanded octet
In a number of compounds of these elements there are more than eight valence electrons around the central atom.Some examples are given below.

Ionic Or Electrovalent Bond
The formation of a positive ion involves ionization, i.e., removal of electrons from the neutral atom and that of the negative ion involves the addition of electron(s) to the neutral atom.

A qualitative measure of the stability of an ionic compound is provided by its enthalpy of lattice formation and not simply by achieving octet of electrons around the ionic species in gaseous state.

Lattice Enthalpy
The Lattice Enthalpy of an ionic solid is defined as the energy required to completely separate one mole of a solid ionic compound into gaseous constituentions.

Bond Parameters

Bond Length
It may be defined as the equilibrium distance between the centres of the nuclei of the two bonded i atoms in a molecule. Bond length are measured by spectroscopic, X-ray diffraction and electron diffraction techniques. It is usually expressed in Angstrom units (A°) or picometres (pm)
1 A° = 10^-10m and 1 pm = 10^-12m

Bond Angle
It is defined as the angle between the orbitals containing bonding electron pairs around the central atom in a molecule.

Bond Enthalpy
It is defined as the amount of energy required to break one mole of bonds of a particular type between two atoms in a gaseous state.

Bond Order:
In the Lewis description of covalent bond, the bond order is given by the number of bonds between the two atoms in a molecule. For example, the bond order in H₂ is one, in O₂ is two and in N₂ is three. Isoelectronic molecules and ions have identical bond orders. For example N₂, CO and NO⁺ have bond order 3. It is found that as the bond order increases, bond enthalpy increases and bond length decreases.

Resonance Structures
It is often observed that a single Lewis structure is inadequate for the representation of a molecule in conformity with its experimentally determined parameters. As in the case of O₃.

O₃ is represented by the above 3 structures. These are called canonical structures.

Experimentally determined oxygen-oxygen bond lengths in the O₃ molecule are same (128 pm). Thus the oxygen-oxygen bonds in the O₃ molecule are intermediate between a double and a single bond. According to the concept of resonance, the canonical structures of the hybrid describes the molecule accurately.

Some of the other examples of resonance structures are provided by the carbonate ion and the carbon dioxide molecule.

Polarity of Bonds
In reality no bond or a compound is either completely covalent or ionic. Even in case of covalent bond between two hydrogen atoms, there is some ionic character. As a result of polarisation, the molecule possesses the dipole moment. Which can be defined as the product of the magnitude of the charge and the distance between the centres of positive and negative charge. It is usually designated by a Greek letter Mathematically, it is expressed as follows: Dipole moment (µ) = change (Q) X distance of separation (r)

In case of polyatomic molecules, the dipole moment not only depend upon the individual dipole moments of bonds known as bond dipoles but also on the spatial arrangement of various bonds in the molecule.lt is due to the shifting of electrons to the side of more eletro negative element. For example,

The shifting of electrons is represented by an arrow. In case of H₂O the resultant dipole moment is given by the following figure:

Fajans Rules:
Just as all the covalent bonds have some partial ionic character, the ionic bonds also have partial covalent character. The partial covalent character of ionic bonds was discussed by Fajans in terms of the following rules:

• The smaller the size of the cation and the larger the size of the anion, the greater the covalent character of an ionic bond.
• The greater the charge on the cation, the greater the covalent character of the ionic bond.
• For cations of the same size and charge, the one, with electronic configuration (n-1)d”ns°, typical of transition metals, is more polarising than the one with a noble gas configuration, ns2 np6, typical of alkali and alkaline earth metal cations.

The cation polarises the anion, pulling the electronic charge toward itself and thereby increasing the electronic charge between the two. This is precisely what happens in a covalent bond, i.e., buildup of electron charge density between the nuclei. The polarising power of the cation, the polarisability of the anion and the extent of distortion (polarisation) of anion are the factors, which determine the per cent covalent character of the ionic bond.

The Valence Shell Electron Pair Repul-Sion (Vspert) Theory
The main postulates of VSEPR theory are as follows:

• The shape of a molecule depends upon the number of valence shell electron pairs (bonded or nonbonded) around the central atom.
• Pairs of electrons in the valence shell repel one another since their electron clouds are negatively charged.
• These pairs of electrons tend to occupy such positions in space that minimise repulsion and thus maximise distance between them.
• The valence shell is taken as a sphere with the electron pairs localising on the spherical surface at maximum distance from one another.
• A multiple bond is treated as if it is a single electron pair and the two or three electron pairs of a multiple bond are treated as a single super pair.
• Where two or more resonance structures can represent a molecule, the VSEPR model is applicable to any such structure.

The repulsive interaction of electron pairs de-crease in the order:
Lone pair (lp) – Lone pair (lp) > Lone pair (lp) – Bond pair (bp) > Bond pair (bp) – Bond pair (bp)

Valence Bond Theory
Valence bond theory was introduced by Heitlerand London (1927) and developed further by Pauling and others. A discussion of the valence bond theory is based on the knowledge of atomic orbitals, electronic configurations of elements, the overlap criteria of atomic orbitals, the hybridization of atomic orbitals and the principles of variation and superposition. First, we consider the formation of H2. When the attractive forces become greater than the repulsive forces, the molecule is formed and the system gets minimum energy. Because energy is released when a bond is formed. The energy so released is called bond enthalpy.

Orbital Overlap Concept
When two atoms approach each other, their atomic orbitals undergo partial interpenetration. This partial interpenetration of atomic orbitals is called overlapping of atomic orbitals. The electrons belonging to these orbitals are said to be shared and this results in the formation of a covalent bond. The main ideas of orbital of overlap concept of formation of covalent bonds are

• Covalent bonds are formed by the overlapping of half filled atomic orbitals present in the valence shell of the atoms taking part in bonding.
• The orbitals undergoing overlapping must have electrons with opposite spins.
  Overlapping of atomic orbitals results in decrease of energy and formation of covalent bond.
• The strength of a covalent bond depends upon the extent of overlapping. The greater the overlapping, the stronger is the bond formed.

The above treatment of formation of covalent bond involving the overlap of half-filled atomic orbitals is called valence bond theory.

Types of Overlapping and Nature of Covalent Bonds
The covalent bond may be classified into two types depending upon the types of overlapping:
(i) Sigma(σ) bond, and
(ii) pi(π) bond

(i) Sigma( σ) bond :
This type of covalent bond is formed by the end to end (hand-on) overlap of bonding orbitals along the internuclear axis. This is called as head on overlap or axial overlap. This can be formed by any one of the following types of combinations of atomic orbitals.
s-s overlapping:
In this case, there is overlap of two half filled s-orbitals along the internuclear axis as shown below:

s-p overlapping:
This type of overlap occurs between half filled s-orbitals of one atom and half filled p-orbitals of another atom.

p-p overlapping :
This type of overlap takes place between half filled p-orbitals of the two approaching atoms.

(ii) pi(π) bond :
In the formation of n bond the atomic orbitals overlap in such a way that their axes remain parallel to each other and perpendicular to the internuclear axis. The orbitals formed due to side wise overlapping consists of two saucer type charged clouds above and below the plane of the participating atoms.

Strength of Sigma and pi Bonds
Basically the strength of a bond depends upon the extent of overlapping. In case of sigma bond, the overlapping of orbitals takes place to a larger extent. Hence, it is stronger as compared to the pi bond where the extent of overlapping occurs to a smaller extent. Further, it is important to note that pi bond between two atoms is formed in addition to a sigma bond. It is always present in the molecules containing multiple bond (double ortriple bonds).

Hybridisation
Hybridisationis defined as the process of intermixing of the orbitals of slightly different energies so as to redistribute their energies, resulting in the formation of new set of orbitals of equivalent energies and shape.

The number of hybrid orbitals is equal to the number of the atomic orbitals that get hybridised.

These hybrid orbitals are stable due to their arrangement which provides minimum repulsion between electron pairs. Therefore, the type of hybridisation indicates the geometry of the molecules.

It is not necessary that only half filled orbitals participate in hybridisation. In some cases, even filled orbitals of valence shell take part in hybridisation.

Types of Hybridisation
There are various types of hybridisation involving s, p and d orbitals. The different types of hybridisation are as under:

(I) sp hybridisation:
This type of hybridisation involves the mixing of one s and one p orbital resulting in the formation of two equivalent sp hybrid orbitals. Each sp hybrid orbitals has 50% s-character and 50% p-character. Such a molecule in which the central atom is sp- hybridised and linked directly to two other central atoms possesses linear geometry.The two sp hybrids point in the opposite direction which provides more effective overlapping resulting in the formation of stronger bonds.

Example of molecule having sp hybridisation BeCl₂:
The ground state electronic.configuration of Be is 1s²2s². In the exited state one of the 2s-electrons is promoted to vacant 2p orbital to account for its divalency. One 2s and one 2p-orbitalsget hybridised to form two sp hybridised orbitals. These two sp hybrid orbitals are oriented in opposite direction forming an angle of 180°. Each of the sp hybridised orbital overlaps with the 2p-orbital of chlorine axially and form two Be-Cl sigma bonds.

II) sp² hybridisation :
In this hybridisation there is involvement of one s and two p-orbitals in orderto form three equivalent sp² hybridised orbitals. For example, in BCl₂ molecule, the ground state electronic configuration of central boron atom is 1s²2s²2p¹. In the excited state, one of the 2s
electrons is promoted to vacant 2p orbital as a result boron has three unpaired electrons.

These three orbitals (one 2s and two 2p) hybridise to form three sp2 hybrid orbitals. The three hybrid orbitals so formed are oriented in a trigonal planar arrangement and overlap with 2p orbitals of chlorine to form three B-Cl bonds. Therefore, in BCl₃ the geometry is trigonal planar with ClBCl bond angle of 120°

III) sp³ hybridisation:
This type of hybridisation can be explained by taking the example of CH₄ molecule in which there is mixing of one s-orbital and three p-orbitals of the valence shell to form four sp³ hybrid orbital of equivalent energies and shape.

There is 25% s-character and 75% p-character in each sp³ hybrid orbital. The four sp3 hybrid orbitals so formed are directed towards the four corners of the tetrahedron.

The angle between sp³ hybrid orbital is 109.5°

The structure of NH₃ and H₂O molecules can also be explained with the help of sp3hybridisation. In NH₃, the valence shell (outer) electronic configuration of nitrogen in the ground state is 2s²\(p_{x}^{1} 2 p_{y}^{1} 2 p_{z}^{1}\) having three unpaired electrons in the sp³ hybrid orbitals and a lone pair of electrons is present in the fourth one. These three hybrid orbitals overlap with 1s orbitals of hydrogen atoms to form three N-H sigma bonds. Due to the force of repulsion, the molecule gets distorted and the bond angle is reduced to 107° from 109.5°. The geometry of such a molecule will be pyramidal.

Other Examples of sp³, sp² and sp Hybridisation:
sp³ Hybridisation in C₂H₆ molecule:
In ethane molecule both the carbon atoms assume sp3 hybrid state. One of the four sp³ hybrid orbitals of carbon atom overlaps axially with similar orbitals of other atom to form sp³-sp³ sigma bond while the other three hybrid orbitals of each carbon atom are used in forming sp³-s sigma bonds with hydrogen atoms Therefore in ethane C-C bond length is 154 pm and each C-H bond length is 109 pm.

sp² Hybridisation in C₂H₄:
In the formation of ethene molecule, one of the sp² hybrid orbitals of carbon atom overlaps axially with sp² hybridised orbital of another carbon atom to form C-C sigma bond. While the other two sp² hybrid orbitals of each carbon atom are used for making sp²-s sigma bond with two hydrogen atoms. The unhybridised orbital (2p_x or 2p_y) of one carbon atom overlaps sidewise with the similar orbital of the other carbon atom to form weak π bond, which consists of two equal electron clouds distributed above and below the plane of carbon and hydrogen atoms.

