1. Find which metal has the highest second ionisation enthalpy

Second ionisation enthalpy means removal of an electron from $M^+$: $M^+ \rightarrow M^{2+} + e^-$

This will be maximum when $M^+$ has an especially stable electronic configuration.

Let us write the ground-state configurations:

• $\mathrm{Sc}: [\mathrm{Ar}]\,3d^1 4s^2$
• $\mathrm{Ti}: [\mathrm{Ar}]\,3d^2 4s^2$
• $\mathrm{V}: [\mathrm{Ar}]\,3d^3 4s^2$
• $\mathrm{Cr}: [\mathrm{Ar}]\,3d^5 4s^1$
• $\mathrm{Mn}: [\mathrm{Ar}]\,3d^5 4s^2$
• $\mathrm{Fe}: [\mathrm{Ar}]\,3d^6 4s^2$

After first ionisation, the $4s$ electron is removed first:

• $\mathrm{Sc}^+: [\mathrm{Ar}]\,3d^1 4s^1$
• $\mathrm{Ti}^+: [\mathrm{Ar}]\,3d^2 4s^1$
• $\mathrm{V}^+: [\mathrm{Ar}]\,3d^3 4s^1$
• $\mathrm{Cr}^+: [\mathrm{Ar}]\,3d^5$
• $\mathrm{Mn}^+: [\mathrm{Ar}]\,3d^5 4s^1$
• $\mathrm{Fe}^+: [\mathrm{Ar}]\,3d^6 4s^1$

Among these, $\mathrm{Cr}^+$ has the especially stable half-filled configuration: $[\mathrm{Ar}]\,3d^5$ Therefore, removing one more electron from $\mathrm{Cr}^+$ requires the maximum energy.

So, $M = \mathrm{Cr}$

2. Find the magnetic moment of $M^+$

Since $M = \mathrm{Cr}$, $M^+ = \mathrm{Cr}^+ = [\mathrm{Ar}]\,3d^5$

For $3d^5$, the number of unpaired electrons is $n = 5$

Spin-only magnetic moment is: $\mu = \sqrt{n(n+2)}\ \text{BM}$

Substitute $n=5$: $\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92\ \text{BM}$

Near integer, $\mu \approx 6\ \text{BM}$

3. Final Answer

The required spin-only magnetic moment is $\boxed{6\ \text{BM}}$