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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?

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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.

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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).

MOs of Mo2

So, to answer your questions:

  1. [Does] Dimolybdenum [...] use its $\ce{4d}$ and $\ce{5s}$ orbitals to form a sextuple bond (2 σσ, 2 ππ, 2 δδ)[?]
    Yes, but the MOs are as described above.

  2. [Are] the $\ce{4s}$ and $\ce{4p}$ orbitals [...] going to hybridize with the $\ce{4d_{z^2}}$ orbital [...] and the $\ce{5s}$[?]
    No, they don’t hybridize the way you imagine it.

  3. [Do] the s and p lone pairs always hybridize, even if they don't form any bonds?
    That question is de facto answered, as there is no hybridization.


(The shown molecular orbitals and the occupation numbers result from a fully relativistic (DKH2) CASSCF(12,12)/def2-QZVP calculation with ORCA 3.0.3; RI was used. The calculation is based on A. C. Borin, J. P. Gobbo, B. O. Roos, Chem. Phys. 2008, 343, 210-216.)

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