Question

Consider the following compounds:
(i) $IF _5$
(ii) $ClI _4^{-}$
(iii) $XeO _2 F _2$
(iv) $NH _2^{-}$
(v) $BCl _3$
(vi) $BeCl _2$
(vii) $AsCl _4^{+}$
(viii) $B ( OH )_3$
(ix) $NO _2^{-}$
(x) $ClO _2^{+}$
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

(i) $ IF _5\left(s p^3 d\right)$
(ii) $ ClI _4^{-}\left(s p^3 d^2\right)$
(iii) $XeO _2 F _2\left(s p^3 d\right)$
(iv) $ NH _2^{-}\left(s p^3\right)$
(v) $BCl _3\left(s p^2\right)$
(vi) $BeCl _2(s p)$
(vii) $AsCl _4^{+}\left(s p^3\right)$
(viii) $B ( OH )_3\left(s p^2\right)$
(ix) $NO _2^{-}\left(s p^2\right)$
(x) $ ClO _2^{+}\left(s p^2\right)$
$x=\left(s p^3\right) 2+s p^3 d(1)+s p^3 d^2(2)=5$
$y=4, z=1 ; \therefore x+y-z=5+4-1=8$