Question

Match Ion given in List I with their Magnetic Moment (Calculated) given in List II.

• A. (A)-(IV), (B)-(II), (C)-(I), (D)-(III)
• B. (A)-(II), (B)-(IV), (C)-(I), (D)-(III)
• C. (A)-(I), (B)-(II), (C)-(III), (D)-(IV)
• D. (A)-(IV), (B)-(I), (C)-(II), (D)-(III)

Hint:

Memorize the common spin-only magnetic moments: \(d^1=1.73\), \(d^3=3.87\), \(d^5=5.92\) BM.

Answer 1

The Correct Option is A

Step 1: Determine configuration of \(\mathrm{Sc^{3+}}\).
\[ Sc=[Ar]3d^14s^2 \] \[ Sc^{3+}=[Ar] \] Number of unpaired electrons \(=0\) \[ \mu=0 \] Therefore \[ (A)\rightarrow(IV) \]

Step 2: Determine configuration of \(\mathrm{Ti^{3+}}\).
\[ Ti^{3+}=[Ar]3d^1 \] \(n=1\) \[ \mu=\sqrt{1(1+2)} =\sqrt3 =1.73 \] Therefore \[ (B)\rightarrow(II) \]

Step 3: Determine configuration of \(\mathrm{V^{2+}}\).
\[ V^{2+}=[Ar]3d^3 \] \(n=3\) \[ \mu=\sqrt{3(5)} =\sqrt{15} =3.87 \] Therefore \[ (C)\rightarrow(I) \]

Step 4: Determine configuration of \(\mathrm{Mn^{2+}}\).
\[ Mn^{2+}=[Ar]3d^5 \] \(n=5\) \[ \mu=\sqrt{5(7)} =\sqrt{35} \approx5.92 \] Thus \[ (D)\rightarrow(III) \]

Step 5: Final matching.
\[ (A)-(IV),\ (B)-(II),\ (C)-(I),\ (D)-(III) \]