JEE Chemistry 2024 Question with Solution

Given: A transition metal M among Sc, Ti, V, Cr, Mn, and Fe has the highest second ionisation enthalpy.

Find: The spin-only magnetic moment of $M^+$ ion as the nearest integer.

From the solution, Cr has the highest second ionisation enthalpy because $\text{Cr}^+$ has the stable half-filled configuration $[\text{Ar}]\,3d^5$, so removing the next electron is difficult.

Electronic configuration:

$$\text{Cr} = [\text{Ar}]\,3d^5 4s^1$$

After losing one electron:

$$\text{Cr}^+ = [\text{Ar}]\,3d^5$$

For a $d^5$ configuration, the number of unpaired electrons is $n = 5$.

The spin-only magnetic moment is:

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

Substituting $n = 5$:

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

Nearest integer:

$$6$$

Therefore, the spin-only magnetic moment value of $M^+$ ion is 6.

Given: Elements are Sc, Ti, V, Cr, Mn, and Fe.

Find: Which element has the highest second ionisation enthalpy and then compute the magnetic moment of $M^+$.

Relevant configurations mentioned in the solution are:

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

The second ionisation enthalpy corresponds to removing an electron from $M^+$. The solution identifies Cr as the correct element because $\text{Cr}^+$ has the especially stable half-filled subshell:

$$\text{Cr}^+ = [\text{Ar}]\,3d^5$$

Removing another electron from this half-filled $d$ subshell is difficult, so the second ionisation enthalpy is highest for Cr.

Now use the spin-only formula:

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

For $\text{Cr}^+$, there are $5$ unpaired electrons, so:

$$\mu = \sqrt{5(7)} = \sqrt{35} \approx 5.92\,\text{BM}$$

Hence, the nearest integer value is 6.