QJEEM24Chb731: JEE Main 2024 Chemistry - Class 12 JEEM24Chb731 Solution: A transition metal ' ' among and has the highest second…

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.