Prasanna

Kerala State Board New Syllabus Plus Two Chemistry Chapter Wise Previous Questions and Answers Chapter 10 Haloalkanes and Haloarenes.

Kerala Plus Two Chemistry Chapter Wise Previous Questions Chapter 10 Haloalkanes and Haloarenes

Question 1.
Most of the organic chlorides, bromides and iodides react with certain metals to give compounds containing carbon-metal bonds. (March – 2010)
i) Give one example for such a compound.
ii) How will you prepare the above compound?
b) Write any two electrophilic substitution reactions of chlorobenzene.
Answer:
i) Grignard reagent; CH₃MgCI (Methyl magnesium chloride)
ii) Grignard reagents are prepared by treating haloalkanes with magnesium metal in dry ether.
\(\mathrm{CH}_{3} \mathrm{Cl}+\mathrm{Mg}_{\frac{\text { dry ether }}{\longrightarrow}} \mathrm{CH}_{3} \mathrm{MgCl}\)

b) Halogenation : When chlorobenzene is treated with chlorine in presence of anhydrous FeCl₃ a mixture of 1,4-Dichlorobenzene (major product) and 1.2-Dichlorobenzene is formed.

Nitration : When chlorobenzene is treated with nitrating mixture (mixture of conc.HNO₃ and cone. H₂SO₄) a mixture of 1-Chloro-4-nitrobenzene (major product) and 1-Chloro-2-nitrobenzene (minor product) is formed.

Question 2.
An organic compound A reacts with metallic sodium in an ether medium to form ethane. A reacts with Magnesium in ether medium to give B, which on hydrolysis gives methane. Identify A and B. Write down the chemical equations for the reactions involved. (Say – 2010)
OR
Bromoethane, when treated with alcoholic KOH, gives ethene, KBr, and H₂O.
a) Identify the type of reaction.
b) Instead of bromoethane, if you take 2- bromobutane, what is the major product obtained? Write down the chemical equation for the reaction.
c) Explain the rule behind the above reaction.
Answer:
A → A CH₃ Cl (Methyl chloride)
B → CH₃Mg Cl (Methyl magnesium chloride)

OR

a) β – elimination (Dehydrohalogenation)

c) Saytzeffs rule – It states that in dehydro-halogenation reactions, the preferred product is that alkene which has the greater number of alkyl groups attached to the doubly bonded carbon atoms.

Question 3.
Haloalkanes and Haloarenes react with metals to give Hydrocarbons or products from which hydrocarbons are obtained easily.
(March – 2011)

Identify the product and name the reaction.

Identify the product and name the reaction.

Identify A & B.
Answer:

This reaction is called the Wurtz-Fitting reaction.

Question 4.
Alkyl halides are the starting materials for the synthesis of a number of organic compounds. How are the following compounds obtained from the alkyl halide CH₃ – CH₂ – Br? (Say – 2011)
a) Ethene
b) Ethanol
c) Butane
d) Ethoxy Ethane
Answer:
a) Conversion of bromoethane to ethene – By β – elimination reaction.
OR

b) Conversion of bromoethane to ethanol – by alkaline hydrolysis.
When bromoethane is treated with aqueous KOH or moist Ag20 ethanol is formed.
OR
\(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{Br}+\mathrm{KOH}_{(\mathrm{aq})} \rightarrow \mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{OH}+\mathrm{KBr}\)
c) Conversion of bromoethane to n-butane – by Wurtz reaction.
When bromoethane is treated with sodium in dry ether n-Butane is formed.
OR

d) Conversion of bromoethane to ethoxyethane – by Williamson’s synthesis.
When bromoethane is treated with sodium ethoxide ethoxyethane is formed.
OR

Question 5.
Nucleophilic substitution reactions are of two types – S_N1 reactions and S_N2 reactions. (March – 2012)
i) Write any two differences between S_N1 and S_N2 reactions.
ii) Write any two reasons for the less reactivity of aryl halides towards nucleophilic substitution reactions.
Answer:
i)

SN1 reaction	Sn2 reaction
1. Molecularity is 1.	1. Molecularity is 2.
2. Rate of reaction is dependent only on the concentration of the alkyl halide.	2. Rate of reaction is dependent on the concentration of the alkyl halide as well as a nucleophile.
3. Mechanism involves two steps – formation of carbocation followed by the nucleophilic attack.	3. Mechanism involves one step via the formation of a transition state.
4. Starting with an optically active alkyl halide results in partial racemization.	4. Starting with an optically active alkyl halide results in a complete inversion of configuration.

ii) Aryl halides are much less reactive than haloalkane or alkyl halides towards nucleophilic substitution reaction due to

1) Resonance effect – in haloalkanes the electron pairs on halogen atom are in conjugation with -IT electrons of the ring and thus the C – X bond acquires a partial double bond character. Asa result the bond cleavage in haloarenes is difficult than in haloalkanes.

2) Difference in hybridization of carbon atom in C – X bond – in haloalkanes, the carbon atom attached to halogen is sp³ hybridized while in case of haloarenes the carbon atom attached to halogen is sp² hybridized which is more electronegative. Hence, the C – X bond length in haloarenes (169 pm) is less than that in haloalkanes (177 pm). It is difficult to break a shorter bond than a longer bond. Therefore, Haloarenes are less reactive towards nucleophilic substitution reaction.

3) InstabilIty of phenyl cation – in case of haloarenes, the phenyl cation formed as a result of self-ionization will not be stabilized by resonance and therefore, S_N1 mechanism Is ruled out.

4) Because of the possible repulsion, it is less likely for the electron-rich nucleophile to approach electron-rich arenes,

Question 6.
Haloarenes undergo electrophilic substitution race fans. Explain the Important electrophilic substitution reactions of chlorobenzene. (Write down the chemical equation) (Say – 2012)
Answer:
Halogenation: When chlorobenzene Is treated with chlorine In presence of anhydrous FeCl₃ a mixture of 1. 4-Dichlorobenzene (major product) and I 2-Dichlorobenzene b formed.

Nitration: When chlorobenzene is treated with nitrating mixture (mixture of conc.HNO₃ and conc. H₂SO₄) a mixture of 1-Chloro-4-nitrobenzene (major product) and 1-Chloro-2-nitrobenzene (minor product) is formed.

Sulphonation: When chlorobenzene is heated with conc.H₂SO₄ a mixture of 4-Chiorobenzene suiphonic acid (major product) and 2-Chiorobenzene sulphonic acid (minor product) is formed.

Question 7.
a) For the preparation of alkyl chlorides from alcohols, thionyl chloride (SOCl₂) is preferred. Give reason. (March – 2013)
b) Haloalkanes undergo b – elimination reaction in presence of alcoholic potassium hydroxide.
i) Which is the major product obtained by the b- elimination of 2-Bromo pentane?
ii) Name the rule, which leads to the product in the above elimination reaction.
c) Write the chemical equation for the preparation of toluene by Wurtz-Fitting reaction.
Answer:
a) S02 and HCI being escapable gases, the reaction of alcohols with thionyl chloride gives pure alkyl chlorides.
R-OH + SOCI₂ → R-CI + SO₂ + HCI
b) i) pent-2-ene is the major product since it has a greater number of alkyl groups attached to the doubly bonded carbon atoms.

Question 8.
Haloarenes undergo nucleophilic and electrophilic substitution reactions. (Say – 2013)
a) Write two examples for ambident nucleophiles.
b) Write one example for the nucleophilic substitution reaction of chlorobenzene.
c) Write any two examples of electrophilic substitution reaction of chlorobenzene.
Answer:
a) Cyanide ion (CN^–) and nitrite ion (NO₂).
b) Chlorobenzene when heated with aqueous sodium hydroxide solution at a temperature of 623 K and a pressure of 300 atmosphere gets converted to phenol by a nucleophilic substitution reaction.

c) Nitration – When chloroform is treated with the nitrating mixture (a mixture of conc, HNO₃ and cone. H₂SO₄), a mixture of 1-Chloro-2-nitrobenzene (minor product) and 1-Chloro-4-nitrobenzene (major product) is obtained.

Friedel-Crafts alkylation – When chloroform is treated with methyl chloride in presence of anhydrous AICI₃ a mixture of 1 – Chloro – 2 – methylbenzene (minor product) and 1-Chloro-4-methylbenzene (major product) are obtained.

Question 9.
a) Most important chemical reactions of haloalkanes are their substitution reactions. (March – 2014)
i) What is S_N1 reaction?
ii) Arrange the four isomeric bromo butanes in the increasing order of their reactivity towards SN1 reaction.
b) How will you prepare chlorobenzene from benzene diazonium chloride?
Answer:
a) i) S_N1 reaction is a unimolecular nucleophilic substitution reaction. It involves the substitution of a weaker nucleophile by a stronger one and follows first-order kinetics, i.e., the rate of reaction depends upon the concentration of only one reactant. It occurs in two steps. In step I, the polarised C – X bond undergoes slow cleavage to produce a carbocation.
ii) S_N1 reaction follows the order 10 alkyl halides < 2° alkyl halides < 3° alkyl halides.
CH₃CH₂CH₂CH₂Br < (CH₃)₂CHCH₂Br < CH₃CH₂CH(Br)CH₃ < (CH₃)₃CBr

b) By Sandmeyer reaction – When freshly prepared benzene diazonium chloride solution is mixed with cuprous chloride the diazonium group is replaced by -Cl to form chlorobenzene.

OR
By Gatterman reaction – When freshly prepared benzene diazonium chloride solution is treated with hydrochloric acid in presence of copper pow- der the diazonium group is replaced by -Cl to form chlorobenzene.

Question 10.
a) i) Write ‘Saytzeff rule’. (Say – 2014)
ii) The products A and B of the following reaction are two isomeric alkenes. Identify Aand B.

b) Identify the main product of the following reactions. Suggest whether the reaction is S_N1 or S_N2.

Answer:
a) i) The Saytzeff rule states that in dehydrohalogenation reactions, the preferred product is that alkene which has a greater number of alkyl groups attached to the doubly bonded carbon atoms.

b) i) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{Br} \stackrel{\mathrm{aqNaOH}}{\longrightarrow}\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{OH}\) This reaction follows S_N1 mechanism. Tertiary alkyl halides undergo S_N1 reaction very fast because of the high stability of 3° carbocations.

ii) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{Br} \longrightarrow \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{OH}\)
This reaction follows S_N1 mechanism. Ben- zylic halides show high reactivity towards the S_N1 reaction because the benzylic carbocation formed is stabilised through resonance.

Question 11.
a) Among the following which one is chlorine-containing insecticide? (March – 2015)
i) DOT
ii) Freon
iii) Phosgene
iv) lodoform

b) Halo arenes undergo Wurtz-Fittig reaction.
i) What is Wurtz-Fittig reaction?

Write the formulae of A and B in the above reaction.
Answer:
a) i) DDT
b) i) A mixture of an alkyl halide and aryl halide gives an alkylarene when treated with sodium in dry ether.

