Question

Give Reasons: ${\mathrm{Cr}}^{2+}$ is a strong reducing agent whereas ${\mathrm{Mn}}^{2+}$ is not. $[\mathrm{Cr}=24,\mathrm{Mn}=25]$

Answer 1

Reducing agent: The reducing agent is a compound that itself gets oxidized(by losing electron/s) while reducing the other substance it is being treated with.

Electronic configuration of ${\mathrm{Cr}}^{2+}$:

• Electronic configuration of Cr is $\left[\mathrm{Ar}\right]3{\mathrm{d}}^{5}4{\mathrm{s}}^1$

• Electronic configuration of ${\mathrm{Cr}}^{2+}$ is $\left[\mathrm{Ar}\right]3{\mathrm{d}}^{4}4{\mathrm{s}}^0$ i.e, one 4s electron and one 3d electron are released as they become outermost shell's electrons.
• Hence, ${\mathrm{Cr}}^{2+}$ is a good reducing agent. It has a ${\mathrm{d}}^4$ configuration.

• While acting as a reducing agent, it gets oxidized to ${\mathrm{Cr}}^{3+}$ and acquires a ${\mathrm{d}}^3$ configuration.
• This ${\mathrm{d}}^3$ configuration can be written as a stable ${\mathrm{t}}_{2\mathrm{g}}$ configuration.

• In the presence asymmetrical field, splitting of the d orbitals takes place. The pattern of splitting depends upon the nature of the crystal field.
• The ${\mathrm{d}}_{\mathrm{xy}},{\mathrm{d}}_{\mathrm{xz}},$ and ${\mathrm{d}}_{\mathrm{yz}}$ orbitals are collectively called the ${\mathrm{t}}_{2\mathrm{g}}$ orbitals, whereas the${\mathrm{d}}_{{\mathrm{z}}^2}$ and ${\mathrm{d}}_{{\mathrm{x}}^2-{\mathrm{y}}^2}$ orbitals are called the ${\mathrm{e}}_{\mathrm{g}}$ orbitals in octahedral geometry.
• So, here in the case of ${\mathrm{Cr}}^{3+}$, all the three electrons enter the low energy ${\mathrm{t}}_{2\mathrm{g}}$ orbital.

Electronic configuration of ${\mathrm{Mn}}^{2+}$:

• Mn is $\left[\mathrm{Ar}\right]3{\mathrm{d}}^{5}4{\mathrm{s}}^2$

• ${\mathrm{Mn}}^{2+}$ is $\left[\mathrm{Ar}\right]3{\mathrm{d}}^{5}4{\mathrm{s}}^0$ i.e, two 4s electrons are released.
• Hence, ${\mathrm{Mn}}^{2+}$ has a stable ${\mathrm{d}}^5$ con uration.

• This has an exactly half-filled d-orbital and has extra stability.
• After reduction, it will become ${\mathrm{Mn}}^{3+}$ has ${\mathrm{d}}^4$ configuration, which is not much stable as ${\mathrm{d}}^5$.
• S, it acts as an oxidizing agent, gets reduced to ${\mathrm{Mn}}^{2+}$, and acquires a stable ${\mathrm{d}}^5$ configuration.

Thus, ${\mathrm{Cr}}^{2+}$ is a strong reducing agent whereas ${\mathrm{Mn}}^{2+}$ is not.