Question

Find the species/elemental atom that has a minimum number of unpaired electrons.

• A. ${Kr}^{+}$
• B. ${Mn}^{2+}$
• C. ${Fe}^{3+}$
• D. $O^{+}$

Answer 1

The correct option is A ${Kr}^{+}$

${Kr}^{+}\to {4s}^2{4p}^5$ $\therefore$ One unpaired electron
${Mn}^{2+}\to {3d}^5{4s}^0$ $\therefore$ $5$ unpaired electron
${Fe}^{3+}\to {3d}^5{4s}^0$ $\therefore$ $5$ unpaired electron
$O^{+}\to {2s}^2{2p}^3$ $\therefore$ $3$ unpaired electron

Hence, option $A$ is correct.