Question 12.
i) State Saytzeff Rule. (Say – 2015)
ii) Identify the major and minor products obtained by the reaction between 2-bromo butane and alcoholic KOH.
iii) Write the product obtained by the reaction between 2-bromo butane and aqueous KOH.
iv) 2-bromo butane exhibit optical isomerism. What is optical isomerism?
Answer:
i) The Saytzeff rule states that in dehydro-halogenation reactions, the preferred product is that alkene which has a greater number of alkyl groups attached to the doubly bonded carbon atoms.

iv) Optical isomerism is the phenomenon in which molecules having the same molecular, as well as structural formulae, differ in the direction of rotation of the plane of plane polarised light.

Question 13.
a) Aryl halides are less reactive in nucleophilic substitution reactions. (March – 2016)
i) Write any two reasons for less reactivity,
ii) Give one example for nucleophilic substitution reactions of aryl halides.
b) Write a method for the preparation of alkyl halides.
c) Which of the following is not a polyhalogen compound?
a) Chloroform
b) Freon
c) Carbon tetrachloride
d) Chlorobenzene
Answer:
i) 1. Resonance Effect- In aryl halides the electron pairs on halogen atom are in conjunction with π-electrons of the ring and different resonating structures are possible, e.g. resonance in chlorobenzene:

The C – X bond acquires a partial double bond character due to resonance. As a result, the bond cleavage in aryl halides is difficult than in alkyl halides and therefore, they are less reactive towards nucleophilic substitution reaction.

2. Difference in hybridisation of carbon atom in C-Xbond: In alkyl halides, the carbon atom attached to halogen is sp³ hybridised while in the case of aryl halides, the carbon atom attached to halogen is sp² hybridised.

The sp² hybridised carbon with a greater s-character is more electronegative and can hold the electron pair of C – X bond more tightly than sp³ hybridised carbon in alkyl halides with less s-character. Thus, the C – X bond length in aryl halides is less than that in alkyl halides. Since it is difficult to break a shorter bond than a longer bond, therefore, aryl halides are less reactive than alkyl halides towards nucleophilic substitution reaction.

3. Instability of phenyl cation – In the case of aryl halides, the phenyl cation formed as a result of self-ionization will not be stabilised by resonance and therefore, S_N1 mechanism is ruled out.

4. Because of the possible repulsion, it is less likely for the electron-rich nucleophile to approach electron-rich arenes. (any two reasons)

ii) Chlorobenzene when heated in aqueous sodium hydroxide solution at a temperature of 623 K and a pressure of 300 atmospheres followed by acidification gets converted to phenol.

b) Alkyl halides can be prepared by treating alcohols with concentrated halogen acid in presence of anhydrous zinc chloride as a catalyst.

c) d) Chlorobenzene

Question 14.
Haloalkanes and haloarenes are compounds containing halogen atom. They undergo many types of reactions. (Say – 2016)
a) Identify the product formed in the following reaction:

b) i) Chloroform is stored in closed, dark coloured bottles completely filled up to the neck. Give reason.
ii) Write any two differences between SN1 and SN² reactions.
Answer:
a) iii) CH₃ – CH₂ = CH₂
b) i) Chloroform is slowly oxidised by air in the presence of light to an extremely poisonous gas, carbonyl chloride, also known as phosgene. It is therefore stored in closed dark coloured bottles completely filled so that air is kept out.
\(2 \mathrm{CHCl}_{3}+\mathrm{O}_{2} \stackrel{\text { light }}{\longrightarrow} 2 \mathrm{COCl}_{2}+2 \mathrm{HCl}\)
ii) Refer March 2012, Question 1 .(i)

Question 15.
a) An ambident nucleophile is …………. (March – 2017)
i) Ammonia
ii) Ammonium ion
iii) Chloride ion
iv) Nitrite ion

b) Haloalkanes and Haloarenes are organohalogen compounds.
i) Suggest a method for the preparation of alkyl chloride.
ii) Aryl halides are less reactive towards Nucleophilic substitution reactions.
Give reasons.
Answer:
a) iv) Nitrite ion
b) i) Alkyl chlorides can be prepared by passing dry hydrogen chloride gas through a solution of alcohol or by heating a solution of alcohol in concentrated aqueous HCI in presence of anhydrous zinc chloride as a catalyst.
\(\mathrm{R}-\mathrm{OH}+\mathrm{HCl} \quad \stackrel{\mathrm{ZnCl}_{2}}{\longrightarrow} \mathrm{R}-\mathrm{Cl}+\mathrm{H}_{2} \mathrm{O}\)

Or, by the action of alcohols with PCl₃, PCI₅ or SOCI₂.

3R-OH + PCl₃ → 3R-CI + H₃PO₃
R-OH + PCl₅ → R-CI + POCl₃ + HCl
R-OH + SOCl₂ → R-CI + SO₂ + HCl

Or, chlorination of hydrocarbons in presence of light or heat.
\(\begin{array}{l}
\text { eg. } \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3} \frac{\mathrm{Cl}_{2} \text { NV ligtt }}{\text { or heat }} \longrightarrow \\
\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Cl}+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}(\mathrm{Cl}) \mathrm{CH}_{3}
\end{array}\)
Or, by the addition of hydrogen chloride to alkenes.
\(\text { eg. } \mathrm{CH}_{2}=\mathrm{CH}_{2}+\mathrm{HCl} \rightarrow \mathrm{CH}_{3}-\mathrm{CH}_{3}\) (Any one method)
ii) Refer March 2012 Question 1 (ii)

Question 16.
On kinetic consideration, nucleophilic substitution in aryl/alkyl halides may be S_N1 or S_N2 mechanisms. (Say – 2017)
a) Briefly explain S_N2 mechanism with an example.
b) In dehydrohalogenation of 2-Bromopentane why Pent-2-ene is major product and Pent-ene is minor product.
Answer:
a) S_N2 mechanism (bimolecular nucleophilic substitution)
1. Takes place in a single step
2. Through the formation of the intermediate transition state
3. Inversion of configuration occurs

b) Saytzeff Rule: In dehydrohalogenation reaction, the more substituted alkene is the major product.

Kerala State Board New Syllabus Plus Two Chemistry Chapter Wise Previous Questions and Answers Chapter 9 Coordination Compounds.

Kerala Plus Two Chemistry Chapter Wise Previous Questions Chapter 9 Coordination Compounds

Question 1.
[Cr(NH₃)₄CI₂] Br is a co-ordination compound. (March – 2010)
a) Identify the central metal ion of the above compound.
b) Name the ligands present in it.
c) What is its coordination number?
d) Write its IUPAC name.
e) Write the Ionization isomer of the above compound.
Answer:
a) Chromium (Cr)
b) Ammine (NH₃), Chloride (Cl- )
c) 6
d) Tetraamminedichloridochromium (lll)bromide
e) [Cr(NH₃)₄CIBr] Cl. By exchanging the ions inside and outside the coordination sphere.

Question 2.
When CuSO₄ is mixed with an excess of NH₃, a deep blue coloured solution is obtained. (Say – 2010)
a) Write the formula of the compound formed.
b) What is the IUPAC name of the compound?
c) What do you understand by the term coordination number and ligand in a coordination compound?
d) Give the oxidation number and coordination number of the central metal atom of the deep blue coloured compound.
Answer:
a) [Cu (NH₃)₄]SO₄
b) Tetraammine copper (ll) sulphate
c) It is the number of ligand atoms to which the metal is directly bonded. Ligands are neutral molecules or ions bounded to the central atom/ion in the coordination entity.
d) Oxidation number of Cu in [Cu(NH₃)₄]SO₄ is +2 Coordination number of Cu in [Cu(NH₃)₄]SO₄ is 4.

Question 3.
The geometry and magnetic properties of complexes can be explained by V.B. Theory. (March – 2011)

The octahedral complex [Co(NH₃]₆]³⁺ is diamagnetic while the octahedral complex [C0F₆]^3- is diamagnetic. Explain using VB. Theory,
Answer:
In the complex, [Co(NH₃)₆]³⁺ the cobalt ion is in +3 oxidation state and has the electronic configuration 3a⁶. It undergoes d²sp³ hybridisation as shown below:

Here, the six pairs of electrons, one from each NH₃ molecule, occupy the six hybrid orbitals. Thus, the complex has octahedral geometry and is diamagnetic be cause of the absence of an unpaired electron.

In [CoF₆]^3- also he cobalt ion is in +3 oxidation state and has the electronic configuration 3d⁶. Since F ion provides a weak ligand field one 4s, three 4p and two 4d orbitaIs hybridise to yield six sp³d² hybrid or bitaIs pointing towards the six ends of an octahedron. The six Fions then donate a pair of electrons to each of these vacant orbitals to have an octahedral geometry. The presence of unpaired electrons in 3d orbit ais makes the complex paramagnetic.

Question 4.
The central ion Ag⁺ with coordination number 2 forms a positive complexion with NH₃ ligand. Also Ag forms a negative complex with CN- ligand.
a) Writetheformula of above positive and negative complexions. Give the IUPAC name of each.
b) Give the denticity of NH₃ and CN- ligands.
c) Write the formula and name of a hexadentate ligand.
Answer:
a) [Ag(NH₃)₂]⁺ – Diamminesilver (l) ion [Ag(CN)₂] – Dicya noargentate(l) ion
b) Denhcìty of NH₃ is 1, since N isthe only donor atom. Densicity of CN is 2, since both C and N atoms can act as donor atoms.
c) (EDTA⁴) ion – Ethylenediaminetetraacetate ion.

Question 5.
Consider the coordination compound [Co(NH₃)₅SO₄] Br. (March – 2012)
a) Write the IUPAC name of the above coordination compound.
b) What is the primary valence and secondary valence of the central metal, cobalt, in the above coordination compound?
c) Which type of structural isomerism is exhibited by the above coordination compound?
Answer:
a) Pentaamminesulphatocobalt (lll) bromide
b) (Co (NH₃)₅ SO₄] Br Primary valency = I Secondary valency =6
c) Ionisation isomerism.

Question 6.
[Cr(NH₃)₅CO₃] Cl is a coordination compound. (Say – 2012)
i) Name the central metal ion of the above compound.
ii) What is its IUPAC name?
iii) Name the ligands present in the above compound.
iv) Whether the ligands present in the above compound are ambdentate ligands? Why?
Write the ionisation isomer of the above compound.
Answer:
i) Chromium (Cr)
ii) Pentamminecarbonatoch romium(lll) chloride
iii) Ammonia, Carbonate ion
iv) No. Both ammonia and carbonate ion are not ambidentate ligands because they have only one donor site to bind with the metal.
v) [Cr(NH₃)₅Cl] CO₃ Pentamminechloridochromium (lll) carbonate

Question 7.
The magnetic behaviour of a complex can be explained on the basis of the Valence Bond (VB) theory. (March – 2013)
a) [CO(NH₃)₆]³⁺ is a diamagnetic complex and [COF₆]^3- is a paramagnetic complex. Substantiate the above statement using VB theory.
b) Classify the above-mentioned complexes into inner orbital and outer orbital complexes.
Answer:
a) Refer (March 2011 Question No.1
b) Inner orbital complex → [Co (NH₃)₆]³⁺ since the innerd orbital (3d) is used for hybridisation in this complex.
Outer orbital complex → [CoF₆]^3- since, the outer d orbital (4d) is used for hybridisation in this complex.

