question_answer

Match Column I with Column II and select the con- answer with respect to hybridisation using the codes gr below:

	Column I (Complex)		Column II (Hybridisation)
(I)	\[{{[Au{{F}_{4}}]}^{-}}\]	(P)	\[ds{{p}^{2}}\]hybridisation
(II)	\[{{[Cu{{(CN)}_{4}}]}^{3-}}\]	(Q)	\[ds{{p}^{3}}\]hybridisation
(III)	\[{{[Cu{{({{C}_{2}}{{O}_{4}})}_{3}}]}^{3-}}\]	(R)	\[s{{p}^{3}}\,{{d}^{2}}\]hybridisation
(IV)	\[{{[Fe{{({{H}_{2}}O)}_{5}}NO]}^{2+}}\]	(S)	\[{{d}^{2}}s{{p}^{3}}\]hybridisation

Codes:

A) (I)\[\to \]Q,         (II)\[\to \]P,        (III)\[\to \]R,       (IV)\[\to \]S

B) (I)\[\to \]P,         (II)\[\to \]Q,        (III)\[\to \]S,       (IV)\[\to \]R

C) (I)\[\to \]P,         (II)\[\to \]Q,        (III)\[\to \]R,       (IV)\[\to \]S

D) (I)\[\to \]Q,         (II)\[\to \]P,        (III)\[\to \]S,       (IV)\[\to \]R

Correct Answer: B

Solution :

[b]

(I) Au in +3 oxidation state with \[5{{d}^{8}}\]configuration has higher CFSE. So complex has \[ds{{p}^{2}}\]hybridisation and is diamagnetic.

(II) Cu is in +1 oxidation state with \[3{{d}^{10}}\]configuration and no \[\left( n-1 \right)d\] orbital is available for \[ds{{p}^{2}}\] hybridisation and complex is diamagnetic.

(III) Co is in +3 oxidation state and \[3{{d}^{6}}\]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.

\[{{[Fe{{({{H}_{2}}O)}_{5}}NO]}^{2+}};\]