Thus, in ethene molecule, the carbon-carbon bond consists of one sp²-sp² sigma bond and one pi (π) bond between p orbitals which are not used in the hybridisation and are perpendicular to the plane of molecule; the bond length is 134 pm. The C-H bond is sp^s sigma with bond length 108 pm. The H-C-H bond angle is 117.6° while the H-C-C angle is 121°.

sp Hybridisation in C₂H₂:
In the formation of ethyne molecule, both the carbon atoms undergo sp- hybridisation having two unhybridised orbital i.e., 2p_y and 2p_x. One sp hybrid orbital of one carbon atom overlaps axially with sp hybrid orbital of the other carbon atom to form C-C sigma bond, while the other hybridised orbital of each carbon atom overlaps axially with the half filled s orbital of hydrogen atoms forming σ bonds. Each of the two unhybridised p orbitals of both the carbon atoms overlaps sidewise to form two π bonds between the carbon atoms. So the triple bond between the two carbon atoms is made up of one sigma and two pi bonds

Hybridisation of Elements involving d-Orbitals
The elements present in the third period contain d orbitals in addition to s and p orbitals. The energy of the 3d orbitals are comparable to the energy of the 3s and 3p orbitals. The energy of 3d orbitals are also comparable to those of 4s and 4p orbitals. As a consequence the hybridisation involving either 3s, 3p, and 3d or 3d, 4s and 4p is possible. However, since the difference in energies of 3p and 4s orbitals is significant, no hybridisation involving 3p, 3d and 4s orbitals is possible.

1. Formation of PCl₅ (sp³d hybridisation):
The ground state and the excited state outer electronic configurations of phosphorus (Z=15) are represented below.

Now the five orbitals (i.eone s, three p, and one d orbitals) are available for hybridisation to yield a set of five sp3d hybrid orbitals which are directed towards the five comers of a trigonal bipyramidal.

Three sigma bond known as equatorial bonds lie in one plane and make an angle of 120° with each other.
The remaining two P-Cl bonds(called axial bonds)-one lying above and the other lying below the equatorial plane, make an angle of 90° with the plane.

As the axial bond pairs suffer more repulsive interaction from the equatorial bond pairs, therefore axial bonds have been found to be slightly longer and hence slightly weaker than the equatorial bonds; which makes PCl₅ molecule more reactive.

2. Formation of SF₆ (sp³d² hybridisation):
In SF₆ the central sulphur atom has the ground state outer electronic configuration 3s²3p⁴. In the exited state the available six orbitals i.e., one s, three p and two d are singly occupied by electrons. These orbitals hybridise to form six new sp³d² hybrid orbitals, which are projected towards the six corners of a regular octahedron in SF₆. These six sp³d² hybrid orbitals overlap with singly occupied orbitals of fluorine atoms to form six S-F sigma bonds. Thus SF₆ molecule has a regular octahedral geometry.

The structure of SF₆ is given below.

Molecular Orbital Theory
Molecular orbital (MO) theory was developed by F. Hund and R.S. Mulliken in 1932. The salient features of this theory are :

• The electrons in a molecule are present in the various molecular orbitals as the electrons of atoms are present in the various atomic orbitals.
• The atomic orbitals of comparable energies and proper symmetry combine to form molecular orbitals.
• While an electron in an atomic orbital is influenced by one nucleus, in a molecular orbital it is influenced by two or more nuclei depending upon the number of atoms in the molecule. Thus, an atomic orbital is monocentric while a molecular orbital is polycentric.
• The number of molecular orbital formed is equal to the number of combining atomic orbitals. When two atomic orbitals combine, two molecular orbitals are formed. One is known as bonding molecular orbital while the other is called antibonding molecular orbital.
• The bonding molecular orbital has lower energy and hence greater stability than the corresponding antibonding molecular orbital.
• Just as the electron probability distribution around a nucleus in an atom is given by an atomic orbital, the electron probability distribution around a group of nuclei in a molecule is given by a molecular orbital.
• The molecular orbitals like atomic orbitals are filled in accordance with the aufbau principle obeying the Pauli’s exclusion principle and the Hund’srule.

Formation of Molecular Orbitals
Linear Combination of Atomic Orbitals (LCAO)
The atomic orbitals of these atoms may be represented by the wave functions ψ_A and ψ_B. The formation of molecular orbitals is the linear combination of atomic orbitals that can take place by addition and by subtraction of wave functions of individual atomic orbitals as shown below.

Energy Level Diagram for Molecular orbitals

The increasing order of energies of various molecular orbitals for O₂ and F₂ is given below:
σ1s < σ*1s < σ2s < σ*2s < σ2p_x <(π2p_x = π2p_y) <(π*2p_x = π*2p_z) < σ*2p_x

This sequence of energy levels of molecular orbitals is not correct for the remaining molecules Li₂, Be₂, B₂, C₂, N₂. For molecules such as B₂, C₂, N₂ etc. the increasing order of energies of various molecular orbitals is
σ1s < σ*1s < σ2s < σ*2s <(π2p_x = π2p_y) < σ2p_x <(π*2p_x = π*2p_z) < σ*2p_z

The important characteristic feature of this order is that the energy of σ 2p_z molecular orbital is higher than that of π2p_x and π2p_y molecular orbitals. If the bonding influence is stronger a stable molecule results and if the antibonding influence is stronger,the molecule is unstable.

Bonding In Some Homonuclear Diatomic Molecules
Bond Order:
Bond order is defined as half of the difference between the number of electrons in the bonding molecular orbitals and that in the antibonding molecular orbitals.
Nh – N
i.e. Bond Order= \(\frac{N_{b}-N_{a}}{2}\). Where N_b is the number of electrons in the bonding molecular orbitals and Na is the number of electrons in the antibonding mo-lecular orbitals.

Significance of bond order:
Bond order conveys the following important informations about a molecule.
i) If the value of bond order is positive, it indicates a stable molecule and if the value of bond order is negative or zero, the molecule is unstable and is not formed.
ii) Bond dissociation energy of a diatomic molecule is directly proportional to the bond order of the molecule. The greater the bond order, the higher is the bond dissociation energy.
iii) Bond order is inversely proportional to the bond length. The higherthe bond ondervalue, smaller is the bond length. For example, the bond length in N₂ molecule (having bond order 3) is less than that in O₂ molecule (having bond order 2).

Magnetic character:
If all the electrons in the mol-ecules of a substance are paired, the substance will be diamagnetic. On the other hand, if there are un-paired electrons in the molecule, the substance will be paramagnetic.

Hydrogen Bonding
Hydrogen bond can be defined as the attractive force which binds hydrogen atom of one molecule with the electronegative atom (F, O orN) of another molecule. When hydrogen is bonded to strongly electronegative element ‘X’, the electron pair shared between the two atoms moves far away from hydrogen atom. As a result the hydrogen atom becomes, highly electropositive with respect to the other atom ‘X’. Since there is displacement of electrons towards X, the hydrogen acquires fractional positive charge (δ⁺) while ‘X’ attain fractional negative charge (δ^–). This results in the formation of a polar molecule having electrostatic force of attraction which can be represented as: H^δ+ – X^δ-

The magnitude of H-bonding depends on the physical state of the compound. It is maximum in the solid state and minimum in the gaseous state. Thus, the hydrogen bonds have strong influence on the structure and properties of the compounds.

Types of Hydrogen Bonds
There are two types of hydrogen bonds

1. Intermolecular hydrogen bond
2. Intramolecular hydrogen bond

1. Intermolecular hydrogen bond:
It is formed between two different molecules of the same or different compounds. For example, H-bond in case of HF molecule, alcohol or water molecules, etc.

2. Intramolecular hydrogen bond:
It is formed when hydrogen atom is in between the two highly electronegative (F, O, N) atoms present within the same molecule. For example, in o-Nitrophenol the hydrogen is in between the two oxygen atoms as shown below:

Students can Download Chapter 11 The p Block Elements Questions and Answers, Plus One Chemistry Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Chemistry Chapter Wise Questions and Answers Chapter 11 The p Block Elements

Plus One Chemistry The p Block Elements One Mark Questions and Answers

Question 1.
The aqueous solution of borax is
a) Acidic
b) Alkaline
c) Neutral
d) Amphoteric
Answer:
b) Alkaline

Question 2.
Say TRUE or FALSE.
Boron in aqueous solution forms B³⁺ ion.
Answer:
False

Question 3.
Which of the halide of group 14 does not exist?
a) CF₄
b) Cl₄
c) SiF₄
d) Pbl₄
Answer:
d) Pbl₄

Question 4.
Orthoboric acid, H₃BO₃ is a
a) Protonic acid
b) Arrhenius acid
c) Lewis acid
d) Bronsted-Lowery acid
Answer:
c) Lewis acid

Question 5.
The zeolite used as a catalyst in petrochemical industries for cracking of hydrocarbons and isomerization is _________ .
Answer:
ZSM-5

Question 6.
Dry ice is __________ .
Answer:
Solid CO₂

Question 7.
Thermodynamically most stable allotrope of carbon is __________ .
Answer:
Graphite

Question 8.
The alkalimetal used in solar cells is __________ .
Answer:
Cs (Caesium)

Question 9.
Acidity is in the order
Answer:
BBr₃ > BCl₃ > BF₃

Question 10.
AlCl₃ fumes in most air because
Answer:
HCl is formed due to hydrolysis in moist air

Plus One Chemistry The p Block Elements Two Mark Questions and Answers

Question 1.
Silicon belongs to the carbon family. Graphite is an important allotrope of carbon. But Si does not form an analogue of graphite. What are the possible reasons?
Answer:
Silicon atom is much bigger in size than carbon. Si-Si bond energy is less than C-C bond energy and also Si does not form compounds in sp² hybrid state.

Question 2.
Boron is an element with atomic number 5.
a) Write down the electronic configuration of boron.
b) Mention any two uses of boron.
c) Write down some compounds of boron.
Answer:
a) 1s²2s²2p¹
b) To increase the hardness of the steel.
Used as the semiconductor in electronic devices.
c) Diborane, borax, orthoboric acid, boron trifluoride

Question 3.
Answer the questions, with the help of the following:

• Hard solid
• Melting point above 450 K.
• Four allotropic forms are known.
• Low electrical conductivity.
• Mass number! 3

1. Which is the element?
2. Write any two uses of this element.
3.
4. Write any three compounds of this element.
Answer:
1. Boron
2. To increase hardness of steel
As semiconductor in electric devices
3. 2B
4. Diborane, Borax, Boric acid

Question 4.
1. Write any two allotropic forms of carbon.
2. C + ½ O₂ → …………….
3. Write any two uses of carbon monoxide.
4. How is carbon dioxide produced?
5. What is the hardest element/form of an element in the world?
Answer:
1. Diamond, Graphite
2. CO (Carbon monoxide)
3. Reducing agent in metallurgy
Used in the manufacture of methanol
4. Carbon dioxide is formed by the complete combustion of carbon.
C + O₂ → CO₂
5. Diamond

Question 5.
From the compounds of group 14 elements write an appropriate example for each of the following:

No.	Type of compound	Name/ Formulae of example
1.	A strong reducing oxide
2.	A giant covalent oxide
3.	A strongly reducing chloride
4.	A covalent chloride not hydrolysed by water

Answer:
1-CO
2-SiO₂
3-SnCl₂
4-CCl₄

Question 6.
Diborane has an unusual structure. Justify the statement with figure.
Answer:
In diborane, each Batom uses sp³ hybrid orbitals for bonding. Out of the four sp³ hybrid orbitals on each B atom, one is without an electron. The terminal B-H bonds are normal 2-centre-2-electron bonds but the two bridge bonds are 3-centre-2-electron bonds. The 3-centre-2-electron bridge bonds are also referred to as banana bonds.

