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Hybridization, shape, and magnetic moment of K3CoCO33 are:

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a

d2sp3, octahedral, 4.9 B.M.

b

sp3d2, octahedral, 4.9 B.M.

c

dsp2, square planar, 4.9B.M

d

sp3, tetrahedral, 4.9 B.M.

answer is B.

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Detailed Solution

K3[Co(CO3)3] +3 d6 configurationweak ligand so no pairing and high spin complex formsso sp3d2 hybridization takes place and octahedral complex formsAlso as it have 4 unpaired electron so magnetic momentμ=n(n+2)=4.9 BMn = no of unpaired electron

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