Molecular Orbitals of Hydrogen Fluoride (HF)
Bearing in mind the energy and symmetry conditions, in the case of $\ce{HF}$ one can construct MOs using the $1s$ AO on H and the $2s$ and $2p$ AOs on F. 



In general, the contribution to MOs is determined by the coefficients in the linear combination. 

Here, one observes that the $1s$ electrons are almost completely localized on the $\ce{F}$ atom. Also, $1 \pi$ electrons are completely localised on the $\ce{F}$ atom because the $2p_x$ and $2p_y$ orbitals on F have a zero net overlap with the $1s$ orbital on $\ce{H}$. 

Electrons in MOs localized on a single atom are referred to as nonbonding electrons. 

Also, I would note that the $3 \sigma$  MO has less bonding character and the $4 \sigma^*$ MO has less anti-bonding character. 

Note that the total bond order is approximately one because the $3 \sigma$ MO is largely localized on the F atom, the $3 \sigma$ MO is not totally bonding, and the $1\pi$ MOs are completely localized on the F atom. 

On account of fluorine being a more electronegative atom, in the $2 \sigma$ bonding orbital the electron density is much greater on the more electronegative fluorine than on the hydrogen. However, in the anti-bonding $4 \sigma^*$ orbital, this polarity is reversed.

Caveat: The preceding paragraph may support your intuition and may be right in a few simple cases, but I wouldn't rely on it too heavily. 

Given below is a diagram showing $2 \sigma$ , $3 \sigma$ and $1 \pi$ MOs in HF