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)	$\left[AuF_{4}\right]^{-}$	(p)	$dsp^{2}$hybridisation
(II)	$\left[Cu\left(CN\right)_{4}\right]^{3-}$	(q)	$sp^{3}$hybridisation
(III)	$\left[Co\left(C_{2}O_{4}\right)_{3}\right]^{3-}$	(r)	$sp^{3}d^{2}$hybridisation
(IV)	$\left[Fe\left(H_{2}O\right)_{5}NO\right]^{2+}$	(s)	$d^{2}sp^{3}$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 $dsp^{2}$ 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 $dsp^{2}$ hybridisation, so $ns$ and $np$ orbitals undergo $sp^{3}$ 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 $d^{2}sp^{3}$ hybridisation.
(IV) $Fe$ is in $+1$ oxidation state and the complex is paramagnetic with three unpaired electrons.
$\left[Fe\left(H_{2}O\right)_{5}NO\right]^{2+}$