I calculated the magnetic moment of $\ce{K4[Cr(CN)6]}$? in the following way:
$\ce{Cr^{+2}}$ has $4$ d electrons. And since $\ce{CN-}$ is a strong field ligand, the electrons will pair up in the $\mathrm{t_{2g}}$. The electronic configuration would be $\mathrm{t_{2g}^4}\;\mathrm{e_g^0}$. Or more specifically, $\mathrm{d}_x^2\; \mathrm{d}_y^1\; \mathrm{d}_z^1\; \mathrm{d}_{z^2}^0\; \mathrm{d}_{x^2-y^2}^0$.
The magnetic moment will be a contribution of both orbital and spin magnetic moments.
$$\mu = \root\of {n(n+2)+L(L+1)},$$
where $n$ will be $2$ as the unpaired electrons are $2$.
$L$ will be $2\times 2 + 1\times 1 +0\times 1 = 5$ corresponding to $\Sigma m_l$.
Solving I get $\mu = 3.74$. But the experimental value is $3.2$.
Why is there this deviation?