Question 8.
Many theories have been put forth to explain the nature of bonding in coordination compounds. (Say – 2013)
a) On the basis of valence bond theory account for the diamagnetic behaviour of [Ni(CN)₄]^2-.
b) What is the shape of the above complex?
c) Arrange the following ligands in the increasing order of their field strengths, (as per the spectrochemical series) CI-, CO, H₂0, OH
Answer:
a) In the complex ion [Ni(CN)₄]^2- the central metal atom Ni is in +2 oxidation state and has the electronic configuration 3d8. Since CN – is a strong field ligand two unpaired electrons are forced to pair up against and give fourdsp2 hybrid orbitals. Each of the hybrid orbitals receive a pair of electrons from a cyanide ion to form [Ni(CN)₄]². Since there is no unpaired electron, the complexion is diamagnetic.

b) Since the hybridisation involved is dsp2 this complex has a square planar geometry.

c) Cl < OH < H₂O < CO

Question 9.
[CO(NH₃)₅SO₂]CI is an octahedral coordination compound. (March – 2014)
a) Write the IUPAC name of the above coordination compound.
b) Write the formula of the ionisation isomer of the above compound.
c) How do ‘d’ orbitals split in an octahedral crystal field?
d) Draw the diagram which indicates the splitting of ‘d’ orbitals in the tetrahedral field.
Answer:
a) Pentaamminesulphatocobalt(lli) chloride
b) [CO(NH₃)₅CI]SO₄

c) In an octahedral crystal field the ligands approach the central metal atom/ion along the coordinate axes. Thus, the d_{x² – y²} and d_z² orbitais which point towards the axes along the direction of the ligand will experience more repulsion and will be raised in energy; and the d_xy, d_yz and d_xz orbitais which are directed between the axes will be lowered in energy relative to the average energy in the spherical crystal filed. The crystal filed splitting in an octahedral crystal field yield three orbitaIs of lower energy (t_2g set) and two orbita Is of higher energy (e_g set). The energy separation is denoted by Δ₀. The energy of the two e_g orbitais will increase by (3/5)Δ₀, and that of the three t_2g will decrease by (2/5)Δ₀. The d-orbitai splitting in an octahedral crystal field is shown below:

d)

Question 10.
a) Valence Bond Theory (VBT) can explain the magnetic behaviour and shape of complexes. Using VBT explain the diamagnetism and square planar shape of [NKCN)₄]^2- (Say – 2014)
b) i) Suggest the shape of the following complexes Ni(CO)₄ and [CoF₆]^3-.
ii) The central ion Co³⁺ with coordination number 6 is bonded to the ligands NH₃ and Br to form a dipositive complexion. Write the formula or IUPAC name of the complexion.
Answer:
a) In the complex ion [Ni(CN)₄]² the central metal atom Ni is in +2 oxidation state and has the electronic configuration 3d⁸. Since CN – is a strong field ligand two unpaired electrons are forced to pair up against Hund’s rule and makes available a vacant 3d, one 4s and two 4p orbitals, which hybridise to give four dsp2 hybrid orbitals. Each of the hybrid orbitals receives a pair of electrons from a cyanide ion to form [Ni(CN)₄]^2-. Since there is no unpaired electron, the complexion is diamagnetic.

Since the hybridisation involved is dsp² this complex has a square planar geometry.

b) i) In [Ni(CO)₄] the central atom Ni undergoes sp³ hybridisation. Hence, it has a tetrahedral geometry. In [CoF₆]³ the central metal ion Co³⁺ is in the sp³d² hybridised state. Hence, it has an octahedral geometry,
ii) [Co(NH₃)₅Br]²⁺
OR
Pentaamminebromidocobalt(lll) ion

Question 11.
Coordination compounds contains central metal atom/ion and ligands.
a) Primary valency of central metal atom/ion in [Co(NH₃)₆] Cl₃ is
i) 3
ii) 6
iii) 4
iv) 9
b) i) What are the postulates of Werner’s theory?
ii) Write the IUPAC names of K₃[Fe(CN)₆], [Co(NH₃)₆] Cl₃.
a) i) 3
b) i)
1) In coordination compounds metals show two types of linkages (valencies) – primary and secondary.
2) The primary valences are normally ionisable and are satisfied by negative ions. ie. oxidation state.
3) The secondary valences are non – ionisable. These are satisfied by neutral molecules or negative ions. The secondary valence is equal to the coordination number and is fixed for a metal.
4) The ions/groups bound by the secondary linkages to the metal have characteristic spatial arrangements corresponding to different coordination numbers.
ii) K₃[Fe(CN)₆] – Potassium hexacyanoferrate(lll) [CO(NH₃)₆]CI₃ – Hexaamminecobalt(lll) chloride

Question 12.
a) Write the IUPAC name of the complex K₃[Cr(C₂O₄)₃]. (Say – 2015)
b) Draw the figure to show the splitting of ‘d’ orbitals in octahedral crystal field.
c) [Fe(H₂O)₆]^3- is strongly paramagnetic, whereas [Fe(CN₆)]^3- is weakly paramagnetic. Write the reason.
Answer:
a) Potassium trioxalatochromate(lll)

c) H₂O is a weaker ligand compared to CN. There is no pairing of electrons in the d-orbital of Fe in [Fe(H₂O)₆]³⁺. But there is the pairing of electrons in the d-orbital of Fe in [Fe(CN₆)]^3-.

Question. 13.
a) Write down the ionization isomer of [Co(NH₃), Cl] SO₄. (March – 2016)
b) Write the IUPAC name of the above compound.
c) [Ni(CO)₄] is diamagnetic while [Nid4]2 is paramagnetic though both are tetrahedral. Why?
Answer:
a) [Co(NH₃)₅SO₄]Cl
b) Pentaamminechlondocobalt(lll) sulphate
c) In [NICI₄]^2- nickel is in +2 oxidation state and the Ni² ion has the electronic configuration 3d⁶. It undergoes sp³ hybridisation.

Each CI ion donates a pair of electrons. The compound ¡s paramagnetic since it contains two unpaired electrons.

[Ni(CO₄)] has tetrahedral geometry but is diamagnetic since nickel is in zero oxidation state and contains no unpaired electron.

Question 14.
Consider the co-ordination compound [CoNH₃)₅Cl]Cl₂ (Say – 2016)
a) Write the IUPAC name of the above coordination compound.
b) i) What is the primary valency and secondary valency of the central metal ion in the above co-ordination compound?
ii) Write the name of isomerism exhibited by the complex [Pt(NH₃)₂Cl₂] Represent the possible isomers.
Answer:
a) Pentaamminechloiidocobalt(lll) chloride
b) i) Primary valency: +3 Secondary valency: 6
ii) Geometric isomerism or cis-trans isomerism

Question 15.
(Co(NH₃)₅SO₄]CI and [Co(NH)₅Cl]SO₄ are coordination compounds. (March – 2017)
a) Identify the isomerism shown by the above compounds.
b) Write the IUPAC names of the above compounds.
c) Identify the ligands ¡ri each of the above compounds.
Answer:
a) Ionisation isomerism
b) [Co(NH₃)₅SO₄]CI – Pentaamm in sulphate cobalt(III) chloride
[Co(NH₃)₅Cl]SO₄ – Pentaamminechloridocobalt(III) sulphate
c) In [Co(NH₃)₅SO₄]CI the ligands are NH₃ and SO₄.
In [Co(NH₃)₅Cl]SO₄ the ligands are NH₃ and Cl.

Question 16.
a) In which of the following, the central atom/ion is in zero oxidation state. (Say – 2017)
i) [Ni(CN)₄]^2-
ii) [NiCl₄]^2-
iii) [Ni(CO)₄]
iv) [Ni(NH₃)₅]²⁺

b) [Ni(CN)₄]^2- has square planar structure and it is diamagnetic.
i) On the basis of valence bond theory explain why [Ni(CN)₄)^2- exhibit these properties.
ii) Identify the ligand in the above-mentioned complex.
Answer:
a) iii) or [NiCo)₄]
b) i) dsp² hybridisation or No unpaired electrons or CN is a strong field ligand.
ii) CN- or cyanide ion

Kerala State Board New Syllabus Plus Two Chemistry Chapter Wise Previous Questions and Answers Chapter 8 The d and f Block Elements.

Kerala Plus Two Chemistry Chapter Wise Previous Questions Chapter 8 The d and f Block Elements

Question 1.
a) Transition elements are ‘d’ block elements. (March – 2010)
i) Write any four characteristic properties of transition elements.
b) Lanthanoids and actinoids are f – block elements.
i) What is the common oxidation state of Lanthanoids?
ii) Name the Lanthanoid with common oxidation state +4.
iii) It is difficult to separate Lanthanoids in the pure state. Explain.
Answer:
a)

• They form coloured compounds.
• They exhibit variable oxidation state.
• They form complex.
• They are good catalysts,

b)

• + 3
• Cerium (Ce)
• Due to lanthanide contraction size is approxi mately equal. So separation is difficult.

Question 2.
Transition metals are widely used as catalysts in industrial processes. (Say – 2010)
a) Name any two industrial processes in which transition elements are used as catalysts.
b) Transition metals exhibit catalytic properties. Why?
c) Why do the transition elements exhibit greater similarity in properties compared to main group elements along the period as well as down the group?
Answer:
a) Fe – Haber’s process forthe manufacture of NH₃ Ni – Hydrogenation of Oil for the manufacture of vanaspati ghee
b) Because of

• variable oxidation state of metals
• ability to form complexes

c) The outer electronic configuration remains almost same and hence they show horizontal similarity.

Question 3.
a) Atomic sizes increase as we come down a group, but in 4th group of the Periodic Table Zr, Hf have almost the same atomic sizes. Why? (March – 2011)
b) E° (standard electrode potential) values generally become less negative as we move across a transition series, but E° values of Ni²⁺/Ni and Zn²⁺/Zn values are exceptions. Justify.
Answer:
a) This is due to Lanthanide contraction. It is the phenomenon of regular decrease in atomic size across the lanthanoid series.
b) Due to the stability of completely filled d¹⁰ configuration of Zn²⁺ its Δ_iHΘ₂ is less. This is responsible for the high negative value of \(E_{z_{n}^{2+} / z_{n}}^{0}\) ( – 0.76 V).

Question 4.
Transition elements are d – block elements, with some exceptions. Usually they are paramagnetic. They show variable oxidation states. They and their compounds show the catalytic property. (Say – 2011)
a) Zn (Atomic number 30) is not a transition element, though it is a d – block element. Why?
b) Which is more paramagnetic Fe²⁺ or Fe³⁺? Why?
c) Why do transition elements show variable oxidation states?.
d) What is the reason for their catalytic property?
Answer:
a) A true transition element is one which has incompletely filled d – orbitals in its ground or in their common oxidation states. The zinc atom has completely filled d – orbitals.
b) Fe³⁺ is more paramagnetic.
The paramagnetic character of a transition metal ion depends on the number of unpaired d – electrons present in it. Fe³⁺ wich 3d⁵ configuration has 5 unpaired d – electrons while Fe²⁺ with 3d⁶ configuration has only 4 unpaired d – electrons.
c) In transition elements the energy difference between (n – 1)d and ns orbitals is very less. Hence, along with ns electrons (n – 1)d electrons can also take part in chemical reactions.
d) Due to their ability to adopt multiple oxidation states and to form complexes.

