Plus Two

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.

Kerala State Board New Syllabus Plus Two Chemistry Chapter Wise Previous Questions and Answers Chapter 5 Surface Chemistry.

Kerala Plus Two Chemistry Chapter Wise Previous Questions Chapter 5 Surface Chemistry

Question 1.
Colloids exhibit certain special properties. (March – 2010)
a) Name the property of colloid involved in the construction of ultramicroscope.
b) Explain the above property.
c) What are the conditions to be satisfied to exhibit Tyndall Effect?
Answer:
a) Tyndall effect

b) When a colloid is viewed at right angles to the passage of light, the path of the beam is illuminated by a bluish light. This effect is called the Tyndall effect and the bright cone of light observed is called the Tyndall cone. Tyndall effect is caused by the scattering of light by colloidal particles in all directions in space.

c) There are two conditions for observing the Tyndall effect:

• The diameter of the dispersed particles is not much smaller than the wavelength of the light used.
• The refractive indices .of the dispersed phase and the dispersion medium differ greatly in magnitude.

Question 1.
In an attempt to prepare ferric hydroxide sol by adding small amount of ferric chloride to water, one person got a precipitate of ferric hydroxide. (Say – 2010)
a) How can you help him to convert Fe(OH)₃ precipitate to Fe(OH)₃ sol?
b) Name the phenomenon behind this.
c) What happens when BaCI₂ is added to Fe(OH)₃ sol?
d) Give reason forthe above.
Answer:
a) By adding a solution of FeCI₃ to the fresh precipitate of ferric hydroxide.
b) Peptization
c) Precipitation of Fe(OH)₃ sol will take place.
d) BaCI₂ being an electrolyte ionises to Ba²⁺ + and Cl ions. The particles of Fe(OH)₃ precipitate adsorb Ba²⁺ ions to their surface and get positively charged. These positively charged particles of the precipitate repel each other and ultimately break up into smaller particles of the size of a colloid. Here BaCI₂ is acting as a peptizing agent.

Question 1.
Physisorption and Chemisorption are 2 types of ad-sorption. (March – 2011)
a) What is the effect of temperature on physisorption and chemisorption?
b) In certain cases physisorption transfers into chemisorption as temperature is increased. Explain with an example.
c) Explain how colloids get coagulated on addition of salts.
Answer:
a) Physical adsorption or physisorption decreases with increase of temperature where as chemical adsorption increases with increase of temperature reaches a maximum value at an optimum temperature and then decreases with increase in temperature.

b) Physisorption of a gas adsorbed at low temperature may transform into chemisorption at a high temperature. This is due to the fact that no activation energy is required for physisorption while chemisorption requires activation energy. For example, dihydrogen is first adsorbed on nickel by van der Waals’ forces. This is physisorption. Molecules of dihydrogen then dissociate to form hydrogen atoms which are held on the surface by chemisorption.

c) When excess of a salt (electrolyte) is added, the colloidal particles interact with ions carrying charge opposite to that present on themselves. (According to Hardy-Schultz rule).

Question 1.
Ferric hydroxide sol can be prepared from freshly prepared ferric hydroxide precipitate. It can also be prepared by adding ferric chloride solution to boiling water. In both cases the sol particles are positively charged. (Say – 2011)
a) Name the above two methods of preparation of ferric hydroxide sol.
b) What happens when an electric potential is applied across two platinum electrodes dipping in ferric hydroxide sol? Explain.
Answer:
a) The method of preparation of ferric hydroxide sol from freshly prepared ferric hydroxide precipitate is called peptization. The method of preparation of ferric hydroxide sol by the addition of ferric chloride solution to boiling water is called hydrolysis,
b) When an electric potential is applied across two platinum electrodes dipped in ferric hydroxide sol electrophoresis will occur. Since ferric hydroxide sol is positively changed the colloidal particles will move towards the platinum electrode which acts as the cathode (negative electrode).

Question 1.
Colloids have many characteristic properties. Among this Tyndall effect is an optical property and coagulation is the process of settling of colloidal particles. (March – 2012)
i) What is the Tyndall effect?
ii) State Hardy Schulze rule which deals with the coagulation of colloids by the addition of an electrolyte.
iii) What is a protective colloid?
Answer:
i) When light passes through colloids the path of light becomes visible. This effect is called Tyndall effect. This is due to scattering of light by colloidal particles.

ii) It state that ‘Thegreaterthe valency of flocculating ion, the greater will be its coagulating, flocculating or precipitating power.

Note:
i) In the coagulation of a positive sol, the flocculating power increases in the order Na⁺ < Ba²⁺ + <AI³⁺
ii) Inthe coagulation of a negative sol, the flocculating power increases in the order Cl < S0²₄ < PO³₄ < [Fe(CN)₆]⁴ Na⁺ < Ba²⁺ < AI³⁺
iii) The lyophilic particles form a layer around the lyophobic particles and thus protect the latter from electrolytes. Such colloids are called protective colloids.

Eg : Gold sol can be protected by adding a little gelatin. Here, gelatin is the protective colloid.

Question 1.
Colloids are widely used in industry and in daily life. (Say – 2012)
i) What are colloids?
ii) Write any four applications of colloids.
Answer:
i) A colloid is a heterogeneous system in which one substance called a dispersed phase is dispersed as very fine particles in another substance called a dispersion medium. The particles in a colloid are larger than simple molecules but small enough to remain suspended. The diameter of colloidal particles ranges between 1nm and 1000 nm.

ii)
1) Electrical precipitation of smoke using Cottrell smoke precipitator-Smoke is a colloid of solid particles such as carbon, arsenic compounds, dust etc., in air. These particles are precipitated using high voltage electrodes.

2) Purification of drinking water – The suspended impurities present in water obtained from natural sources is coagulated by adding alum and is made fit for drinking purposes.

3) Medicines – Most of the medicines are colloidal in nature. Colloidal medicines are more effective because they have large surface area and are therefore easily assimilated.

4) In the rubber industry – Rubber latex is a colloid of negatively charged rubber particles which is coagulated to rubber by adding formic acid.

Question 1.
a) The accumulation of molecular species at the surface rather than in the bulk of a solid or liquid is termed adsorption. (March – 2013)
i) What is adsorption isotherm?
ii) Write the mathematical expression of Freundlich adsorption isotherm.
b) Enzymes are known as biochemical catalysts. Write any two important characteristics of enzyme catalysis.
Answer:
a) i) A plot between the amount of gas adsorbed pergram of adsorbent (x/m) and the pressure of the adsorbate at constant temperature is called Adsorption isotherm.

\(\frac{x}{m}=k p^{1 / n}\)
OR
\(\log \frac{x}{m}=\frac{1}{n} \log P+\log k[latex]

x → Amount of gas adsorbed by ‘m’ gram of the adsorbent at a pressure P. ‘k’ and ‘n’ are constants.

b)

• Enzymes are highly specific in nature.
• Enzymes are highly efficient.
• They are highly active under an optimum temperature.

Question 1.
There are mainly two types of adsorption of gases on solids (Say – 2013)
a) What are the two types of adsorption of gases on solids?
b) Write any two characteristics of each of the above two types of adsorption.
Answer:
a) Physical adsorption or physisorption and chemical adsorption or chemisorption,
b)

Physical Adsorption	Chemical Adsorption
1) Adsorbate mole­cules are held on the surface of the adsor­bent by weak van der Waals’ forces	1) The adsorbate mole­cules are held on the surface of the adsor­bent by chemical bonds
2) Not specific in nature	2) Highly specific in nature
3) Low temperature is favourable for adsor­ption	3) High temperature is favourable for adsor­ption
4) Enthalpy of adsor­ption is low (20 – 40 kJ mol-1)	4) Enthalpy of adsor­ption is high (80 – 240kJ mol-1)

Question 1.
Sols are colloidal systems in which dispersion medium is liquid and dispersed phase is solid. (March – 2014)
a) Write any four differences between lyophilic sols and lyophobic sols.
b) Peptisation is a method .of preparation of sols. Write a general procedure for peptisation.
Answer:
a)

Lyophilic Sols	Lyophobic Sols
1. Liquid-loving	1. Liquid-hating
2. Reversible	2. Irreversible
3. Stable	3. Not stable
4. Cannot be easily coagulated and no stabilising agent re­quired	4. Can be easily co­agulated and need a stabilising agent for preservation

b) Peptization is the process of converting a precipitate into colloidal sol by shaking it with a dispersion medium in the presence of small amount of electrolyte. The electrolyte used for this purpose is called peptizing agent. This method is applied to convert a freshly prepared precipitate into a colloidal sol. During peptization, the precipitate adsorbs one of the ions of the electrolyte on its surface. This causes the development of positive or negative charges on precipitates, which ultimately break up into smaller particles of the size of a colloid.

Question 1.
a) ‘Adsorption’ has many applications. Write any two applications of adsorption. (Say – 2014)
b) Physisorption and chemisorption are the two types of adsorption. Write any four differences between them.
Answer:
a) 1) In production of high vacuum
2) In gas masks
b)

Physical Adsorption	Chemical Adsorption
1) Adsorbate mole­cules are held on the surface of the adsor­bent by weak van der Waals’ forces	1) The adsorbate mole­cules are held on the surface of the adsor­bent by chemical bonds
2) Not specific in nature	2) Highly specific in nature
3) Low temperature is favourable for adsor­ption	3) High-temperature is favourable for adsor­ption
4) Enthalpy of adsor­ption is low (20 – 40 kJ mol-1)	4) Enthalpy of adsor­ption is high (80 – 240kJ mol-1)

Question 1.
a) Which of the following is Lyophobic colloid? (March – 2015)
1) Starch in water
ii) Gum in water
iii) Soap in water
iv) Gold sol

b) Write four applications of colloids.
Answer:
a) iv) Gold sol
b)

• Electrical precipitation of smoke
• Purification of drinking water
• Medicines
• Tanning
• The cleansing action of soaps and detergents
• Photographic plates and films
• Coagulation of rubber latex etc. (Any four)

Question 1.
a) Which of the following is an example of absorption? (Say – 2015)
i) Water on silica gel
ii) Water on CaCI₂
iii) Hydrogen on finely divided Nickel
iv) Oxygen on the metal surface

b) Write any two differences between absorption and adsorption.
OR
Based on particles of the dispersed phase, colloidal systems are classified into multimolecular, macromolecular, and associated colloids.
a) Which of the following colloidal system is an example of the multimolecular system?
i) Starch water
ii) Soap solution
iii) Ferric hydroxide in water
iv) Polyvinyl alcohol in water

b) Associated colloids are also known as micelles. How are they formed?
Answer:
a) ii) Water on CaCl₂
b)

Absorption	Adsorption
1. It involves the uniform distribution of the molecular species throughout the bulk.	1. It involves the unequal distribution of the molecular species in bulk and at the surface.
2. It occurs through­out the body of material.	2. It is a surface phenomenon.

OR
a) iii) Ferric hydroxide in water.

b) Associated colloids or micelles are formed by the aggregation of ions of an electrolyte above a particular concentration and temperature. Soap is an example of associated colloid. It is sodium or potassium salt of higher fatty acids and can be represented as RCOO-Na⁺ or RCOO K⁺. When dissolved in water, it dissociates into RCOO and Na⁺ or K³ ions. The RCOO ions consist of two parts – a long hydrocarbon chain R (non-polar ‘tail’) which is hydrophobic (water-repelling), and a polar group COO (polar ionic ‘head’) whic is hydrophobic (water-loving). Therefore, the RCOO ions are present on the surface with their COO groups in water and the hydrocarbon chains (R) staying away from it and remain at the surface. At critical micelle concentraion, the anions are pulled into the bulk of the solution and aggregate to form a spherical shape with their hydrocarbon cains pointing towards the centre of the sphere with COO part remaining outward on the surface of the sphere. An aggregate thus formed is known as ionic micelle.