Question 7.
Match the following:

A	B
CO	i. A semi conductor in electronic devices
H2	ii. Reducing agent in metallurgy
O3	iii. Thermal decomposition of Ammonia
Boron	iv. Used as a chemical reagent in organic chemistry

Answer:
CO – ii,
N₂ – iii,
O₃ – iv,
Boron – i

Question 8.
BCl₃ fumes in moist air.
1. Give reason.
2. Write down a balanced equation that can reveal the answer.
Answer:
1. When water of moist air reacts with boron halide hydrogen chloride is formed which causes fumes,

2. BCl₃ + 3H₂O → H₃BO₃ + 3HCl

Question 9.
‘Generally, non-metal oxides are basic.’
1. Do you agree?
2. What do you meant by oxides?
3. Which are the different types of oxides?
4. Give examples for each type of oxides.
Answer:
1. No. Generally, non metallic oxides are acids.
2. The binary compounds formed by the combination of oxygen with metals or nonmetals are called oxides.
3. Acidic oxides, basic oxide, amphoteric oxide, neutral oxide.
4. Acidic oxide → SO₂, NO₂
Basic oxide → MgO
Amphoteric oxide → Al₂O₃, ZnO
Neutral oxide → CO, N₂O

Question 10.
Match the following:
1. Borane – H₃BO₃
2. Boric acid – Na₂B₄O₇.10H₂O
3. Borax – Amphoteric oxide
4. Al₂O₃ – Boron hydride
Answer:
1. Borane – Boron hydride
2. Boric acid – H₃BO₃
3. Borax – Na₂B₄O₇.10H₂0
4. Al₂O₃ – Amphoteric oxide

Question 11.
How diborane reacts with
1. Oxygen?
2. Water?
Answer:
1. Diborane catches fire spontaneously upon exposure to air. It burns in oxygen releasing an enormous amount of energy.
B₂H₆ + 3O₂ → B₂O₂ + 3H₂O; ∆_rH^Θ -1976 kJ mol^-1

2. Diborane is readily hydrolysed by water to give boric acid.
B₂H₆(g) + 6H₂O(l) → 2B(OH)₃(aq) + 6H₂(g)

Question 12.
1. CCl₄ cannot be hydrolysed. Give reason.
2. Draw the structure of the dimer of AlCl₃.
Answer:
1. Carbon has no d orbitals to accommodate the lone pair of electrons from oxygen atom of H₂0.
2.

Question 13.
1. Draw the structure of boric acid.
2. Starting from borax how will you prepare boric acid? (Write the chemical equation).
Answer:
1.

2. Na₂B₄O₇ + 2HCl + 5H₂O → 2NaCl + 4B(OH)₃.

Question 14.
CO₂ is a gas but SiO₂ is a solid. Give reason.
Answer:
CO₂ exists as discrete molecules due to the formation of pπ -pπ double bond between carbon and oxygen. But SiO₂ has a three dimensional network structure in which each Si atom is covalently bonded in a tetrahedral manner to four oxygen atoms.

Question 15.
1. What are zeolites?
2. What is ZSM-5?
Answer:
1. Zeolites are aluminosilicates with three-dimensional network structure in which Al atoms replace few Si atoms. Cations such as Na⁺, K⁺ or Ca²⁺ balance the negative charge of aluminosilicate anion.

2. ZSM-5 is a zeolite catalyst used in petrochemical industry to convert alcohols directly into gasoline.

Plus One Chemistry The p Block Elements Three Mark Questions and Answers

Question 1.
Explain the following:
1. Allotropy
2. Coke and Charcoal
Answer:
1. The phenomenon of existence of an element in two or more forms, which have different physical properties but almost similar chemical properties, is called allotropy and the different forms are called allotropes.

2. Coke and Charcoal are amorphous forms of carbon. Coke is formed-by the destructive distillation of coal. It is used as a fuel and also as a reducing agent in metallurgy.

Question 2.
Give justifications.
1. The first ionisation enthalpy of carbon is greater than that of boron, whereas the reverse is correct for the second ionisation enthalpy.
2. Graphite is a better lubricant on moon than that on earth.
Answer:
1. This is because carbon has greater nuclear charge. For second ionisation enthalpy, an electron is to be removed from 2p of carbon while in boron the second electron is in 2s orbital. Removal of a 2s electron is more difficult due to the high penetrating power of 2s orbital.

2. Graphite contains hexagonal sheets of carbon held together by weak van der Waals’ forces and low density. In the moon, the different layers experience less weight due to less gravitational force and the density is still reduced. The lighter layers can easily slide over another to make graphite more lubricating on the moon than on earth.

Question 3.
1. Identify the compound with the following structure:

2. Mention the angles a & b.
3. Write down the corresponding elements in the figure.
1……………..
2……………..
3……………..
4……………..
5……………..
6……………..
Answer:
a) Diborane(B₂H₆).
b) a-97°
b-120°
c) 1-H, 2-H, 3-B, 4-B, 5-H, 6-H

Question 4.
The simplest boron hydride is diborane.
1. Write down the molecular formula of diborane.
2. Distinguish between terminal hydrogen and bridging hydrogen atoms of diboran.
Answer:
1. B₂H₆

2. In diborane, four hydrogen atoms and two boron atoms are present in a single plane. These hydrogen atoms are called terminal atoms. The terminal B-H bonds are regular two centre-two electron bonds. The remaining two hydrogen atoms above and below this plane are called bridging hydrogen atoms. The two bridge bonds (B-H-B) are three centre-two electron bonds.

Question 5.
1. How is orthoboric acid prepared?
2. Account for the acidic nature of orthoboric acid.
Answer:
1. Orthoboric acid is prepared by acidifying an
aqueous solution of borax.
Na₂B₄O₇ + 2HCl + 5H₂O → 2NaCl + 4B(OH)₃

2. Orthoboric acid is a weak nonobasic acid. It acts as a Lewis acid by accepting electrons from a hydroxyl ion.
B(OH)₃ + 2HOH → [B(OH)₃]^– + H₃O⁺

Question 6.
1. How is diborane prepared in the laboratory?
2. BCl₃ is a good Lewis acid. Why?
Answer:
1. Diborane can be conveniently prepared in the laboratory by the oxidation of sodium borohydride with iodine.
2NaBH₄ + l₂ → B₂H₆ + 2Nal + H₂

2. In BCl₃, the central boron atom contains only six electrons. Hence, it has the tendency to accept electrons and acts as a Lewis acid.

Question 7.
1. Name the allotropes of carbon.
2. Carbon monoxide is highly poisonous. Do you agree? Justify.
Answer:
1. Graphite, diamond and fullerene.

2. I agree with this statement. Because, CO has strong and reversible binding with haemoglobin resulting in the formation of carboxyhaemoglobin. This reduces the amount of haemoglobin available in blood for oxygen transport. This causes laboured respiration, muscle weakness and even death.

Question 8.
1. Diamond is hard and non conducting while graphite is soft and conducting. Why?
2. Explain the action of heat on boric acid.
3. What is inorganic benzene? How is it formed?
Answer:
1. In diamond, all the carbon atoms are in sp³ hybridised state. Due to this closely packed arrangement, it is hard. Since, there are no free electrons it is an insulator. But in graphite, the carbon atoms are in sp² hybridised state. It is a good conductor due to the presence of delocalised electrons between the layers. Graphite cleaves easily between the layers. Therefore, It is very soft and slippery.

2. On heating, orthoboric acid above 370 K forms metaboric acid, HBO₂ which on further heating yields boric oxide, B₂O₃.

3. B₃N₃H₆ is called Inorganic benzene. It is obtained by heating diborane with ammonia.

Question 9.
1. Explain the difference in properties of diamond and graphite on the basis of their structures.
2. How do you explain the lower atomic radius of gallium as compared to aluminium?
Answer:
1. In diamond, all the carbon atoms are in sp³ hybridised state. As a result of this hybridisation, a closely packed arrangement is present in diamond and it explains the hardness of diamond. Due to the absence of electrons, diamond is an insulator.

In graphite, the carbon atoms are in sp² hybridised state. As a result of this hybridisation, graphite has a layered structure with hexagonal rings. Each layer can slide over the other which explains the lubricating property of graphite.

2. This can be explained on the basis of the variation in the inner core of the electronic configuration. The presence of additional 10 d-electrons offer only poor screening effect for the outer electrons from the increased nuclear charge in gallium. Consequently, the atomic redius of gallium (135 pm) is less than that of aluminium (143 pm).

Question 10.
1. Boron resembles silicon in many of its properties. What is this resemblance generally known as?
2. What is dry ice? What is it used for?
3. What are silicones? How do they differ from silicates?
Answer:
1. Diagonal relationship

2. Carbon dioxide can be obtained as a solid in the form of dry ice, by allowing the liquified CO₂ to expand rapidly.
Dry ice is used as a refrigerant for ice-cream and frozen food.

3. Silicones are a group of organosilicon polymers containing R₂SiO repeating units.
Silicates are minerals with SiO₄^4- as the basic structural unit. In silicates either the discrete unit is present or a number of such units are joined together via corners by sharing 1, 2, 3 or 4 oxygen atoms per silicate units.

Plus One Chemistry The p Block Elements Four Mark Questions and Answers

Question 1.
Match the following:

Answer:
1 – c – f
2 – a – h
3 – b – g
4 – d – e

Question 2.
a) Elements of group 13 are
1) Al, Cr, Cd, Ga, Ti
2) B, C, Si, Ga, Ti
3) B, Al, Ga, Ti, In
4) B, Al, Cr, Ca, Ti
b) Which of the following is not a mineral of boron? (borax, bauxite, colemanite, tincal)
c) In the reaction between NH₃ and BF₃ ammonia acts as
(Lewis base, Lewis acid, brownsted base)
d) Diamond and Graphite are carbon’s
(isotropic forms, allotropic forms, isotopic forms, amorphous form)
Answer:
a) 3) B, Al, Ga, TI, In
b) Bauxite
c) Lewis base
d) allotropic forms

Question 3.
The hydrides of boron are called boranes.
1. How diborane reacts with ammonia?
2. Account for the exceptional hardness of diamond.
Answer:
1. Diborane reacts with ammonia to give B₂H₆.2NH₃
initially, which can be formulated as [BH₂(NH₃)₂]⁺[BH₄]^–. Further heating gives borazine, B₃N₃H₆ known as ‘inorganic benzene’ in view of its ring structure with alternate B-H and N-H groups,

2. In diamond, all the carbon atoms are in sp3 hybridised state. As a result of this hybridisation, a rigid three dimensional network of carbon atom is generated with directional covalent bonds throughout the lattice. It is very difficult to break this network structure.

Plus One Chemistry The p Block Elements NCERT Questions and Answers

Question 1.
Why does boron trifluoride behave as a Lewis acid? (2)
Answer:
The B atom in BF₃ has only 6 electrons in the valence shell and thus needs two more electrons to complete its octet. Therefore, it easily accepts a pair of electrons from nucleophiles such as F^–, NH₃, (C₂H₅)₂O, RCH₂OH etc. and thus behaves as a Lewis acid.

Question 2.
Explain why is there a phenomenal decrease in ionisation enthalpy from carbon to silicon? (2)
Answer:
Due to increase in atomic size and screening effect the force of attraction of the nucleus for the valence electron decreases considerably in Si as compared to C. As a result, there is a phenomenal decrease in ionisation enthalpy from carbon to silicon.

Question 3.
Consider the compounds, BCl₃ and CCl₄. How will they behave with water? (2)
Answer:
The B atom in BCl₃ has only six electrons in the valence shell and hence is an electron-deficient molecule. It easily accepts a pair of electrons donated by water and hence BCl₃ undergoes hydrolysis to form boric acid (H₃BO₃) and HCl.
BCl₃ + 3H₂O → H₃BO₃ + 3HCl

In contrast, C atom in CCl₄ has 8 electrons in the valence shell. Therefore, it is an electron-precise molecule. As a result, it neither accepts nor donates a pair of electrons. In simple words, it does not accept a pair of electrons from H₂O molecule and hence CCl₄ does not undergo hydrolysis in water.

