$[N i C l_{4}]^{2 -}$ is paramagnetic while $[N i (C O)_{4}]$ is diamagnetic though both are tetrahedral. Why?

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Solution

In $[N i C l_{4}]^{2 -}$ oxidation state of $N i$ is $+ 2$ and it have

$↿⇂ ↿⇂ ↿⇂ ↿ ↿$

$8$ electrons in d shell which is arranged as above, it has $2$ unpaired electrons, so it is paramagnetic.

In $[N i (C O)_{4}]$ oxidation state of $N i$ is $0$ and two electron from $4 s$ is pushed into $3 d$ oribtal.

$↿⇂ ↿⇂ ↿⇂ ↿⇂ ↿⇂$

It has fully filled d-orbital.

$∴ [N i (C O)_{4}]$ is diamagnetic

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(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)

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

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

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

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