Question 1.
i) Catalysis can be classified into two groups homogenous and heterogeneous. (March – 2016)
a) What do you mean by homogenous catalysis?
b) Write one example for heterogeneous catalysis.
ii) Which of the following is an emulsifying agent?
a) Milk
b) Butter
c) Gum.
d) Lampblack
Answer:
i) a) When the reactants and the catalyst are in the same phase (i.e., liquid or gas), the process is said to be homogenous catalysis.
b) e.g. Oxidation of sulphur dioxide into sulphur trioxide in the presence of Pt or V₂O₅.
[latex]2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \stackrel{\mathrm{Pt}(\mathrm{s})}{\longrightarrow} 2 \mathrm{SO}_{3}(\mathrm{~g})\)
In this process the reactants, sulphur dioxide and dioxygen are in the gaseous state while the catalyst Pt or V₂O₅ is in the solid state.
ii) c) gum (for O/W emulsions)
d) lamp black (for W/O emulsions)

Question 1.
Dispersed phase and dispersion medium are two phases of the colloidal system (Say – 2016)
a) Name the colloid in which dispersed phase is liquid and dispersion medium is solid.
i) Sol
ii) foam
iii) Emulsion
iv) Gel

b) Physisorption and Chemisorption are two types of adsorption. Write any four differences between them.
Answer:
a) iv) Gel
b) Refer March 2017 Question No 1 (a)

Question 1.
There are mainly two types of adsorption. They are physisorption and chemisorption. (March – 2017)
a) Differentiate between physisorption and chemisorption.
b) Write any two applications of adsorption.
Answer:
a) Ant two

Physisorption	Chemisorption
1) It is arises because of van der Waals force.	1) It is caused by chemical bond formation.
2) It is not specific in nature.	2) It is highly specific in nature.
3) It is reversible in nature.	3) It is irreversible.
4) More easily liquefiable gases are adsorbed readily.	4) Gases which can react with the adsorbent show chemisorption.
5) Enthalpy of adsorption is low (20-40 kJ mol-1) in this case.	5) Enthalpy of adsorption is high (80-240 kJ mol-1) in this case.
6) Low temperature is favourable for adsor­ption. It decreases with increase of temperature.	6) Hig temperature is favourable for adsorption. It increases with the increase of temperature.
7) No appreciable activation energy is needed.	7) High activation energy is sometimes needed.
8) It increases with an increase of surface area.	8) It too increases with an increase of surface area.

b) Production of high vacuum, gas masks, control of humidity, removal of colouring matter from solutions, heterogeneous catalysis, separation of inert gases, in curing diseases, Froth floatation process, adsorption indicators, chromatographic analysis etc. (Any two applications required).

Question 1.
a) Which among the following Is not an electrical property of colloids? (Say – 2017)
i) Electrophoresis
ii) Electro osmosis
iii) Coagulation
iv) Tyndal effect

b) Freundlich adsorption Isotherm is
x/m = a kp^1/n where n > 1
Answer the following questions based on Freundlich adsorption isotherm:
i) What Is adsorption isotherm?
ii) Explain the terms In the above equation.
Answer:
iv) Tyndaleffed
b) i) lt is a curve obtained by plotting extent of adsorption against pressure at a constant temperature.

ii) x = mass of gas adsorbed
m = mass of adsorbent
p = pressure
k, n are constant n > 1

Kerala State Board New Syllabus Plus Two Chemistry Chapter Wise Previous Questions and Answers Chapter 4 Chemical Kinetics.

Kerala Plus Two Chemistry Chapter Wise Previous Questions Chapter 4 Chemical Kinetics

Question 1.
The order of a chemical reaction can be zero and even a fraction but molecularity cannot be zero or a non-integer. (March – 2010)
i) What do you mean by the order of a reaction?
ii) What is the molecularity of a reaction?
iii) The conversion of molecules ‘A’ to ‘B’ follows second order kinetics. If the concentration of A is increased to three times, how will it affect the rate of formation of ‘B’?
Answer:
i) It is sum of powers of the concentration – terms of the reactants in the rate law expression.
ii) The number of reacting species in an elementary reaction.
iii) Increases by 9 times.
Rate = k[R]²
Rate’= k[3R]² = 9 [R]²; i.e., Rate = 9 Rate

Question 2.
In a class room discussion about order and molecularity of a chemical reaction, Ramu argued that “there are reactions which appear to be of higher order but actually follow first order kinetics”. (Say – 2010)
a) How far is his statement true? Give your opinion in this regard. Justify your answer using suitable example.
b) List out any three important differences between order and molecularity.
Answer:
a) The reaction appears to be of higher order but actually follows a lower order kinetics. Such reactions are called pseudo order reactions. For example, hydrolysis of ethyl acetate. This reaction appears to be of second order but actually follows first order kinetics.

\(\begin{array}{l}
\mathrm{CH}_{3} \mathrm{COOC}_{2} \mathrm{H}_{5}+\mathrm{H}_{2} \mathrm{O}_{\longrightarrow} \mathrm{H}^{+}, \mathrm{CH}_{3} \mathrm{COOH}+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH} \\
\text { Rate }=\mathrm{k}\left[\mathrm{CH}_{3} \mathrm{COOH}\right]\left[\mathrm{H}_{2} \mathrm{O}\right]
\end{array}\)

Here the concentration of water does not get altered much during the course of the reaction.
Hence, [H₂O] can be taken as a constant. The equation thus becomes
Rate = k[CH₃CQOH] where k = k [H₂O]
Thus, the reaction behaves as a first-order reaction.

(b)

Order	Molecularity
i. Experimental	i. Theoretical
ii Can be zero	ii Can not be zero
iii Can be a fraction	iii Cannot be a fraction

Question 3.
The hydrolysis of an ester in acid medium is a first-order reaction. (March – 2011)
a) What do you mean by a first-order reaction?
b) What is the relation between Rate Constant and Half-Life Period of a reaction?
c) Half-Life Period of a first-order reaction is 20 seconds. How much time will it take to complete 90% of the reaction?
Answer:
a) When the sum of the powers of the concentra¬tion terms in the rate expression is one, that reaction is called a first-order reaction.
Rate = k[A]¹

b) In the case of the first-order reaction,

Question 4.
The value of rate constant K of a reaction depends on temperature. From the values of K at two different temperatures, the Arrhenius parameters E_a, and A can be calculated. (Say – 2011)
a) The rate constant of a reaction at 600K and 900K are 0.02 s^-1 and 0.06 s^-1 respectively. Find the values of E_a and A.
b) Write the unit of rate constant ‘K’ of a reaction if the concentration is in mol L^-1 and time in s. (order of the reaction is two)
Answer:

Question 5.
Rate of a reaction is the change in concentration of any one of the reactants or any one of the products in unit time. (March – 2012)
i) Express the rate of the following reaction in terms of reactants and products:
2H I → H₂ + l₂
ii) If rate expression for the above reaction is, rate = k[H I]², What is the order of the reaction?
iii) Define order of a reaction.
iv) Whether the molecularity and the order of the above reaction are the same? Give reason.
Answer:
i) In terms of reactants

iii) Order is the sum of the powers of the concentration terms in the rate law.
rate = K [A]^x [B]^y
∴ Order = x + y

iv) Yes.
2H I = H₂ + l₂
rate = K [H I]² ∴ Order = 2
Molecularity is the number of reacting species taking part in an elementary reaction which must collide simultaneously in order to bring about a chemical reaction.

∴ Here molecularity = 2

Question 6.
For a first-order reaction, the half-life period (t_1/2) is independent of initial concentration of its reacting species. (Say – 2012)
i) What is meant by half-life period of a reaction?
ii) By deriving the equation for t_1/2 of first-order reaction, prove that it is independent of initial concentration of its reacting species.
[Hint: Fora first ortler reaction, \(\left.\mathrm{k}=\frac{2.303}{\mathrm{t}} \log \frac{[\mathrm{R}]_{0}}{[\mathrm{R}]}\right]\)
Answer:
i) The half-life period of a reaction is the time in which the concentration of a reactant is reduced to one half of its initial concentration,
ii) For a first order reaction,

Thus, for a first order reaction, half-life period is constant, i.e., it is independent of initial concentration of the reacting species.

Question 7.
a) Zero-order reaction means that the rate of a reaction is independent of the concentration of reactants. (March – 2013)
i) Write an example for a zero-order reaction.
ii) Write the integral rate expression for the zero-order reaction, R→ P.

b) The temperature dependence of the rate of a chemical reaction can be accurately explained by the Arrhenius equation. With the help of the Arrhenius equation calculate the rate constant for the first-order reaction C₂H₅l(g) → C₂H₄(g) + Hl(g) at 700K. Energy of activation (E_a) for the reaction is 209 kJ moh1 and rate constant at 600 K is 1.60 x 10^-5 S^-1 [Universal gas constant R = 8.314 JK^-1 mol^-1]
Answer:
a) i) Zero-order reaction

where ‘I is the constant of integration.
At t = 0 1 = [R]₀
[R]₀ → InitiaI concentration of the reactant
[R] → concentration at time
∴ Equation (1) becomes. [R] = ^–kt + [R]₀

Question 8.
The conversion of molecule A to B follows second order kinetics. (Say – 2013)
a) If the concentration of A is increased to 4 times, how will it affect the formation of B?
b) Indicate the order and molecularityofthe reaction given below.
\(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}+\mathrm{H}_{2} \mathrm{O} \stackrel{\mathrm{H}}{\longrightarrow} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}+\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\)
Answer:
a) Rate r,= k[A]²
If the concentration of A is increased by four times, the new rate, r2 = k[4A]²
or r₂ = 16 k [A]²
or r₂ = 16 r₁
i.e., rate is increased by 16 times.

b) This is a pseudo first order reaction.
Order = 1, Molecularity = 2

Question 9.
a) Consider a general reaction aA + bB → cC + dD. The rate expression for the reaction is Rate = K[A]*[B]^y (March – 2014)
i) Establish the significance of ‘(a+b)’ and ‘(x+y)’ in terms of order and molecularity.
ii) Write any two differences between order and molecularity.
b) “Reactions with zero order is possible, but zero molecularity is not”. Justify the statement.
Answer:
a) i) (a + b) – Molecularity of the reaction
(X + y) – Order of the reaction

ii)

Order	Molecularity
1. Sum of the pow­ers of the concen­tration of the reac­tants in the rate law expression	1. No. of reacting species taking part in an elementary reaction, which must collide simultaneously to bring about a chemical reaction
2. Experimental quantity	2. Theoretical concept

b) The order of a reaction can be zero which means that the rate of the reaction is independent of the concentration of the reactants. But, molecularity of a reaction cannot be zero which means that there is no reacting species and hence no reaction is possible.

Question 10.
a) Unit of rate constant (K) of a reaction depends on the order of the reactions. (Say – 2014)
Values of ‘K’ of two reactions are given below. Find the order of each reaction.
i) K = 3 x 10^-2 molL^-1 s^-1
ii) K = 5 x 10^-3 mol^-1 Ls^-1
b) i) Write integrated rate equation for a first order reaction.
ii) Write the relation between half life (t^1/2) and rate constant (K) of a first order reaction.
iii) Rate constant (K) of a reaction is 5 x 10.2 s^-1.
Find the half life (t_1/2) of the reaction.
Answer:
a) i) Zero-order reaction by analysing the unit.
ii) Second-order reaction

Question 11.
The terms order and molecularity are common in chemical kinetics. (March – 2015)
a) What do you mean by order and molecularity?
b) i) Write two factors influencing rate of a reaction.
ii) WnteArrhenius equation.
Answer:
a) Order of a chemical reaction is the sum of powers of the concentration of the reactants in the rate law expression.
Molecularity of a reaction is the number of reating species (atoms, ions or molecules) taking part in an elementary reaction, which must collide simultaneously in order to bring about a chemical reaction.

b) i) Temperature, Nature of the reactant, Concentration of the reactant (Pressure in the case of gases). Presence of catalyst, Presence of radiation/light, Surface area etc. (any one)

\(\text { ii) } k=A e^{\frac{-E_{0}}{R T}} \text { OR } \ln k=-\frac{E_{a}}{R T}+\ln A\)

Question 12.
Integrated rate expression for rate constant of first-order reaction is given by \(\mathrm{K}=\frac{2.303}{\mathrm{t}} \log \frac{[\mathrm{R}]_{0}}{[\mathrm{R}]}\), for a general reaction R → R (Say – 2015)
i) Derive an expression for half life period of first order reaction.
ii) A first order reaction has a rate constant 1.15 x 10^-3s^-1. How long will 5g of the reactant take to reduce 3 g?
Answer:

Question 13.
i) The molecularity of the reaction 2NO + O₂ → 2NO₂ is, (March – 2016)
a) 5
c) 2
c) 3
d) O

ii) a) What do you mean by rate oía reaction?
b) What will be the effect of temperature on rate of a reaction?
iii) A first order reaction is found to have a rate constant, k = 5.5 x 10^-14 s^-1. Find out the half-life of the reaction.
Answer:
c) 3

ii) a) The rate of a reaction is defined as the change in concentration of any one of the reactants or products ¡n unit time. i.e.,

b) The rate of most of the chemical reactions (endothermic reactions) increase with increase in temperature. For a chemical reaçtion with rise ¡n temperature by 100, the rate constant is nearly doubled. The temperature dependence of the rate of a chemical reaction can be accurately explained by Arrhenius equation, \(k=A e^{\frac{-E_{0}}{R T}}\)

iii) For a first order reaction, half-life period,

Question 14.
Rate of a reaction is the change in concentration of any one of the reactants or any one of the products in unit time (Say – 2016)
a) Express the rate of the following reaction in terms of reactants and products \(2 \mathrm{NO}_{(g)}+\mathrm{O}_{2(g)} \rightarrow 2 \mathrm{NO}_{2(g)}\).
b) i) \(\mathrm{N}_{2} \mathrm{O}_{5(g)} \rightarrow 2 \mathrm{NO}_{2(g)}+\frac{1}{2} \mathrm{O}_{2(g)}\) is a first order reaction. Find the unit of K.
ii) Calculate the time required for the completion of 90% of a first order reaction. (K = 0.2303s^-1)
Answer:

Question 15.
a) Plot a graph showing variation in the concentration of reactants against time for a zero-order reaction. (March – 2017)
b) What do you mean by zero-order reaction?
C) The initial concentration of the first-order reaction, \(\mathrm{N}_{2} \mathrm{O}_{5(\mathrm{~g})} \rightarrow 2 \mathrm{NO}_{2(\mathrm{~g})}+1 / 2 \mathrm{O}_{2(g)}\) was 1 24 x 10^-2 mol L^-1 at 300 K. The concentration of N₂O₅ after ‘1’ hour was 0.20 x 10^-2 mol L^-1. Calculate the rate constant of the reaction at 300 K.
Answer:

b) It is a reaction for which the rate of the reaction is proportional to zero power of the concentration of reactants, i.e, order is zero.
Or, this is a reaction for which the rate of the reaction is independent of the concentration of the reactants.