Question 1.
a) Potassium dichromate (K₂ Cr₂ O₇) is an important compound of chromium. Describe the method of preparation of potassium dichromate from chromite ore. (March – 2012)
b) The gradual decrease in the size of lanthanoid elements from lanthanum to lutetium is known as lanthanoid contraction. Write anyone consequence of lanthanoid contraction.
Answer:
a) K₂ Cr₂ O₇ is prepared from chromate one Fe Cr₂ O₄.
Step I: The powdered ore is heated with molten alkali in free access of air to form soluble sodium chromate.
4 Fe Cr₂ O₄ + 16 NaOH + 7O₂ → 8Na₂CrO₄ + 2Fe₂O₃ + 8H₂O

Step II: Sodium chromate (Na₂Cr O₄) is filtered and acidified with dil. H₂SO₄ to form sodium dichromate.
2Na₂CnO₄ + H₂SO₄ → Na₂Cn₂O₇ + Na₂SO₄ + H₂O

Step III: Na₂Cn₂O₇ solution is treated with KCI to form K₂Cr₂O₇.
Na₂Cn₂O₇ + 2 KCI → K₂Cn₂O₇ + 2NaCI

b) Consequences of lanthanoid contraction ane

• Difficulty in separation of lanthanoids due to similanity in chemical properties.
• The similarity in size of elements belonging to same group of second & third transition series.

Question 1.
Assume that you are going to present a seminar on transition elements. Prepare a seminar paper by stressing any four important properties of transition elements. (Say – 2012)
Answer:
The transition elements are the elements in groups 3 – 12 of the periodic table in which the d – orbitals are progressively filled.
1) Magnetic properties : Most of transition elements show paramagnetism due to the presence of unpaired electrons. The magnetic moment \((\mu) \mu=\sqrt{n(n+2)}\)
2) Formation of coloured ions : Transition elements form coloured compounds due to the presence of unpaired d – electrons, which can take part in d – d transition.
3) Formation of complex compounds: Transition metals form a large number of complex compounds. This is due to the comparatively smaller sizes of the metal ions, their high ionic charges and the availability of d – orbitals for bond formation, e.g., K₄[Fe(CN)₆], [Co(NH₃)₆]CI₃
4) Catalytic properties : Transition elements and their compounds act as good catalysts. This is attributed to their ability to adopt show multiple oxidation states because of and to form complexes due to the presence of partially filled d – orbitals, e.g., Finely divided Fe is used as a catalyst in Haber’s process.

Question 1.
Account for the following trends in atomic and ionic radii of transition metals. (March – 2013)
i) Ions of the same charge in a given series (3d, 4d or 5d) show progressive decrease in radii with icreasing atomic number.
ii) The atomic radii of elements in 4d series are more than that of corresponding elements in 3d series.
iii) The atomic radii of the corresponding elements in ‘4d’ series and ‘5d’ series are virtually the same.
Answer:
i) This is because each time a new electron enters a ‘d’ orbital the nuclear change increases by unity. The shielding effect of a ‘d’ electron is not that effective. Hence, the net electrostatic attraction between the nuclear charge and the outermost electron increases and the ionic radius decreases.

ii) The effect of addition of new shells in 4d series overtakes the effect of increase in nuclear charge.
Thus, electrostatic attraction between nucleus and valence electron decreases and hence atomic size increases.

iii) This phenomenon is due to the intervention of the 4f – orbitals which must be filled before the 5d series of elements begins. The filling of 4f before 5d – orbitals results in a regular decrease in atomic radii called Lanthanoid Contraction due to imperfect shielding of intervening 4f – orbital electrons which compensate for the expected increase in atomic size on moving down the group, with increasing atomic number. The net result of Lanthanoid Contraction is that the second and the third d series exhibit similar radii.

Question 1.
i) d – Block elements belong to group 3 – 12 in the periodic table, in which the d orbitals are progressively filled. (Say – 2013)
a) What is their common oxidation state?
b) Name two important compounds of transition elements.
c) Transition elements form a large number of complex compounds, why?
ii) What is mischmetal?
Answer:

1. a) (n – 1)d^1-10 ns^1-2
   b) Potassium dichromate, Potassium permanganate. KMnO₂
   c) This is due to the following two factors: Cations of transition metals are very
   • small in size
   • high effective nuclear
   • have vacant d – orbitals
2. ‘Misch metal’ is an alloy which consists of a lanthanoid metal (- 95%) and iron (- 5%) and traces of S, C, Ca and Al. It is used in Mg-based alloy to produce bullets, shell and lighter flint.

Question 1.
Potassium dichromate is an orange coloured crystal and is an important compound used as an oxidant in many reactions. (March – 2014)
a) How do you prepare K₂Cr₂O₇ from chromite ore?
b) How will you account for the colour of potassium dichromate crystals?
Answer:
a) The chromite ore is fused with sodium carbonate in free access of airto get sodium chromate.
\(\begin{aligned}
4 \mathrm{FeCr}_{2} \mathrm{O}_{4}+8 \mathrm{Na}_{2} \mathrm{CO}_{3}+& 7 \mathrm{O}_{2} \longrightarrow \\
& 8 \mathrm{Na}_{2} \mathrm{CrO}_{4}+2 \mathrm{Fe}_{2} \mathrm{O}_{3}+8 \mathrm{CO}_{2}
\end{aligned}\)
The yellow solution of sodium chromate is filtered and acidified with sulphuric acid to get sodium dichromate.
\(2 \mathrm{Na}_{2} \mathrm{CrO}_{4}+2 \mathrm{H}^{+} \rightarrow \mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{4}+2 \mathrm{Na}^{+}+\mathrm{H}_{2} \mathrm{O}\)
The sodium dichromate solution is treated with potassium chloride to get potassium dichromate.
\(\mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}+2 \mathrm{KCl} \rightarrow \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}+2 \mathrm{NaCl}\)
Orange crystals of potassium dichromate crystallise out.

b) This is due to charge transfer spectra i.e., Chromium being a transition element has vacant d – orbitals.

Question 1.
Potassium permanganate and potassium dichromate are two transition metal compounds. (Say – 2014)
a) Write any four characteristics of transition metals.
b) i) Write any two uses of potassium permanganate.
ii) Draw the structure dichromate ion.
Answer:
a)

• Variable oxidation states.
• Formation of coloured ions in aqueous solution.
• Formation of complex compounds.
• Formation of interstitial compounds,

b) i)

• Lab reagent
• For bleaching of wool, cotton, silk and other textile fibres.

ii) The dichromate ion consists of 2 tetrahedra sharing one comer with Cr – O – Crbond angle of 126°.

Question 1.
Fourteen elements following Lanthanum are called Lanthanoids: (March – 2015)
a) What is Lanthanoid contraction? Give reason for it.
b) KMnO₄ is a purple coloured crystal and it acts as an oxidant. How will you prepare KMnO₄ from MnO₂?
Answer:
a) The overall decrease in atomic and ionic radii from lanthanum to lutetium is called lanthanoid contraction.

Lanthanoid contraction is caused by the imperfect shielding of one 4f electron by another in the same set of orbitals. The shielding of one 4f electron by another is less than that of one d electron by another. Hence, as the nuclear charge increases along the lanthanoid series, there is fairly regular derease in the size of the entire 4fn orbitals.

b) MnO₂ is fused with an alkali metal hydroxide and an oxidising agent like KNO₃ to get dark green potassium manganate, K₂MnO₄.
2MnO₂ + 4KOH + O₂ → K₂MnO₄ + 2H₂O

Potassium manganate disproportionates in a neutral or acidic solution to give potassium permanganate.
3MnO^2-₄ + 4H⁺ → 2MnO^4- + MnO₂ + 2H₂O

Question 1.
a) Which of the following oxidation state is common for lanthanids? (Say – 2015)
i) +2
ii) +3
iii) +4
iv) +5
b) Drawthe structures of chromate and dichromate ions.
c) Zirconium (Zr) belongs to ‘4d’ and Hafnium (Hf) belongs to ‘5d’ transition series. It is difficult to separate them. Explain.
Answer:
a) ii) +3


c) It is a consequence of lanthanoid contraction, a cumulative effect of the contraction of atomic radii of the lanthanoid series caused by the imperfect shielding of one electron by another in the 4f sub-shell.

Question 1.
a) Which of the following oxidation state is not shown by Maganese? (March – 2016)
a) +1
b) +2
c) +4
d ) +7

b) Represent the structure of dichromate ion.
c) Potassium permanganate (KMnO₄) is a strong oxidizing agent. Write any two oxidizing reactions of KMnO₄.
Answer:
a) +1

c) 1. Permanganate ion oxidises iodide to iodine in acid medium:
\(\left.10\right|^{+}+2 \mathrm{MnO}_{4}+16 \mathrm{H}^{+} \rightarrow 2 \mathrm{Mn}^{2+}+8 \mathrm{H}_{2} \mathrm{O}+5 \mathrm{I}_{2}\)

2. Permanganate ion oxidises Fe2+ ion (green) to Fe³⁺ ion (yellow) in acid medium:
\(5 \mathrm{Fe}^{2+}+\mathrm{MnO}_{4}+8 \mathrm{H}^{+} \rightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_{2} \mathrm{O}+5 \mathrm{Fe}^{3+}\)

Question 1.
Transition elements are d-block elements and inner transition elements are f-block elements. (Say – 2016)
i) Write any two properties of transition elements.
ii) Name a transition metal compound and write one use of it.
iii) What is Lanthanoid Contraction?
iv) Write any two consequences of Lanthanoid Contraction.
Answer:
i) Transition elements are metals, have high melting points and high enthalpy of atomisation, exhibit variable oxidation states, show paramagnetism, form coloured compounds, form complex compounds, show catalytic properties, form interstitial compounds, form alloys etc. (any two properties)
ii) Fe to makes steal
iii) The overall decrease in atomic and ionic radii from lanthanum to lutetium, caused by the poor shielding of one 4f electron by another is called lanthanoid contraction.
iv)
1. The atomic radii of second row of transition elements are almost similar to those of third row of transition elements.
2. The almost identical radii of Zr (160 pm) and Hf (159 pm).
3. All the lanthanoids have quite similar properties and due to this they are difficult to be separated.
4. The basic strength of hydroxides decreases from La(OH)₃ to Lu(OH)₃ due to decrease in size of M³⁺ ions and consequent increase in the covalent character of M – OH bond.

