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VBT Tetrahedral vs Square Planar

Colorless, te...

Question

Colorless, tetrahedral complexes among the following are :

_{3}

[Cu(CN)

_{4}

]

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_{2}

[NiCl

_{4}

]

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K[BF

_{4}

]

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Ni(CO)

_{4}

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Solution

The correct options are
A K $_{3}$ [Cu(CN) $_{4}$ ]
C Ni(CO) $_{4}$
D K[BF $_{4}$ ]
K $_{3}$ [Cu(CN) $_{4}$ ]
$C u^{+} = d^{10}$ configuration, so no unpaired electron and colorless.
K[BF $_{4}$ ]
no unpaired electron so colorless.
Ni(CO) $_{4}$
$d^{10}; s p^{3}$ hybridization so no unpaired electron and colorless.

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Similar questions

_{4}

_{4}

^{2 -}

_{4}^{2 -}

[N i C l_{4}]

[N i C O_{4}]

Among $T i F_{6}^{2 -}, C o F_{6}^{3 -}, C u_{2} C l_{2} a n d N i C l_{4}^{2 -}$ the colorless species are

_{3}

_{4}

_{3}

^{+}

_{4}

_{3}

_{6}

^{3 +}

^{-}

_{3}

_{6}

\begin{matrix} C o l u m n I & C o l u m n 2 & C o l u m n 3 (I) & [N i (C O)_{4}] & (i) & s p^{3} & (P) & D i a m a g n e t i c (H i g h s p i n) (I I) & [P t C l_{4}]^{2 -} & (i i) & s p^{3} d^{2} & (Q) & P a r a m a g n e t i c (H i g h s p i n) (I I I) & [N i (C O)_{4}] & (i i i) & d s p^{2} & (R) & D i a m a g n e t i c (L o w s p i n) (I V) & K_{2} [C o (C N)_{4}] & (i v) & d^{2} s p^{3} & (S) & P a r a m a g n e t i c (L o w s p i n) \end{matrix}

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