Or, this is a reaction for which rate of the reaction,

Question 16.
The effect of temperature on rate of reaction is given by Arrhenius equation. (Say – 2017)
i) Write Arrtenius equation.
ii) Define activation energy (Ea).
iii) Rate constant K₂ of a reaction at 310 K is two times of its rate constant K., at 300 K. Calculate activation energy of the reaction. (1og2 0.3010, log 1=0)
Answer:
\(\begin{array}{l}
\text { I) } \mathrm{K}=\mathrm{A} \mathrm{e}^{-\mathrm{Ea} / \mathrm{RT}} \text { or } \\
\log \mathrm{K}=\log \mathrm{A} \frac{-\mathrm{Ea}}{2.303 \mathrm{RT}}
\end{array}\)

ii) Activation energy is the energy required to form an activated complex or It is the energy difference between the activated complex and the reactant molecules.

Kerala State Board New Syllabus Plus Two Chemistry Chapter Wise Previous Questions and Answers Chapter 3 Electrochemistry.

Kerala Plus Two Chemistry Chapter Wise Previous Questions Chapter 3 Electrochemistry

Question 1.
From the position of elements in the electrochemical series, copper (Cu) can displace silver (Ag) from silver nitrate solution. (March – 2010)
a) Represent the cell constructed with silver and copper electrodes.
b) Write down the reaction taking place at the anode.
c) Write down the reaction taking place at the cathode.
d) Write the Nernst equation for the above cell reaction.
Answer:

Question 2.
In a class room, the teacher has explained the quantitative aspects of electrolysis by stating the Faraday’s laws of electrolysis. (May – 2013)

a) State the Faraday’s laws of electrolysis.
b) Explain the term electrochemical equivalent.
c) Calculate the quantity of electricity required to deposit 0.09 g of Aluminium during the following electrode reaction:
Al³⁺ + 3e → Al (Atomic mass of Al = 27)
Answer:
a) First law : The amount of a substance which is deposited or liberated at any electrode during ectrolysis is directly proportional to the quantity of electricity flowing through the electrolyte.

Second law: If the same quantity of electricity is passed through different electrolytes the amount of substances formed is directly proportional to their chemical equivalent weights.

b) It is the quantity of a substance formed when one-ampere current is passed through an electrolyte for one second.

c) Quantity of electricity required to deposit 27 g of
Al = 3F = 3 x 96500 C = 289500 C
∴ the quantity of electricity required to deposit 0.09
\(g \text { of } A l=\frac{289500 \times .09}{27}=965 C\)

Question 3.
The limiting molar conductivity of an electrolyte is to Aained by adding the limiting molar conductivities of cation and anion of the electrolyte. (March – 2011)
a) Name the above law.
b) What is meant by limiting molar conductivity?
c) Explain how conductivity measurements help to determine the ionization constant of a weak electrolyte like Acetic Acid.
d)Explain the change of conductivity and molar conductivity of a solution with dilution.
Answer:
a) Kohlrauschs law.
b) It is the conductivity of an electrolyte when the concentration of the solution approaches zero (or at infinite dilution).
c) The limiting molar conductivity of acetic acid \(\left(\Lambda_{\mathrm{Gh}_{3} \mathrm{COOH}}\right)\) isdeterrnined by app’ying Kohlrauschts law \(\left(\Lambda_{\mathrm{CH}_{3} \mathrm{COOH}}^{0}=\Lambda_{\mathrm{CH}_{3} \mathrm{COONa}}^{0}+\Lambda_{\mathrm{HCl}}^{0}-\Lambda_{\mathrm{NaCl}}^{0}\right)\). Then, degree of dissociaijon,OE is determined using the re lation, \(a=\frac{\Lambda_{\mathrm{CH}_{3} \mathrm{COOH}}^{\mathrm{C}}}{\Lambda_{\mathrm{CH}_{3} \mathrm{COOH}}^{\mathrm{O}}}\)

From α, the diissociation constant can be deter mined using the relation, \(K_{a}=\frac{c a^{2}}{(1-\alpha)}\)

Substituting for CL we get,

d. Conductivity decreases with dilution because the number of ions per unit volume that carry the current in a solution decreases on dilution. The vanation of molar conductivity with dilution is different for strong and weak electrolytes. For strong electrolytes molar conductivity in creases steadily with increase in dilution due to decrease in interionic attraction. Thus, a straight line is obtained when ∧_m is plotted against C^1/2. For weak electrolytes molar conductivity increases with dilution steadily intially and shows a steep increase, especially at lower concentration due to increase in degree of dissociation.

Thus, a plot of ∧_m against C^1/2 gives a curve,

Question 4.
The standard electrode potentials of some electrodes are given below: (May – 2011)
E°_(zn2+,zn) = – 0.76V
E°_{(cu2+, Cn)} = + 0.34V
E°_{(Ag+, Ag)} = + 080V
E°_{(H+, H2)} = 0V

a) Can CuSO₄ solution be kept in silver vessel?
b) Zinc or Copper which can displace hydrogen from dii. H₂SO₄?
c) What is the reaction taking place at SHE when its connected to Ag/Ag electrode to form a galvanic cell?
d) Find the value of K_C (equillibnum constant) in the Daniel cell at 298k.
Answer:
a) Yes.
Since the standard reduction potential of silver is more than that of copper it is less reactive than copper and hence cannot react with CuSO₄.

b) Zinc with negative standard electrode potential is more active than hydrogen and hence can displace hydrogen from dil.H₂SO₄. But copper with a positive standard electrode potential is less active than hydrogen and hence cannot displace hydrogen from dil. H₂SO₄.

c) Since silver is less active than hydrogen, when silver electrode is connected with S.H.E, silver will act as cathode and S.H.E will act as anode. At the anode H is oxidiseci to H⁺. Hence, the reaction taking place at S.H.E is

Question 5.
Daniell cell is a galvanic cell made of zinc and copper electrodes. (March – 2012)
i) Write anode and cathode reactions in Daniell cell.
ii) Nernst equation for the electrode reaction
Mⁿ⁺ + + ne- 2 M is
\(\mathrm{E}_{\left(\mathrm{M}^{n+} / \mathrm{M}\right)}=\mathrm{E}_{\left(\mathrm{M}^{n+} / \mathrm{M}\right)}^{0}-\frac{2.303 \mathrm{RT}}{\mathrm{nF}} \log \frac{1}{\left[\mathrm{M}^{\mathrm{n}^{+}}\right]}\)
Derive Nernst equation for Daniel cell.
OR
Leclariche cell, Lead storage cell and Fuel cell are galvanic cells having different uses.

i) Among these, the Leclanche cell is a primary cell and Lead storage cell is a secondary cell. Write any two differences between primary cells and secondary cells.
ii) What is a Fuel cell?
iii) Write the overall cell reaction in H₂ – O₂ Fuel cell.
Answer:
i) Anode : Zn → Zⁿ²⁺ + 2e- (oxidation half reaction)
Cathode : Cu²⁺ + 2e^– → Cu (reduction half reaction)
Overall cell reaction is Zn + Cu²⁺ → Zn²⁺ + Cu Theceilcanberepresentedas: Znl Zn²⁺llCu²⁺lCu
Anode : Zn I ZnSO₄
Cathode : Cu I CuSO₄

which is the Nernst equation for Daniell cell.
OR

1) Primary cell	Secondary cell
a) Electrode reaction cannot be reversed.	a) Electrode reaction can be reversed by an external electric energy source.
b) Reaction occur only once & after use they become dead; not chargeable.	b) Reaction can occur many times in both directions Rechargeable

ii) Fuel cells are galvanic cells in which chemical energy from fuels like H₂, CO, CH₄ etc are converted to electrical energy.

Question 6.
Innumerable number of galvanic cells can be constructed on the pattern of Daniell cell by taking combination of different half cells. (May – 2012)
i) What is galvanic cell?
ii) Name the anode and cathode of the Daniell cell.
iii) Write the name of the half-cell represented by Pt_(S)/H_2(g)/H⁺_(aq).
iv) What is the potential of the above half-cell at all temperate res?
Answer:
i) It is a device for converting chemical energy released into electrical energy.
ii) Anode – Zn rod dipped in ZnSO₄
Cathode – Cu rod dipped on CauSO₄
iii) Standard Hydrogen Electrode (S.H.E)
iv) S.H.E is assigned a zero potential at all temperatures.

Question 7.
With decrease in concentration of an electrolytic solution, conductivity (K)decreases and molar conductivity (∧_m) increases. (March – 2013)
i) Write the equation showing the relationship between conductivity and molar conductivity.
ii) How will you account for the increase in molar conductivity with decrease in concentration?
iii) Limiting molar conductivity (L°_m) of a strong electrolyte can be determined by graphical extrapolation method. Suggest a method for the determination of limiting molar conducivity of a weak electrolyte, taking acetic acid (CH₃COOH) as example.
Answer:
i) Molar conductivity \(\left(\Lambda_{m}\right)=\frac{\kappa \times 1000}{M}\)
where M → Molanty and K → Conductivity
OR
\(\lambda_{\mathrm{m}}=\frac{\mathrm{K}}{\mathrm{C}}\) Concentration of where solution in mol/litre.

ii) Molar conductivity increase with decrease in con centration or increase in dilution as number of ions as well as mobility of ions increases with dilution.

a) For strong electrolytes, the number of ions do not increase appreciably on dilution and only mobility of ions increases due to decrease in interionic attraction.

∴ increases a little as shown in the above graph.

b) For weak electrolytes, the number of ions, as well as mobility of ions, increases on dilution. Hence, increases steeply on dilution, especially near lower concentrations as shown in graph.

iii) Using Kohlrauschs law

Thus the molar conductivity of CH3COQH at infinite dilution can be determined from the knowledge Of \(\lambda_{\mathrm{m}\left(\mathrm{CH}_{3} \mathrm{COONa}\right)}^{0}, \lambda_{\mathrm{m}(\mathrm{HCl})}^{0}, \lambda_{\mathrm{m}(\mathrm{NaCl})}^{0}\)

Question 8.
We can construct innumerable number of galvanic cells on the pattern of Daniell cell by taking combi nation of different half cells. (May – 2013)
a) What is a galvanic cell?
b) Name the cathode and anode used in the Daniell cell.
c) Name the cell represented by Pt_(S), H_2(g)/H⁺_(aq).
d) According to convention what is the potential of the above cell at all temperatures?
e) Write the use of the above cell.
Answer:
a) It is a device for converting chemical energy into electrical energy. The decrease in free energy in a spontaneous chemical process appears as elec trical energy. e.g., Daniell cell,
b) A zinc rod dipped in 1 M solution of ZnSO₄ acts as the anode. Here oxidation takes place. A copper rod dipped in 1 M solution of CuSO₄ acts as the cathode. Here reduction takes place.
c) This represents the andard Hydrogen Electrode (S.H.E), when it acts as the anode.
d) According to convention, S.H.E is assigned a zero potential at all temperatures
e) It is used as a primary reference electrode for determining the standard electrode potential of an unknown electrode. The electrode whose standard potential is to be determined is coupled with a reference electrode of known potential i.e., S.H.E to get a galvanic cell. The potential of the resulting galvanic cell is determined experimentally. E = E – E Knowing the potential of one electrode that of the other can be calculated.

