# How would you calculate the formal charge of each atom in the following compounds: NO, AlH_4^- ?

For $\text{tetrahydroaluminate anion}$, we know that there is a negative charge on the ion. The charge is formally centred on the aluminum nucleus, because it is associated with 4 bonding electrons rather than 3. Another way to look at the aluminate ion is as the alkali metal hydride adduct of $A l {H}_{3}$, i.e. $L i H \cdot A l {H}_{3}$. Lithium aluminum hydride can still potentially deliver up to 4 equiv of hydride.
For $N O$, this has a problematic Lewis representation with an odd electron. $N = O$ is reasonable. We have $5 + 6$ valence electrons to distribute. Around oxygen there are 2 lone pairs; around nitrogen there is one lone pair, and one single electron. Each centre is thus neutral.