Question

Correct order of reducing agent, in $V^{2+}$ or $C{o}^{2+}$ is:

• A. $V^{2+}$ = $C{o}^{2+}$
• B. $V^{2+}$ < $C{o}^{2+}$
• C. $V^{2+}$ > $C{o}^{2+}$
• D. none of these

Answer 1

Answer: (B)

$V^{2+}$ < $C{o}^{2+}$

Option $\left(B\right)$ is correct.

$C{o}^{2+}$ is more reducing agent than $V^{2+}$.

$C{o}^{2+}$ has $3d^7$ as outer electronic configuration. So it can easily lose electron to form stable $3d^5$ configuration (half filled orbitals); Hence it is reducing agent. On the other hand, $V^{2+}$ has $3d^3$ as outer electronic configuration So it can easily gain electron to form stable $3d^5$ configuration ( as it has half filled level). Hence it is oxidising agent.