Coordination CompoundHard
Question
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
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
Options
A.(I) → q
(II) → p
(III) → r
(IV) → s
(II) → p
(III) → r
(IV) → s
B.(I) → p
(II) → q
(III) → s
(IV) → r
(II) → q
(III) → s
(IV) → r
C.(I) → p
(II) → q
(III) → r
(IV) → s
(II) → q
(III) → r
(IV) → s
D.(I) → q
(II) → p
(III) → s
(IV) → r
(II) → p
(III) → s
(IV) → r
Solution
(I) Au in +3 oxidation state with 5d8 configuration has higher CFSE. So complex has dsp2 hybridisation and is diamagnetic.
(II) Cu is in +1 oxidation state with 3d10 configuration and no (n -1)d orbital is available for dsp2 hybridisaiton, so ns and np orbitals undergo sp3 hybridisation and complex is diamagnetic.
(III) Co is in +3 oxidation state and 3d6 configuration has higher CFSE. So complex is diamagnetic and has d2sp3 hybridisation.
(IV) Fe is in +1 oxidation state and the complex is paramagnetic with three unpaired electrons.
[Fe(H2O)5NO]2+ ;
(II) Cu is in +1 oxidation state with 3d10 configuration and no (n -1)d orbital is available for dsp2 hybridisaiton, so ns and np orbitals undergo sp3 hybridisation and complex is diamagnetic.
(III) Co is in +3 oxidation state and 3d6 configuration has higher CFSE. So complex is diamagnetic and has d2sp3 hybridisation.
(IV) Fe is in +1 oxidation state and the complex is paramagnetic with three unpaired electrons.
[Fe(H2O)5NO]2+ ;

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