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Q.

Which one of the following is outer orbital complex

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a

CoNH36+3

b

MnCN6-3

c

FeCN64−

d

NiNH36+2

answer is D.

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Detailed Solution

The hybridization of metal ion in octahedral complexes will be sp3d2 (or) d2dp3.In the ions with configuration d8,d9,d10 there is no availability of two vacant d-orbitals in (n-1) d substance.Moreover they cannot rearrange. Thus for the octahedral complexes having the metal ions with d8, d9, d10 configuration the hybridization is always sp3d2. They form outer orbital complexes without undergoing rearrangement.
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