Solved Problem

Question 1: Lithium cannot be kept in Kerosene other like alkali metals e.g. Na,K etc- Why?

Solution: This is because it floats to the surface of kerosene due to its low density and is usually wrapped in paraffin wax.

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Question 2:  The chemistry of Lithium is very much similar to that of magnesium even though they are placed in different groups - Explain.

Solution: The ratio of their charge to size is nearly same by which they show the diagonal relationship .

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Question 3: Which is the weakest base among NaOH, Ca(OH)₂, KOH and Be(OH)₂

Solution: Be(OH)₂ is weakest base , because alkali metal hydroxides are more stronger base than alkaline earth metal hydroxides. Also basic character of hydroxides of alkaline earth metals increases down the group. So Be(OH)₂ is the weakest one.

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Question 4: NaHCO₃ and NaOH cannot exist together in solution- Why?

Solution: NaHCO₃ is an acid salt which must react with NaOH which is strong base. The reaction   is as follows:

NaHCO₃ + NaOH → Na₂CO₃ + H₂O

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Question 5: A mixture of Al(OH)3 and Fe(OH­)3 is given to you. How would youseparate it?

Solution:It can be easily separated by the use of conc. NaOH solution which react only with Al(OH)3  and forms soluble sodium aluminate. Fe(OH)3 will remain in the residue.

Al(OH)₃ + NaOH → NaAlO₂ +2H₂O

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Question 6: When SnCl₂ solution is added excess into a solution of HgCl₂,a white ppt is turning grey. Why?

Solution: When SnCl2 solution is added to  HgCl2 solution , initially white ppt, of Hg2Cl2 is formed which further reacts with SnCl2 forming metallic grey Hg.

2HgCl₂  + SnCl₂ →  SnCl₄ + Hg₂Cl₂ (white)

Hg₂Cl₂ + SnCl₂ →  SnCl₄  + Hg₂ (grey)

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Question 7: (NH₄)­₂SO₄,FeSO₄ 6H_­2O, Is this compound alum or not?

Solution: The general formula of alum is

Since it has no resemblance with the general formula it is not an alum.

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Question 8: Give reason for decreasing order of conductivity of following

Cs⁺ > Rb⁺   > k⁺ > Na⁺ > Li⁺

Solution:  Ions are hydrated in solution since Li is very small it is heavily hydrated. This make the radius of the hydrated ions large and hence it move only slowly (although Li⁺ is very small) and the radius of hydrated Cs⁺ ion is smaller than the radius of hydrated Li⁺.

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Question 9: BaO₂ is a peroxide but PbO₂ is not a peroxide why?

Solution: Metallic oxides which on treatment with dilute acids produce hydrogen peroxide are called peroxides. All peroxides contain a peroxide ion (O­₂)^2– having the structure – O – O – PbO₂ does not contain a peroxide ion (O₂)₂ and it can not be called as peroxides.

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Question 10: Statues coated with white lead on long exposure to atmosphere turn black and the original colour can be restered in treatment with H2O2 why?

Solution: On long exposure to atmosphere, white lead is converted into black PbS due to the action of H2S present in the atmosphere.  As a result statues turn black.

PbO₂ + 2H₂S →  PbS + 2H₂O

On treatment of these blackened statues with H2O2, the black PbS gets oxidised to white PbSO4 and the colour is restored

PbS + 4H₂O₂ →  PbSO₄ + 4H₂O

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Question 11: Epson salt on strong heating gives

(A)  MgSO4 × 6H2O                         (B)  MgSO4

(C) MgO                                         (D)  MgO + SO3

Solution: 

MgSO₄×7H­₂O →  MgO + SO₃ + 7H₂O

Epsom salt (D)

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Question 12  The species that does not contain peroxide ions is

(A)  PbO₂                                        (B)  H₂O₂

(C) SrO₂                                          (D)  BaO₂

Solution:

(A) PbO₂ contains O^–2 ions and O.S. of Pb is +4, whereas (B), (C) and (D) contains [O – O]^–2 ions (peroxide ions)

(A)

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Question 13: The following compounds have been arranged in the order of increasing thermal stabilities. Identify the correct order.

K₂CO₃(I), MgCO₃ (II), CaCO₃ (III),  BeCO₃ (IV)

(A)  I < II < III < IV                                 (B)  IV < II < III < I

(C) IV < II < I < III                                 (D)  II < IV < IIII < I

Solution: Carbonates of group 2 decompose to give oxides. Thermal stabilities of carbonates increase from Mg to Ba. Beryllium carbonate is unstable and is kept only in the atmosphere of CO₂.

Group 1 carbonate (and also hydroxides) except lithium do not decompose on heating. In this problem K₂CO₃  remains unchaged on heating.

 (B)

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Question 14: When zeolite, which is hydrated sodium aluminium silicate, is treated with hard water, the sodium ions are exchanged with

(A)  H⁺ ions                                           (B)  Ca⁺² ions

(C) SO₄^–2 ions                                       (D)  Mg⁺² ions

Solution:  (B) and (D), Na⁺ ions are exchanged with Ca⁺² and Mg⁺² ions, but not with H⁺ ion, because of its smaller size than Ca⁺² or Mg⁺² ions.

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Question 15: Setting of plaster of paris involves

(A)  Oxidation by atmospheric oxygen

(B)  Reaction with atmospheric carbon dioxide

(C) Dehydration

(D)  Hydration to yield another hydrate

Solution: 

CaSO₄ × 1/2H₂O → CaSO₄×2H₂O

When mixed with water, it forms a plastic mass, within about half an hour, sets to a hard solid mass consisting of interlaced crystals of gypsum.          

 (D)

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Question 16:  Which is deliquescent?

(A)  MgCl₂                                       (B)  NaOH

(C) CaCl₂                                        (D)  All

Solution: Because MgCl₃, NaOH and CaCl₂ all absorbs water from the atmospheric moisture.

(D)

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Question 17: AlCl₃ fumes in air because of

(A)  hydrolysis                                (B)  dehydrates

(C) reduction                                  (D)  oxidation

Solution: It undergoes hydrolysis and gives fumes of HCl

AlCl₃ + 3H₂O →  Al(OH)₃ + 3HCl

(A)

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Question 18 : Carbon dioxide is isostructural with

(A)  HgCl₂                                        (B)  SnCl₂

(C) C₂H₆                                          (D)  NO₂

Solution:

 HgCl₂ →  linear structure

CO₂ →  linear structure

(A)

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Question 19 : Which of the following oxides is neutral

(A)  CO                                           (B)  SnO₂

(C) SiO₂                                          (D)  ZnO

Solution:

SiO₂ is acidic, ZnO and SnO₂ are amphoteric only CO is neutral.

(A)

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Question 20 : Red lead on reaction with dil. HNO₃ form

(A)  PbO                                         (B)  PbO₂

(C) PbO + Pb(NO₃)₂                         (D)  PbO₂ + Pb(NO­₃)₂

Solution: 

Pb₃O₄ + 4HNO₃ →  2Pb(NO₃)₂ + PbO₂ + 2H₂O

(D)

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Related Resources

• Look here for past year papers of IIT JEE  to get an idea about the type of questions asked in exam

• You should also refer to the syllabus of chemistry for IIT JEE which would help you to plan your study

• Knowing about the important books for preparing IIT JEE would help you alot.

To read more, Buy study materials of S- Block elements comprising study notes, revision notes, video lectures, previous year solved questions etc. Also browse for more study materials on Chemistry here.

 

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