(any two consequences required)

Question 1.
a) Transition elements are ‘d’ block elements. (March – 2017)
i) Write any four characteristic properties of transition elements.
ii) Cr²⁺ and Mn³⁺ have d⁴ configuration. But Cr²⁺ is reducing and Mn³⁺ is oxidising. Why?
b) Which of the following is not a lanthanoid element?
i) Cerium
ii) Europium
iii) Lutetium
iv) Thorium
Answer:
a) i) Transition elements are metals, have high melting points and high enthalpy of atomisation, exhibit variable oxidation states, show paramagnetism, form coloured compounds, form complex compounds, show catalytic properties, form interstitial compounds, form alloys etc. (any four properties)

ii) For Cr²⁺ to Cr³⁺ configuration changes from d⁴ to d³ For Mn³⁺ to Mn²⁺ d⁵ configuration results in extra stability due to half-filled configuration,

b) iv) Thorium

Kerala State Board New Syllabus Plus Two Chemistry Chapter Wise Previous Questions and Answers Chapter 7 The p Block Elements.

Kerala Plus Two Chemistry Chapter Wise Previous Questions Chapter 7 The p Block Elements

Question 1.
Elements in the groups 13 to 18 in the Periodic table constitute the ‘p’ block elements. (March – 2010)
i) Name the most important oxo acid of Nitrogen.
ii) How will you prepare the above oxo acid on large scale?
iii) In general, noble gases are least reactive. Why?
Answer:
i) Nitric acid (HNO₃)
ii) HNO₃ can be prepared on a large scale by Gstwald process. It involves three steps.
a) NH₃ is oxidised catalytically by atmospheric oxygen.

\(4 \mathrm{NH}_{3 \mathrm{gg}}+5 \mathrm{O}_{2(\mathrm{~g})} \frac{\mathrm{Pt} / \mathrm{Rh} \text { gangue catalyst }}{500 \mathrm{~K}, 9 \mathrm{bar}}, 4 \mathrm{NO}_{(\mathrm{g})}+6 \mathrm{H}_{2} \mathrm{O}_{(9)}\)

b) Nitric oxide thus formed combines with oxy gen giving NO₂.

\(2 \mathrm{NO}_{(g)}+\mathrm{O}_{2(g)} \rightleftharpoons 2 \mathrm{NO}_{2(g)}\)

C) Nitrogen dioxide so formed dissolves in water to give HNO₃.

\(3 \mathrm{NO}_{2(g)}+\mathrm{H}_{2} \mathrm{O}_{(1)} \rightarrow 2 \mathrm{HNO}_{3(\mathrm{aq})}+\mathrm{NO}_{(9)}\)

iii)

• The noble gases except helium (1s²) have completely filled ns2np6 electronic configuration in their valence shell.
• They are octet completed and stable so they are inert.

Question 2.
After a discussion about the structures of hydrides of the group – 15 elements, Neethu wrote the order of bond angles as NH₃ < PH₃ < AsH₃ (Say – 2010)
a) is this the correct order?
b) Justify your answer.
c) Give the hybridization and shape of these hydrides.
d) Also arrange the above hydrides in the increasing order of their thermal stability. Justify your answer.
Answer:
a) No.
b) From top to bottom in the group the bond angle of group 15 hydrides decreases. As the electronegativity of the central atom decreases on moving down the group, the bond pair-bond pair repulsion decreases. Hence the bond angle decreases ¡n the order NH₃ > PH₃ > AsH₃.

C) In all hydrides the central atom is sp³ hybridised. The molecules assume trigonal pyramidal geometry with a lone pair on the central atom.

d) Thermal stability of the group 15 hydrides increases BiH₃ < AsH₃ < PH₃ < NH₃ This is due to the fact that moving up the group the EH bond dissociation enthalpy (‘E’ is a group 15 element) increases due to a decrease in size of the central atom and the molecules will decompose only at higher temperatures.

Question 3.
The Discovery of Haber’s process for the manufacture of Ammonia is considered to be one of the principal discoveries of the twentieth century. (March – 2011)
a) Which is the promoter used ¡ri the earlier process when Iron was used as a catalyst?
b) What is the temperature condition for the maximum yield of Ammonia? Justify.
c) Explain how can you convert NH₃ to HNO₃, on a large scale commercially.
Answer:
a) Molybdenum (Mo)

b) By Le – Chaltiers principle, the rate of an exothermic reaction increases with a decrease in temperature of 500°C to get a good yield of the product.

C) NH₃ is converted to HNO₃ commercially by Ostwald’s process. It involves three steps.
i) NH₃ is oxidised catalytically by atmospheric oxygen.

ii) Nitric oxide thus formed combines with oxygen giving NO₂.
\(2 \mathrm{NO}_{(g)}+\mathrm{O}_{2(g)} \rightleftharpoons 2 \mathrm{NO}_{2(g)}\)

iii) Nitrogen dioxide so formed dissolves in water to give HNO₃.
\(3 \mathrm{NO}_{2(g)}+\mathrm{H}_{2} \mathrm{O}_{(1)} \rightarrow 2 \mathrm{HNO}_{3(\mathrm{aq})}+\mathrm{NO}_{(g)}\)

Question 4.
The phosphorus of group 15 and sulphur of group 16 are two industrially important ‘p’ block elements. Their compounds are also industrially important. (Say – 2011)
a) \(4 \mathrm{H}_{3} \mathrm{PO}_{3} \stackrel{\text { heat }}{\longrightarrow} 3 \mathrm{H}_{3} \mathrm{PO}_{4}+\mathrm{PH}_{3}\) show that this is a disproportionation reaction.
b) PCl₃ fumes in moisture. Give reason.
c) Sulphuric acid can be manufactured from sulphur using V₂O₅ as a catalyst.
i) Give the name of the method.
ii) Outline the principle.
Answer:
a) Disproportionation reactions are a special type of redox reaction in which an element in one oxidation state is simultaneously oxidised and reduced. In phosphorous acid (H₃PO₃) phosphorus is in the intermediate oxidation state of +3. It is increased to +4 (oxidation) in phosphoric acid (H₃PO₄) and decreased to – 3 (reduction) in phosphine (PH₃).

b) PCI₃ undergoes hydrolysis in presence of moisture giving fumes of HCI. PCI₃ + 3H₂O → H₃PO₃ + 3HCI

c) i) Contact Process

Question 5.
a) Important allotropic forms of phosphorus are white phosphorus, red phosphorus and black phosphorus. Among these which allotropic form is more reactive? Why? (March – 2012)
b) In the manufacture of sulphuric acid (H₂SO₄) the final product obtained is oleum.
i) What is Oleum?
ii) Write chemical equation for the conversion of oleum to sulphuric acid.
c) Name the halogen which forms only one oxoacid and also write the formula of the oxo acid of that halogen.
d) Which element among inert gases form a maximum number of compounds? Write the formula of one of the compounds formed by the element.
Answer:
a) White Qhosfhorus It consists of discrete tetrahedral P4 molecules.
b) i) Pyrosuiphuncacid
ii) H₂S₂O₇ + H₂O → 2H₂SO₄
c) Flounne or HOF or hypoflourous acid
d) Compounds – Xe F₂, XeF₄ Xe O₃, Xe OF₂ etc.,

Question 6.
i) What are the products obtained when copper reacts with concentrated nitric acid? (Say – 2012)
ii) Name two important xenon fluorides.
iii) Interhalogen compounds are compounds formed by the combination of different halogen atoms. Which are more reactive, halogens or interhalogen compounds? Give reason.
Answer:
i) Copper reacts with concentrated nitric acid to give copper nitrate and nitrogen dioxide.
Cu + 4HNO_3(conc) + Cu(NO₃)₂ + 2NO₂ + 2H₂O
ii) Two important xenon fluorides are XeF₂ and XeF₄.
iii) Interhalogen compounds are more reactive than halogens (except fluorine). Because, the bond between different halogen atoms (X – X’) in interhalogen compounds is weakerthan the bond between similar halogen atoms. Due to the difference in size and electronegativity. The F – F bond is the weakest due to interelectronic repulsion.

Question 7.
a) Nitrogen forms a number of oxides in the different oxidation stats. Write the names and structural formulae of any four oxides of nitrogen. (March – 2013)
b) Boiling point of H₂O (373 K) is very much higher than that of H₂S (213k). Give reason.
c) Suggest a method for the quantitative estimation of ozone (O₃).
Answer:

b) Molecules of water are highly associated through hydrogen bonding resulting in its high boiling point. Hydrogen bonding is not possible ¡n H₂S.
c) When O₃ reacts with an excess of Kl solution buffered with a borate buffer (pH = 9.2) 12 is liberated, which can be titrated against a standard solution of sodium thiosulphate. 2Kl+ H₂O + O₃ → O₂ + KOH + l₂

Question 8.
a) Name the products obtained when the copper reacts with concentrated nitric acid. (Say – 2013)
b) Write down the chemical reaction between concentrated nitric acid and aluminium.
c) What is the basicity of H₃PO₃?
d) How do you account for the basicity of H₃PO₃?
e) Write down the three steps involved in the manufacture of sulphuric acid by the Contact Process.
f) Write any two important uses of noble gas elements.
Answer:
a) Copper nitrate, Nitrogen dioxide and Water. [Cu + 4HNO_3(conc) → Cu(NO₃) + 2NO₂ + 2H₂O]
b) Aluminium does not dissolve in concentrated nitric acid because it is rendered passive due to the formation of a thin protective layer of metal oxide on the surface of the metal which cuts off the further action.
c) T’ do.
d) The basicity of oxo acids of phosphors is deter mined by the number of P – OH bonds, because only those H atoms which are attached with oxygen ¡n P – OH form are ionisable and cause the basicity. H₃PO₃ has two P – OH bonds.

e) i) Preparation of suiphurdioxide by burning sulphur. S(s) + O₂(g) → SO₂(g)
ii) Oxidation of sulphur dioxide to suphurtrioxide catalytically with atmospheric oxygen.

iii) Preparation of oleum by absorbing sulphur trioxide in sulphuric acid. It is diluted with enough water to get sulphuric acid of desired concentration.
\(\begin{array}{l}
\mathrm{SO}_{3}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{I}) \rightarrow \mathrm{H}_{2} \mathrm{~S}_{2} \mathrm{O}_{7}(\mathrm{I}) \\
\mathrm{H}_{2} \mathrm{~S}_{2} \mathrm{O}_{7}(\mathrm{I})+\mathrm{H}_{2} \mathrm{O}(\mathrm{I}) \rightarrow \mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})
\end{array}\)

Orthophosphorous acid Neon is used in discharge tubes and fluorescent bulbs for advertisement display purposes. Argon is used to provide an inert atmosphere in high-temperature metallurgical processes.