Question 9.
a) The cell reactìon in Daniell cell is Zn_(s) + CU²⁺_(aq) → Zn²⁺_(aq) + CU_(s) and Nernst equation for single electrode potential for general electrode reaction \(\mathrm{M}^{\mathrm{n}+}{ }_{(\mathrm{aq})}+\mathrm{ne}^{-} \longrightarrow \mathrm{M}_{(\mathrm{s})}\) is \(E_{M^{n+} M}=E_{M^{n+} / M}^{0}-\frac{2.303 R T}{n F} \log \frac{[M]}{\left[M^{n+}\right]}\) Derive Nernst equation for Daniell cell. (March – 2014)
b) Daniell cell is a primary cell while lead storage cell is a secondary cell. Write any one difference between primary cells and secondary cells.
Answer:
a) In Daniell cell, the electrode potential for any given concentration of Cu²⁺ and Zn²⁺ ions, we can write, For Cathode:

b) In primary cells the reaction occurs only once and after use over a period of time cell becomes dead and cannot be reused again. Here the cell reaction is irreversible. e.g., dry cell Secondary cells after use can be recharged by passing current through them in the opposite direction so that they can be used again. Here the cell reaction is reversible. e.g., Lead storage cell.

Question 10.
Fuel cells are special types of galvanic cells. (May – 2014)
a) i) Whataregalvaniccells?
ii) Write any two advantages of fuel cells.
b) Write the electrode reactions is H₂ – O₂ fuel cells.
a) i) These are devices for converting chemical energy into electrical energy. The decrease in free energy in a spontaneous chemical process appears as electrical energy. e.g., Daniell oeil.
ii) 1) Fuel cells are pollution free.
2) Fuel cells are highly efficient (about 70%) compared to thermal plants (about 40%)
3) Fuel cells run continuously as long as the reactants are supplied. (any two)

b) At cathode:
O₂(g) + 2H₂O(l) + 4e → 4OH(aq)
At anode:
2H₂(g) + 4OH(aq) → 4H₂O(l)

Question 11.
You are supplied with the following substances: Copper rod, Zinc rod, Salt bridge, two glass bea kers, a piece of wire, 1 M CuSO₄ solution, 1 M ZnSO₄ solution. (March – 2015)
a) Represent the cell made using the above materials.
b) i) Write the Nemst equation for the above cell.
ii) Calculate the standard EMF of the cell if
\(\begin{array}{l}
E_{\left(\mathrm{Zn}^{2+} \mid \mathrm{Zn}\right)}=-0.76 \mathrm{~V} \\
\mathrm{E}_{\left(\mathrm{Cu}^{2+} \mid \mathrm{Cu}\right)}=+0.34 \mathrm{~V}
\end{array}\)
Answer:
a) Zn(s)|Zn²⁺(1 M)||Cu²⁺(1 M)|Cu(s)
OR

Question 12.
a) Conductance (G), conductivity (K) and molar conductivity (∧_m) are terms used in electrolytic conduction. (May – 2015)
i) Write any two factors on which conductivity depends on.
ii) How do conductivity and molar conductivity vary with concentration of electrolytic solution?
b) Write any one difference between primary cell and secondary cell.
Answer:
a) i) 1. the nature of the electrolyte added
2. size of the ions produced and their solvation
3. the nature of the solvent and its viscosity
4. concentration of the electrolyte
5. temperature (any two factors)

ii) Conductivity always decreases with decrease in concentration both for weak and strong electrolytes. This can be explained by the fact that the number of ions per unit volume that carry the current in a solution decreases on dilution.

Molar conductivity, ∧_m = _kV

∧_m increases with decrease in concentration. This is because the total volume (V) of the solution containing one mole of electrolyte also increases. The decrease in K on dilution is more than compensated by increase in its volume.

In the case of strong electrolytes Am increases slowly with dilution and can be represented by the equation:
\(\Lambda_{m}=\Lambda_{m}^{0}-A c^{1 / 2}\)

where ‘c’ is the molar concentration, ‘A’ is a constant (equato to -ve of slope) and ∧_m° is the limiting molar conductivity. Here, the plot of ∧_m against ‘C^1/2‘ will be a straight line. In the case of weak electrolytes ∧_m increases steeply on dilution, especially near lower concentrations due to increase in degree of dissociation.

b) Pnmary cell – Cell in which the reaction occurs only once and after use over a period of time the cell becomes dead and cannot be reused again. Secondary cell – Cell which can be recharged after use by passing current through it in the opposite direction so that it can be used again.

Question 13.
a) Which of the following is a secondary cell? (March – 2016)
a) Dry cell
b) Leclanche cell
C) Mercury cell
d) None of these

b) What is the relationship between resistance and conductance?
c) One of the fuel cells uses the reaction of hydrogen and oxygen to form water. Write down the cell re action taking place in the anode and cathode of that fuel cell.
Answer:
d) None of these
b) Resistance is inversely proportional to conductance. Or, Conductance is the inverse of resistance.

Question 14.
Galvanic cells are classified into primary and secondary cells (May – 2016)
a) Write any two differences between primary cell and secondary cell.
b) i) What is a fuel cell?
ii) Write the overall cell reaction in H₂ – O₂ fuel cell.
Answer:
a) Primary cell
Cell reaction cannot be reversed and hence can not be recharged, cannot be reused again. e.g. Dry cell, Mercury cell

Secondary cell
Cell reaction can be reversed and hence can be recharged, can be resued again. e.g. Lead storage battery, nickel-cadmium cell

b) i) Fuel cell is a galvanic cell that is designed to convert the energy of combustion of fuels directly into electncal energy.
ii) 2H₂(g) + O₂(g) → 2H₂O(l)

Question 15.
a) Represent the galvanic cell based on the cell reaction given below (March – 2017)
\(\mathrm{Cu}_{(\mathrm{s})}+2 \mathrm{Ag}_{(\mathrm{aq})} \rightarrow \mathrm{Cu}_{(\mathrm{aq})}^{2+}+2 \mathrm{Ag}_{(\mathrm{s})}\)
b) Write the half cell reaction of the above cell.
c) ∧_m⁰ for NaCI. HCI and NaAc are 126.4, 425.9 and 91.0 S cm² mol^-1 respectively. Calculate ∧_m⁰ for HAc.
Answer:

Question 16.
a) Identify the weak electrolyte from the following: (May – 2017)
i) KCl
ii) NaCl
iii) KBr
iv) CH₃COOH

b) Kohlrausch’s law helps to determine the degree of dissociation of a weak electrolyte at a given concentration.
i) State Kohlrausch’s law.
ii) The molar conductivity ∧_m of .001 M acetic acid is 4.95 x 10^-5 S cm² mol^-1. Calculate the degree of dissociation (α) at this concentra tion if limiting molar conductivity \(\wedge_{m}^{0}\) for H+ is 340 x 10^-5 S cm² mol^-1 and for CH₃COO is 50.5 x 10^-5 S cm² mol^-1.
Answer:
a) iv) CN₃COOH
b) i) Kohlrausch’s law: Limiting molar conductivity of an electrolyte is the sum of individual contnbutions of anion and cation respectively.

Kerala State Board New Syllabus Plus Two Chemistry Chapter Wise Previous Questions and Answers Chapter 2 Solutions.

Kerala Plus Two Chemistry Chapter Wise Previous Questions Chapter 2 Solutions

Question 1.
Colligative properties are properties of solutions which depend on the number of solute particles irrespective of their nature. (March – 2010)
a) Name the four important colligative properties.
b) What happens to the colligative properties when ethanoic acid is treated with benzene? Give reason.
Answer:
a) 1) Relating lowering of vapour pressure of the solvent
2) Depression of freezing point of the solvent
3) Elevation of Boiling point of the solvent
4) Osmotic pressure of the solution

b) Molecules of ethanoic add (acetic acid) dimenses in benzene due to hydrogen bonding. As a result of dimerisation, the actual number of solute particles in the solution is decreased. As colligative property decreases molecular mass increases.

Question 2.
a) Mr. Raju has determined the molecular masses of different solutes in different solvents by osmotic pressure measurements and presented them in the following table. Please help him to complete the table. (May – 2010)

Solute	Solvent	Theoretical Molecular Mass	Experimental Molecular Mass
NaCI Benzoic acid Urea Acetic acid CaCI2 Glucose Ai2(So4)3	Water Benzene Water Benzene Water Water Water	A B C D E F G	A/2 — — — — — —

b) The extent of deviation from ideal behaviour of a solution is explained by van’t Hoff factor, i. What is meant by van’t Hoff factor?
Answer:
a) 2B, C, 2D, E/3, F, G/5.
b) 2B and 2 D are due to association
\(\mathrm{i}=\frac{\text { Normal molar mass }}{\text { Abnormal molar mass }}\)
OR
\(\mathrm{i}=\frac{\text { Observed colligative property }}{\text { Calculated colligative property }}\)

Question 3.
Colligative properties can be used to determine the molecular mass of solutes in solutions. (March – 2011)
a) What do you mean by ‘Colligative Property’?
b) Fordeterminingthe molecular mass of polymers, osmotic pressure is preferred to other properties. Why?
c) For intravenous injections only solutions with freezing point depression equal to that of 0.9% NaCI solution is used. Why?
Answer:
a) Colligative properties are those properties which depend upon no. of particles in the solution.

b) i) Pressure measurement can be done around the room temperature.
ii) Molarity of the solution is used instead of molality.
iii) Its magnitude is large compared to other colligative properties.
iv) Polymers have poor solubility.

c) This is because the osmotic pressure associated with the fulid inside the blood cell is equivalent to that of 0.9 (m/w) sodium chloride solution, i.e., blood is isotonic with this solution.

Question 4.
Relative lowering of vapour pressure, elevation of boiling point, depression of freezing point, and osmotic pressure are important colligative properties of dilute solutions. (May – 2011)
a) Relative lowering of vapour pressure of an aqueous dilute solution of glucose is 0.018. What is the mole fraction of glucose in the solution?
b) An aqueous dilute solution of a non-volatile solute boils at 373.052 K. Find the freezing point of the solution.
For water K_b = 0.52K Kg mol^-1
For water K_f = 1.86 K Kgmol^-1
Normal boiling point of water = 373K
Normal freezing point of water = 273K
Answer:
a) 0.018
Because according to Raoult’s law for solutions containing non-volatile solutes, the relaive lower-ing of vapour pressure is equal to the mole fraction of the non-volatile solute.

Question 5.
Vapour pressure of a solution is different from that of pure solvent. (March – 2012)
i) Name the law which helps us to determine partial vapour pressure of a volatile component in solution.
ii) State the above law.
iii) Vapour pressure of chloroform (CHCI₃) and dichloro methane (CH₂CI₂) at 298 K are 200 mm and 415 mm of Hg respectively. Calculate the vapour pressure of solution prepared by mixing 24 g of chloroform and 17 g of dichloro methane at 298 K. [At. Mass : H -1, C – 12, Cl – 35.5]
Answer:
i) Raoult’s law.

ii) It states that at a given temperature for a solution of volatile liquids, the partial v.p. of each component in the solution is directly proportional to its mole fraction.

Question 6.
Colligative properties are properties of solutions which depend on the number of solute particles in the solution. (May – 2012)
i) Write the names of four important colligative properties.
ii) The value of van’t Hoff factor, ‘i’ for aqueous KCI solution is close to 2, while the value of ‘i’ for ethanoic acid in benzene is nearly 0.5. Give reason.
i) The Colligative Properties are:
1) Relative lowering of vapour pressure
2) Elevation of boiling point
3) Depression of freesing point
4) Osmotic pressure

ii) This is caused by dissociation in the case of KCI and association in the case of acetic acid.

KCI in aqueous solution undergoes dissociation as KCI → K⁺ + Cl^–

Thus, if complete ionisation occurs the number of particles in solution becomes double and hence van’t Hoff factor (i) for aqueous KCI solution is close to 2.

In the case of ethanoic acid (acetic acid) association (dimerisation) occurs in benzene through in- termolecular hydrogen bonding. Thus, if complete association occurs the number of particles in solution becomes half and hence van’t Hoff factor (i) for ethanoic acid in benzene is nearly 0.5.

Question 7.
Elevation of boiling point is a colligative property, (March – 2013)
i) What are colligative properties?
ii) Elevation of boiling point (D T_b) is directly proportional to molality (m) of solution.
Thus, DT_b = K_bm, K_b is called the molal elevation constant.
From the above relation derive an expression to obtain molar mass of the solute,
iii) The boiling point of benzene is 353.23K. When 1.80 g of a non-volatile solute was dissolved in 90g of benzene, the boiling point is raised to 354.11K. Calculate the molar mass of the solute. K_b for benzene is 2.53K kg mol^-1.
Answer:
i) Colligative properties are those properties which depend upon the number of solute particles irrespective of their nature relative to the total number of particles present in the solution.

