Atomization enthalpies of transition elements

Question

So, my book says that transition elements have higher enthalpies of atomization than other elements (say s- or p- block) because of stronger metallic bonding, primarily due to large number of unpaired electrons in the d-orbitals.

Now, if that is the case, then the atomization energy for $\ce{Ni}$ should be less than for $\ce{Fe}$ and $\ce{Co}$, as $\ce{Ni}$ has only 2 unpaired electrons compared to $\ce{Fe}$ and $\ce{Co}$ (4 and 3 respectively).

Actually, it is just the opposite.

$$\begin{array}{cc} \text{Metal} & \Delta_\text{atom}H(\ce{M})\text{ / }\mathrm{kJ~mol^{-1}} \\ \hline \ce{Ni} & 430 \\ \ce{Co} & 425 \\ \ce{Fe} & 416 \end{array}$$

Why could this be?

Answer 1

These sorts of generalities always sweep aside other considerations.

$\ce{^26Fe}$ has a ​body-centered cubic crystal structure.

$\ce{^27Co}$ has a ​hexagonal close-packed crystal structure.

$\ce{^28Ni}$ has a ​face-centered cubic crystal structure.

Here is a nice table showing the overall trends.

Answer 2

According to J.D.Lee's "Concise Inorganic Chemistry", the extra electrons in the metals of the 2nd half of the transition series, like Fe, Ni etc.. are excited to the p level .. and then the unpaired electrons in the d, s and p levels participate in metallic bonding .. so Ni has the max. unpaired electrons, followed by Co and then Fe ... and hence strength of metallic bonding follows same order