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

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

Accepted Answer

Sol. [Ni(CN)4]2– ⇒ Ni+2 = 3d8(CN– →SFL), dsp2, square planarμ = 0 Bm[Mn Br4]–2 ⇒ Mn+2 = 3d5(Br– → WFL) ⇒ sp3 tetrahedral⇒ number of unpaired electrons = 5μ = $\sqrt{5(5 + 2)} = \sqrt{35}$ = 5.9 Bm[Fe F6]–4 ⇒ Fe+2 = 3d6(F– → WFL) ⇒ sp3d2 ⇒ octahedralμ = $\sqrt{4(4 + 2)}$ = 4.9 Bm

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