Question 4.
Aluminium trifluoride is insoluble in anhydrous HF but dissolves on addition of NaF. Aluminium trifluoride precipitates out of the resulting solution when gaseous BF₃ is buddled through. Give reasons. (3)
Answer:
Anhydrous HF is a covalent compound and is strongly H-bonded. Therefore, it does not give free F^– ions and hence AlF₃ does not dissolve in HF. In contrast, NaF is an ionic compound and hence F^– ions are easily available. As a result, it combines with AlF₃ to form the soluble complex.
3NaF + AlF₃ → Na₃[AlF₃]

On bubbling gaseous BF₃, AlF₃ is precipitated. It is because BF₃ is a stronger Lewis acid than AlF₃. This is attributed to smaller size and higher electronegativity of Boron. As a result of this B has much higher tending to form complexes than Al. Therefore, when BF₃ is added to the above solution, AlF₃ gets precipitated.
Na₃[AlF₆] + 3BF₃ → 3Na[BF₄] + AlF₃(s)

Question 5.
In some of the reactions, thallium resembles aluminium, whereas in others it resembles with group 1 metals. Support this statement by giving some evidences. (3)
Answer:
Aluminium shows a uniform oxidation state of+3 in its compounds. Like aluminium, thallium also shows +3 oxidation state in some of its compounds like TlCl₃, Tl₂O₃, etc. Al is known to form octahedral complexes like [AlF₆]^3-. Similarly, Tl also forms octahedral complexes as [TlF₆]^3-.

Thallium also resembles group 1 metals. Like group 1 metals which show a stable oxidation state of +1 in their compounds Tl, due to inert pair effect, also shows +1 oxidation state in some of its compounds such as Tl₂O, TlCl, TlClO₄, etc. Similarly, like group 1 oxides, Tl₂O is strongly basic.

Students can Download Chapter 1 Some Basic Concepts of Chemistry Notes, Plus One Chemistry Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Chemistry Notes Chapter 1 Some Basic Concepts of Chemistry

INTRODUCTION
Chemistry is the branch of science which deals with the composition, properties and transformation of matter. These aspects can be best understood in terms of basic constituents of matter: atoms and molecules. That is why chemistry is called the sci-ence of atoms and molecules.

IMPORTANCE OF CHEMISTRY
Chemistry plays an important role in almost all walks of life. In recent years chemistry has tackled with a fair degree of success some of the pressing aspects of environmental degradation. Safer alternatives to environmentally hazardous refrigerants like CFCs, responsible for ozone depletion in the stratosphere, have been successfully synthesised.

NATURE OF MATTER
Matter is anything which has mass and occupies space. Matter can exist in three physical states: solid, liquid and gas.

At the macroscopic or bulk level, matter can be clas-sified as mixtures or pure substances.
A material containing only one substance is called a pure substance. Materials containing more than one substance are called mixtures. Pure substances are further classified into two types: elements and com¬pounds. Mixtures are also of two types: homoge¬neous mixtures and heterogeneous mixtures. A mix¬ture is said to be homogeneous if it has same com¬position throughout. Some examples of homoge¬neous mixtures are air, gasoline, kerosene, milk, alloys, etc. Heterogeneous mixtures are the mixtures which have different composition in different parts. Some examples of heterogeneous mixtures are iron and sulphur, muddy water, etc.

PROPERTIES OF MATTER AND THEIR MEASUREMENT
Every substances has characteristic properties. These are classified into two categories – physical properties and chemical properties.
Physical properties are those properties which can be measured or observed without changing the iden¬tity or the composition of the substance. Some ex¬amples of physical properties are colour, odour, melt¬ing point, boiling point, density, etc. chemical properties are characteristic reactions of different substances; these include acidity or basic¬ity, combustibility, etc.

THE INTERNATIONAL SYSTEM OF UNITS (SI UNITS)
A unit may be defined as the standard of reference chosen to measure any physical quantity. There are many different systems of units.
The improved metric system of units accepted inter-nationally is called International System of Units or SI units. The SI system has seven base units from which all other units are derived.

UNCERTAINTY IN MEASUREMENT
Scientific measurements involving some measuring devices have some degree of uncertainty. The magnitude of uncertainty depends on the accuracy of the measuring device and also on the skill of its operator.
The closeness of a set of values obtained from identical measurements of a quantity is known as precision of the measurement.
The term accuracy is defined as the closeness of a measurement or a set of measurements to its true value.

SIGNIFICANT FIGURES
The total number of digits in a measurement is called the number of significant figures. It includes the num¬ber of figures that are known with certainty plus the last uncertain digit, beginning with the first non-zero digit.
1) All non-zero digits are significant. For example, in 285 cm, there are three significant figures and in 0.25 mL, there are two significant figures.

2) Zeros preceding to first non-zero digit are not significant. Such zero indicates the position of decimal point. Thus, 0.03 has one significant figure and 0.0052 has two significant figures.
3) Zeros between two non-zero digits are significant. Thus, 2.005 has four significant figures.

4) Zeros at the end or right of a number are significant provided they are on the right side of the decimal point. For example,0.200 g has three significant figures. But, if otherwise, the zeros are not significant. For example, 100 has only one significant figure.

5) Exact numbers have an infinite number of significant figures. For example, in 2 balls or 20 eggs, there are infinite significant figures as these are exact numbers and can be represented by writing infinite number of zeros after placing a decimal i.e.,2 = 2.000000 or 20 = 20.000000

LAWS OF CHEMICAL COMBINATIONS
Law of Conservation of Mass
It states that matter can neither be created nor destroyed.
This law was put forth by Antoine Lavoisier in 1789.

Law of Definite Proportions
This law was given by, a French chemist, Joseph Proust. He stated that a given compound always contains exactly the same proportion of elements by weight. It is sometimes also referred to as Law of definite composition.

Law of Multiple Proportions
This law was proposed by Dalton in 1803. According to this law, if two elements can combine to form more than one compound, the masses of one element that combine with a fixed mass of the other element, are in the ratio of small whole numbers.

Gay Lussac’s Law of Gaseous Volumes
This law was given by Gay Lussac in 1808. He observed that when gases combine or are produced in a chemical reaction they do so in a simple ratio by volume provided all gases are at same temperature and pressure.

Avogadro Law
In 1811, Avogadro proposed that equal volumes of gases at the same temperature and pressure should contain equal number of molecules Avogadro made a distinction between atoms and molecules which is quite understandable in the present times.

DALTON’S ATOMIC THEORY
The main postulates of the theory are:-

1. Matter is made up of extremely small, indivisible particles called atoms.
2. Atoms of the same element are identical in all respects i.e. size and mass.
3. Atoms of different elements are different, i.e., they possess different sizes, shapes, masses, and chemical properties.
4. Atoms of different elements may combine with each other in a simple whole number ratio to form compound atoms or molecules.
5. Atoms can neither be created nor destroyed, i.e., atoms are indestructible.

ATOMIC AND MOLECULAR MASSES
ATOMIC MASS
The mass of an atom is extremely small. These are very inconvenient for calculations. This difficulty was overcome by expressing atomic masses as relative masses, i.e., with respect to the mass of an atom of a standard substance. The scale in which the relative masses are expressed is called atomic mass unit scale or amu scale. One atomic mass unit (amu) is equal to one-twelfth (1/ 12) the mass of an atom of carbon-12. Recently the symbol ‘u’ (unified mass) is used in place of amu.

AVERAGE ATOMIC MASS
The atomic masses of many elements have fractional values because they exist as mixture of isotopes.
In the case of such elements, the atomic mass is taken as the average of the atomic masses of the various isotopes. For example, ordinary chlorine is a mixture of two isotopes with atomic masses 35 u and 37 u and they are present in the ration 3:1. Therefore,
Atomic mass of Chlorine = \(\frac{35 \times 3+37 \times 1}{3+1}\) = 35.5u

MOLECULAR MASS
Molecular mass is the sum of atomic masses of the elements present in a molecule.
For example, molecular mass of water
= 2 x atomic mass of hydrogen + 1 x atomic mass of oxygen
= 2 × (1.008u) + 1 × 16.00 u
= 18.02 u

FORMULA MASS
In ionic compounds, we use formula mass instead of molecular mass. Formula mass of an ionic compound is the sum of the atomic masses of all atoms in a formula unit of the compound.

MOLE CONCEPT AND MOLAR MASSES
‘Mole’ was introduced as the seventh base quantity for the amount of substance in SI system. One mole is the amount of a substance that contains as many particles or entities as there are atoms in exactly 12 g (or 0.012 kg) of the ¹²C isotope. This number is known as ‘Avogadro constant’ (N_A = 6.022 × 10²³). The mass of one mole of a substance in grams is called its molar mass. The molar mass is numerically equal to atomic/ molecular/ formula mass in u.

PERCENTAGE COMPOSITION
One can check the purity of a sample by analysing its mass percentage
Mass % of an element in a compound =

EMPIRICAL FORMULA FOR MOLECULAR FORMULA
An empirical formula represents the simplest whole number ratio of various atoms present in a compound whereas the molecular formula shows the exact number of different types of atoms present in a molecule of a compound.

Problem 1.2
A compound contains 4.07 % hydrogen, 24.27 % carbon and 71.65 % chlorine. Its molar mass is 98.96 g. What are its empirical and molecular formulae?
Solution:
Step 1. Conversion of mass per cent to grams.
Since we are having mass per cent, it is convenient to use 100 g of the compound as the starting material. Thus, in the 100 g sample of the above compound, 4.07g hydrogen is present, 24.27g carbon is present and 71.65 g chlorine is present.

Step 2. Convert into number moles of each element
Divide the masses obtained above by respective atomic masses of various elements.

Step 3. Divide the mole value obtained above by the smallest number
Since 2.021 is smallest value, division by it gives a ratio of 2:1:1 for H:C:CI.
In case the ratios are not whole numbers, then they may be converted into whole number by multiplying by the suitable coefficient.

Step 4. Write empirical formula by mentioning the numbers after writing the symbols of respective elements.
CH₂Cl is, thus, the empirical formula of the above compound.

Step 5. Writing molecular formula
a) Determine empirical formula mass. Add the atomic masses of various atoms present in the empirical formula.
ForCH₂Cl, empirical formula mass is 12.01 + 2 1.008 + 35.453 = 49.48 g

b) Divide Molar mass by empirical formula mass

c) Multiply empirical formula by ‘n’ obtained above to get the molecular formula
Empirical formula = CH₂Cl, n = 2.
Molecular formula = 2 × [CH₂Cl]=C₂H₄Cl₂

STOICHIOMETRY AND STOICHIOMETRIC CALCULATIONS
‘Stoichiometry’ deals with the calculation of masses (sometimes volumes also) of the reactants and prod¬ucts involved in a chemical reaction. The coefficients of reactants and products in a balanced chemical equation is called the stoichiometric coefficients.

LIMITING REAGENT
The amount of the product obtained in a chemical reaction is determined by the amount of the reactant that is completely consumed in the reaction. This reactant is called the limiting reagent. Thus, limiting reagent may be defined as the reactant which is com¬pletely consumed in a reaction containing two or more reactants.

REACTIONS IN SOLUTIONS
1. Mass per cent :
It is obtained by the following relation:

2.Mole Fraction :
It is the ratio of number of moles of a particular component to the total number of moles of the solution. If a substance ‘A’ dissolves in substance ‘B’ and their number of moles are nA and nB respectively; then the mole fractions of A and B are given as

3. Molarity :
It is the most widely used unit and is denoted by M. It is defined as the number of moles of the solute in 1 litre of the solution.

4. Molality :
It is defined as the number of moles of solute present in 1 kg of solvent. It is denoted by ‘m’.

Note :
Molarity of a solution changes with temeprature. But molality of a solution does not change with temperature since mass remains unaffected with temperature.

