Nitrogen dioxide is paramagnetic in a gaseous state but diamagnetic in solid because:

1.	NO2 exists as a dimer in a gaseous as well as solid state.
2.	NO2 exists as a monomer in a gaseous as well as solid state.
3.	NO2 exists as a dimer in a gaseous state and it becomes a monomer in solid state.
4.	NO2 exists as a monomer in a gaseous state and it becomes dimerized in solid state.

The correct statement regarding the bond angle of H₂O and H₂S is:

1. Both have the same bond angle.

2. Bond angle of H₂O is larger than that of H₂S

3. Bond angle of H₂O is smaller than that of H₂S 

4. None of the above.

PCl₅ reacts with finely divided silver on heating and a white silver salt is obtained, which dissolves on adding excess aqueous NH₃ solution gives "A". Compond A is : 

1. \([Ag(NH_{3})_{2}]Cl\)
2. \([Ag(NH_{3})_{2}]Cl_{2}\)
3. \([Ag(NH_{3})_{4}]Cl_{3}\)
4. \([Ag(NH_{3})_{}]Cl_{2}\)

Phosphorus forms a number of oxoacids. Out of these oxoacids, phosphinic acid has strong reducing property. its structure and reaction is 

1. and 3AgNO₃ + H₃PO₄ → Ag₃PO₄ + 3HNO₃

2. and 2AgNO₃ + H₃PO₂ + H₂O  2Ag + H₃PO₃ + 2HNO₃

3. and $4{\mathrm{AgNO}}_3+2{\mathrm{H}}_2\mathrm{O}+ {\mathrm{H}}_3{\mathrm{PO}}_2 \to 4\mathrm{Ag}\downarrow + 4{\mathrm{HNO}}_3 + {\mathrm{H}}_3{\mathrm{PO}}_4$

4. and H₄P₂O₇ + 4AgNO₃ → Ag₄P₂O₇ + 4HNO₃

An amorphous solid“A” burns in the air to form a gas “B” which turns lime water milky. The gas is also produced as a by-product during the roasting of sulphide ore. This gas decolourises acidified aqueous KMnO₄ solution and reduces Fe³⁺ to Fe²⁺. The solid "A" and the gas "B" are:

1. 'A' = S₈: ‘B' = SO₂

2. 'A' = S₈: ‘B' = SO₃

3. 'A' = S₁₀: ‘B' = SO₂

4. Not predictable

On heating lead (II) nitrate gives a brown gas “A” The gas “A” on cooling changes to colourless solid “B”. Solid “B” on heating with NO changes to a blue solid ‘C’.

Identify ‘A’, ‘B’ and ‘C and also write reactions involved and draw the structures of ‘B’ and ‘C’.

1. A=NO_2​, B=N₂​O₄​ and C=N_2​O₃​.

2. A=N₂​O₄​_​, B=NO₂ and C=N_2​O₃​.

3. A=N₂​O₄​_​, B=N_2​O₃ and C=NO₂ 

4. None of above

On heating compound (A) gives a gas (B) which is a constituent of air. This gas when treated with 3 moles of hydrogen (H₂) in the presence of a catalyst gives another gas (C) which is basic in nature. Gas C on further oxidation in moist condition gives a compound (D) which is a part of acid rain. Compounds (A) and (D) are :
 

1. A = N₂ ;      D = HNO₃

2. A = NH₄NO₂ ;      D = HNO₃

3. A = NH₄NO₂ ;     D = NH₃

4. A = NH₄NO₂ ;     D = HNO₂

The disproportionation reaction of ${\mathrm{H}}_3{\mathrm{PO}}_3$ is as follows:

${\mathrm{H}}_3{\mathrm{PO}}_3$ $\stackrel{∆}{\to }$ $\mathrm{X}$ $+$ $\mathrm{Y}$

 X and Y in the above reaction are:

1. H₃PO₂, HPO₂  

2. H₃PO₄, PH₃

3. H₂PO₆, PO₃

4. None of the above.