1. Magnetic moment depends on number of unpaired electrons

For first-row transition metal ions, the magnetic moment is usually estimated by the spin-only formula:

$\mu = \sqrt{n(n+2)}\; \text{BM}$

where $n$ is the number of unpaired electrons.

So, the ion with the maximum number of unpaired electrons will have the maximum magnetic moment.

2. Find electronic configuration of each ion

Recall the atomic numbers:

• $Ti: 22$
• $Cr: 24$
• $Mn: 25$
• $Fe: 26$

Also, electrons are removed first from $4s$ before $3d$.

(A) $Mn^{2+}$

Neutral Mn:

$Mn = [Ar]3d^5 4s^2$

So,

$Mn^{2+} = [Ar]3d^5$

This has 5 unpaired electrons.

(B) $Fe^{2+}$

Neutral Fe:

$Fe = [Ar]3d^6 4s^2$

So,

$Fe^{2+} = [Ar]3d^6$

For free ion/high spin $3d^6$, number of unpaired electrons = 4.

(C) $Ti^{2+}$

Neutral Ti:

$Ti = [Ar]3d^2 4s^2$

So,

$Ti^{2+} = [Ar]3d^2$

This has 2 unpaired electrons.

(D) $Cr^{2+}$

Neutral Cr:

$Cr = [Ar]3d^5 4s^1$

So,

$Cr^{2+} = [Ar]3d^4$

This has 4 unpaired electrons.

3. Compare magnetic moments

Using $\mu = \sqrt{n(n+2)}$:

• $Ti^{2+}$: $n=2$ $\mu = \sqrt{2(2+2)} = \sqrt{8}$
• $Cr^{2+}$: $n=4$ $\mu = \sqrt{4(4+2)} = \sqrt{24}$
• $Fe^{2+}$: $n=4$ $\mu = \sqrt{24}$
• $Mn^{2+}$: $n=5$ $\mu = \sqrt{5(5+2)} = \sqrt{35}$

Clearly,

$\sqrt{35} > \sqrt{24} > \sqrt{8}$

So, $Mn^{2+}$ has the maximum magnetic moment.

4. Correct option

$\boxed{A\; (Mn^{2+})}$