Among Sc, Mn, Co and Cu, identify the element with highest enthalpy of atomisation. The spin only magnetic moment value of that element in its +2 oxidation state is ________ BM (in nearest integer).

The enthalpy of atomisation ($$ΔH_\text{atom}$$) of the 3d transition elements first rises from Sc to about Ni and then falls towards Cu and Zn. This behaviour is linked to the strength of metallic bonding, which in turn depends mainly on the number of 3d electrons available for delocalised metallic bonding.

Approximate experimental values (kJ mol−1) illustrate the trend: Sc ≈ 330, Mn ≈ 280, Co ≈ 425, Cu ≈ 335. Hence, Co exhibits the largest enthalpy of atomisation among Sc, Mn, Co and Cu.

For the magnetic moment we consider Co in its +2 oxidation state.

Atomic number of Co = 27, so neutral Co is $$[Ar]\,3d^7\,4s^2$$.

Co2+ is formed by losing the two 4s electrons: $$[Ar]\,3d^7$$.

In an aqueous or other weak-field environment the 3d orbitals remain high-spin. A 3d7 high-spin configuration has the distribution $$t_{2g}^5\,e_g^2$$, giving

Number of unpaired electrons, $$n = 3$$.

The spin-only formula is $$\mu_\text{so} = \sqrt{n(n+2)}\; \text{BM}$$.

Substituting $$n = 3$$:

$$\mu_\text{so} = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \text{ BM} \approx 4 \text{ BM (nearest integer)}.$$

Therefore, the element is Co and its spin-only magnetic moment in the +2 state is 4 BM.