Question 8.
Liquid solutions can be classified into ideal and non-ideal solutions on the basis of Raoult’s law. (May – 2013)
a) State Raoult’s law.
b) What are ideal solutions?
c) Write two important properties of ideal solutions.
d) What type of deviation is shown by a mixture of chloroform and acetone? Give reason.
Answer:
a) Raoult’s law states that for a solution of volatile liquids, the partial vapour of each component in the solution is directly proportional to its mole fraction. Or The partial vapour pressure of any volatile component of a solution is equal to the product of the vapour pressure of pure component and mole fraction of that component in the solution.

b) Ideal solutions are solutions which obey Raoult’s law over the entire range of concentration.

c) i) Enthalpy of mixing of the pure components to form the solution is zero, i.e.; Δ_mixH = 0
ii) Volume of mixing is zero, i.e.; Δ_mixV = 0

d) Negative deviation. This is because chloroform molecule is able to form hydrogen bond with acetone molecule.

Question 9.
Osmotic pressure is a colligative property and it is proportional to the molarity of the solution. (March – 2014)
a) What is osmotic pressure?
b) Molecular mass of NaCI determined by osmotic pressure measurement is found to be half of the actual value. Account for it.
c) Calculate the osmotic pressure exerted by a solution prepared by dissolving 1.5 g of a polymer of molar mass 185000 in 500 mL of water at 37°C. [R = 0.0821 L atm K^-1 mol^-1].
Answer:
a) Osmotic pressure is the extra pressure applied on the solution to stop osmosis
b) NaCI, as a strong electrolyte dissociates to form Na+ and Ch ions. The number of particles doubles colligative property also doubles. Observed molecular mass will be half.

Question 10.
Molarity (M), molality (m) and mole fraction (x) are some methods for expressing concentrations of solutions. (May – 2014)
a) Which of these are temperature independent?
b) i) Define ‘molefraction’.
ii) A mixture contains 3.2 g methanol (molecular mass = 32 u) and 4.6 g ethanol, (molecular mass = 46 u) Find the molefraction of each ’ component)
Answer:
a) Molality and Mole fraction are temperature independent because these are mass to mass relationships. Mass is independent of temperature,
b) i) Mole fraction is defined as the ratio of the number of moles of that component to the total number of moles of all the components in solution.

Question 11.
a) Among the following which is not a colligative property? (March – 2015)
i) Osmotic pressure
ii) Elevation of boiling point
iii) Vapour pressure
iv) Depression of freezing point

b) i) 200 cm³ of an aqueous solution of a protein contains 1.26 g of protein. The osmotic pressure of solution at 300 K is found to be 8.3 x 10^-2 bar. Calculate the molar mass of protein. (R = 0.083 L bar K^-1 mol^-1)
ii) What is the significance of van’t Hoff factor?
Answer:
a) iii) Vapour pressure
b) i) Osmotic pressure, n = 8.3 x 10^-2 bar
Volume of the solution, V = 200 cm³ = 0.200 L
Temperature, T = 300 K
R = 0.083 L bar mol^-1 K^-1

Thus, in the case of association, the value of ‘i’ is less than unity while for dissociation ‘i’ is greater than unity. If i =1 it means that there is no association or dissociation of the solute particles in solution.

Question 12.
a) Draw a vapour pressure curve, by plotting vapour pressure against mole fraction of an ideal solution of two volatile components A and B (not to scale). Indicate partial vapour pressure of A and B (P_A and P_B) and total vapour pressure (P_total) (May – 2015)
b) What is an ideal solution?
c) Modify the above plot for non-ideal solution showing positive deviation. (Draw the above plot once again and modify).
Answer:

b) A solution which obeys Raoult’s law over the entire range of concentration is known as an ideal solution.

Question 13.
a) Number of moles of the solute per kilogram of the solvent is (March – 2016)
a) Mole fraction
b) Molality
c) Molarity
d) Molar mass

b) The extent to which a solute is dissociated or associated can be expressed by Van’t Hoff factor.’ Substantiate the statement.
c) The vapour pressure of pure benzene at a certain temperature is0.850 bar. Anon volatile, non-electrolyte solid weighing 0.5 g when added to 39 g of benzene (molar mass 78 g mol^-1), the vapour pressure becomes 0.845 bar. What is the molar mass of the solid substance?
Answer:
a) b) Molality
b) b) The van’t Holf factor,

When i = 1 ⇒ there is no association or dissociation of solute particles.
When i < 1 ⇒ there is association of solute particles. When i > 1 ⇒ there is dissociation of solute particles.

Question 15.
Osmotic pressure is a colligative property. (May – 2016)
a) What is osmotic pressure?
b) 1.00 g of a non-electrolyte solute dissolved in 50g of benzene lowered the freezing point of benzene by 0.40 k. The freezing point depression constant of benzene is 5.12 K kg/mol. Find the molar mass of solute.
Answer:
a) Osmotic pressure is the extra pressure applied on the solution side to just stop osmosis i.e., the flow of the solvent from its side to solution side through the semipermeable membrane,

Question 16.
a) Henry’s law is related to solubility of a gas in liquid. (March – 2017)
i) State Henry’s law.
ii) Write any two applications of Henry’s law.
b) 1000cm³ of an aqueous solution of a protein contains 1.26 gm of the protein. The osmotic pressure of such a solution at 300 K is found to be 2.57 x 10^-3 bar. Calculate molar mass of the protein. (R = 0.083 L bar mol^-1 K^-1)
Answer:
a) i) Henry’s law states that at constant tempera-ture, the solubility of a gas in a liquid is directly proportional to the pressure of the gas. p = K_H x where, p is the partial pressure of the gas in vapour phase, K_H is the Henry’s law constant and x is the mole fraction of the gas in the solution.

ii) 1. To increase the solubility of CO₂ in soft drinks and soda water, the bottle is sealed under high pressure.

2. To avoid bends and the toxic effects of high concenration of nitrogen in blood, the tanks used by scuba divers are filled with air diluted with helium.

3. Low partial pressure of oxygen at high altitudes leads to low concentration of oxygen in the blood and tissues of people living at high altitudes or climbers and causes anoxia. (Any two applications)

Question 17.
a) The mole fraction of water in a mixture containing equal number of moles of water and ethanol is (May – 2017)
i) 1
ii) 0.5
iii) 2
iv) 0.25

b) The following are the vapour pressure curves of a pure solvent and a solution of a non-volatile solute in it.

Based on the above curves answer the following questions:
i) What do the curves A and B indicate?
ii) Explain why the value of TB is greater than that of Tb⁰.
Answer:
a) ii) 0.5
b) i) A-Vapour pressure curve of solvent Vapour pressure curve of solution
ii) Due to the presence of a non-volatile solute vapour pressure of solution is less than solvent and the boiling point is increased.

Kerala State Board New Syllabus Plus Two Chemistry Chapter Wise Previous Questions and Answers Chapter 1 The Solid State.

Kerala Plus Two Chemistry Chapter Wise Previous Questions Chapter 1 The Solid State

Plus Two Chemistry The Solid State 4 Marks Important Questions

Question 1
a) Schottky defects and Frenkel defects are two stoichiometric defects shown by crystals. (March – 2010)
i) Classify the following crystals into those showing Schottky defects and Frenkel defects:
NaCI, AgCI, CsCI, CdCI₂
ii) Name a crystal showing both Schottky defect and Frenkel defect.
b) Schematic alignment of magnetic moments of ferromagnetic, antiferromagnetic and ferrimagnetic substances are given below. Identify each of them.
i) ↑↓↑↓↑↓↑↓
ii) ↑↑↓↑↓↑↑↑
iii) ↑↑↑↑↑↑↑↑
Answer:
a) i) Schottchydefect – NaCI, CsCI Frenkel defect – AgCI, CdCI₂
ii) AgBr

b) i) Antifero magnetism
ii) Ferrimagnetism
iii) Ferromagnetism

Question 2.
Based on the nature of order present in the arrangement of the constituent particles, solids are classified into two, crystalline and amorphous. (May – 2010)
a) List out any four points of difference between crystalline and amorphous solids.
b) A list of solids are given below:
Quartz, glass, iodine, ice.
From this, identify crystal (s)
i) having sharp melting point.
ii) which is/are isotropic
Answer:

Crystalline	Amorphous
i) Long-range order ii) Sharp melting point iii) Newly formed surface is smooth iv)  Anisotropic	i) Short-range order ii) Range of melting point iii) Newly formed surface is rough. iv) Isotropic

b) i) Quartz, Iodine, Ice
ii) Glass

Question 3.
Cristal defects give rise to certain special properties in the solids. (March – 2011)
a) What is meant by Frenkel Defect?
b) Why does LiCI not exhibit Frenkel Defect?
c) Explain the pink colour of LiCI when heated in . the vapours of Li.
Answer:
a) The dislocation of a cation from its original site to an interstitial site. It creates a vacancy defect at its original site and an interstitial defect at its new location.
b) The size of the cation is bigger than the void.
c) Due to F – centre. It is an electron trapped anion vacancy.

Question 4.
A cubic unit cell is characterized by a = b = c and α = β = γ = 90° (May – 2011)
a) Name three important types of cubic unit cells and calculate the number of atoms in one unit cell in the above three cases.
b) A metal forms cubic crystals. The mass of one unit cell of it is M/NA gram, where M is the atomic mass of the metal and NA is Avogardo Number. What is the type of cubic unit cell possessed by the metal?
Answer:
a) Simple cubic unit cell or Primitive unit cell, Body – centred cubic unit cell (bcc) and Face – centred cubic unit cell (fee).
b) Primitive cubic unit cell: This unit cell has atoms only at its comers. There are 8 corners for a cube.
Contribution by atom at the corner = 1/8
Total number of atoms in one unit cell \(=8 \times \frac{1}{8}=1\) atom

Body – centred cubic unit cell:
This unit cell has an atom at each of its corners and also one atom at its body centre.
8 corners \(\times \frac{1}{8}\) per corner atom \(=8 \times \frac{1}{8}=1\) atom
1 body centre atom = 1 x 1 = 1 atom
∴ Total number of atoms per unit cell = 2 atoms

Face – centred cubic unit cell:
This unit cell contains atoms at all the corners and at the centre of all the faces of the cube.
8 corners \(\times \frac{1}{8}\) per corner atom \(=8 \times \frac{1}{8}=1\) atom Contribution by an atom at the face centre = \(\frac{1}{2}\)

6 face – centred atoms x \(\frac{1}{2}\) atom per unit cell \(=6 \times \frac{1}{2}=3\) atoms

∴ Total number of atoms per unit cell = 4 atoms

c) Mass of one unit cell = \(\frac{\mathrm{M}}{\mathrm{N}_{\mathrm{A}}} \mathrm{g}\)

Mass of 1 mole of unit cells \(\mathrm{N}_{\mathrm{A}} \times \frac{\mathrm{M}}{\mathrm{N}_{\mathrm{A}}}\) = M gram = Gram atomic mass It means that 1 unit cell contains one atom of the metal. Hence, the type of unit cell is primitive cubic or simple cubic.

Question 5.
Solids can be classified into three types on the basis of their electrical conductivities. (March – 2012)
i) Name three types of solids classified on the basis of electrical conductivities.
ii) How will you explain such classification based on Band theory?
Answer:
i) Conductors, insulators & semiconductors
ii) In conductors, the valance band overlaps with Ir e conduction band or no energy gap exists between the valance band and conduction band.

∴ The electrons can easily go into the conduction band and hence metals are good conductors. In insulators, the energy gap between valance band and conduction band is very large. Hence electrons from valance band cannot move into the conduction band.

Semi conductors have energy gap between conductors and insulators. At room temperature, these are not good conductors. But with an in crease in temperature electrons acquire sufficient energy to move from valance band into conduction band resulting in an increase in conductivity.

Question 6.
Schottky and Erenkel defects are stoichiometric defects. (May – 2012)
i) Write any two differences between Schottky defect and Frenkel defect.
ii) When pure NaCI (Sodium Chloride) crystal is heated in an atmosphere of sodium vapours, it turns yellow. Give reason.
Answer:
i)

Schottky defect	Frenkel defect
1. Vacancy defect which arises due to the messing of equal number of cations and anions from the lattice sites.	Interstitial defector dislocation defect which arises when the smaller ion, usually cation is dislocated from its normal site to an interstitial site.
3. Cation and anion in are of almost similar sizes	there is a large difference size of ions
4. The density of the crystal is lowered	It does not affect the density of the crystal

ii) It is caused by metal excess defect due to anion vacancies. When crystals of NaCI are heated in a atmosphere of sodium vapour, the sodium atoms are deposited on the surface of the crystal. The Cl- ions diffuse to the surface of the crystal and combine with Na atoms to give NaCI. The electrons released from Na atoms diffuse into the crystal and occupy anionic sites to form Fcentres, which impart yellow colourto the crystal. The colour results by excitation of these electrons when they absorb energy from the visible light falling on the crystals.