Students can Download Chapter 13 Hydrocarbons Questions and Answers, Plus One Chemistry Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Chemistry Chapter Wise Questions and Answers Chapter 14 Environmental Chemistry

Plus One Chemistry Environmental Chemistry One Mark Questions and Answers

Question 1.
The greatest affinity for haemoglobin is shown by
a) NO
b) CO
c) O₂
d) CO₂
Answer:
b) CO

Question 2.
London smog is ___________ in nature.
Answer:
reducing

Question 3.
Addition of phosphate fertilizers into water leads to
a) Increased growth of decomposers
b) Reduced algal growth
c) Increased algal growth
d) Nutrient enrichment
Answer:
d) Nutrient enrichment

Question 4.
Which of the following is a viable particulate?
a) Algae
b) Smoke
c) Mist
d) Fumes
Answer:
a) Algae

Question 5.
Which of the following is not a greenhouse gas?
a) CO₂
b) CH₄
c) O₂
d) Water vapour
Answer:
c) O₂

Question 6.
Methyl isocyanate is prepared by the action of ___________ on methyl amine
Answer:
COCl₂

Question 7.
In a photo chemical smog the gas that causes eye irritation is ___________ .
a) CO₂
b) CH₄
c) PAN
d) Acrolein
e) both (c) and (d)
Answer:
PAN

Question 8.
Super sonic jet planes can contribute to zone depletion by introduction of the gas straight to the stratosphere
Answer:
NO

Question 9.
The difference between the amounts of dissolved oxygen in a sample of water saturated with oxygen and that after incubation for a period of 5 days is known as.
Answer:
BOD

Question 10.
Excess of nitrate ions in drinking water causes.
Answer:
Methenoglobinemia

Plus One Chemistry Environmental Chemistry Two Mark Questions and Answers

Question 1.
Write the equation for the combustion of ethane.
a) Complete combustion
b) Incomplete combustion
Answer:
a) 2C₂H₆(g) +7O₂ → 4CO₂(g) + 6H₂O(v)
b) 2C₂H₆(g)+5O₂ → 4CO(g) + 6H₂O(v)

Question 2.
Write the three common pollutants.
Answer:
1. Gases such as CO and SO₂
2. Compounds of metals like lead, mercury, zinc.
3. Radioactive substance

Question 3.
Three equations are given below. After studying it, write about Acid rain.
(1) CO₂(g) + H₂O(l) → 2H+ (aq) + CO₃^2-(aq)
(2) 2SO₂(g) + O₂(g) +2H₂O(l) → 2H₂SO₄(aq)
(3) 4NO₂(g) + O₂(g) + 2H₂O(l) → 2HNO₃ (aq)
Answer:
Oxides of Sulphur and Nitrogen, mist of HCI and Phosphoric acid, etc. present in polluted airdissolve in rain water making it more acidic. This is known as acidic rain. SO₂ and NO₂ present in polluted air are the major contributors of acid rain.
2SO₂(g) +O₂(g) + 2H₂O(l) → 2H₂SO₄(aq)
4NO₂(g) +O₂(g) + 2H₂O(l) → 2HNO₃(aq)

Question 4.
Marble of Taj Mahal reacts with acid rain. What is its result?
Answer:
The marble with which Taj Mahal is made is continuously eaten away by acid rain. It is due to the presence of chemical factories in Agra. The acids present in acid rain react with marble and making the marble dull and rough.
CaCO₃ + H₂SO₄ → CaSO₄ +H₂O + CO₂

Question 5.
Why usage of chlorofluorocarbons being discouraged? or Explain ozone layer depletion?
Answer:
The decomposition of CFC’s destroying ozone. CF₂Cl₂(g) + hv → Cl(h) + CF₂Cl(g)
The reactive Cl atom reacts with O₃ to form ClO radical. Cl(g) + O₃ → ClO(g) + O₂(g)
Thus each Cl atom produced, can destroy many O₃ molecules. This leads to ozone depletion.

Question 6.
What is meant by BOD?
Answer:
It is biochemical oxygen demand. It is the amount of dissolved oxygen required by micro-organisms to oxidise organic and inorganic matter present in polluted water.

Question 7.
1. How we can control air pollution?
2. How does the soil pollution occur?
Answer:
1. Air pollution can be controlled by the following ways.
• Exhaust to oxidize CO to CO₂.
• CO₂ level can be maintained by planting trees.
• Hydrogen gas is looked upon as pollution less future fuel.
• The large amount of nitrogen oxides emitted from power plant can be removed by scrubbing it with H₂SO₄.
2. It occurs due to
• Indiscriminate use of fertilizers, pesticides etc.
• Dumping of waste materials.
• Deforestation

Question 8.
What do you mean by greenhouse effect?
Answer:
Global warming is caused by the excess amount of CO₂ in the atmosphere. When CO₂ accumulate due to the decrease of trees it will increase the temperature of earth. This leads to melting of ice. So the sea level rises.

Question 9.
What do you mean by global warming?
Answer:
As more and more infrared radiations are trapped, the atmosphere becomes hotter and the global temperature rises up. This is known as global warming. There has been a marked increase in the levels of CO₂ in the atmosphere due to severe deforestation and burning of fossil fuels.

Question 10.
Give two examples in which green chemistry has been applied.
Answer:
1) Oxidative cracking process in the formation of ethylene is a significant step.
2) Fuel cells for cellular phones have been developed. This cell last for the full life time of the phone.

Question 11.
Oxygen play a key role in the troposphere while ozone in the stratosphere.
i) How is ozone formed in the atmosphere?
ii) What are the causes of depletion of ozone layer?
Answer:
i) By electric discharge of O₂ during lightning.
ii) UV radiations enter to the earth surface and it causes skin diseases.

Question 12.
No new industries are allowed in thickly populated cities by Government order. Why such an order is issued?
Answer:
The city will be destroyed due to acid rain caused by industrial pollution. The Govt, order was to protect the environment (orthe city).

Question 13.
Discuss the importance of dissolved oxygen in water.
Answer:
It helps living organisms in water. They inhale dissolved oxygen in water.

Question 14.
Define environmental chemistry.
Answer:
Environmental chemistry is defined as the branch of science which deals with the chemical processes occurring in the environment. It involves the study of origin, transport, reactions, effects and the fates of chemical species in the environment.

Question 15.
Explain tropospheric pollution.
Answer:
Tropospheric pollution occurs due to the presence of gaseous and the particulate pollutants.
• Gaseous air pollutants. These include mainly oxides of sulphur (SO₂, SO₃), oxides of nitrogen (NO, NO₂) and oxides of carbon (CO, CO₂) in addition , to hydrogen sulphide (H₂S), Hydrocarbons, ozone and other oxidants.
• Particulate pollutants. These include dust, mist, fumes, smoke, smog, etc.

Question 16.
Which gases are responsible for greenhouse effect? List some of them.
Answer:
The main gas responsible for greenhouse effect is CO₂. Other greenhouse gases are methane, nitrous oxide, water vapours, chlorofluorocarbons (CFC’s) and ozone.

Question 17.
A large number of fish are suddenly found floating dead on a lake. There is no evidence of toxic dumping but you find an abundance of phytoplankton. Suggest a reason for the fish kill.
Answer:
Excessive phytoplankton (organic pollutants such as leaves, grass, trash, etc) present in water is biodegradable. A large population of bacteria decomposes this organic matter in water. During this process they consume the oxygen dissolved in water. Water has already limited dissolved oxygen (≈10 ppm) which gets is further depleted. When the level of dissolved oxygen falls below 6 ppm, the fish cannot survive. Hence, they die and float dead in water.

Question 18.
How can domestic waste be used as mannure?
Answer:
Domestic waste comprises two types of materials, biodegradable such as leaves, rotten food, etc, and non-biodegradable such as plastics, glass, metal scrap, etc. The non-biodegradable waste is sent to industry for recycling. The biodegradable waste should be deposited in the land fills. With the passage of time, it is converted into compost manure.

Question 19.
For your agricultural field or garden, you have developed a compost producing pit. Discuss the process in the light of bad odour, flies and recycling of wastes for a good produce.
Answer:
The compost producing pit should be set up at a suitable place or in a tin to protect ourselves from bad odour and flies. It should be kept covered so that flies cannot make entry into it and the bad odour is minimized. The recyclable material like plastics, glass, newspapers, etc., should be sold to the vendor who further sells it to the dealer. The dealer further supplies it to the industry involved in recycling process.

Plus One Chemistry Environmental Chemistry Three Mark Questions and Answers

Question 1.
Fish grow in cold water as well as in warm water.
1. Do you agree? What is the reason?
2. pH of rain water is 5.6. Is it true? What is the reason?
Answer:
1. No. Fish grow in cold water. Warm water contains less dissolved O₂ than cold water,

2. Normally rain water contains dissolved CO₂ and hence it shows acidic pH of 5.6.
H₂O + CO₂ \(\rightleftarrows \) H₂CO₃ or 2H⁺ + CO₂
When pH of rain water became below 5.6, it becomes acid rain.

Question 2.
Write the mechanism of formation of photochemical smog.
Answer:
At high temp, the petrol and diesel engines, N₂ & O₂ combine to form NO which is emitted into atmosphere. NO is then oxidised in airto form NO₂ which absorbs sunlight and form NO and free O atom.

The O atoms being reactive and combine with O₂ to form O₃.
O₂ (g) + O(g) → O₃(g)
The O₃ react with NO formed by the photochemical decomposition of NO₂.
NO(g) + O₃(g) → NO₂(g) + O₂(g)
NO₂ and O₃ are good oxidising agents and they react with unburnt hydrocarbons in the polluted airto form substances such as acrolein and formaldehyde. These are the main substances of photochemical smog.

Question 3.
1. What are the major pollutants of water?
2. What is meant by eutrophication?
Answer:
1. Micro-organism present in domestic sewages, organic wastes, plant nutrients, toxic metals, sediments, pesticides and radioactive substances,

2. Addition of P to water as PO₄^3- ion encourages the formation of algae which reduces the dissolved oxygen content of water. This is called eutrophication.

Question 4.
1. Acid rain causes extensive damage to vegetation and aquatic life. )
i) What do you mean by acid rain?
ii) Name the chemicals responsible for acid rain.
2. List gases which are responsible for greenhouse effect.
Answer:
1. i) When the pH of the rain water drops below 5.6, it is called acid rain.
ii) Oxides of nitrogen and sulphur mist of hydrochloric acid and phosphoric acid etc.

2. Such as carbon dioxide, methane, ozone, chlorofluorocarbon compounds (CFC).

Question 5.
Statues and monuments in India such as Tajmahal are affected by acid rain. How?
Answer:
The statues of monuments in India are affected by acid rain. For example, the air in the vicinity of Taj Mahal contains very high levels of oxides of sulphur and nitrogen. This results in acid rain which reacts with marble of Taj Mahal causing pitting.
CaCO₃ + H₂SO₄ → CaSO₄ + H₂O + CO₂
CaCO₃ + 2HNO₃ → Ca(NO₃)2 +H₂O + CO₃.
As a result, the monument is being slowly eaten away and the marble is getting decolourised and lustreless. Thus, acid rain is considered as a threat to Taj Mahal.

Question 6.
1. Writethree major consequences of air pollution.
2. Write any two suitable methods to control air pollution.
Answer:
1. We cannot get pure oxygen for inspire.
Air pollution causes global warming.
It leads to acid rain.
It leads to diseases.

2. Plant trees.
Reduce the use of motor vehicles.
Do not burn plastics.

Question 7.
Carbon monoxide gas is more dangerous than carbon dioxide gas. Why?
Answer:
CO combines with haemoglobin to forms a complex entity, carboxyhaemoglobin which is about 300 times more stable than oxy-haemoglobin. In blood, when the concentration of carboxyhaemoglobin reaches 3 -4%, the oxygen-carrying capacity of the blood is significantly reduced. In other words, the body becomes oxygen-starved. This results into headache, nervousness, cardiovascular disorder, weak eyesight etc.

