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Q. Consider the following three compounds (i) $AX^{n-}_{2n}$, (ii) $AX_{3n}$ and (iii) $AX^{n+}_{4n}$, 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 $''x^{3}+y^{2}+z''$ (Where x, y and z are total number of lone pair at central atom in compound (i), (ii) and (iii) respectively

Chemical Bonding and Molecular Structure

Solution:

$n=1$, then $X=H; A=N$
(i) $NH^{-}_{2} $ (ii) $NH_{3}$ (iii) $NH^{+}_{4}$
lone pair $x=2, y=1, z=0$
$x^{3}+y^{2}+z=(2)^{3}+(1)^{2}+(0)=9$