Select the correct statement about elements of group ${15}^{th}$.

(A) The order of stability of oxidation state for +3 is $B{i}^{3+}>S{b}^{3+}>A{s}^{3+}$ and for +5 is $B{i}^{5+}<S{b}^{5+}<A{s}^{5+}$

(B) In the case of nitrogen, all oxidation states from +1 to +4 tend to disproportionate in acid solution.

(C) There is considerable increase in covalent radius from N to P but from As to Bi only a small increase in covalent radius is observed.

(D) All of these.

Which of the following compounds does not give nitrogen on heating?

(1) $Na{N}_3$                                                     

(2) ${\left(N{H}_4\right)}_{2}S{O}_4$

(3) $N{H}_{4}N{O}_2$                                               

(4) $\left(N{H}_4\right)2C{r}_2{O}_7$

A brown-coloured mixture of two gases is obtained by reducing 6 N nitric acid with metallic copper. This mixture on cooling condenses to a blue liquid which on freezing (-30 ^oC) gives a blue solid. The correct statement for the blue liquid or solid is: 

Which of the following statements is false?  

$N{H}_{4}Cl{O}_4+HN{O}_3\left(dilute\right)\to HCl{O}_4+\left[X\right]$

$\left[X\right]\stackrel{∆}{\to }Y\left(g\right)$

$\left[X\right]$ and $\left[Y\right]$ are respectively.

(a) $N{H}_{4}N{O}_3 and N_{2}O$       (c) $N{H}_{4}N{O}_2 and N_2$

(b) $HN{O}_4 and O_2$                (d) None

One of the products of the following reaction $KCNO+{\left(NH\right)}_{4}S{O}_4$ is :

(A) $N{H}_{4}N{O}_2$                              (B) $N{H}_{4}N{O}_3$

(C) $N{H}_2-CO-N{H}_2$                  (D) $N_2$ and $N{O}_2$ both

Which of the following orders is correct?

(1) $Sb{H}_3>N{H}_3>As{H}_3>P{H}_3$ – Boiling Point

(2) $N{H}_3>P{H}_3>As{H}_3>Sb{H}_3$  – Thermal Stability

(3) $N{H}_3>P{H}_3>As{H}_3>Sb{H}_3$  – Basic Character

(4) $N{H}_3>P{H}_3>As{H}_3>Sb{H}_3$  – Bond Angle

(A) (1), (2) and (3) only                             (B) (2), (3) and (4) only

(C) (1), (3) and (4) only                             (D) (1), (2), (3) and (4).

Which of the following is correct?

(1) $S_3{O}_9$  – contains no S - S linkage.

(2) $S_2{O}_6^{2-}$   – contains – O - O linkage.

(3) ${\left(HP{O}_3\right)}_3$  – contains P – P linkage.

(4) $S_2{O}_8^{2-}$ - contains S – S linkage.

The oxidation states of the S atom in $S_4{O}_6^{2-}$  from left to right, respectively, are:

1. + 6, 0, 0, and +6                         

2. +5, +1, +1, and +5

3. +5, 0, 0, and +5                           

4. +3, +1, +1, and +3

One gas bleaches the coloured matter by reduction temporary while the other one by oxidation
permanently. The gases are respectively:

(1) $S{O}_2 and C{l}_2$                             

(2) $O_3 and C{l}_2$

(3) $H_{2}S and C{O}_2$                             

(4) $C{l}_2 and S{O}_2$