(a) Write down the ionization isomer of $[C o {(N H_{3})}_{5} C l] S O_{4}$ .
(b) Write the IUPAC name of the above compound.
(c) $[N i {(C O)}_{4}]$ is diamagnetic while ${[N i {C l}_{4}]}^{2 -}$ is paramagnetic though both are tetrahedral. Why?

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Solution

(a) I UPAC name - pentaaminechlorocobalt(III)sulphate.

(b) $[C O (N H_{3})_{5} C l] S O_{4}$ and $[C O (N H_{3})_{5} S O_{4}] C l$ both are ionisation isomers.

(c) Though both $[N i C l_{4}]^{2 -}$ and $N i C O_{4}]$ are tetrahedral, their magnetic character is different due to the difference in their ligands. $C l^{-}$ is weak field ligand so it does not cause the pairing of 3d electrons which are unpaired in their ground state. Thus $[N i C l_{4}]^{2 -}$ is paramagnetic. (Ref. image (i)]
In $N i (C O)_{4}$ , $N i$ is in zero oxidation state with $3 d^{8} 4 s^{2}$ (Ref. image (ii)] configuration.
$C O$ is a strong field ligand which cause pairing of $3 d$ electrons and to shift $4 s$ electrons in $3 d$ orbital. so no unpaired are present and compound is diamagnetic.

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Similar questions

[NiCl₄]²⁻ is paramagnetic while [Ni(CO)₄] is diamagnetic though both are tetrahedral. Why?

[N i C l_{4}]

[N i C O_{4}]

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

d^{4}

Δ_{0} > P .

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

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

(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 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

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