### Related Formula Spin-only magnetic moment expression: mu = sqrtn(n+2)mathrm\ B.M.$\mu = \sqrt{n(n+2)}\mathrm{\ B.M.}$ ### Core Logic Evaluating the oxidation states and stability profiles: - In TiO$TiO$: Ti^2+$Ti^{2+}$ - In VO$VO$: V^2+$V^{2+}$ - In Mn_2O_3$Mn_2O_3$: Mn^3+$Mn^{3+}$ Mn^3+$Mn^{3+}$ possesses a very high reduction potential (E^circ_Mn^3+/Mn^2+ = +1.57mathrm\ V$E^\circ_{Mn^{3+}/Mn^{2+}} = +1.57\mathrm{\ V}$), making it an exceptionally strong oxidizing agent because it easily gains an electron to form stable Mn^2+$Mn^{2+}$ (d^5$d^5$ configuration). ### Step 1: Calculate the Magnetic Moment of Mn(III) Electronic configuration of Mn^3+$Mn^{3+}$: Mn^3+ = [Ar]3d^4 implies n = 4text unpaired electrons$Mn^{3+} = [Ar]3d^4 \implies n = 4\text{ unpaired electrons}$ Calculating the spin-only magnetic moment: mu = sqrt4(4+2) = sqrt24 approx 4.89mathrm\ B.M.$\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.89\mathrm{\ B.M.}$ ### Step 2: Rounding to Nearest Integer Rounding 4.89mathrm\ B.M.$4.89\mathrm{\ B.M.}$ to the nearest integer gives 5$5$. ### Pattern Recognition High reduction potentials are strongly tied to manganese in its +3$+3$ oxidation state. To quickly estimate magnetic moments, remember that a system with n$n$ unpaired electrons always results in a value of 'n.textsomething$n.\text{something}$' B.M. Thus, 4$4$ unpaired electrons rightarrow 4.89mathrm\ B.M.$\rightarrow 4.89\mathrm{\ B.M.}$, which rounds up to 5$5$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: d- and f-Block Elements