Question 9.
Compounds of nitrogen, phosphorus and sulphur such as ammonia, phosphoric acid and sulphuric acid are used in the fertilizer industry. (March – 2014)
a) Describe Haber process for the manufacture of ammonia.
b) Write the chemical equation forthe preparation of phosphoric acid (H₃PO₄) from ortho phosphorous acid (H₃PO₃).
c) Describe contact process for the manufacture of sulphuric acid.
Answer:
a) On a large scale, ammonia is manufactured by Haber’s process. In this process nitrogen gas reacts with hydrogen gas to form ammonia gas as per the reaction:
\(\mathrm{N}_{2(g)}+3 \mathrm{H}_{2(g)}=2 \mathrm{NH}_{3(g)} ; \Delta_{f} \mathrm{H}^{\circ}=-46.1 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

According to Le Chatelier’s principle, high pressure favours the formation of ammonia. The optimum conditions for the production of ammonia are a pressure of 200 x 10⁵ Pa (about 200 atm), a temperature of 700 K and the use of catalysts such as iron oxide with small amounts of K₂O and Al₂O₃ to increase the rate of attaintment of equilibrium. The flow chart for the production of ammonia is shown below:

b) Orthophosphorous acid on heating disproportionates to give orthophosphoric acid and phosphine.
4H₃PO₃ → 3H₃PO₄ + PH₃
c) i) Bunning of sulphur or sulphide ores in air to generate SO₂. S_(s) + O_2(g) → SO_2(g))
ii) Conversion of SO₂ to SO₃ by the reaction with oxygen in the presence of V₂O₅ catalyst.
\(\begin{array}{l}
2 \mathrm{SO}_{2(\mathrm{~g}}+\mathrm{O}_{2(\mathrm{~g})} \stackrel{\mathrm{V}_{2} \mathrm{O}_{3}}{\longrightarrow} 2 \mathrm{SO}_{3(\mathrm{~g})} \Delta_{\mathrm{r}} \mathrm{H}^{\circ} \\
=-196.6 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{array}\)
A pressure of 2 bar and a temperature of 720 K are applied.

iii) Absorption of SO₃ gas in H₂SO₄ to give oleum (H₂S₂O₇) SO₃ + H₂SO₄ → H₂S₂O₇ Dilution of oleum with water gives H2S04 of the desired concentration. H₂S₂O₈ + H₂O → 2H₂SO₄

Question 10.
Ammonia and Nitric acid are two industrially mportant compounds. (Say – 2014)
a) Write any two uses of ammonia.
b) Complete the following equations. (Balancing is not required)
i) NH₃ + O₂ > \(\frac{\mathrm{Pt}}{500 \mathrm{~K}, 9 \mathrm{ber}}]\)
ii) Cu + Con. HNO₃ →
iii) Zn + dil. HNO₃ →
iv) NH₃ + excess Cl₂ →

OR

a) Phosphorus forms a number of oxoacids. Write the name or formulae of any two dibasic oxoacids of phosphorus.
b) Account for the following:
i) PCl₃ fumes in moist air.
ii) Nitrogen does not form a pentahalide.
iii) Boiling point of PH₃ is less than that of NH₃
iv) NO₂ undergone dimerisation.
Answer:
a)

• To produce various nitro geneous fertilisers.
• In the manufacture of some inorganic nitrogen compounds like nitric acid.

OR

a) i) Orthophosphorous acid (H₃ PO₃)
ii) Pyrophosphorous acid (H₄P₂O₅)
b) i) PCI₃ hydrolyses in the presence of moisture giving fumes of HCI. PCl₃ + 3H₂O → H₃PO₃ + 3HCI
ii) It does not have ‘d’ orbitais to expand its covalence beyond four. That is why it does not form pentahalide.
iii) Unlike NH₃, PH₃ molecules are not associated through hydrogen bonding in liquid state. That is why the boiling point of PH₃ s lower than that of NH₃.
iv) NO₂ contains odd number of valence electrons. It behaves as a typical odd electron molecule. On dimerisation, it is converted to stable N₂O₄ molecule with even number of electrons.

Question 11.
Some elements in p – block shows allotropy. (March – 2015)
a) What are the allotropic forms of sulphur?
b) i) How will you manufacture Sulphuric Acid by contact process?
ii) What are interhalogen compounds?
Answer:
a) Rhombic sulphur (α – sulphur) and Monoclinic sulphur (β – sulphur)
b) i) The manufacture of sulphuric acid by contact process involves three steps:
1) Burining of sulphur or sulphide ores in air to generate SO₂. S(s) + O_2(g) → SO_2(g)
2) Conversion of SO₂ to SO₃ by the reaction with ozygen in the presence of V₂O₅ catalyst.
\(2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \frac{\mathrm{V}_{2} \mathrm{O}_{5}}{2 \mathrm{~S} \mathrm{O}_{3}(\mathrm{~g}) \mathrm{\Delta}_{\mathrm{r}} \mathrm{H}^{\circ}}=-196.6 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

This reaction is exothermic, reversible and the forward reaction leads to a decrease in volume. Therefore, low temperature and high pressure are the favaourable conditions for maximum yield. In practice, a pressure of 2 bar and a temperature of 720 K are applied.

3) Absorption of SO₃ gas in H₂SO₄ to give oleum (H₂S₂O₇) SO₃ + H₂SO₄ → H₂S₂O₇ Dilution of oleum with water gives H2S04 of the desired concentration. H₂S₂O₈ + H₂O → 2H₂SO₄

The flow diagram for the manufacture of sulphuric acid by Contact Process is

ii) Compounds formed by the reaction between two different halogens are called interhalogen compounds. They can be assigned general compositions as XX’, XX₃’, XX₅ and XX₇’ where X is halogen of larger size and X’ of smaller size and X is more electropositive than X’.

Question 12.
a) Name two oxoacids of Sulphur.
b) i) How will you manufacture ammonia by Haber process?
ii) Write any two uses of inert gases.
Answer:
a) Sulphurous acid (H₂S₂O₅), Sulphuric acid (H₂SO₄), Peroxodisuiphuric acid (H₂S₂O₅), Pyrosulphunc acid or Oleum (H₂S₂O₇) – Any two.
b) i) Ammonia is manufactured by Haber’s process. In this process nitrogen gas reacts with hydrogen in the ratio 1:3 to form ammonia: \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g}) ; \Delta_{f} \mathrm{H}^{\odot}=-46.1 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
High pressure favours the formation of ammonia 200 atm, a temperature of 700 K and the use of catalysts such as iron oxide.

ii) Helium is used for filling balloons for meteorological observations, Neon is used in discharge tubes and fluorescent bulbs. Argon is used to provide an inert atmosphere in high-temperature metallurgical processes and for filling electric bulbs, Xenon and Krypton are used in light bulbs designed for special purposes. (Any two)

Question 13.
a) What are interhalogen compounds? Write any two examples. (Say – 2015)
b) Write a method of preparation of phosphine from white phosphorus.
c) Write the name or formula of oxo acid of chlorine, in which chlorine possess oxidation number +7. Draw the structure of XeO₃ and XeF₂.
Answer:
a)These are compounds formed by the reaction of two different halogens. They can be assigned general compositions as XX’, XX, XX and XX where X is halogen of larger size and X’ of smaller size and X is more electropositive than X’. e.g. dF, dF₃, BrF₅, IF₇ (any two)

b) Phosphifle is prepared by heating white phosphorus with concentrated NaOH solution in an inert atmosphere of CO₂.
P₄ + 3NaOH + 3H₂O → 4 PH₃ + 3NaH₂PO₂

c) Perchloric acid or Chionc (VII) acid (HOCIO₃)

d)

Question 14.
a) Account for the following: (March – 2O16)
i) NH₃ acts as a Lewis base.
ii) PCI₃ fumes ¡n moist air.
iii) Fluorine shows only – 1 oxidation state.

b) i) SuggestanytwofluoridesofXenon.
ii) Write a method to prepare any one of the above mentioned Xenon fluorides.
OR
a) Account for the following:
i) H₂O is a liquid while H₂S is a gas.
ii) Noble gases have very low boiling points.
iii) NO₂ dimerises to N₂O₄.
b) i) What are interhalogen compounds?
ii) Suggest any two examples of interhalogen compounds.
Answer:
a) i) Nitrogen atom in NH₃ has one lone pair of electrons which is available for donation. There fore, it acts as a Lewis base.
ii) PCI₃ hydrolyses in the presence of moisture giving fumes of HCI. PCI₃ + 3H₂O → H₃PO₃ + 3HCI
iii) Fluorine is the most electronegative element and cannot exhibit any positive oxidation state. Fluorine atom has no d orbitals in its valence shell and therefore cannot expand its octet.

b) i) Xenon ditluoride, XeF₂
Xenon tetrafluoride, XeF₄
Xenon hexafluonde, XeF₆ (any two)
ii) XeF₂ is prepared by treating Xe with excess fluorine at 673 Kandl bar.
\(\mathrm{Xe}(\mathrm{g})+\mathrm{F}_{2}(\mathrm{~g}) \longrightarrow 673 \mathrm{~K}, 1 \text { bar } \quad>\mathrm{XeF}_{2}(\mathrm{~s})\)
Or, XeF₄ is prepared by treating Xe with excess fluorine in 1: 5 ratio at 873 K and 7 bar.
\(\mathrm{Xe}(\mathrm{g})+2 \mathrm{~F}_{2}(\mathrm{~g}) \longrightarrow \mathrm{B} 73 \mathrm{~K}, 7 \mathrm{bar} \quad \rightarrow \mathrm{XeF}_{4}(\mathrm{~s})\)
Or, XeF₆ is prepared by treating Xe with excess fluorine in 1 :20 ratio at 573 K and 60 – 70 bar.
\(\mathrm{Xe}(\mathrm{g})+3 \mathrm{~F}_{2}(\mathrm{~g}) \longrightarrow 573 \mathrm{~K}, 60-70 \mathrm{bar} \quad \longrightarrow \mathrm{XeF}_{6}(\mathrm{~s})\)

OR

a) i) Due to small size and high electronegativity of oxygen it is capable of forming hydrogen bond. Thus, water molecules can associate through intermolecular hydrogen bonds and hence ¡t exists as a liquid.

Due to big large and low electronegativity of sulphur t is not capable of forming hydrogen bond. So hydrogen sulphide molecules cannot associate through intermolecular bonds and hence it exists as a gas.

ii) Noble gases being monoatomic have no interatomic forces except weak dispersion forces and therefore, they are liquefied at very low temperatures. Hence, they have low boiling points.

iii) NO₂ contains odd numberof valence elecrons. It behaves as a typical odd electron molecule. On dimensation, it is converted to stable N₂O₄ molecule with even number of electrons.
b) i) These are compounds formed by the reaction between two different halogens.
ii) dF, BrF, IF, BrCI, ICI, dF₃, BrF₃, IF₃, ICI₃, IF₅,
BrF₅, dF₅, IF₇ (any two)

Question 15.
Nitrogen shows different oxidation states in different oxides. (Say – 2016)
a) In which of the fof lowing oxides, nitrogen is in + 4 oxidation state?
a) NO
ii) N₂O
iii) N₂O₃
iv) NO₂
b) Prepare a short write upon Nftric acid highlighting its structure, manufacture and any two properties.
OR
Phosphorous forms oxoacids
a) In which of the following phosphorous is in + 1 oxidation state?
i) H₃PO₂
ii) H₃PO₃
iii) H₄P₂O₇
iv) H₃PO₄
b) Prepare a short write up on Ammonia highlighting its structure, manufacture and properties.
Answer:
a) iv) NO₂
b) Nitric acid is the most important oxoacid of nitrogen. HNO₃ exists as planar molecule.

Manufacture of nitric acid: On a large scale, nitric acid is prepared mainly by Ostwald’s process. This method is based upon catalytic oxidation of NH₃ by atmospheric oxygen.