Question 7.
a) NaCI has fcc structure. Calculate the number of NaCI units in a unit cell of NaCI. (March – 2013)
b) Calculate the density of NaCl, if edge length
of NaCI unit cells is 564pm. [Molar mass of
NaCI =58.5g/mol].
Answer:
a) Number of Na ions = 12 (at edge centres) x \(\frac{1}{4}\) +
= 1(at body centre) x 1
= 3 + 1 = 4
Number of Cl- ions = 8 (at the corners) x \(\frac{1}{8}\) +
= 6 (at face centres) x \(\frac{1}{2}\)
= 1 + 3 = 4
∴ Number of NaCI units per unit cell (z) = 4

Question 8.
Unit cells can be broadly classified into 2 categories primitive and centred unit cells. (May – 2013)
a) What is a unit cell?
b) Name the three types of centred unit cells.
c) The unit cell dimension of a particular crystal system is a = b = c, α = β = γ = 90°. ldentify the crystal system.
d) Give one example for the above crystal system.
Answer:
a) The smallest repeating unit of a crystal.
b) Body centred unit cell, Face centred unit cell and End centred unit cell.
c) Cubic crystal system.
d) NaCI (Rock salt struãture)

Question 9.
a) Every substance has some magnetic properties associated with it. How will you account for the following magnetic properties? (March – 2014)
i) Paramagnetic property
ii) Ferromagnetic property

b) A compound is formed by two elements P and Q. Atoms of Q (as anions) make hep lattice and those of the element P (as cations) occupy all the tetrahedral voids. What is the formula of the compound?
Answer:
a) i) Paramagnetic substances are weakly attracted by a magnetic field. They are magnetised in a magnetic field in the same direction. They lose their magnetism in the absence of magnetic field. Paramagnetism is due to presence of one or more unpaired . electrons, e g., O₂, Cu²⁺

ii) Ferromagnetic substances are strongly attracted by a magnetic field. They are permanantly magnetised. When a ferromagnetic substance is placed in a magnetic field all the magnetic domains get oriented in the direction of the magnetic field and a strong magnetic effect is produced, e.g,, Fe, Co

For hep lattice, No. of particles per unit cell = 6

∴ Number of anions of Q in the unit cell = 6
Number of tetrahedral voids = 2 x N = 2 x 6 = 12
∴ Number of cations of P in the unit cell = 12
Hence, formula of the compound = P₁₂Q₂ = P₂Q

Question 10.
a) Crystalline solids are ‘anisotropic’. What is ‘anisotropy’? (May – 2014)
b) Copper crystals have fee unit cells.
i) Compute the number of atoms per unit cell of copper crystals.
ii) Calculate the mass of a unit cell of copper crystals. (Atomic mass of copper = 63.54 u)
Answer:
a) Anisotropy means physical properties shows different values along different directions, eg. refrative index electrical resistance,
b) i) Face centred cubic unit cell (fee) – It contains atoms at all the corners and at the centre of all the faces of the cube.
8 corners x \(\frac{1}{8}\) per corner atom

Question 11.
Unit cells can be divided into two categories, primitive and centred unit cells. (March – 2015)
a) Differentiate between Unit Cell and Crystal Lattice.
b) Calculate the number of atoms per unit cell in the following:
i) Body centred cubic unit cell (bcc)
ii) Face centred cubic unit cell (fee)
Answer:
a) Unit Cell – It is the smallest portion of a crystal lattice which, when repeated in different directions, generates the entire lattice.

Crystal Lattice – It is the regularthree dimensional arrangement of points in space.

b) i) Body centred cubic unit cell (bcc) – It has atoms at each of its corners and one atom at its body centre.
8 corners x \(\frac{1}{8}\) per corner atom = 8 x \(\frac{1}{8}\) = 1 atom
1 body centre atom = 1 x 1 = 1 atom
∴ Total number of atoms per unit cell = 2 atoms

ii) Face centred cubic unit cell (fee) – It contains atoms at all the corners and at the centre of all the faces of the cube.

8 corners x \(\frac{1}{8}\) per corner atom
\(=8 \times \frac{1}{8}=1\) atom

6 face – centred atoms x \(\frac{1}{2}\) atom per unit cell \(=6 \times \frac{1}{2}=3\) atoms
∴ Total number of atoms per unit cell = 4 atoms

Question 12.
a) Which of the following is not a characteristic of a crystalline solid? (May – 2015)
i) Definite heat of fusion
ii) Isotropic nature
iii) A regular ordered arrangement of constituent particles
iv) A true solid

b) Frenkel defect and Shottky defects are two stoichiometric defects found in crystalline solids.
i) What are stoichiometric defects?
ii) Write any two differences between Frenkel defect and Schottky defect.
Answer:
a) ii) Isotropic nature
b) i) Stoichiometric defects are those point defects which do not disturb the stoichiometry of the solid.

ii)

Schottky defect	Frenkel defect
1. Cation and anion in are of almost similar sizes	there is a large difference size of ions
2. The density of the crystal is lowered	It does not affect the density of the crystal

Question 13.
a) Which of the following is a molecular solid? (March – 2016)
a) Diamond
b) Graphite
c) Ice
d) Quartz

b) Unit cells can be classified into primitive and centered unit cells. Differentiate between primitive and centered unit cells.
c) Presence of excess Sodium makes NaCI crystal coloured. Explain on the basis of crystal defects.
Answer:
a) Ice
b) In primitive unit cell constituent particles are present only at the corner positions. Unit cells in which one constituent particles are present at the centres of a faces in addition to those at corners.

c) Such anionic sites occupied by unpaired electrons are called F – centres (colour centres). They impart yellow colourto the crystals of NaCI. The colour results by excitation of these electrons when they absorb energy from the visible light falling on the crystals.

Question 14.
A unit cell is a term related to crystal structure. (May – 2016)
a) What do you mean by unit cell?
b) Name any two types of cubic unit cells.
c) Calculate the number of atoms in each of the above – mentioned cubic unit cells.
d) Identify the substance which shows Frenkel defect:
i) NaCI
ii) KCI
iii) ZnS
iv) AgBr
Answer:
a) Unit cell is the smallest portion of a crystal lattice which, when repeated in different directions, generates the entire lattice.
b) Simple cubic unit cell, body centred cubic (bcc) unit cell, face centred cubic (fee) unit cell (any two)
c) Number of atoms per unit cell
i) Simple cubic unit cell:
Total number of atoms in one unit cell \(=8 \times \frac{1}{8}=1\) atom

ii) Body centred cubic (bcc) unit cell: Total number of atoms in one unit cell
\(=\left(8 \times \frac{1}{8}\right)+(1 \times 1)=1+1=2\) atoms

iii) Face centred cubic (fee) unit cell: Total number of atoms in one unit cell \(=\left(8 \times \frac{1}{8}\right)+\left(6 \times \frac{1}{2}\right)=1+3=4\)
(any two required)

d) ZnS or AgBr
(AgBr shows both Schottky and Frenkel defects)

Question 15.
a) Identify the non – stoichiometric defect (March – 2017)
i) Schottky defect
ii) Frenkel defect
iii) Interstitial defect
iv) Metal deficiency defect
b) What type of substance could make better permanent magnets – ferromagnetic or ferrimagnetic? Justify your answer.
c) In terms of Band theory write the differences between conductor and insulator.
Answer:
a) iv) Metal deficiency defect

b) Ferromagnetic substances could make better permanent magnets because when these substances are placed in a magnetic field all the magnetic moments (domains) get oriented in the direction of the magnetic field and a strong magnetic effect is produced. This ordering of domains persist even when the magnetic field is removed and the ferromagentic substance becomes a permanent magnet.

The schematic alignment of magnetic moments in a ferromagnetic substance is as shown below:

c) In conductors the valence band is either partially filled or it is overlaped with a higher energy unoccupied conduction band so that the electrons can flow easily under an applied electric field. Whereas in insulators the energy gap between the filled valence band and the next higher unoccupied conduction band is large so that electrons cannot jump to it.

Question 16.
a) From the following choose the incorrect statement about crystalline solids. (May – 2017)
i) Melt at sharp temperature.
ii) They have definite heat of fusion.
iii) They are isotropic
iv) They have long range order.

b) Cubic unit cells are divided into primitive, bcc and fee.
i) Calculate the number of atoms in a unit cell of each of the following:
* bcc
* fcc

ii) Write two examples for covalent solids.
a) iii) They are isotropic
b) i) \(\begin{array}{l}
\text { bcc }-2\left(8 \times \frac{1}{8}+1=2\right) \\
\text { fCc }-4\left(8 \times \frac{1}{8}+6 \times \frac{1}{2}=4\right)
\end{array}\)
ii) Graphite, Diamond

Kerala State Board New Syllabus Plus Two Chemistry Previous Year Question Papers and Answers.

Kerala Plus Two Chemistry Previous Year Question Paper Say 2018 with Answers

Board	SCERT
Class	Plus Two
Subject	Chemistry
Category	Plus Two Previous Year Question Papers

Time: 2 Hours
Cool off time: 15 Minutes
Maximum: 60 Score

General Instructions to candidates:

• There is a ‘cool off time’ of 15 minutes in addition to the writing time of 2 hrs.
• Use the ‘cool off time’ to get familiar with the questions and to plan your answers.
• Read questions carefully before you answering.
• Read the instructions carefully.
• Calculations, figures, and graphs should be shown in the answer sheet itself.
• Malayalam version of the questions is also provided.
• Give equations wherever necessary.
• Electronic devices except non-programmable calculators are not allowed in the Examination Hall.

Answer all questions from 1 to 7. Each question carries 1 score. (7 × 1 = 7)

Question 1.
If N spheres are there in a close packing, what is the total number of tetrahedral and octahedral voids present in it?
Answer:
Tetrahedral 2N, octahedral N

Question 2.
What is the order of a reaction, if its half life is independent of initial concentration?
Answer:
1^st order

Question 3.
What is the magnetic moment of an atom having d configuration.
Answer:
Zero

Question 4.
Gabriel synthesis of used forthe preparation of which type of amines?
i) Primary
ii) Secondary
iii) Tertiary
iv) Quaternary
Answer:
i) Primary

Question 5.
Which vitamin is responsible for blood clotting?
Answer:
Vitamin K

Question 6.
Name the linear polymer formed during the condensation polymerization between phenol and formaldehyde.
Answer:
bakelite

Question 7.
Which is the chemical substance discovered by Paul Ehlrich for the treatment of syphilis?
Answer:
Salvarsan or Arephenamine

II. Questions from 8 to 20 carry 2 score each. Answer any 10 questions. (10 × 2 = 20)

Question 8.
Draw the vapour pressure-mole fraction curve for a non-ideal solution having positive deviation, if A and B are the two volatile components.
Answer:

Question 9.
Calculate the depression in freezing point of a 0.2 molal solution if kg for water is 1.86 K kg mol^-1.
Answer:
ΔT_f = k_jm
= 1.86 × 0.2
= 0.372K

Question 10.
Suppose you are given a sample of NaCl salt. How will you prepare chlorine gas in laboratory using the above sample? (Write balanced chemical equations)
Answer:
By the electrolysis of sodium chloride solution Cl₂ gas can be prepared
2NaCl + 2H₂O → 2Na⁺ + 2OH^– + H₂⁺_(g) + Cl_2(g)

Question 11.
Give one use each of Freon 12, DDT, CCl4and CHl3.
Answer:
Freon – 12-Refrigerant
DDT – insecticide
CCl₄ – For the manufacture of refrigerant
CHl₃ – Antiseptic

Question 12.
Write equations showing Wurtz-Fittig reaction and Fittig reaction.
Answer:
Fitting reaction

Question 13.
Identify A and B in the following equations:

Answer:

Question 14.
How the conversion of carbon dioxide to carboxylic acid can be effected using. Grignard reagent?
Answer:

Question 15.
Complete the following equations:


Answer:

Question 16.
Describe primary and secondary structure of proteins.
Answer:
Structure of proteins:
1) Primary structures – amino acids are arranged in sequence.
2) Secondary structure:
a) α – helix – polypeptide chains are coild to form a helical structure eg. Myosine
b) β-pleated structure: amino acid chains lie side by side and bonded by hydrogen bonds, eg. Keratine.

Question 17.
Explain homopolymers and copolymers with examples.
Answer:
Polymers formed by polymerisation of one type of monomer are called homopolymer, eg. Polythene Polymers formed by polymerisation 0 two or more different monomers are called copolymers eg. SBR, Rubber, Nylon 6, 6.

Question 18.
Briefly explain different types of artificial sweetening agents.
Answer:
Commonly used artificial sweetener is saccharin. It is 550 times sweeter than sucrose. Alitame is 1000 times sweet as cane sugar.
Aspartame, monolellin etc are other sweetening agent.

Question 19.
Write the IUPAC names of the following compounds:
a) [Ni(CO)₄]
b) K₃[Fe(C₂O₄)₃]
Answer:
a) Ni(CO)₄ Tetra carbonyl nickel (0)
b) K₃ [Fe(C₂O₄)₃] Potassium tris oxalate ferrate iii

Question 20.
Distinguish Ferromagnetism and Ferrimagnetism.
Answer:
Ferromagnetic substance – Magnetic moments are in one direction.

