question_answer

Match Column-1 with Column-II and select correct answer the respect to hybridisation using the codes given below:

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

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-s

Correct Answer: B

Solution :

(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 3d10 configuration and no (n - 1)d orbital is available for \[ds{{p}^{2}}\] hybridsation, so ns and np orbitals undergo \[s{{p}^{3}}\] hybridisation and complex is diamagnetic. (III) Co is in +3 oxidation state with 3d6 configuration has higher CFSE. So complex has 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.