Total number of complex compound which have low spin as well as Diamagnetic.
[Fe(CN)₆]³⁻, [Co(H₂O)₆]³⁺, [NiF₆]²⁻, [Cr(H₂O)₆]³⁺, [Fe(CN)₆]⁴⁻, [Co(C₂O₄)₃]³⁻, [Ni(en)₃]²⁺, [PtCl₄]²⁻

Question

Total number of complex compound which have low spin as well as Diamagnetic.

[Fe(CN)₆]³⁻, [Co(H₂O)₆]³⁺, [NiF₆]²⁻, [Cr(H₂O)₆]³⁺, [Fe(CN)₆]⁴⁻, [Co(C₂O₄)₃]³⁻, [Ni(en)₃]²⁺, [PtCl₄]²⁻

TestHub · Accepted Answer

The total number of low-spin, diamagnetic complexes is 5.

$d^{6}$ octahedral complexes with strong splitting: $[C o (H_{2} O)_{6}]^{3 +}$ , $[F e (CN)_{6}]^{4 -}$ , and $[C o (C_{2} O_{4})_{3}]^{3 -}$ have a $t_{2 g}^{6} e_{g}^{0}$ configuration with zero unpaired electrons.

High oxidation state $d^{6}$ complex: In $[N i F_{6}]^{2 -}$ , the high $+ 4$ charge on nickel causes strong crystal field splitting, pairing all electrons into $t_{2 g}^{6}$ .

$5 d$ square planar complex: $[PtC l_{4}]^{2 -}$ is a $d^{8}$ square planar complex with very large splitting, causing all 8 electrons to pair up completely.

Chemistry - Coordination Chemistry Question with Solution | TestHub

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