Question

Match Column I with Column II and select the correct answer with respect to hybridisation using the codes given below:

Column I (Complex)		Column II (Hybridisation)
(I)	${\left[Au{F}_4\right]}^{-}$	(p)	$ds{p}^2$hybridisation
(II)	${\left[Cu{\left(CN\right)}_4\right]}^{3-}$	(q)	$s{p}^3$hybridisation
(III)	${\left[Co{\left(C_2{O}_4\right)}_3\right]}^{3-}$	(r)	$s{p}^3{d}^2$hybridisation
(IV)	${\left[Fe{\left(H_{2}O\right)}_{5}NO\right]}^{2+}$	(s)	$d^{2}s{p}^3$hybridisation

• A. $(I)-(q),(II)-(p),(III)-(r),(IV)-(s)$
• B. $(I)-(p),(II)-(q),(III)-(s),(IV)-(r)$
• C. $(I)-(p),(II)-(q),(III)-(r),(IV)-(s)$
• D. $(I)-(q),(II)-(p),(III)-(s),(IV)-(r)$

Answer 1

Answer: (B)

(I) $Au$ in $+3$ oxidation state with $5d^8$ configuration has higher $CFSE$. So complex has $ds{p}^2$ hybridisation and is diamagnetic.
(II) $Cu$ is in $+1$ oxidation state with $3d^{10}$configuration and no $\left(n-1\right)d$ orbital is available for $ds{p}^2$ hybridisation, so $ns$ and $np$ orbitals undergo $s{p}^3$ hybridisation and complex is diamagnetic.
(III) $Co$ is in $+3$ oxidation state and $3d^6$ configuration has higher $CFSE$. So complex is diamagnetic and has $d^{2}s{p}^3$ hybridisation.
(IV) $Fe$ is in $+1$ oxidation state and the complex is paramagnetic with three unpaired electrons.
${\left[Fe{\left(H_{2}O\right)}_{5}NO\right]}^{2+}$