Plus One Chemistry Environmental Chemistry Four Mark Questions and Answers

Question 1.
Classify the following pollutants:
(a) Carbon monoxide (CO)
(b) Detergents
(c) Plastic
(d) DDT
(e) Sewage
(f) Cigarette smoke
Answer:

Type of Pollution
Air	Water	Soil
CO Cigarette smoke	DDT Detergent Sewage	DDT Plastic

Question 2.
As a result of stratospheric pollution, the ozone layer is destructed.
a) What is ozone layer?
b) How is it useful to us?
c) Write a harmful effect of ozone layer depletion.
Answer:
a) The layer of ozone seen in stratosphere is called ozone layer.
b) Ozone layer plays a significant role in protecting earth from UV rays.
c) Due to ozone layer depletion agricultural crops were found to give reduced yields. Small aquatic organisms which are very sensitive are destroyed due to the increase in the level of UV radiation.

Question 3.
a) What is smog?
b) What are the adverse effects of photochemical smog?
c) Write any two methods to control photochemical smog.
Answer:
a) Smog is a mixture of smoke and fog. This is the most common example of air pollution that occurs in many cities throughout the world. There are two types of smog:
1) Classical smog
2) Photochemical smog

b) Adverse effects of photochemical smog:
• Eye irritants
• Nose irritation
• Head ache
• Chest pain
• Dryness of throat
• Cough
• Difficulty in breathing

c) 1) Use catalytic converters in automobiles.
2) Plant certain plants (e.g. Pinus) which can metabolise nitrogen oxide.

Students can Download Chapter 3 Classification of Elements and Periodicity in Properties Notes, Plus One Chemistry Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Chemistry Notes Chapter 3 Classification of Elements and Periodicity in Properties

Introduction
The systematic classification of elements made the study of elements easy. In this unit, we will study the historical development of the periodic table and also learn how elements are classified.

Genesis Of Periodic Classification
While Dobereiner initiated the study of periodic relationship, it was Mendeleev who was responsible for publishing the Periodic Law for the first time. It states as follows:

The properties of the elements are a periodic function of their atomic weights.
Mendeleev arranged elements in horizontal rows and vertical columns of a table in order of their increasing ‘ atomic weights. Elements with similar properties occupied the same vertical column or group. He realized that some of the elements did not fit in with his scheme of classification if the order of atomic weight was strictly followed. He ignored the order of atomic weights, thinking that the atomic measurements might be incorrect, and placed the elements with similar properties together.

At the same time, keeping his primary aim of arranging the elements of similar properties in the same group, he proposed that some of the elements were still undiscovered and, therefore, left several gaps in the table. He left the gap under aluminium and a gap under silicon, and called these elements Eka-Aluminium and Eka-Silicon. Mendeleev predicted the existence of gallium and germanium, and their general physical properties. These elements were discovered later.

Modern Periodic Law And The Present Form Of The Periodic Table
Modem periodic law states that “The physical and chemical properties of the elements are periodic functions of their atomic numbers”. Atomic number is equal to the nuclear charge and the elements are arranged in the increasing order of atomic number.

The period number correspond to the highest principal quantum number (n) of the elements.

Nomenclature Of Elements With Atomic Number Greater Than 100
The names (IUPAC) are derived directly form the atomic number using numerical roots for zero and numbers 1 to 9. The roots are linked together in the order of digits and ‘ium’ is added at the end. The roots for 0,1, 2 9 are nil, un, bi, tri, quad, pent, hex, sept, oct and enn respectively. For example, the element with atomic number 110 will have the name Ununnilium (Un+ un+nil + ium), The element with atomic number 114 has the name Ununquadium (un + un + quad + ium) and the element with atomic number 120 will be Unbinilium (un + bi + nil + ium).

Electronic Configurations And Types Of Elements: s, p, d, f- Blocks

The s-Block Elements
The elements of Group 1 (alkali metals) and Group 2 (alkaline earth metals) which have ns1 and ns2 outermost electronic configuration belong to the s-Block Elements. They are all reactive metals with low ionization enthalpies.

They lose the outermost electron(s) readily to form 1+ ion (in the case of alkali metals) or 2+ ion (in the case of alkaline earth metals). The metallic character and the reactivity increase as we go down the group. Because of high reactivity they are never found pure in nature.

The compounds of the s-block elements, with the exception of those of lithium and beryllium are predominantly ionic.

The p-Block Elements
The p-Block Elements comprise those belonging to group 13 to 18 and these together with the s-Btock Elements are called the Representative Elements or Main Group Elements. The outermost electronic configuration varies from ns2np1 to ns2np6in each period. At the end of each period is a noble gas element with a closed valence shell ns2np6 configuration. All the orbitals in the valence shell of the noble gases are completely filled by electrons and it is very difficult to alter this stable arrangement by the addition or removal of electrons. The noble gases thus exhibit very low chemical reactivity. Preceding the noble gas family are two chemically important groups of non-metals. They are the halogens (Group 17) and the chalcogens (Group 16). These two groups of elements have high negative electron gain enthalpies and readily add one or two electrons respectively to attain the stable noble gas configuration. The non-metallic character increases as we move from left to right across a period and metallic character increases as we go down the group.

The d-Block Elements (Transition Elements)
These are the elements of group 3 to 12 in the centre of the Periodic Table. These are characterised by the filling of inner d orbitals by electrons and are therefore referred to as d-Block Elements. These elements have the general outer electronic configuration (n-1) d1-10ns^2. They are all metals. They mostly form coloured ions, exhibit variable valence (oxidation states), paramagnetism and oftenly used as catalysts. However, Zn, Cd and Hg which have the electronic configuration, (n-1) d10ns2 do not show most of the properties of transition elements. In a way, transition metals form a bridge between the chemically active metals of s-block elements and the less active elements of groups 13 and 14 and thus take their familiar name “Transition Elements”.

The f-Block Elements (Inner-Transition Elements)
The two rows of elements at the bottom of the Periodic Table, called the Lanthanoids, Ce(Z = 58) -Lu(Z = 71) and actinoids, Th(Z = 90)-Lr(Z= 103) are characterised by the outer electronic configuration (n-2)f1-14 (n-1 )d°-1ns2. The last electron added to each element is filled in f- orbital. These two series of ‘ elements are hence called the Inner Transition Elements (f-Block Elements). They are all metals. Within each series, the properties of the elements are quite similar. The elements after Uranium are called Transuranium Elements.

Metals, Non-metals and Metalloids. In addition to displaying the classification of elements into s, p, d and f-blocks, they can be divided into Metals and Non-Metals. Metals usually have high melting and boiling points. They are good conductors of heat and electricity. They are malleable (can be flattened into thin sheets by hammering) and ductile (can be drawn into wires). In contrast, non-metals are located at the top right hand side of the Periodic Table.

In fact, in a horizontal row, the property of elements change from metallic on the left to non-metallic on the right. Non-metals are usually solids or gases at room temperature with low melting and boiling points (boron and carbon are exceptions). They are poor conductors of heat and electricity. Most nonmetallic solids are brittle and are neither malleable nor ductile. The elements become more metallic as we go down a group; the nonmetallic character increases as one goes from left to right across the Periodic Table. The elements (e.g., silicon, germanium, arsenic, antimony and tellurium) running diagonally across the Periodic Table show properties that are characteristic of both metals and nonmetals. These elements are called Semi-metals or Metalloids.

Periodic Trends In Properties Of Elements
Most of the properties such as atomic radius, ionic radius, Ionisation enthalpy, electron gain enthalpy and electron negativity are directly related to electronic configuration of their atoms. They show variation with change in atomic number within a period or a group.

Trends In Physical Properties

1. Atomic Radius :
lt is defined as the distance from the centre of the nucleus of an atom to the outermost shell of electrons. Electron cloud surrounding the atom does not have a sharp boundary since, the determination of the atomic size cannot be precise. Hence it is expressed in terms of different types of radii. Some of these are covalent radius and metallic radius. Covalent radius is defined as one half of the distance between the centres of nuclei of two similar atoms bonded by a single covalent bond. Metallic radius may be defined as half of the internuclear distance between two adjacent atoms in the metallic crystal.

2. Ionic Radius:
The removal of an electron from an atom results in the formation of a cation, whereas gain of an electron leads to an anion. The ionic radii can be estimated by measuring the distances between cations and anions in ionic crystals. When we find some atoms and ions which contain the same number of electrons, we call them isoelectronic species. For example, O2-, F~, Na+ and Mg2+ have the same number of electrons (10). Their radii would be different because of their different nuclear charges.

3. Ionization Enthalpy:
A quantitative measure of the tendency of an element to lose electron is given by its Ionization Enthalpy. It represents the energy required to remove an electron from an isolated gaseous atom (X) in its ground state. To understand the trends in ionization enthalpy, we have to consider two factors: (i) the attraction of electrons towards the nucleus, and (ii) the repulsion of electrons from each other. The effective nuclear charge experienced by a valence electron in an atom will be less than the actual charge on the nucleus because of “shielding” or “screening” of the valence electron from the nucleus by the intervening core electrons.

The first ionization enthalpy of boron (Z = 5) is slightly less than that of beryllium (Z = 4) even though the former has a greater nuclear charge. It is because, s-electron is attracted to the nucleus more than a p-electron. In beryllium, the electron removed during the ionization is an s-electron whereas the electron removed during ionization of boron is a p-electron. The penetration of a 2s-electron to the nucleus is more than that of a 2p-electron; hence the 2p electron of boron is more shielded from the nucleus by the inner core of electrons than the 2s electrons of beryllium. Therefore, it is easier to remove the 2p-electron from boron Compared to the removal of a 2s-electron from beryllium.

Thus, boron has a smaller first ionization enthalpy than beryllium. Another “anomaly” is the smaller first ionization enthalpy of oxygen compared to nitrogen. This arises because in the nitrogen atom, three 2p-electrons reside in different atomic orbitals (Hund’s rule) whereas, in the oxygen atom, two of the four 2p-electrons must occupy the same 2p-orbital resulting in an increased electron-electron repulsion. Consequently, it is easier to remove the fourth 2p-electron from oxygen than it is, to remove one of the three 2p-electrons from nitrogen.

4. Electron Gain Enthalpy :
When an electron is added to a neutral gaseous atom (X) to convert it into a negative ion, the enthalpy change accom-panying the process is defined as the Electron Gain Enthalpy (∆_eg H).

5. Electronegativity:
A qualitative measure of the ability of an atom in a chemical compound to attract shared electrons to itself is called electronegativity. Unlike ionization enthalpy and electron gain enthalpy, it is not a measurable quantity. However, a number of numerical scales of electronegativity of elements viz., Pauling scale, Mulliken-Jaffe scale, Allred-Rochow scale have been developed.

Trends In Chemical Properties
1. Oxidation State :
The atomic property, valency is better explained in terms of oxidation state. It is the charge which an atom of element has or appears to have when present in the combined state. Electronegative elements generally acquire negative oxidation states while electropositive elements acquire positive oxidation states.

2. Anomalous properties of second-period elements:
The first element of each group in s and p block differs in many respects from the remaining members of the respective groups. This is due to their small size, high charge/ radius ratio, high electronegativity and availability of less valence orbitals. The first member has only 4 valence orbitals (2s, 2p) whereas the second member of the same group will have nine valence orbitals (3s, 3p, 3d) for bonding. B can form only (BF₄)^– while Al forms (AlF₆)^3-

In group 1 only Li forms covalent compounds and in many respects, Li resembles Mg of group 2. Similarly, Be resembles Al of group 13. This type of similarity in properties is known as diagonal relationship.

Chemical Reactivity
Across a period ionisation enthalpy increases and electron gain enthalpy becomes more negative. Thus elements at the extreme left show lower ionisation enthalpies (more electropositive nature) and those at the right (excluding nobel gases) show larger negative electron gain enthalpies (more electronegative). Therefore high chemical reactivity is found with elements at the two extremes compared to those at the centre. Electropositivity leads to metallic behaviour and electronegativity leads to non-metallic behaviour.

