# What is the hybridisation in dimolybdenum?

Dimolybdenum's $\mathrm{4s}$ and $\mathrm{4p}$ orbitals are full. It is going to use its $\mathrm{4d}$ and $\mathrm{5s}$ orbitals to form a sextuple bond (2 $\sigma$, 2 $\pi$, 2 $\delta$). I am wondering if the $\mathrm{4s}$ and $\mathrm{4p}$ orbitals are going to hybridize with the $\mathrm{4d}_{z^2}$ orbital (the one forming the first $\sigma$ bond) and the $\mathrm{5s}$, given the differences in energy. The hybridisation would be $\mathrm{ds^2p^3}$, which is certainly uncommon.

If so, do the $\mathrm{s}$ and $\mathrm{p}$ lone pairs always hybridize, even if they don't form any bonds?

The electronic configuration of $\ce{Mo2}$ is given in Bursten, B.E.; Cotton, F.A.; Hall, M.B. Dimolybdenum: nature of the sextuple bond. J. Am. Chem. Soc. 1980, 102 (20), 6348–6349:

$$\ldots\mathrm{ 9\sigma_g^{1.88} 5\pi_u^{3.78} 2\delta_g^{3.42} 10\sigma_g^{1.92} 9\sigma_u^{0.08} 2\delta_u^{0.58} 5\pi_g^{0.22} 10\sigma_u^{0.12} }$$

Which can be simplified as:

$$\ldots\mathrm{ (9\sigma_g)^2 (5\pi_u)^4 (2\delta_g)^4 (10\sigma_g)^2 (9\sigma_u)^0 (2\delta_u)^0 (5\pi_g)^0 (10\sigma_u)^0 }$$

• $9\sigma_{\mathrm{g}}$ is not degenerate; it is $\mathrm{4d_{z^2}+4d_{z^2}}$.
• $5\pi_{\mathrm{u}}$ is doubly degenerate; it is $\mathrm{4d_{xz}+4d_{xz}}$ and $\mathrm{4d_{yz}+4d_{yz}}$.
• $2\delta_{\mathrm{g}}$ is doubly degenerate; it is $\mathrm{4d_{xy}+d_{xy}}$ and $\mathrm{d_{x^2-y^2}+d_{x^2-y^2}}$.
• $10\sigma_{\mathrm{g}}$ is not degenerate; it is $\mathrm{5s+5s}$.

Now, let us have a look at the beautiful 4d orbitals, kindly provided by our user Philipp:

From here, you can see that no hybridization is necessary, because the $\mathrm{4d}$ and the $\mathrm{5s}$ orbitals can all form bonds within themselves.

In $\ce{Mo2}$ the six bonding and six anti-bonding $\sigma(s), \sigma(d_{z^2}), \pi(d_{xz,yz})$ and $\delta(d_{xy})$ molecular orbitals with occupation numbers of less than 2 and more than 0, the active space, are only formed by $\ce{5s}$ and $\ce{4d}$ atomic orbitals (see DHMO’s answer for the $\ce{4d}$ atomic orbitals).

1. [Does] Dimolybdenum [...] use its $\ce{4d}$ and $\ce{5s}$ orbitals to form a sextuple bond (2 σσ, 2 ππ, 2 δδ)[?]
2. [Are] the $\ce{4s}$ and $\ce{4p}$ orbitals [...] going to hybridize with the $\ce{4d_{z^2}}$ orbital [...] and the $\ce{5s}$[?]