Match Column-I with Column-II and select the correct answer with respect to hybridisation using the codes given below :
Column - I | Column - II | ||
(Complex) | (Hybridisation) | ||
(I) | [Au F4]– | (p) | dsp2 hybridisation |
(II) | [Cu(CN)4]3– | (q) | sp3 hybridisation |
(III) | [Co(C2O4)3]3– | (r) | sp3d2 hybridisation |
(IV) | [Fe(H2O)5NO]2+ | (s) | d2sp3 hybridisation |
Codes :
(I) (II) (III) (IV)
Text Solution
Verified by ExpertsB
(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 (n –1)d orbital is available for dsp 2 hybridisaiton, 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.
[Fe(H 2 O) 5 NO] 2+ ; 
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