4NH₃(g) + \(5 \mathrm{O}_{2}(\mathrm{~g}) \frac{\text { PUR } \text { guage catalyst }}{500 \mathrm{~K}, \text { bar }}\) 4NO_(g) + 6H₂O_(g)

Nitric oxide thus formed combines with oxygen giving NO₂.

2NO_(g) + O_2(g) \(\rightleftharpoons\) 2NO_2(g)

Nitrogen dioxide so formed, dissolves in water to give HNO₂.

3NO_2(g) + H₂O_(l) → 2HNO_3(aq) + NO_(g)

Properties of nitric acid: It is a colourless liquid. In aqueous solution nitric acid behaves as a strong acid giving hydronium and nitrate ions.

HNO_2(g) + H₂O₍₁₎ → H₃O_+(aq) + NO_3(aq)

Concentrated nitric acid is a strong oxidising agent and attacks most metals except noble metals such as gold and platinum.

OR

a) i) H₃PO₂
b) Structure of ammonia: The ammonia molecule is trigonal pyramidal with the nitrogen atom at the apex. It has three bond pairs and one lone pair of electrons.

Manufacture of ammonia: On a large scale ammonia is manufactured by Haber’s process.

\(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g}) ;=-46.1 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

High pressure would favour the formation of ammonia. (about 200 atm), a temperature of —700 K and the use of catalyst such as iron oxide with small amounts of K₂O and Al₂O₃.

Ammonia gas is highly soluble in water. Its aqueous solution is weakly basic due to the formation of OH ions.

\(\mathrm{NH}_{3(\mathrm{~g})}+\mathrm{H}_{2} \mathrm{O}_{(0)} \rightleftharpoons \mathrm{NH}_{4^{+}(\mathrm{aq})}+\mathrm{OH}_{-(\mathrm{aq})}\)

It forms salts with acids. It precipitates the hydroxides of many metals from their salt solutions. The presence of a lone pair of electrons on the nitrogen atom of the ammonia molecule makes it a Lewis base.

Question 16.
Nitrogen forms a number of oxides and oxoacids. (March – 2017)
a) Which of the following is a neutral oxide of Nitrogen.
i) N₂O
ii) N₂O₅
iii) NO₂
iv) N₂O₄

b) Prepare a short write – up on Nitric acid high lighting its laboratory preparation, chemical properties and uses.

OR

Phosphorous forms a number of compounds.

a) The gas liberated when calcium phosphide is treated with die. HCl is
i) Cl
ii) H₂
iii) PH₃
iv) All the above

b) Prepare a short write up on PCl₃ and PCI₅ highlighting the preparation and chemical properties of PCl₃ and structure of PCl₅.
Answer:
a) i) N₂O

b) Laboratory preparation: In the laboratory, nitric acid is prepared by heating KNO₃ or NaNO₃ and concentrated H₂SO₄ in a glass retort.

NaNO₃ + H₂SO₄ → NaHSO₄ + HNO₃

Uses: in the manufacture of ammonium nitrate for fertilisers and other nitrates for use in explosives and pyrotechnics; for the preparation of nitroglycerin, trinitrotoluene and other organic nitro compounds; in the pickling of stainless steel etching of metals and oxidiser in rocket fuels.

OR

a) iii) PH₃
b) Preparation of PCI₃: By passing dry chlorine over heated shite phosphorus.

P₄ + 6Cl₂ → 4PCl₃

Or, by the action of thionyl chloride with white phosphorus.

P₄ + 8SOCl₂ → 4PCI₃ + 4SO₂ + 2S₂Cl₂

Properties of PCI₃: It is a colourless oily liquid and hydrolyses in the presence of moisture. Hence, it fumes in moist air.

PCI₃ + 3H₂O → H₃PO₃ + 3HCI

Structure of PCI₅: In gaseous and liquid phases, PCI₅ has a trigonal bipyramidal structure. The three equational P – CI bonds are equivalent, while the two axial bonds are longer than equatorial bonds. This is due to the fact that the axial bond pairs suffer more repulsion as compared to equatorial bond pairs.

Question 17.
a) Identify the most acidic compound from the following (Say – 2017)
i) H₂O
ii) H₂S
iii) H₂Se
iv) H₂Te
b)
i) Explain step P and Q.
ii) Give a reaction which indicates dehydration property of conc. H₂SO₄.
iii) Write any two uses of sulphuric acid.
OR
a) Identify the least basic compound among the following:
i) NH₃
ii) PH₃
iii) AsH₃
iv) SbH₃

b) i) Halogens have maximum negative electron gain enthalpy in the respective periods. Give reason.
ii) Draw the structure of Perch bric acid (HClO₄)
iii) Write the formulae of any two interhalogen compounds.
Answer:
a) iv or H₂Te
b) i)

ii) Charring action of cane sugar to carbon
C₁₂H₂₂O₁₁ + Con H₂SO₄ → 12 C+ 11 H₂O

iii) Dehydrating agent, laboratory reagent
OR
a) iv) SbH₃
b) i) by getting one electron octet ¡s completed. So electronegative ¡s very high

Kerala State Board New Syllabus Plus Two Chemistry Chapter Wise Previous Questions and Answers Chapter 6 General Principle and Processes of Isolation of Elements.

Kerala Plus Two Chemistry Chapter Wise Previous Questions Chapter 6 General Principle and Processes of Isolation of Elements

Question 1.
Analyse the table given below: (March – 2010)

Metal	Ore
Copper	Copper pyrites, Copper glance, Cuprite
Zinc	Zinc blende, Calamine, Zincite
Aluminium	Bauxite, Diaspore
Iron	Haematite, Magnetite, Iron pyrites

a) Which of the ores mentioned in the above table can be concentrated by magnetic separation method? Justify your answer.
b) Identify the ores that can be concentrated by leaching.
c) What do you mean by leaching?
Answer:
a) Haematite, Magnetite, Iron pyrites Magnetic separation is based on difference in magnetic properties of the ore components. If either the ore or the gangue (one of these two) is capable of being attraced by a magnetic field, then such separations are carried out. In the case of iron ores mentioned above the ore particles are magnetic while the impurities are non-magnetic. Thus, when the ground ore is carried on a conveyer belt which passes over a magnetic roller, the ore particles are attracted towards the magnetic roller while the non-magnetic particles are collected away from the magnetic roller,

b) Bauxite

c) It is a chemical method used for the concentration of ore. For example, the ore of Al, bauxite is concentrated by leaching (Baeyer’s process). Bauxite is heated with NaOH solution. As a result of this reaction, sodium meta aluminate is formed. The aluminate solution is neutralised by passing CO₂ solution and hydrated Al₂O₃ is precipitated by seeding with freshly prepared samples of hydrated Al2O₃. Hydrated alumina is filtered, dried and heated to obtain pure Al2O₃.

Question 2.
You are provided with samples of some impure metals such as Titanium and Nickel. (Say – 2010)
a) Which method would you recommend for the purification of each of these metals?
b) Briefly explain each method.
Answer:
a) Ti – van Arkel method
Ni – Monds process

b) The crude metal is heated in an evacuated vessel with iodine. The metal iodide is formed. The metal iodide is decomposed on a tungsten filament, at about 1800 K. The pure metal is ’ deposited on the filament.

Mond process: In this method impure Ni is heated in a stream of carbon monoxide forming a volatile complex, nickel tetracarbonyl.

The carbonyl is subjected to higher temperature so that it is decomposed to give pure metal.

Question 3.
The concept of AG° of coupled reactions are used to explain reductions in metallurgy. (March – 2011)
a) Explain the above statement.
b) In the blast furnace for manufacturing iron, most of the reduction is carried out by CO rather than C(Coke). How can you account for this?
Answer:
a) The reduction of a metal oxide which is not feasible (\(\Delta G^{\ominus}\) positive) is coupled with the oxidation of a suitable reducing agent, which is usually a highly feasible reaction (\(\Delta G^{\ominus}\) highly negative) so that the \(\Delta G^{\ominus}\) of the overall reaction (coupled reaction) becomes negative and the reduction process occurs spontaneously.This is in accordance with Ellingham diagram. The \(\Delta G^{\ominus}\) value should be negative to make the reaction feasible.

b) CO is a good reducing agent at low temperature than coke. This is because, in the Ellingham diagram, at low temperature, the CO → CO₂ line is below the C → CO line. Thus, oxidation of CO is more feasible than that of coke. Hence, in the blast furnace, CO reduces Fe₂O₃ even at the lower tem perature range. At high-temperature coke reacts with CO₂ to form carbon monoxide. Thus CO is the actual reducing agent.

Question 4.
Bauxite, Al₂O₃, xH₂O, is an important ore of aluminium. It is concentrated by leaching. Explain the method. (Say – 2011)
Answer:
Leaching of alumina from bauxite (Baeyer’s process) The powered bauxite ore is diagested with a concentrated solution of NaOH at 473 – 523 K and 35 – 36 bar pressure. AI₂O₃ is leached out as sodium aluminate along with SiO₂ as sodium silicate. Other impurities like iron oxides and TiO₂ are left behind.

AI₂O_3(s) + 2NaOH_(aq) + 3H₂O_(l) → 2Na[AI(OH)₄]_(aq)

The solidum aluminate in solution is neutralised by passing CO₂ gas and hydrated Al₂O₃ is precipitated. At this stage, the solution is seeded with freshly prepared samples of hydrated Al₂O₃ which induces the precipitation of hydrated Al₂O₃.

2Na [AI(OH)₄]_(aq) + CO_2(g) → AI₂O₃. XH₂O_(s) + 2NaHCO_3(aq)

The sodium silicate remains in the solution and hy-drated alumina is filtered, dried and heated to give back pure Al₂O₃.

\(\mathrm{Al}_{2} \mathrm{O}_{3} \cdot \mathrm{xH}_{2} \mathrm{O}_{(\mathrm{s})} \stackrel{1470 \mathrm{~K}}{\longrightarrow} \mathrm{Al}_{2} \mathrm{O}_{3(\mathrm{~s})}+\mathrm{xH}_{2} \mathrm{O}\)

Question 5.
a) All ores are minerals, but all minerals are not ores. Why? (March – 2012)
b) Carbonate ores are usually subjected to calcination, while sulphide ores are subjected to roasting. Comment on the statement.
Answer:
a) The naturally occuring materials in which the metals are present either in the native or in the combined state are called minerals.

The minerals from which the metals can be extracted economically are called ores.

Hence all ores are minerals but all minerals are not ores.

b) Calcination is the proœss of heating the ore in a limited supply of air below its metting point. This removes the volatile impurities and moisture from the ore. Oxygen is not used up during calcination.

\(\mathrm{ZnCO}_{3(\mathrm{~s})} \stackrel{\Delta}{\longrightarrow} \mathrm{ZnO}_{(\mathrm{s})}+\mathrm{O}_{2(g)}\)

Roasting is the process of heating the concentrated ore in a regular supply of air in a furnace below the melting point of the metal. It is usually employed in the concentration of sulphide ores.

