NEET 2020 Chemistry Question 103 Solution

Chromium has atomic number $$Z = 24$$. Its exceptional, half-filled stable ground state electronic configuration is: $$\text{Cr} = [\text{Ar}] 3d^5 4s^1$$

During ionization to form $$Cr^{2+}$$, electrons are removed first from the outermost $$4s$$ subshell, followed by the $$3d$$ subshell: $$\text{Cr}^{2+} = [\text{Ar}] 3d^4 4s^0$$ In the five degenerate $$3d$$ orbitals, according to Hund's rule of maximum multiplicity, all 4 electrons occupy separate orbitals with parallel spins: $$\uparrow, \uparrow, \uparrow, \uparrow, \_$$. Therefore, the number of unpaired electrons is $$n = 4$$.

The spin-only magnetic moment ($$\mu_s$$) is calculated using the formula: $$\mu_s = \sqrt{n(n + 2)} \text{ BM}$$ Substituting $$n = 4$$: $$\mu_s = \sqrt{4(4 + 2)} = \sqrt{4 \times 6} = \sqrt{24} \text{ BM}$$ $$\sqrt{24} = 4.8989... \approx 4.90 \text{ BM}$$

- $$n = 3 \implies \mu = \sqrt{15} = 3.87\text{ BM}$$ (e.g. $$Cr^{3+}$$) - $$n = 5 \implies \mu = \sqrt{35} = 5.92\text{ BM}$$ (e.g. $$Fe^{3+}, Mn^{2+}$$) - $$n = 2 \implies \mu = \sqrt{8} = 2.84\text{ BM}$$ (e.g. $$Ni^{2+}$$).