Question

Some octahedral complex ions have the same d-orbital splitting diagrams whether they are high spin or low spin. For which of the following transition metal ions is this true?

1. ${\mathit{V}}^{\mathbf{3}\mathbf{+}}$
2. ${\mathbf{Ni}}^{\mathbf{2}\mathbf{+}}$
3. ${\mathbf{Ru}}^{\mathbf{2}\mathbf{+}}$
   

Short answer

${\mathrm{V}}^{3+},{\mathrm{Ni}}^{2+}$have the same d-orbital splitting diagram.

Step-by-step solution

Step:1

${\mathrm{V}}^{3+}$has electronic configuration is$3d^2$ . So for both high spin or low spin electrons will remain in${\mathrm{t}}_{2\mathrm{g}}$ orbital.${\mathrm{Ni}}^{2+}$ has electronic configuration$3{\mathrm{d}}^8$. So for both high spin or low spin 6 electrons are there in role="math" localid="1664046353183" ${\mathrm{t}}_{2\mathrm{g}}$ and 2 electrons are there in role="math" localid="1664046362068" ${\mathrm{e}}_{\mathrm{g}}$orbital. Same d-orbital splitting is there in both ions.

Step:2

But$R{u}^{+2}$has electronic configuration is$4{\mathrm{d}}^8$. So for high spin 4 electrons will remain in localid="1664047044412" ${\mathrm{t}}_{2\mathrm{g}}$ orbital and for low spin 6 electrons will remain in orbital. So same d-orbital splitting diagram is not there in ${\mathrm{Ru}}^{+2}$.