Coordination CompoundHardBloom L3

Question

Geometry, hybridisation and magnetic moment of the ions [Ni(CN)4]2–, [MnBr4]2– and [FeF6]4– respectively are:

Options

A.tetrahedral, square planar, octahedral: sp3, dsp2 , sp3d2 : 5.9, 0, 4.9
B.tetrahedral, square planar, octahedral: dsp2, sp3 , sp3d2 : 0, 5.9, 4.9
C.square planar, tetrahedral, octahedral: dsp2, sp3 , d2sp3 : 5.9, 4.9, 0
D.square planar, tetrahedral, octahedral: dsp2, sp3 , sp3d2 : 0, 5.9, 4.9

Solution

Sol. [Ni(CN)4]2– ⇒ Ni+2 = 3d8

(CN– →SFL), dsp2, square planar

μ = 0 Bm

[Mn Br4]–2 ⇒ Mn+2 = 3d5

(Br– → WFL) ⇒  sp3 tetrahedral

⇒ number of unpaired electrons = 5

μ = $\sqrt{5(5 + 2)} = \sqrt{35}$ = 5.9 Bm

[Fe F6]–4 ⇒ Fe+2 = 3d6

(F– → WFL) ⇒ sp3d2 ⇒ octahedral

μ = $\sqrt{4(4 + 2)}$ = 4.9 Bm

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