Question

Consider the following compounds:
(i) $I{F}_5$
(ii) $Cl{I}_4^{-}$
(iii) $Xe{O}_2{F}_2$
(iv) $N{H}_2^{-}$
(v) $BC{l}_3$
(vi) $BeC{l}_2$
(vii) $AsC{l}_4^{+}$
(viii) $B(OH{)}_3$
(ix) $N{O}_2^{-}$
(x) $Cl{O}_2^{+}$
The value of " $x+y-z$ " is
where $x,y$ and $z$ are total number of compounds in given compounds in which central atom used their all three porbitals, only two p-orbitals and only one p-orbital in hybridisation respectively.

Answer 1

Answer: 8

(i) $I{F}_5\left(s{p}^{3}d\right)$
(ii) $Cl{I}_4^{-}\left(s{p}^3{d}^2\right)$
(iii) $Xe{O}_2{F}_2\left(s{p}^{3}d\right)$
(iv) $N{H}_2^{-}\left(s{p}^3\right)$
(v) $BC{l}_3\left(s{p}^2\right)$
(vi) $BeC{l}_2(sp)$
(vii) $AsC{l}_4^{+}\left(s{p}^3\right)$
(viii) $B(OH{)}_3\left(s{p}^2\right)$
(ix) $N{O}_2^{-}\left(s{p}^2\right)$
(x) $Cl{O}_2^{+}\left(s{p}^2\right)$
$x=\left(s{p}^3\right)2+s{p}^{3}d(1)+s{p}^3{d}^2(2)=5$
$y=4,z=1;\therefore x+y-z=5+4-1=8$