Find the element with highest enthalpy of atomisation

For transition elements, enthalpy of atomisation generally depends on the strength of metallic bonding, which is related to the number of unpaired electrons available for metal-metal bonding.

Among $\mathrm{Sc}, \mathrm{Mn}, \mathrm{Co}$ and $\mathrm{Cu}$:

• $\mathrm{Sc}$: relatively low
• $\mathrm{Mn}$: lower than expected because of stable half-filled $3d^5 4s^2$ configuration
• $\mathrm{Co}$: high enthalpy of atomisation
• $\mathrm{Cu}$: lower than Co

Hence, the element with the highest enthalpy of atomisation among these is: $\boxed{\mathrm{Co}}$

Write the electronic configuration of $\mathrm{Co}$ and $\mathrm{Co}^{2+}$

Atomic number of Co is $27$.

Neutral Co: $\mathrm{Co}: [\mathrm{Ar}]\,3d^7 4s^2$

For $\mathrm{Co}^{2+}$, remove two electrons from $4s$ first: $\mathrm{Co}^{2+}: [\mathrm{Ar}]\,3d^7$

Find number of unpaired electrons in $3d^7$

For a free ion $3d^7$ configuration, the number of unpaired electrons is $3$.

So, $n = 3$

Calculate spin-only magnetic moment

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

Substituting $n=3$: $\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87\ \text{BM}$

Nearest integer: $\boxed{4}$

Compare with stored answer

Derived answer = $4$

Stored correct answer = $4$

Therefore, the answer agrees.