They are strongly attracted my magnetic field, eg Fe, Co
Ferrimagneticsubstances: Magnetic moments are unequal and in opposite direction.

eg. Fe₃O₄, MgFe₂O₄

III. Questions from 21 to 29 carry 3 score each. Answer any 7 questions. (7 × 3 = 21)

Question 21.
Silver atoms are arranged in CCP lattice structure. The edge length of its unit cell is 408 pm. Calculate the density of silver. (Atomic mass of silver is 108.4)
Answer:

Question 22.
The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction assuming that it does not change with temperature.
Answer:

Question 23.
Explain any three chemical methods for the preparation of Lyophobic colloids with suitable examples.
Answer:
1) Oxidation methods: Oxidation of aqueous solution of H₂S with SO₂
SO₂ + 2H₂S → 3S + 2H₂O
2) Reduction method : Reduction of AuCl₃ solution using SnCl₂
2AuCl₃ + 3SNCl₂ → 3SnCl₄ + 2Au
3) Hydrolysis. By adding a saturated solution of ferric .chloride dropwise to a large excess of boiling water
FeCl₃ + 3H₂O → Fe(OH)₃ + 3HCl

Question 24.
Explain the following refining processes:
a) Distillation
b) Vapour phase refining
c) Zone refining
Answer:
a) Distillation: The impure metal is heated to form pure metals as distillate it is collected and condensed impurities are left behind, eg Zn and Hg.
Only metals with low boiling point can apply this method.

b) Vapour phase refining
i) Van Arkel method – eg Titanium, Zirconium, Thorium etc.

Mond process: Nickel is strongly heated with carbon monoxide to form Nickel tetra carbonyl this is again heated strongly to get pure nickel.

c) The impure metal bar is heated at one end with moving circular heater. The heater is now slowly moved along the rod. The pure metal recrystallises from the melt while impurities remain in the melt. Finally the end where impurities have collected is cut off. The impure metal bar is heated at one end with moving circular heater. The heater is now slowly moved along the rod. The pure metal recrystallises from the melt while impurities remain in the melt. Finally the end where impurities have collected is cut off.

Question 25.
A solution of CuSO₄ is electrolysed for 20 minutes with a current of 1.5 amperes. What is the mass of copper deposited at cathode?
(Atomic mass of copper – 63)
Answer:
Cu²⁺ + 2e → Cu
Q = It   1.5 × 20 × 60 = 1800 C
Mass of Cu deposited by 1800 C
\(\frac{63.5 \times 1800}{2 \times 96500}\) = 0.5875 g

Question 26.
Briefly explain the manufacture of sulphuric acid by contact process.
Answer:
Contact Process
1) Sulphur is burnt in air to form Sulphur Dioxide
S + O₂ → SO₂
2) Sulphur Dioxide is again oxidised to SO₃ with atmospheric oxygen in the presence of

3) SO₃ is treated with Sulphuric acid to get Oleum
SO₃ + H₂SO₄ → H₂S₂O₇
4) Oleum is diluted to get H₂SO₄
H₂S₂O₇ + H₂O → 2H₂SO₄

Question 27.
Explain with the help of equations, preparation of Xenon fluorides.
Answer:
Xe + F₂ → XeF₂
Xe + 2F₂ → XeF₄
Xe + 3F₂ → XeF₆

Question 28.
Describe lanthanoid contraction. Write any two consequences of it.
Answer:
The steady but slow decrease in the size of atoms or ions of the lanthanoids with increase in atomic number is called Lanthanoid Contraction.

Consequences:

1. As the size of the Lanthanoid ions decreases from La to Lu. The covalent character of hydroxides increases and hence the basic strength decreases.
2. The change in ionic radii of lanthanoids is very small their properties are almost similar. This makes the separation of lanthanoids are very difficult.

Question 29.
How the conversion of an aldehyde to acetal can carried out?
(Write chemical equations)
Answer:

IV. Questions from 30 to 33 carry 4 score each. Answer any 3. (3 × 4 = 12)

Question 30.
Predict the products of electrolysis of the following substances at anode and cathode using suitable chemical equations.
a) Aqueous NaCl
b) H₂SO₄ solution
Answer:
Electrolysis of aqeous NaCl
a) At Cathode H⁺ + 1e^– → \(\frac{1}{2}\)H₂
As the standard reduction potential for H⁺ ions are more it is easily reduced at cathode.
At anode Ch ions are oxidised Cl^– → Cl + e
2Cl → Cl_2(g)

b) Electrolysis of Sulphuric acid (dilute)
At anode
2H₂O → O₂ + 4H⁺ + 4e^–
At cathode
H⁺ + 1e → H
H + H → H₂
Electrolysis of concentrated H₂SO₄
At cathode
H⁺ + 1e → H
H + H → H₂
At anode
2SO\(\mathrm{O}_{4}^{2-}\) → S₂O\(\mathrm{O}_{8}^{2-}\) + 2e

Question 31.
Draw a diagram depicting crystal field splitting in an octahedral environment of d-orbitals. Label the diagram properly. Calculate the crystal field stabilization energy for a d³ configuration.
Answer:
Crystal field splitting in octahedral field.

CFSE for d³ configuration in octa hedral field. CFSE for 3 unpaired electrones
0 – \(\frac{2}{5}\)Δ₀ × 3 = \(\frac{-6}{5}\)Δ₀

Question 32.
a) Predict the products A and B.

b) How methanol is prepared industrially?
Answer:
2CH₃ – CH = CH₂ + (BH₃)₂ → (CH₃ – CH₂ – CH₂)₃ B → CH₃ – CH₂ – CH₂ – OH
By the catalytic hydrogenation of carbon monoxide in presence of a catalyst at 573K and under 200 to 300 atmospheric pressure to form methanol

Question 33.
a) Symbolically represent standard hydrogen electrode, when it acts as an anode and as cathode.
b) Write Nernst equation for a Daniel cell.
(Assume activity of metals is unity)
Answer:

Kerala State Board New Syllabus Plus Two Chemistry Previous Year Question Papers and Answers.

Kerala Plus Two Chemistry Previous Year Question Paper March 2019 with Answers

Board	SCERT
Class	Plus Two
Subject	Chemistry
Category	Plus Two Previous Year Question Papers

Time: 2 Hours
Cool off time: 15 Minutes
Maximum: 60 Score

General Instructions to candidates:

• There is a ‘cool off time’ of 15 minutes in addition to the writing time of 2 hrs.
• Use the ‘cool off time’ to get familiar with the questions and to plan your answers.
• Read questions carefully before you answering.
• Read the instructions carefully.
• Calculations, figures, and graphs should be shown in the answer sheet itself.
• Malayalam version of the questions is also provided.
• Give equations wherever necessary.
• Electronic devices except non-programmable calculators are not allowed in the Examination Hall.

I. Answer all questions from 1 to 7. Each carries 1 score. (7 × 1 = 7)

Question 1.
The monomeric unit of natural rubber is …………
Answer:
Isoprene or 2-methyl 1, 3 butadiene or

Question 2.
The weakest reducing agent among the hydrides of group 15 elements is …………
Answer:
Ammonia or NH₃.

Question 3.
The reaction in which an amide is converted into a primary amine by the action of Br₂ and alcoholic NaOH is known as ………….
Answer:
Hoffmann bromamide reaction.

Question 4.
\(\mathrm{MnO}_{4}^{-}\) and ……… are formed by the disproportionation of \(\mathrm{MnO}_{4}^{2-}\) in acidic medium.
Answer:
MnO₂ or Mn⁴⁺.

Question 5.
In a solution of components‘A’ and ‘B’, at molecular level, A – B interactions are weaker than those between A – A or B – B interactions. Then the type of deviation shown by this solution is called ……………
Answer:
Positive deviation.

Question 6.
Identify the co-ordination compound which can exhibit linkage isomerism, among the following.
a) [Pt(NH₃)2Cl₂)
b) [Co(NH₃)₅(SO₄)]Br
c) [CO(NH₃)₅(NO₂)]Cl₂
d) [Cr(NH₃)₆][CoF₆]
Answer:
c) [CO(NH₃)₅(NO₂)]Cl₂

Question 7.
For the reaction, 2NO_(g) + O_2(g) → 2NO_2(g), the rate law is given as,
Rate = k[NO]² [O₂], The order of the reaction with respect to O₂ is …………
Answer:
one

II. Answer any ten questions from 8 to 20. Each carries 2 scores. (10 × 2 = 20)

Question 8.
Write the chemical equation representing Reimer-Tiemann reaction.
Answer:

Question 9.
What is reverse osmosis? Write any one of its applications.
Answer:
If the pressure applied is larger than osmotic pressure the direction of osmosis gets reversed. It is called reverse osmosis.
Aplication:

1. Desalination of sea water.
2. Purification of drinking water.

Question 10.
Identify the products and give the name of the following reaction:

Answer:
Cannizzaro reaction:

Question 11.
Explain Haloform reaction.
Answer:
Compounds with CH₃ – CO or CH₃ – CH – OH gp only gives haloforms such compounds when react with sodium hypohalite or a mixture of halogen and NaOH gives haloform.

Question 12.
What is meant by step growth polymerisation? Give an example.
Answer:
Step growth polymers are condensation polymers. The eliminate simple molecules like water or ammonia on addition, eg. nylon 6,6

Question 13.
An element crystallizes in F.C.C. manner. What is the length of a side of the unit cell, if the atomic radius of the element is 0.144 nm?
Answer:
a = 2\(\sqrt{2}\)r
r = 0.144 nm
= 2\(\sqrt{2}\) × 0.144 = 0.407 nm

Question 14.
Draw the structure of H₃PO₂ and account for its reducing character.
Answer:

Due to the presence of two P – H bonds it is a strong reducing agent.

Question 15.
2-Bromobutane is optically active. Explain the stereo¬chemical aspect of SN¹ reaction of 2-Bromobutane with OH- ions.
Answer:

Question 16.
Briefly explain the different types of emulsions and give examples for each.
Answer:
Emulsion: Both dispersed phase & dispersion medium are liquids:
1) Oil in water type, e.g. milk, vanishing cream.
2) Water in oil type, butter.

Question 17.
Give the structural formula and IUPAC name of the product formed by the reaction of propanone with CH₃MgBr in dry ether, followed by hydrolysis.
Answer:

Question 18.
Examine the graph given below. Identify the integrated rate equation and the order of the reaction corresponding to it.

Answer:
Order – zero
Rate equation for zero order K = \(\frac{[\mathrm{Ro}]-[\mathrm{R}]}{\mathrm{t}}\)

Question 19.
How is primary amine distinguished from a secondary amine using a chemical test?
Answer:
A) Carbyl amine test: Only primary amine gives carbyl amine test. Secondary amine does not give this test, (foul smelling gas).
(OR)
B) Hinsberg test: Primary amine react with benzene sulphonyl chloride to form a precipitate which is soluble in alkali. But the precipitate formed by the secondary ammine is insoluble in alkali.
(OR)

Question 20.
Predict the products obtained by the reaction of 2-methoxy-2-methyl propane with HI.
Answer:

III. Answer any seven questions from 21 to 29. Each carries 3 scores. (7 × 3 = 21)

Question 21.
Explain the terms, Zeta potential, electrophoresis and electro-osmosis.
Answer:
Zeta potential: The potential difference between the fixed layer and the diffused layer of an electrical, double layer.
Electrophoresis: The movement of colloidal particles under the influence of an electric field.
Electro osmosis: The movement of particles of dispersion median under the influence of electric field.

Question 22.
The rate constant of a reaction at 293 K is 1.7 × 10⁵ s^-1. When the temperature is increased by 20K, the rate constant is increased to 2.57 × 10⁶ s^-1. Calculate Ea and A of the reaction.
Answer:

Question 23.
Identify A, B and C in the following sequence of reactions:

Answer:
A = CH₃CONH₂
B = CH₃-COOH
C = CH₂Br – COOH

Question 24.
Briefly explain different types of neurologically active drugs and give exmaple for each type.
Answer:
Tranquilizers: Medicines or chemicals used for mental-stress, e.g. Equanil, barbituric acid, veronal, luminal, seconal (anyone).
Analgesics are painkillers and are also neurologically active drug, e.g. paracetamol, morphine, heroine.

