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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.


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 .


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.


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


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


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)


Question 7: (NH42SO4,FeSO4 6H­2O, 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.


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+.


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.


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


Question 11: Epson salt on strong heating gives

(A)  MgSO4 × 6H2O                         (B)  MgSO4

(C) MgO                                         (D)  MgO + SO3


MgSO4×7H­2O →  MgO + SO3 + 7H2O

Epsom salt (D)


Question 12  The species that does not contain peroxide ions is

(A)  PbO2                                        (B)  H2O2

(C) SrO2                                          (D)  BaO2


(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)



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.



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.


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


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.          



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.



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



Question 18 : Carbon dioxide is isostructural with

(A)  HgCl2                                        (B)  SnCl2

(C) C2H6                                          (D)  NO2


 HgCl2 →  linear structure

CO2 →  linear structure



Question 19 : Which of the following oxides is neutral

(A)  CO                                           (B)  SnO2

(C) SiO2                                          (D)  ZnO


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



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

(A)  PbO                                         (B)  PbO2

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


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



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