What is the hybridisation in dimolybdenum?
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.