Answer: Option A. -> $N{H}_3$
:
A
We can look at the $p{K}_a$ table and see that $N{H}_3$ is more stable than $H_{2}N-N{H}_2$.

Answer: Option D. -> Metal Nitrite
:
D
Nitrates on heating with lead decompose to give the corresponding nitrites and Lead (II) Oxide
$NaN{O}_3+Pb\to NaN{O}_2+PbO$ 

Answer: Option B. -> False
:
B
Sulphuric acid reacts with ammonia to form ammonium sulphate
$H_{2}S{O}_4+2N{H}_3\to (N{H}_4{)}_{2}S{O}_4$
Therefore, CaO is used to dry ammonia gas
$CaO+H_{2}O\to Ca(OH{)}_2$

Answer: Option A. -> True
:
A
$HN{O}_2$ (Nitrous acid) is monobasic acid because it can donate only a single proton.
Basicity is the number of protons that an acid can give. From the structure,

We can see that there is only one H atom and so only one proton can be abstracted from a $HN{O}_2$ molecule. What can you say about the basicity of nitric acid?

Answer: Option A. -> True
:
Let us take a look at the structure of $H_{3}P{O}_4$:
As we can see from the above structure, the Oxygen atom shown to have double bond character acts as a site to accept Hydrogen bonds while the other Hydrogen atoms attached via a single bond to Oxygen atoms can participate in Hydrogen bond. So in all, for every molecule there are 3 donor sites and one acceptor site.

Answer: Option D. -> Heating $N{H}_{4}N{O}_2$ carefully
:
D
Heating $N{H}_{4}N{O}_3$ carefully produces $N_{2}O$ but heating $N{H}_{4}N{O}_2$ liberates $N_2$ gas.

Answer: Option D. -> Nitrogen does not have any other allotropes under any circumstances
:
A, B, and C
Nitrogen atom is small in size and has the tendency to form $p\pi -p\pi$ multiple bonds. In fact, the Nitrogen – Nitrogen triple bond in $N_2$ molecule has very high thermodynamic stability with a dissociation energy of $942kJ.mo{l}^{-1}$. This and other molecular orbital considerations account for the lack of allotropes of Nitrogen under normal conditions. However, it would be wrong to say that Nitrogen doesn’t have allotropes at all.
Gaseous nitrogen condenses into the $\beta$-hcp form at $-{210.01}^{\circ }C$. On freezing further, at temperatures below $-{237.6}^{\circ }C$ Nitrogen assumes the cubic α form. At room temperature and atmospheric pressure, gaseous $N_2$ does not have any other allotropes.
 

Answer: Option C. -> N<P<As<Sb<Bi
:
C
This increase in atomic radii from N to Bi is due to the corresponding increase in the highest principal quantum number for each succeeding element. Thus option C is the correct one.

Answer: Option A. -> True
:
A
Cyanamide is an important organic chemical, which is used in agriculture and in the production of various pharmaceuticals and other organic compounds. It can be thought of as a nitrile group attached to an amine group.
Calcium Cyanamide is CaNCN, an inorganic compound that is extensively used as a fertilizer. It is synthesized by the famous Frank-Caro process (also known as the cyanamide process) where Calcium Carbide is heated with $N_2$ gas in a reactor at 1273 K. The reaction is exothermic:
$Ca{C}_2$    +  $N_2$                       $\to$          CaNCN          + C
( Calcium          (Nitrogen)                     ( Calcium                 (Carbon) 
  Carbide)                                             Cyanamide)
$CaC{N}_2$       +    $4H_{2}O$                $\to$       $Ca(OH{)}_2$    +    $C{O}_2$       +            $2N{H}_3$   
                                                           (Calcium hydroxide)                                                               (Ammonia)

Answer: Option D. -> All of these
:
D
Lead Heavy metal nitrates like copper (II) Nitrate or Lead (II) Nitrate decompose on heating to give the corresponding metal (II) oxide, nitrogen dioxide gas and dioxygen gas.
$Pb(N{O}_3{)}_2\to PbO+O_2+N{O}_2$