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Square Planar Complexes

a What type o...

Question

(a) What type of isomerism is shown by the complex $[C o (N H_{3})_{5} (S C N)]^{2 +}$ ?
(b) Why is $[N i C l_{4}]^{2 -}$ paramagnetic while $[N i (C N)_{4}]^{2 -}$ is diamagnetic? (Atomic number of Ni = 28)
(c) Why are low spin tetrahedral complexes rarely observed?

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Solution

(a) Linkage isomerism.
(b) Because $C N^{-}$ is stronger ligand than $C l^{-}$ .
(c) As the energy gap between $e_{g}$ and $t_{2 g}$ in tetrahedral complexes is very low, electrons go to higher states to avoiding pairing.

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Similar questions

[C o (N H_{3})_{5} (S C N)]^{2 +}

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

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

N i = 28

[C o (e n)_{2} C l_{2}]^{+} .

d^{4}

Δ_{0} > P .

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

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

[C r {(N H_{3})}_{4} {C l}_{2}] C l

{[C o {(e n)}_{3}]}^{3 +}

e n =

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

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

C r = 24, C o = 27, N i = 28

(i) Draw the geometrical isomers of complex $[C o (e n)_{2} C l_{2}]^{+}$ .

(ii) On the basis of crystal field theory, write the electronic configuration for d⁴ ion if $Δ_{0} > P$ .

(iii) $[N i C l_{4}]^{2 -}$ is paramagnetic, while $[N i (C O)_{4}]$ is diamagnetic, though both are tetrahedral. Why? (Atomic number of Ni = 28)

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