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Kerala Plus One Chemistry Notes Chapter 2 Structure of Atom

Introduction
The atomic theory of matter was first proposed by John Dalton. His theory, called Dalton’s atomic theory, regarded the atom as the ultimate particle of matter.

Sub-Atomic Particles
Discovery Of Electron
The experiments of Michael Faraday in discharge tubes showed that when a high potential is applied to a gas taken in the discharge tube at very low pres-sures, certain rays are emitted from the cathode. These rays were called cathode rays.

The results of these experiments are summarised below:
1. The cathode rays start from cathode and move towards the anode.

2. In the absence of electrical or magnetic field, these rays travel in straight lines.ln the presence of electrical or magnetic field, they behave as negatively charged particles, i.e.,they consist of negatively charged particles, called electrons.

3. The characteristics of cathode rays (electrons) do not depend upon the material of electrodes and the nature of the gas present in the cathode ray tube.Thus, we can conclude that electrons are the basic constituent of all the atoms.

Charge To Mass Ratio Of Electron
In 1897, the British physicist J.J. Thomson measured the ratio of electrical charge (e) to the mass of electron (m_e) by using cathode ray tube and applying electrical and magnetic field perpendicular to each other as well as to the path of electrons.

From the amount of deviation of the particles from their path in the presence of electrical or magnetic field, the value of e/m was found to be 1.75882 × 10¹¹ coulomb per kg or approximately 1.75288 × 10⁸ cou-lomb per gram. The ratio e/m was found to be same irrespective of the nature of the gas taken in the dis-charge tube and the material used as the cathode.

Charge Of The Electron
Millikan (1868-1953) devised a method known as Oil drop experiment (1906-14), to determine the charge on the electrons. He found the charge on the electron to be – 1.6 × 10-¹⁹C.

Mass of the electron (m)

Discovery Of Protons And Neutrons
Electrical discharge earned out in the modified cathode ray tube led to the discovery of canal rays. The characteristics of these positively charged particles are listed below:

• unlike cathode rays, the e/m ratio of the particles depend upon the nature of gas present in the cathode ray tube.
• Some of the positively charged particles carry a multiple of the fundamental unit of electrical charge.
• The behaviour of these particles in the magnetic or electrical field is opposite to that observed for cathode rays.

The smallest and lightest positive ion was obtained from hydrogen and was called proton. Later, electrically neutral particles were discovered by Chadwick (1932) by bombarding a thin sheet of beryllium by α – particles when electrically neutral particles having a mass slightly greater than that of the protons was emitted. He named these particles as neutrons.

Atomic Models

Thomson Model Of Atom
J.J. Thomson was the first to propose a model of the atom. According to him, the atom is a sphere in which positive charge is spread uniformly and the electrons are embedded in it so as to make the atom electrically neutral. This model is also known as “plumpudding model’. But this model was soon discarded as it could not explain many of the experimental observations.

Rutherford’s Nuclear Model of Atom
Rutherford and his students (Hans Geiger and Ernest Marsden) bombarded very thin gold foil with α – particles. The experiment is known as α -particle scattering experiment. On the basis of the observations, Rutherford drew the following conclusions regarding the structure of atom :

1. Most of the space in the atom is empty as most of the α -particles passed through the foil undeflected.

2. A few α – particles were deflected. Since the α – particles are positively charged, the deflection must be due to enormous repulsive force showing that the positive charge of the atom is not spread throughout the atom as Thomson had presumed. The positive charge has to be concentrated in a very small volume that repelled and deflected the positively charged α – particles.

3. Calculations by Rutherford showed that the volume occupied by the nucleus is negligibly small as compared to the total volume of the atom.

On the basis of above observations and conclusions, Rutherford proposed the nuclear model of atom (after the discovery of protons). According to this model:
1.The positive charge and most of the mass of the atom was densely concentrated in extremely small region. This very small portion of the atom was called nucleus by Rutherford.

2. The electrons move around the nucleus with a very high speed in circular paths called orbits. Thus, Rutherford’s model of atom resembles the solar system in which the nucleus plays the role of sun and the electrons that of revolving planets.

3. Electrons and the nucleus are held together by electrostatic forces of attraction.

Atomic Numberand Mass Number
’ Knowing the atomic number Z and mass number A of an element, we can calculate the number of protons, electrons and neutrons present in the atom of the element.
Atomic Number (Z) = Number of protons = Number of electrons
Mass Number (A) – Atomic number (Z) = Number of neutrons

Isotopes, Isobars And Isotones
Isotopes are atoms of the same element having the same atomic number but different mass numbers. They contain different number of neutrons. For ex-ample, there are three isotopes of hydrogen having mass numbers 1,2 and 3 respectively. All the three isotopes have atomic number 1. They are represented as \(_{ 1 }^{ 1 }{ H }\), \(_{ 1 }^{ 2 }{ H }\) and \(_{ 1 }^{ 3 }{ H }\) and named as hydrogen or protium, deuterium (D) and tritium (T) respectively. Isobars are atoms of different elements which have the same mass number. For example, \(_{ 6 }^{ 14 }{ C }\) and \(_{ 7 }^{ 14 }{ N }\) are isobars.
Isotones may be defined as atoms of different elements containing same number of neutrons. For example \(_{ 6 }^{ 13 }{ C }\) and \(_{ 7 }^{ 14 }{ N }\) are isotones.

Developments Leading To The Bohr’S Model Of Atom
Neils Bohr improved the model proposed by Rutherford. Two developments played a major role in the formulation of Bohr’s model of atom. These were:

1. electromagnetic radiation possess both wave like and particle like properties(Dual character)
2. Experimental results regarding atomic spectra which can be explained only by assuming quantized electronic energy levels in atoms.

Wave Nature Of Electromagnetic Ra-Diation
Light is the form of radiation and it was supposed to be made of particles known as corpuscules.
As we know, waves are characterised by wavelength (λ), frequency (υ) and velocity of propagation (c) and these are related by the equation
c = vλ or v = \(\frac { c }{ \lambda } \)

The wavelengths of various electro magnetic radia-tions increase in the order.
γ rays < X-rays< uv rays < visible < IR < Microwaves < Radio waves

Particle Nature Of Electro Magnetic Radiation: Planck’S Quantum Theory
Planck suggested that atoms and molecules could emit (or absorb) energy only in discrete quantities and not in a continuous manner, a belief popular at that time. Planck gave the name quantum to the smallest quantity of energy that can be emitted or absorbed in the form of electromagnetic radiation. The energy (E) of a quantum of radiation is proportional to its frequency (υ) and is expressed by the equation E = hυ

Photoelectric Effect
When a metal was exposed to a beam of light, electrons were emitted. This phenomenon is called photoelectric effect. Obseravations of the photoelectric effect experiment are the following:

• There is no time lag belween the striking of light beam and the ejection of electrons from the metal surface.
• The number of electrons ejected is proportional to the intensity or brightness of light.
• For each metal, there is a characteristic minimum frequency, u₀ (also known as threshold frequency) below which photoelectric effect is not observed. At a frequency u>u₀, the ejected electrons come out with certain kinetic energy.

The kinetic energies of these electrons increase with the increase of frequency of the light used.

Using Plank’s quantum theory Einstein explained photoelectric effect. When a light particle, photon with sufficient energy strikes an electron instantaneously to the electron during the collision and the electron is ejected without any time lag. Greater the energy of photon greater will be the kinetic energy of ejected electron and greater will be the frequency of radiation.

If minimum energy to eject an electron is hv0 and the photon has an energy equal to hv. Then kinetic en-ergy of photoelectron is given by, hv=hv₀ + 1/2 m_ev² where m_e is the mass of electron and hv₀ is called the work function.

Duel Behaviour Of Electromagnetic Ra-Diation
Light has dual behaviour that is it behaves either as a wave or as a particle. Due to this wave nature, it shows the phenomena of interference and diffraction.

Evidence For The Quantized Electronic Energy Levels : Atomic Spectra
It is observed that when a ray of white light is passed through a prism, the wave with shorter wavelength bends more than the one with a longer wavelength. Since ordinary white light consists of waves with ail the wave-lengths in the visible range, a ray of white light is spread out into a series of coloured bands called spectrum. In a continuous spectrum light of different colours merges together. For example violet merges into blue, blue into green and soon.

Emission and absorption spectra
The spectrum of radiation emitted by a substance that has absorbed energy is called an emission spectrum. Atoms, molecules or ions that have absorbed radiation are said to be “excited”.

A continuum of radiation is passed through a sample which absorbs radiation of certain wavelengths. The missing wavelength which corresponds to the radiation absorbed by the matter, leave dark spaces in the bright continuous spectrum. The study of emission or absorption spectra is referred to as spectroscopy Line spectra or atomic spectra is the spectra where emitted radiation is identified by the appearance of bright lines in the spectra.

Line spectrum of Hydrogen
The hydrogen spectrum consists of several series of lines named after their discoverers. Balmershowed in 1885 on the basis of experimental observations that if spectral lines are expressed in terms of wavenumber (\(\overline { v } \)), then the visible lines of the hydrogen spectrum obey the following formula :
\(\overline { v } \) = 109,677 \(\left[\frac{1}{2^{2}}-\frac{1}{n^{2}}\right] \mathrm{cm}^{-1}\)
where n = 3, 4, 5, ………….
The series of lines described by this formula are called the Balmer series.

The value 109,677cm^-1 is called the Rydberg constant for hydrogen. The first 5 series of lines correspond to n₁ = 1, 2, 3, 4, 5 are known as Lyman, Balmer, Paschen, Bracket and Pfund series respectively. Line specrum becomes more complex for heavier atoms.

Bhor’S Model For Hydrogen Atom
Bhors model for hydrogen atom says that
1. the energy of an electron does not change with time.
The diagram shows the Lyman, Balmer and Paschen series of transitions for hydrogen atom.

2. The frequency of radiation absorbed or emitted when transition occurs between two stationary states that differ in energy by ∆E, is given by :
\(v=\frac{\Delta E}{h}=\frac{E_{2}-E_{1}}{h}\)
E₁ and E₂ are the energies of the lower and higher allowed energy states respectively.
The angular momentum of an electron in a given stationary state can be expressed as in equation,

Bohr’s theory for hydrogen atom:
1. The stationary states for electron are numbered n = 1,2,3. These integral numbers are known as Principal quantum numbers.
2. The radii of the stationary states are expressed as:
r_n = n² a₀
where a₀ = 52.9 pm

3. The most important property associated with the electron, is the energy of its stationary state. It is
given by the expression, \(E_{n}=-R_{H}\left(\frac{1}{n^{2}}\right)\)
where R_H is called Rydberg constant and its value is 2.18 × 10^-18 J. The energy of the lowest state, also called as the ground state, is
E₁ = -2.18 × 10^-18 \(\left(\frac{1}{1^{2}}\right)\) = -2.18 × 10^-18 J. The energy of the stationary state for n = ∝, will be :
E₂ = -2.18 × 10^-18 J\(\left(\frac{1}{2^{2}}\right)\) = -0.545 × 10^-18 J.

When the electron is free from the influence of nucleus(n = ∞), the energy is taken as zero. When the electron is attracted by the nucleus and is present in orbit n, the energy is emitted and its energy is lowered. That is the reason for the presence of negative sign and depicts its stability relative to the reference state of zero energy and n = ∞

4. Bohr’s theory can also be applied to the ions containing only one electron, similar to that present in hydrogen atom. For example, He⁺ Li²⁺, Be³⁺ and so on. The energies of the stationary states associated with these hydrogen-like species are given by the expression,

Explanation of Line Spectrum of Hydrogen
The frequency (v) associated with the absorption and emission of the photon can be evaluated by using equation,

Limitations of Bohr’s Model
Bohr’s model was too simple to account for the following points:
1. It fails to account for the finer details (doublet, that is two closely spaced lines) of the hydrogen atom spectrum. This model is also unable to explain the spectrum of atoms other than hydrogen Further, Bohr’s theory was also unable to explain the splitting of spectral lines in the presence of magnetic field (Zeeman effect) or an electric field (Stark effect).
2. It could not explain the ability of atoms to form molecules by chemical bonds.

