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Isomerism in Coordination Compounds

Geometry of ...

Question

Geometry of $[N i (C N)_{4}]^{2 -}$ is tetrahedral.
If true 1 else 0.
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Solution

The correct option is A 0
Geometry of $[N i (C N)_{4}]^{2 -}$ is square planar due to $d s p^{2}$ hybridization.

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Similar questions

[N i (C N)_{4}]^{2 -}

is tetrahedral.

Write the given statement is true (or) false.

[N i C l_{4}]^{2 -}

is paramagnetic but when

C N^{-}

ions are added, the product

[N i (C N)_{4}]^{2 -}

is diamagnetic because in

[N i C l_{4}]^{2 -}

C l^{-}

is a weak ligand and thus, electrons are unpaired in

N i^{2 +}

making it paramagnetic,

s p^{3}

hybridised and tetrahedral.
If true enter 1, else enter 0.

Q. The geometry of

[N i (C N)_{4}]^{2 -}

The geometry of

[N i (C N)_{4}]^{2 -}

is tetrahedral.

Q. What is the geometry of

[N i (C N)_{4}]^{2 -}, [N i C l_{4}]^{2 -}

[N i (C O)_{4}]

?

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