BITSAT 2024 Shift 1 Paper

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 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:$\mu_s=\sqrt{n(n+2)}\,\mathrm{BM}$Substituting $n=5$ into the formula:$\mu_s=\sqrt{5(5+2)}=\sqrt{35}$Calculating further gives:$\mu_s\approx 5.85\,\mathrm{BM}$Therefore, the spin-only magnetic moment of $\mathrm{Fe}^{3+}$ is approximately 5.85 Bohr Magnetons (BM), which can be rounded to about 6 BM .