Question

Consider the following three compounds (i) $A{X}_{2n}^{n-}$, (ii) $A{X}_{3n}$ and (iii) $A{X}_{4n}^{n+}$, where central atom $X$ is $15th$ group element and their maximum covalency is $3n$. If total number of proton in surrounding atom $X$ is $n$ and value of $n$ is one, then calculate value of ${}^{\prime \prime }{x}^3+y^2+z^{\prime \prime }$ (Where x, y and z are total number of lone pair at central atom in compound (i), (ii) and (iii) respectively

Answer 1

Answer: 9

$n=1$, then $X=H;A=N$
(i) $N{H}_2^{-}$ (ii) $N{H}_3$ (iii) $N{H}_4^{+}$
lone pair $x=2,y=1,z=0$
$x^3+y^2+z=(2{)}^3+(1{)}^2+(0)=9$