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(B) The electronic configuration of $Ti$ $(Z=22)$ is $[Ar] 3d^2 4s^2$. Thus,$Ti^{2+}$ is $[Ar] 3d^2$,which has $n=2$ unpaired electrons.The electronic

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Same number of unpaired electrons

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(B) The electronic configuration of $Ti$ $(Z=22)$ is $[Ar] 3d^2 4s^2$. Thus,$Ti^{2+}$ is $[Ar] 3d^2$,which has $n=2$ unpaired electrons.The electronic configuration of $Ni$ $(Z=28)$ is $[Ar] 3d^8 4s^2$. Thus,$Ni^{2+}$ is $[Ar] 3d^8$. In $3d^8$,there are $2$ unpaired electrons $(n=2)$.Therefore,both $Ti^{2+}$ and $Ni^{2+}$ have the same number of unpaired electrons.

$I. \ Zn^{2+}$	$(i) \ d^8 \ \text{configuration}$
$II. \ Cu^{2+}$	$(ii) \ \text{Colourless}$
$III. \ Ni^{2+}$	$(iii) \ \mu = 1.73 \ \text{BM}$

$Ti^{2+}$ and $Ni^{2+}$ have .....

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