2 Zns + 3O₂ → 2ZnO + 2SO₂
2 PbS + 3O₂ → 2PbO + 2SO₂
2 Cu2S + 3O₂ → 2Cu₂O + 2SO₂

Question 6.
Concentrated ore of iron, coke and limestone are fed into a blast furnace from the top. (Say – 2012)

i) Write down the reason for adding limestone along with the concentrated ore of iron.
ii) Write down the reactions taking place at the higher temperature range in the blast furnace.
OR
Metals are extracted from their chief ore.
i) Name the pencil pal ore of aluminium.
ii) Write the equations for the reactions taking place at the anode and at the cathode during the extraction of aluminium by the electrolytic process.
Answer:
i) Lime stone is added to blast furnace morder to remove acidic impurities (gangue) like silica (SiO₂). At high temperature, lime stone decomposed to form calcium oxide, which acts as a basic flux and removes acidic silica gaunge as calcium silicate slag.

\(\begin{array}{l}
\mathrm{CaCO}_{3} \stackrel{\text { Heat }}{\longrightarrow} \mathrm{CaO}+\mathrm{CO}_{2} \\
\mathrm{CaO}+\mathrm{SiO}_{2} \rightarrow \mathrm{CaSiO}_{3}
\end{array}\)

ii) The following reaction take place at the higher temperature range (900 K – 1500 K) in the blast furnace:

\(\mathrm{C}+\mathrm{O}_{2} \rightarrow \mathrm{CO}_{2}, \mathrm{FeO}+\mathrm{CO} \rightarrow \mathrm{Fe}+\mathrm{CO}_{2}\)

OR

i) The principal ore of Al is Bauxite (Al₂O₃ 2H₂O). In Hall-Heroult process for the electrolytic extration of Al, purified Al₂O₃ mixed with Na₃ALF₆ or CaF₂ acts as the electrolyte, steel cathode and graphite anode. The following reactions take place during electrolysis:

At cathode: Al₃(melt) + 3e → 4 Al(l)

At anode : The oxygen liberated at anode reacts with the carbon of anode producing CO and CO₂.

\(\begin{array}{l}
\mathrm{C}_{(\mathrm{s})}+\mathrm{O}_{(\mathrm{melt})}^{2 .} \rightarrow \mathrm{CO}_{(9)}+2 \mathrm{e} \\
\mathrm{C}_{(\mathrm{s})}+2 \mathrm{O}_{(\mathrm{melt})}^{2 .} \rightarrow \mathrm{CO}_{2(\mathrm{~g})}+4 \mathrm{e}^{-}
\end{array}\)

The overall reaction is

\(2 \mathrm{Al}_{2} \mathrm{O}_{3}+3 \mathrm{C} \rightarrow 4 \mathrm{Al}+3 \mathrm{CO}_{2}\)

Question 7.
a) Match the items of Column I with hems of Column II. (March – 2013)

Column 1	Column II
i) Aluminium	a) Malachite
ii) Iron	b) Bauxite
iii) Copper	c) Limestone
iv) Zinc	d) Haematite
	e) Calamine

b) The reduction of the metal oxide is easier if the metal formed is in liquid state, at the temperature of reduction. Give reason.
Answer:
a)

Column 1	Column II
i) Aluminium	Bauxite
ii) Iron	Haematite
iii) Copper	Malachite
iv) Zinc	Calamine

b) The entropy ¡s higher if the metal is in liquid state than when it is in solid state. So the value of entropy change or the reduction process will be more positive. Thus, the value of AGe becomes more negative and the reduction becomes easier.

Question 8.
The scientific and technological processes used for isolation of the metal from its ore is known as metallurgy. (Say – 2013)
a) Name the method used for removing gangue from suiphide ores.
b) Explain the above method.
c) Give two examples for alloy steel.
Answer:
a) Froth floatation method

b) It is used for removing gangue from suiphide ores. The mineral particles become wet by oils while the gangue particles by water. Finely powdered ore is agitated with water containing little frothing agent (e.g. pine oil) and froth stabilizers (e.g. cresols) by passing a forceful current of air. Heavier gangue particles are left to the bottom. The froth is skimmed off and then dried.

c)

• Stainless steel (Fe-74%, Cr-18%, Ni-8%)
• Nickel steel (Fe-96%, Ni-4%)

Question 9.
a) Calcination and roasting are pre-treatments in metallurgy before metal extraction. Differentiate between calcination and roasting. (March – 2014)
b) Match the items of Column I with items of Column II.

Column I	Column II
i) Distillation	a) Ge
ii) Liquation	b) Ni
iii) Zone refining	c) Cu
iv) Vapour phase refining	d) Zn
	e) Sn

a) Calcination is the process of heating the ore in the absence or limited supply of air when the volatile matter escapes leaving behind the metal oxide. Here oxygen ¡s not consumed, It is applied to hydrated oxides, hydroxides and carbonates.

Roasting is the process of heating the ore in a regular supply of air in a furnace at a temperature below the melting point of the metal. Here oxygen is consumed. it is applied to suiphide ores.

b)

Column I	Column II
i) Distillation	d) Zn
ii) Liquation	e) Sn
iii) Zone refining	a) Ge
iv) Vapourphasereflning	b) Ni

Question 10.
Sulphide ores are concentrated by froth floatation process. (Say – 2014)
a) Write the name or formula of any two sulphide ores of copper.
b) Explain froth floatation process.
Answer:
a)

• Copper pyrites (CuFeS₂)
• Copper glance (Cu₂S)

b) This method is used for removing gangue from sulphide ores and is based on the principle of preferential wetting of solid surface by vanousliq ulds. i.e., the mineral particles become wet by oils while the gangue particles by water. Finely powered ore is agitated with water containing collectors (e.g. pine oils, fatty acids, xanthates, etc.) and froth stabilizers (e.g. cresols, aniline) by passing a forceful current of air. The collectors enhance non-wettability of the mineral particles and froth stabilisers stabilise the froth. The froth which is formed at the surface of water carries up lighter ore particles and heavier gangue particles are left to the bottom. The froth is skimmed off and then dried for recovery of the ore particles.

Question 1.
a) Name two metals which can be refined by van Arkel Method. (March – 2015)
b) Match the items of Column I with items of Column II.

Column I	Column II
i) Bauxite	a) Zinc
ii) Malachite	b) Iron
iii) Calamine	c) Copper
iv) Magnetite	d) Aluminium
	e) Lead

Answer:
a) Zirconium (Zr) or Titanium (Ti)
b) i – d;
ii – c;
iii – a;
iv – b

Question 11.
The process involved in metallurgy are the concentration of the ore, isolation of the metal from its concentrated ore and purification of the metal. (Say – 2015)
a) Froth floatation method is an ore concentration method. What is the principle behind the process?
b) What is the role of limestone (CaCO₃) in the extraction of iron?
c) Monds process is used for refining of Ni and Van Arkel method is used for refining Zr (Zirconium). Write one similarity between these processes.
Answer:
a) The principle behind froth floatation is adsorption. The ore particles are preferentially wetted by the oil and are carried to the surface by the froth. The gangue material wetted by water sinks to of the tank. The froth is light and is skimmed off.

b) Inside the blastfumace lime stone is decomposed to CaO which acts as a basic flux and removes the silicate impurity (acidic gangue) of the ore as calcium silicate slag. The slag is in molten state and separates our from iron.

\(\begin{array}{l}
\mathrm{CaCO}_{3} \stackrel{\text { Heat }}{\longrightarrow} \mathrm{CaO}+\mathrm{CO}_{2} \\
\mathrm{CaO}+\mathrm{SiO}_{2} \rightarrow \mathrm{CaSiO}_{3}
\end{array}\)

c) Vapour phase refining techniques. The metal is converted into its volatile compound and collected elsewhere. It is then deèomposed to give pure metal.

Question 12.
a) Which of the following is the ore of zinc? (March – 2016)
a) Bauxite
b) Magnetite
c) Malachite
d) Calamine

b) There are several methods for refining metals. Explain a method for refining Zirconium.
Answer:
a) d) Calamine
b) Van Arkel method. The crude metal is heated in an evacuated vessel with iodine. The metal iodide is formed.

Zr+ 2l₂ → Zrl₄

The metal iodide is decomposed on a tungsten filament, at 1800 K. Pure metal is deposited on the filament.

Zrl₄ → Zr + 2l₂

Question 13.
Metals are extracted from their ores (Say – 2016)
a) Among the following which metal is extracted from bauxite:
i) Zinc
ii) Iron
iii) Aluminium
iv) Copper

b) Suiphide ores are subjected to roasting while carbonate ores are subjected to calcination. Comment on the statement.
Answer:
a) iii) Aluminium

b) In roasting, the ore is heated in a regular supply of air in a furnace at a temperature below the melting point of the metal. Here, oxygen is consumed. The sulphide ores need to be converted to oxides. Hence these are subjected to roasting.

e.g. 2ZnS + 3O₂ → 2ZnO + 2SO₂

In calcination, the ore ¡s heated in a limited supply fair below its melting point morder to remove he volatile matter. Here, oxygen is not consumed. Hence, the carbonate ores are subjected to calcination.

e.g. ZnCO₃(s) → ZnO(s) + CO₂(g)

Question 14.
Leaching is a process of concentration of ores. Explain the leaching of alumina from bauxite. (March – 2017)
Answer:
Bauxite usually contains silica (SiO₂), iron oxides and titanium oxide (TiO₂) as impurities. The powdered ore is digested with a concentrated solution of NaOH at 473 – 523 K and 35 – 36 bar pressure. Al₂O₃ is leached out as sodium aluminate leaving the impurities behind. (SiO₂ is also leached as sodium silicate).

Al₂O₃(s) + 2NaOH_(aq) + 3H₂O(l) → 2Na[Al(OH)₄](aq)

The aluminate in solution is neutralised by passing CO₂ gas and hydrated Al₂O₃ is precipitated. At this stage, the solution is seeded with freshly prepared samples of hydrated Al₂O₃ to induce the precipitation.

2Na[Al(OH)₄](aq) + CO₂(g) → Al₂O₃ xH₂O + 2NaHCO₃(aq)

The sodium silicate remains in the solution and hydrated alumina is filtered, dried and heated to give back pure Al₂O₃.

\(\mathrm{AI}_{2} \mathrm{O}_{3} \cdot \mathrm{xH}_{2} \mathrm{O} \stackrel{1470 \mathrm{~K}}{\longrightarrow} \mathrm{Al}_{2} \mathrm{O}_{3}(\mathrm{~s})+\mathrm{xH}_{2} \mathrm{O}(\mathrm{g})\)

Question 15.
a) Which of the following is not an Ore of Iron? (Say – 2017)
i) Haematite
ii) Magnetite
iii) Malachite
iv) Sidenote

b) Explain froth floatation process for the concentration of Ore.
Answer:
a) i) Haematite

b) It is used for removing gangue from sulphide ores. The mineral particles become wet by oils while the gangue particles by water. The finely powdered ore is agitated with water containing little frothing agent (e.g. pine oil) and froth stabilizers (e.g. cresols) by passing a forceful current of air. Heavier gangue particles are left to the bottom. The froth is skimmed off and then dried.