d and f Block Elements

We calculate the number of unpaired electrons in each ion,

(i) $\mathrm{Cr}^{3+} \rightarrow[\mathrm{Ar}] 3 d^3 \rightarrow 3$ unpaired electrons

(ii) $\mathrm{Fe}^{3+} \rightarrow[\mathrm{Ar}] 3 d^5 \rightarrow 5$ unpaired electrons

(iii) $\mathrm{Mn}^{2+} \rightarrow[\mathrm{Ar}] 3 d^5 \rightarrow 5$ unpaired electrons

(iv) $\mathrm{Co}^{2+} \rightarrow[\mathrm{Ar}] 3 d^7 \rightarrow 3$ unpaired electrons

(v) $\mathrm{Ni}^{2+} \rightarrow[\mathrm{Ar}] 3 d^8 \rightarrow 2$ unpaired electrons

Now compare:

• $\mathrm{Cr}^{3+}$ and $\mathrm{Co}^{2+} \rightarrow$ both have 3 unpaired electrons.

• $\mathrm{Fe}^{3+}$ and $\mathrm{Mn}^{2+} \rightarrow$ both have 5 unpaired electrons.

Two distinct pairs have same number of unpaired electrons.

$\mathrm{Fe}^{3+}$ is $3 d^5$, all electrons unpaired (high-spin $d^5$ ): Spin-only moment

$ \begin{aligned} & =\sqrt{[n(n+2)]}=\sqrt{[5 \times 7]} \\ & =\sqrt{35} \approx 5.92 \mathrm{BM} \approx 6 \mathrm{BM} \end{aligned} $

Given the species, we need to determine which have the same magnetic moment by counting their unpaired electrons.

For the species $ \text{Cr}^{2+} $ and $ \text{Mn}^{3+} $, both have the same number of unpaired electrons:

Number of unpaired electrons, $ n = 4 $

The magnetic moment $\mu$ is calculated using the formula:

$ \mu = \sqrt{n(n+2)} = \sqrt{4(4+2)} = \sqrt{24} \, \text{BM} $

Thus, $ \text{Cr}^{2+} $ and $ \text{Mn}^{3+} $ have the same magnetic moment.

The spin-only magnetic moment of the $ \mathrm{Fe}^{3+} $ ion is determined by its electronic configuration and the number of unpaired electrons.

Electronic Configuration:

The neutral iron atom ($_{26} \mathrm{Fe}$) has the electronic configuration:

$ [\mathrm{Ar}] 3d^6 4s^2 $

When it forms a $ \mathrm{Fe}^{3+} $ ion, it loses three electrons. The configuration becomes:

$ [\mathrm{Ar}] 3d^5 $

Number of Unpaired Electrons:

In the $ \mathrm{Fe}^{3+} $ ion, there are 5 electrons in the 3d subshell. All these electrons are unpaired, hence:

$ n = 5 $

Calculating the Spin-Only Magnetic Moment:

The formula for the spin-only magnetic moment ($\mu_s$) is:

$ \mu_s = \sqrt{n(n+2)} \, \text{BM} $

Substituting $ n = 5 $ into the formula:

$ \mu_s = \sqrt{5(5+2)} = \sqrt{35} $

Calculating further gives:

$ \mu_s \approx 5.85 \, \text{BM} $

Therefore, the spin-only magnetic moment of $ \mathrm{Fe}^{3+} $ is approximately 5.85 Bohr Magnetons ($\mathrm{BM}$), which can be rounded to about 6 $\mathrm{BM}$.

Option (d) does not contain a pair of mixed oxides. Fe₂O₃ is not mixed oxide.

While, Mn₃O₄ (2MnO + MnO₂); Co₃O₄ (CoO + Co₂O₃); Fe₃O₄ (Fe₃O₃ + FeO) and Pb₃O₄ (2PbO + PbO₂), all are mixed salts.