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Give reasons for the following.
Co³⁺ readily loses one electron whereas Mn³⁺ readily gains one electron. (Atomic numbers of Co = 27…
$\text{Co}^{3+}$ readily loses one electron whereas $\text{Mn}^{3+}$ readily gains one electron.
(Atomic numbers of $\text{Co} = 27$ and $\text{Mn} = 25$)
Answer
Answer
$\mathrm{Mn^{3+}}$ is $3d^4$: by gaining one electron it becomes $\mathrm{Mn^{2+}}$, $3d^5$, a half-filled d subshell that is extra stable, so $\mathrm{Mn^{3+}}$ readily gains an electron (it is a strong oxidising agent).
The cobalt statement fits $\mathrm{Co^{2+}}$ (the printed "$\mathrm{Co^{3+}}$" appears to be a misprint): $\mathrm{Co^{2+}}$ ($3d^7$) readily loses one electron to form $\mathrm{Co^{3+}}$ ($3d^6$), which with strong-field ligands has the very stable low-spin $t_{2g}^6$ arrangement (high crystal field stabilisation energy), so Co(II) is easily oxidised to Co(III) in its complexes.