5
$\begingroup$

I am familiar with the double hump trend for atomisation of transition metals, and I can explain the unusually low atomisation of $\ce{Mn}$ with the exchange principle, specifically that in the solid state the spins are more localised whereas for the gas/vapour the spins are able to match and hence there is a gain in exchange energy (stabilising).

However, what I am struggling to understand is (ignoring the middle drop due to exchange):

Why there is a hump in the first place?

I have come across a few ideas, some suggesting that it is a bonding/antibonding situation, whereby, first you fill a d-band gap up to $\ce{Mn}$ and then an antibonding contribution comes in.

Is there a more eloquent way of explaining this trend?

$\endgroup$
0

1 Answer 1

6
$\begingroup$

The hump is apparent because of the dip. To be more clear, it is good to know that all these attributes follow the same trends:

  • Metallic character
  • Strength of metallic bonding
  • Enthalpy of atomisation
  • Melting point
  • Density

If you plot a graph with any of these attributes on the $y$ axis, you'll notice that more or less in each of these cases, the graph rises first reaches a maxima just before the middle of the series, undergoes a dip in the middle again rises and then finally decreases till the end of the series.

It can be somewhat explained (these observations are made on an experimental basis) by the strength of metallic bonding. As the number of unpaired electrons increase, the metallic bonding becomes stronger. Note that $\ce{Cr}$ has the highest number of unpaired $e^-$s: $6$. This explains the hump.

$\ce{Mn}$ on the other hand has a half-filled electronic configuration. Thus, it is more stable. Stability order on the grounds of electronic configuration follows the order:

Fully filled > Half filled > Partially filled

A lesser strength of metallic bonding, plus the stability collectively account for the dip.

Also, as a quick roundup of densities, density of $\ce{Os}$ is the highest: $\pu{22.59 gm cm^{-3}}$ and $\ce{Ir}$ stands quite close $\pu{22.56 gm cm^{-3}}$. Another fun fact, gold and mercury are also quite dense: $\pu{\rho_{Au} = 19.32 gm cm^{-3}}$, $\pu{\rho_{Hg} = 13.56 gm cm^{-3}}$. As a comparison (as to how dense I actually mean): $\pu{\rho_{Pb} = 11.34 gm cm^{-3}}$.

The low melting points of $\ce{Mn}$ and $\ce{Tc}$ are also mapped to half-filled stable configurations.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.