"A transition metal '$\mathrm{M}$' among $\mathrm{Sc}, \mathrm{Ti}, \mathrm{V}, \mathrm{Cr}, \mathrm{Mn}$ and $\mathrm{Fe}$ has the highest second ionisation enthalpy. The spin-only magnetic moment value of $\mathrm{M}^{+}$ ion is _______ BM (Near integer) (Given atomic number $$\mathrm{Sc}: 21, \mathrm{Ti}: 22, \mathrm{~V}: 23, \mathrm{Cr}: 24, \mathrm{Mn}: 25, \mathrm{Fe}: 26$$)"

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Identify which metal (M) has the highest second ionization enthalpy.

We are comparing the elements Sc, Ti, V, Cr, Mn, and Fe in terms of their second ionization enthalpy (IE₂). Recall:

The second ionization enthalpy (IE₂) is the energy required to remove one electron from the singly charged ion $ \mathrm{M}^{+} $ to form $ \mathrm{M}^{2+} $.

For most 3d transition metals, the first electron is removed from the 4s orbital.

Particularly large values of IE₂ often occur if, after removal of the first electron, one is forced to remove an electron from a stable configuration (e.g., half-filled d-orbital).

Let us outline the ground-state (neutral atom) electron configurations and the removal of the first and second electrons:

Sc $\bigl[ \mathrm{Ar} \bigr] 3d^1\,4s^2 $

$\mathrm{Sc} \rightarrow \mathrm{Sc}^{+}$: remove 1 electron from 4s

$\mathrm{Sc}^{+} \rightarrow \mathrm{Sc}^{2+}$: remove the second 4s electron

Final: $\mathrm{Sc}^{2+} = [\mathrm{Ar}]\,3d^1$

Ti $\bigl[ \mathrm{Ar} \bigr] 3d^2\,4s^2 $

1st electron from 4s → $\mathrm{Ti}^{+} = [\mathrm{Ar}]\,3d^2\,4s^1$

2nd electron from 4s → $\mathrm{Ti}^{2+} = [\mathrm{Ar}]\,3d^2$

V $\bigl[ \mathrm{Ar} \bigr] 3d^3\,4s^2 $

1st electron from 4s → $\mathrm{V}^{+} = [\mathrm{Ar}]\,3d^3\,4s^1$

2nd electron from 4s → $\mathrm{V}^{2+} = [\mathrm{Ar}]\,3d^3$

Cr $\bigl[ \mathrm{Ar} \bigr] 3d^5\,4s^1 $

1st electron from 4s → $\mathrm{Cr}^{+} = [\mathrm{Ar}]\,3d^5$ (a stable half-filled d$^{5}$ configuration)

2nd electron now must come from the 3d shell → $\mathrm{Cr}^{2+} = [\mathrm{Ar}]\,3d^4$

Because removing an electron from a half-filled d$^{5}$ orbital costs a lot of energy, $\mathrm{Cr}$ generally has a notably high second IE.

Mn $\bigl[ \mathrm{Ar} \bigr] 3d^5\,4s^2 $

1st electron from 4s → $\mathrm{Mn}^{+} = [\mathrm{Ar}]\,3d^5\,4s^1$

2nd electron from 4s → $\mathrm{Mn}^{2+} = [\mathrm{Ar}]\,3d^5$ (thus still d$^{5}$ in the end)

The second ionization for Mn is large but less dramatic than for Cr because Mn$^{2+}$ ends with a half-filled d$^{5}$. You are not removing from a half-filled d$^{5}$ to get Mn$^{2+}$.

Fe $\bigl[ \mathrm{Ar} \bigr] 3d^6\,4s^2 $

1st electron from 4s → $\mathrm{Fe}^{+} = [\mathrm{Ar}]\,3d^6\,4s^1$

2nd electron from 4s → $\mathrm{Fe}^{2+} = [\mathrm{Ar}]\,3d^6$

From experimental data (and the arguments above), $\mathrm{Cr}$ indeed has the highest second ionization enthalpy among these six metals.

Determine the electronic configuration of $\mathbf{Cr}^{+}$ and its spin-only magnetic moment.

Neutral $\mathrm{Cr}$: $\bigl[ \mathrm{Ar} \bigr]\,3d^5\,4s^1$

$\mathrm{Cr}^{+}$: Removal of the 4s electron ⇒ $\bigl[ \mathrm{Ar} \bigr]\,3d^5$

The $3d^5$ configuration has 5 unpaired electrons.

The spin-only magnetic moment $\mu$ is given by

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

where $n$ = number of unpaired electrons. Here $n = 5$, so

$ \mu = \sqrt{5(5+2)} \;=\; \sqrt{35} \;\approx\; 5.92 \;\text{BM}. $

This is often rounded to about 5.9 or 6.0 BM.

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