Question 25.
Write any three applications of d-block and f-block elements.
Answer:

1. They act as catalyst.
2. They form alloys.
3. Cu, Ag Au are coinage metals.
4. They have different oxidation states.

Question 26.
Give the open chain and ring structures of glucose and account for the existence of glucose in two anomeric forms.
Answer:

Question 27.
A 5% solution (by mass) of cane sugar (C₁₂H₂₂O₁₁) in water has a freezing point of 271 K. Calculate the freezing point of 5% (by mass) solution of glucose (C₆H₁₂O₆) in water. Freezing point of pure water is 273.15 K.
Answer:
ΔTf = \(\frac{1000 \mathrm{~K}_{\mathrm{f}} \cdot \mathrm{W}_{2}}{\mathrm{~W}_{1} \cdot \mathrm{M}_{2}}\)
for 5% suger solution
W₂ = 5
W₁ = 100 – 5 = 95
M₂ = 342
Tf = 271 k
T°f = 273.15
ΔTt= T°f – Tf = 2.15 K

For 5% gulcose solution W₂ = 5, W₁ = 100 – 5 = 95g, M₂ = 180

Question 28.
Explain the steps involved in the vapour phase refining of Ni and Zr.
Answer:
Ni by Monds process.
Ni is strongly heated at a temperature of 330 – 350K to form nickel tetra carbonyl. It is again heated strongly 450 – 470 to decompose it.

Zr by Van Arkel method.
Zirconium is treated with iodine to form zirconium tetra iodide. It is electrically heated to about 1800K with tungsten filament. Zr I₄ is decomposed
Zr + Zl₂ → Zrl₄
Zrl₄ → Zr + 2l₂

Question 29.
What are inter halogen compounds? Which interhalogen compound is used to fluorinate Uranium? How is it prepared?
Answer:
Inter halogen compounds are formed by the direct combination of halogen.

CIF₃ or Br F₃ are used to Fluorinate Uranium.
They are compounds of two or more halogen atoms.

IV. Answer any three questions from 30 to 33. Each carries 4 scores. (3 × 4 = 12)

Question 30.
How can the following conversions be effected?
i) Ethanol-Fluroethane
ii) But-l-ene → But-2-ene
Answer:

Question 31.
Diagrammatically represent H₂ – O₂ fuel cell and write the half cell reactions taking place in this cell.
Answer:

Reaction at cathode
O₂g + 2H_2(l)O + 4\(\bar{e}\) → 4O\(\bar{H}\)_(ag)
Reaction at anode
2H_2(g) + 4O\(\bar{H}\)(aq) → 4H₂O_(I)+ 4\(\bar{e}\)
2H₂ + O₂ → 2H₂O

Question 32.
What are point defects? Explain the non-stoichiometric point defects in ionic crystals.
Answer:
Point defect: The irregularities from ideal arrangement around a point or an atom in crystalline substance.
Non-stoichiometric defect: The defect which do not disturb the stoichiometry of the crystalline substance.
I) Metal excess defect:

• Due to anion vacancies. Electron trapped anion vacancies are called F centre which gives colour.
• Due to extra cation at interstitial site.

II) Metal deficiency defect:

• Due to cation vacancies.

Question 33.
i) With the help of a diagram, give the splitting of d-orbitals of Mn²⁺ ion and octahedral crystal field.
ii) On the basis of crystal field theory, explain why [Mn(H₂O)₆]²⁺ contains five unpaired electrons while [Mn(CN)₆]^4- contains only one unpaired electron.
Answer:
i) Splitting of d orbital in octahedral system.

ii) In [Mn (H₂O)₆]²⁺ Mn²⁺ have five electrons.
As (H₂O) is a weak ligand, i.e., Δ < P no pairing occurs and electronic configuration t₂g³ eg².

But in [Mn(CN)₆]^4- Mn²⁺ has five electrons and are in paired state as CN^– is a strong ligend and Δ > P pairing occurs the electronic configuration is t₂g⁵ eg⁰.

Kerala State Board New Syllabus Plus Two Zoology Chapter Wise Previous Questions and Answers Chapter 8 Biodiversity and Conservation.

Kerala Plus Two Zoology Chapter Wise Previous Questions Chapter 8 Biodiversity and Conservation

Question 1.
Gopalan cultivated a variety of fruit crops and plants in this field and Raman destroyed the fruit crops and plants and cultivated rubber trees which are double in number. (MARCH-2010)
a) In your opinion which method is suitable for the ecosystem? Give reason.
b) Mention any three factors for the extinction of species.
Answer:
Method by Gopalan
Biodiversity depends on variety of species.  Extinction of species is due to

1. Over exploitation
2. Habitat loss and fragmentation
3. Co-extinction

Question 2.
a) The “Evil Quartet” is the nickname used to describe the causes of biodiversity losses. Explain the reason leading to accelerated rates of extinction of flora and fauna.  (MAY-2010)
b) Philosophically or spiritually, we need to realize that every plant or animal species has an intrinsic value and we have a moral duty to protect them. Justify the statement and write down the protective measures.
Answer:
a) (1) Habitat loss and fragmentation
(2) Over-exploitation
(3) Alien species invasions
(4) Co-extinction
b) Agreed with the statement as we have a moral duty to care for their well-being and pass on our biological legacy in good order to future generations.
Protective measures-
Insitu conservation – Biosphere reserve, National park, Wild Life Sanctuary etc.
Exsitu conservation – Zoo, Botanical gardens, seed bank etc.

Question 3.
The year 2010 has been declared as the International Biodiversity Year by United Nations (UN)  (MARCH-2011)
a) Point out the levels of diversity in nature.(1 Score)
b) Give a brief description of The Evil Quartet.
Answer:
a) Genetic, Species, Ecosystem level diversity.
b) i) Habitat loss and fragmentation
ii) Overexploitation
iii) Alien species Invasion
iv) Co-extinction etc.

Question 4.
The given graph shows the distribution of insects in different latitudes of earth.  (MARCH-2012)

a) What is your observation?
b) List the three reasons for greater biodiversity in tropical regions.
c) Write two causes of biodiversity lossess.
Answer:
a) Species richnes decreases from equator to poles.
b) Climate is constant and predictable Glaciations were absent. Tropical region get more solar energy.
c) Habital loss and fragmentation Invansion of alien species.

Question 5.
Last twenty years alone have witnessed the disappearance of 27 animal species from earth.  (MAY-2012)
a) Name an animal disappeared recently.
b) What may be the causes for this loss?
c) How can we conserve biodiversity?
Answer:
a) Out of Syllabus
b) i) Habitat loss and fragmentation
ii) Over-exploitation
iii) Alien species invasions
iv) Co-extinctions
c) It is through in situ (on site) conservation and ex situ (offsite) conservation

Question 6.
Gopalan cultivated a variety of fruit crops and plants . in this field and Raman destroyed the fruit crops and plants and cultivated rubbertrees which are double in number.  (MARCH-2013)
a) In your opinion which method is suitable for the ecosystem? Give reason.
b) Mention any three factors for the extinction of species.
Answer:
Method by Gopalan
Biodiversity depends on variety of species. Extinction of species is due to
1. Over exploitation
2. Habitat loss and fragmentation
3. Co-extinction

Question 7.
While preparing the species area relationship graph of 4 areas, the following z values are obtained.  (MAY-2013)
Area A = 0.1 Area B = 0.8 Area C = 1.2 Area D = 0.3
a) Which area shows maximum species richness?
b) What are the expected reasons for the loss of biodiversity in areas with low species richness?
Answer:
a) Area a = 0.1
b) (i) Habitat loss and fragmentation
(ii) Over-exploitation
(iii) Alien species invasions
(iv) Co-extinctions

Question 8.
“Nature provides all for the need of man but not for his greed.” (MARCH-2014)
a) Do you agree with this statement? Justify your answer.
b) Distinguish between two types of biodiversity conservations.
Answer:
a) yes, For example forest is used for some basic needs of a man but not for clearing of trees .
b) Exitu conservation – conservation of flora and fauna outside the natural habitat
Eg- Botanical garden
Insitu conservation – conservation of flora and fauna in the natural habitat
Eg-wild life sanctuaries and national parks

Question 9.
a) Variety of species are present around us, what they constitute and comment.  (MAY-2014)
b) Comment on in situ conservation and ex situ conservation.
c) In these aspects explain biodiversity hot spots with example – give importance to recent issues with regard to Western Ghats.
Answer:
a) Biodiversity
b) Exitu conservation – conservation of flora and fauna outside the natural habitat Insitu conservation- conservation.of flora and fauna in the natural habitat
c) Biodiversity hotspots are regions with very high levels of species richness and high degree of endemism Western ghats is the hotspots where accelerated habitat loss occurs.

Question 10.
We have a moral responsibility to take good care of earth’s biodiversity and pass it on in good order to next generation.  (MARCH-2015)
a) Define Biodiversity.
b) Write causes for biodiversity losses.
c) Name two types of biodiversity conservation.
Answer:
a) It is the variety within and between all species of plants, animals and micro-organisms and the ecosystems within which they live and interact.
b) i) Habitat loss and fragmentation
ii) Over-exploitation
iii) Alien species invasions
iv) Coextinctions
c) Insitu conservation, Exitu conservation

Question 11.
Two approaches for the conservation of biodiversity is shown as A and B.  (MAY-2015)

a) Identify the type of biodiversity conservation shown in A and B.
b) Write the difference between the two, types of biodiversity conservation shown in A and B.
c) Which of the above approach is more desirable’ when there is an urgent need to save a species?
Answer:
a) A-Insitu conservation B-Exitu conservation
b) A- It is the conservation of animals in natural habitat
B- It is the conservation of plants outside the natural habitat
c) exitu conservation

Question 12.
Observe the concept diagram of the Evil Quartet of biodiversity loss. (MARCH-2016)

a) Write A and B
b) What is Co-Extinction?
Answer:
a) A- Habitat loss and fragmentation
B -Alien species invasions
b) When a species becomes extinct, the plant and animal species associated with it also become extinct

Question 13.
Read the statement and choose the correct option:   (MARCH-2016)
A: Sacred grooves are examples of in situ conservation
B: Biodiversity hotspots have low degree of endemism.
C: Biodiversity increases when number of organisms in a particular species increases.
a) Statement ‘A’ alone is correct
b) Statement ‘A’ and ‘B’ are correct
c) Statement ‘A’ and ‘C’ are correct.
d) Statement ‘C’ alone is correct
Answer:
Statement ‘A’ alone is correct.

Question 14.
a) “When we conserve and protect the whole ecosystem, its biodiversity at all levels is protected.” Based on this statement explain the strategies of biodiversity conservation.  (MAY-2016)
OR
b) “When need turns to greed, it leads to biodiversity loss.” Substantiate this statement by explaining two causes of biodiversity loss.
Answer:
a) Insitu conservation- It is the conservation of plant and animal sp. in their natural habitat.
Eg-biosphere reserves, national parks and sanctuaries.
Exitu conservation- It is the conservation of plant and animal sp outside the natural habitat.
Eg- Zoological parks, botanical gardens and wildlife safari parks.
OR
b) i) Habitat loss and fragmentation: The degradation of many habitats by pollution affects the survival of many species. It results large habitats are broken up into small fragments.
Amazon rain forest is cleared for cultivating soya beans or for conversion to grasslands for raising beef cattle
ii) Over-exploitation : It leads to the over-exploitation of natural resources. For example the extinction of Steller’s sea cow, passenger pigeon was due to humans.
iii) Alien species invasions : The introduction of foreign species cause the reduction or extinction of indigenous species.
The Nile perch introduced into Lake Victoria in east Africa led to the extinction of more than 200 species of cichlid fish in the lake
iv) Co-extinctions : If two species are in obligatory relationship the extinction of one species affect the other.
Eg – coevolved plant-pollinator mutualism where extinction of one species leads to the extinction of the other.

Question 15.
Z – values of a frugivorous bat species are given below. Which value is not applicable to continents?  (MARCH-2017)
1) 0.6
2) 0.65
3) 0.20
4) 0.68
Answer:
Incorrect option

Question 16.
Distinguish in situ conservation from ex situ conservation with one example each.  (MARCH-2017)
Answer:
Insitu conservation- Ms the conservation of plant and animal sp. in natural habitat.
Eg- biosphere reserves, national parks and sanctuaries.
Exitu conservation- It is the conservation of plant and animal sp outside the natural habitat.
Eg- Zoological parks, botanical gardens and wildlife safari parks.

Question 17.
What are the advantages of biofertilizers over chemical fertilizers? Give an example for biofertilizer.  (MARCH-2017)
Answer:
a) 1) It prevents pollution
2) It improves soil structure and function
b) biofertilizer- Mycorrhiza

Question 18.
Explain the three levels of biodiversity. (MAY-2017)
OR
Explain different types of biodiversity conservation with example.
Answer:
1. Genetic diversity – A single species might show high diversity at the genetic level over its distributional range
2. Species diversity – Diversity at species level
3. Ecological diversity – Diversity at ecosystem level
OR
In situ conservation – the species are protected in their natural habitat.
example: National Park, Wildlife sanctuaries etc.
ex situ conservation – threatened animals and plants are protected outside their natural habitat.
example: Zoological Park, botanical gardens etc.