P BLOCK ELEMENTS CLASS 12

On addition of conc. H2SO4 to a chloride salt, colourless fumes are evolved

but in case of iodide salt, violet fumes come out. This is because

• (i) H2SO4 reduces HI to I2

• (ii) HI is of violet colour

• (iii) HI gets oxidised to I2

• (iv) HI changes to HIO3

In qualitative analysis when H2S is passed through an aqueous solution of

salt acidified with dil. HCl, a black precipitate is obtained. On boiling the

precipitate with dil. HNO3, it forms a solution of blue colour. Addition of excess

of aqueous solution of ammonia to this solution gives _________.

• (i) deep blue precipitate of Cu (OH)2

• (ii) deep blue solution of [Cu (NH3)4]2+

• (iii) deep blue solution of Cu(NO3)2

• (iv) deep blue solution of Cu(OH)2.Cu(NO3)2

On heating with concentrated NaOH solution in an inert atmosphere of CO2,

white phosphorus gives a gas. Which of the following statement is incorrect

about the gas?

• (i) It is highly poisonous and has smell like rotten fish.

• (ii) It’s solution in water decomposes in the presence of light.

• (iii) It is more basic than NH3.

• (iv) It is less basic than NH3.

On heating lead nitrate forms oxides of nitrogen and lead. The oxides formed are ______.

• (i) N2O, PbO

• (ii) NO2, PbO

• (iii) NO, PbO

• (iv) NO, PbO2

Maximum covalency of nitrogen is ______________.

• 3

• 5

• 4

• 6

Which of the following statements is wrong?

• (i) Single N–N bond is stronger than the single P–P bond.

• ii) PH3 can act as a ligand in the formation of coordination compound with transition elements.

• (iii) NO2 is paramagnetic in nature.

• (iv) Covalency of nitrogen in N2O5 is four.

A brown ring is formed in the ring test for NO3– ion. It is due to the formation of

• (i) [Fe(H2O)5 (NO)]2+

• (ii) FeSO4.NO2

• (iii) [Fe(H2O)4(NO)2]2+

• (iv) FeSO4.HNO3

Which of the following are peroxoacids of sulphur?

• (i) H2SO5 and H2S2O8

• (ii) H2SO5 and H2S2O7

• (iii) H2S2O7 and H2S2O8

• (iv) H2S2O6 and H2S2O7

In the preparation of compounds of Xe, Bartlett had taken O2⁺PtF6^– as a base compound. This is because

• (i) both O2 and Xe have same size.

• (ii) both O2 and Xe have same electron gain enthalpy.

• (iii) both O2 and Xe have almost same ionisation enthalpy.

• (iv) both Xe and O2 are gases.

Which of the following elements can be involved in pπ–dπ bonding?

• Carbon

• Nitrogen

• Phosphorus

• Boron