Towards Quantum Mechanical Model Of The Atom
Two important developments which contributed significantly in the formulation of a more suitable and general model for atoms were:

1. Dual behaviour of matter
2. Heisenberg uncertainty principle

Dual Behaviour of Matter
The French physicist, de Broglie proposed that matter, like radiation, should also exhibit dual behaviour i. e., both particle and wavelike properties. This means that just as the photon, electrons should. also have momentum as well as wavelength. de Broglie, from this analogy, gave the following relation between wavelength (λ) and momentum (p) of a material particle.
\(\lambda=\frac{h}{m v}=\frac{h}{p}\)

Heisenberg’s Uncertainty Principle
Werner Heisenberg a German physicist in 1927, stated uncertainty principle which is the consequence of dual behaviour of matter and radiation. It states that it is impossible to determine simultaneously, the exact position and exact momentum (or velocity) of an electron. Mathematically, it can be given as in equation,

∆x is the uncertainty in position and ∆p_x (or ∆v_x) is the uncertainty in momentum (or velocity) of the particle. If the position of the electron is known with high degree of accuracy (∆x is small), then the velocity of the electron will be uncertain ∆v_x is large]. On the other hand, if the velocity of the electron is known precisely ( ∆v_x is small), then the position of the electron will be uncertain (∆x will be large). Thus, if we carry out some physical measurements on the electron’s position or velocity, the outcome will always depict a fuzzy or blur picture.

Significance of Uncertainty Principle
Heisenberg Uncertainty Principle rules out existence of definite paths or trajectories of electrons and other similar particles. The trajectory of an object is determined by its location and velocity at various moments. If we know where a body is at a particular instant and if we also know its velocity and the forces acting on it at that instant, we can tell where the body would be sometime later. We, therefore, conclude that the position of an object and its velocity fix its trajectory. The effect of Heisenberg Uncertainty Principle is significant only for motion of microscopic objects and is negligible for that of macroscopic objects.

Reasons for the Failure of the Bohr Model
In Bohr model, an electron is regarded as a charged particle moving in well defined circular orbits about the nucleus. The wave character of the electron is not considered in Bohr model. Further, an orbit is a clearly defined path and this path can completely be defined only if both the position and the velocity of the electron are known exactly at the same time. This is not possible according to the Heisenberg uncertainty principle. Bohr.model of the hydrogen atom, therefore, not only ignores dual behaviour of matter but also contradicts Heisenberg uncertainty principle. There was no point in extending Bohr model to other atoms. In fact, an insight into the structure of the atom was needed which could account for wave-particle duality of matter and be consistent with Heisenberg uncertainty principle. This came with the advent of quantum mechanics.

Quantum Mechanical Model Of Atom
Quantum mechanics is a theoretical science that deals with the study of motions of microscopic objects such as electrons.

In quantum mechanical model of atom, the behaviour of an electron in an atom is described by an equation known as Schrodinger wave equation. Fora system, such as an atom or molecule whose energy does not change with time, the Schrodinger equation written as Hψ = Eψ where H is a mathematical operator, called Hamiltonian operator, E is the total energy and ψ is the amplitude of the electron wave called wave function.

Hydrogen Atom And The Schrodinger Equation
The wave function ψ as such has no physical significance. It only represents the amplitude of the electron wave. However ψ² may be considered as the probability density of the electron cloud. Thus, by determining ψ² at different distances from the nucleus, it is possible to trace out or identify a region of space around the nucleus where there is high probability of locating an electron with a specific energy.

According to the uncertainty principle, it is not possible to determine simultaneously the position and momentum of an electron in an atom precisely. So Bohr’s concept of well defined orbits for electron in an atom cannot hold good. Thus, in quantum mechanical mode, we speak of probability of finding an electron with a particular energy around the nucleus. There are certain regions around the nucleus where probability of finding the electron is high. Such regions are called orbitals. Thus an orbital may be defined as the region in space around the nucleus where there is maximum probability of finding an electron having a specific energy.

Orbitals and Quantum Numbers
Orbitals in an atom can be distinguished by their size, shape and orientation. An orbital of smaller size means there is more chance of finding the electron near the nucleus. Similarly, shape and orientation mean that there is more probability of finding the electron along certain directions than along others. Atomic orbitals are precisely distinguished by what are known as quantum numbers. Each orbital is designated by three quantum numbers labelled as n, l and m_l

The principal quantum number n’ is a positive integer with value of n= 1, 2, 3 ……………

The principal quantum number determines the size and to large extent the energy of the orbital.

The principal quantum number also identifies the shell. With the increase in the value of ‘n’, the number of allowed orbital increases and are given by ‘n²’ Ait the orbitals of a given value of ‘n’ constitute a single shell of atom and are represented by the following letters
n= 1 2 3 4 ………………
Shell = K LM N ………………

Size of an orbital increases with increase of principal quantum number ‘n’. Since energy of the orbital will increase with increase of n.

Azimuthal quantum number, ‘F is also known as orbital angular momentum or subsidiary quantum number. It defines the three-dimensional shape of the orbital. For a given value of n, l can have n values ranging from 0 to (n – 1), that is, for a given value of n, the possible value of l are: l = 0, 1, 2, ……….. (n – 1)

Each shell consists of one or more subshells or sub-levels. The number of subshells in a principal shell is equal to the value of n. For example h the first shell (n = 1), there is only one sub-shell which corresponds to l = 0. There are two sub-shells (l= 0, 1) in the second shell (n = 2), three l= 0, 1, 2) and so on. Each sub-shell is assigned an azimuths! quantum number (l). Sub-shells corresponding to different values of l are represented by the following symbols.
l : 0 1 2 3 4 5 …………….
Notation for sub-shell : s p d f g h …………….

Magnetic orbital quantum number. ‘m_l’ gives information about the spatial orientation of the or bital with respect to standard set of co-ordinate axis. For any sub-shell (defined by T value) 21+ 1 values of m,are possible and these values are given by:
m, = -l, -(l-1), (l-2)… 0, 1… (l-2), (l-1), l Thus for l = 0, the only permitted value of m,= 0, [2(0) + 1 = 1, one s orbital].

Electron spin ‘s’:
George Uhlenbeck and Samuel Goudsmit proposed the presence of the fourth quantum number known as the electron spin quantum number (m_s). Spin angular momentum of the electron — a vector quantity, can have two orientations relative to the chosen axis. These two orientations are distinguished by the spin quantum numbers ms which can take the values of +½ or -½. These are called the two spin states of the electron and are. normally represented by two arrows, ↑ (spin up) and ↓ (spin down). Two electrons that have different m_s values (one +½ and the other -½) are said to have opposite spins. An orbital cannot hold more than two electrons and these two electrons should have opposite spins.

Shapes of Atomic Orbitals
The orbital wave function or V for an electron in an atom has no physical meaning. It is simply a mathematical function of the coordinates of the electron.

According to the German physicist, Max Bom, the square of the wave function (i.e., ψ²) at a point gives the probability density of the electron at that point.

For 1 s orbital the probability density is maximum at the nucleus and it decreases sharply as we move away from it. The region where this probability I density function reduces to zero is called nodal surfaces or simply nodes. In general, it has been found that ns-orbital has (n – 1) nodes, that is, number of nodes increases with increase of principal quantum number n.

These probability density variation can be visualised . in terms of charge cloud diagrams.

Boundary surface diagrams of constant probability density for different orbitals give a fairly good representation of the shapes of the orbitals. In this representation, a boundary surface or contour surface is drawn in space for an orbital on which the value of probability density |ψ|² is constant. Boundary ‘ surface diagram for a s orbital is actually a sphere centred on the nucleus. In two dimensions, this sphere looks like a circle. It encloses a region in which probability of finding the electron is about 90%. The s-orbitals are spherically symmetric, that is, the probability of finding the electron at a given distance is equal in all the directions.

unlike s-orbitals, the boundary surface diagrams of p orbitals are not spherical. Instead, each p orbital consists of two sections called lobes that are on either side of the plane that passes through the nucleus. The probability density function is zero on the plane where the two lobes touch each other. The size, shape and energy of the three orbitals are identical. They differ, however, in the way the lobes are oriented. Since the lobes may be considered to lie along the x, y or z-axis, they are given the designations 2p_x, 2p_y, and 2p_z. It should be understood, however, that there is no simple relation between the values of m, (-1, 0 and+1) and the x, y and z directions. For our purpose, it is sufficient to remember that, because there are three possible values of m, there are, therefore, three p orbitals whose axes are mutually perpendicular. Like s orbitals, p orbitals increase in size and energy with increase in the principal quantum number

The number of nodes are given by (n -2), that is number of radial node is 1 for 3p orbital, two for 4p orbital and so on.

For l = 2, the orbital is known as d-orbital and the minimum value of principal quantum number (n) has to be 3 as the value of l cannot be greater than n-1. There are five m; values (-2, -1, 0, +1 and +2) for l = 2 and thus there are five d orbitals. The five d-orbitals are designated as d_xy, d_yz, d_xz, d_x²-y² and d_z². The shapes of the first fourd-orbitals are similarto each other, where as that of the fifth one, d_z², is different from others, but all five 3d orbitals are equivalent in energy. The d orbitals for which n is greater than 3 (4d, 5d…) also have shapes similar to 3d orbital, but differ in energy and size.

Besides the radial nodes (i.e., probability density function is zero), the probability density functions for the np and nd orbitals are zero at the plane (s), passing through the nucleus (origin). For example, in case of p_z orbital, xy-plane is a nodal plane, in case of d_xy orbital, there are two nodal planes passing through the origin and bisecting the xy plane containing z-axis. These are called angular nodes and number of angular nodes are given by T, i.e., one angular node for p orbitals, two angular nodes for cf orbitals and so on. The total number of nodes are given by (n-1), i.e., sum of I angular nodes and (n-l-1) radial nodes.

Energies Of Orbitals
The order of energy of orbitals in single electron sys-tem are given below:
1s < 2s = 2p < 3s = 3p = 3d < 4s = 4p = 4d = 4f The orbitals having same energy are called degenerate.

Filling Of Orbitals In Atom
Aufbau principle: According to this principle in the ground state of an atom, an electron will occupy the orbital of lowest energy and orbitals are occupied by electrons in the order of increasing energy.

Pauli’s exclusiohn principle : Pauli’s exclusion principle states that ‘no two electrons in an atom can have the same values for all the four quantum numbers’

Since the electrons in an orbital must have the same n, I and m quantum numbers, if follows that an orbital can contain a maximum of two electrons provided their spin quantum numbers are different. This is an important consequence of Pauli’s exclusion principle which says that an orbital can have maximum two electrons and these must have opposite spins.

Hund’s rule of maximum multiplicity :
This rule states that electron pairing in orbitals of same energy will not take place until each available orbital of a given subshell is singly occupied (with parallel spin).
The rule can be illustrated by taking the example of carbon atom. The atomic number of carbon is 6 and its electronic configuration is 1s²2s²2p². The two electrons of the 2p subshell can be distributed in the following three ways.

According to Hund’s rule, the configuration in which the two unpaired electron occupying 2px, and 2py orbitals with parallel spin is the correct configuration of carbon.

Exceptional configurations of chromium and copper
The electronic configuration of Cr (atomic number 24) is expected to be [Ar] 4s² 3d⁴, but the actual configuration is [Ar] 4s¹ 3d⁵. Similarly, the actual configuration of Cu (At. No. 29) is [Ar] 4s¹ 3d¹⁰ instead of the expected configuration [Ar] 4s² 3d⁹.

This is because of the fact that exactly half filled or completely filled orbitals (i.e., d⁵, d¹⁰, f⁷, f¹⁴) have lower energy and hence have extra stability.