For #"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 #AlH_3#, i.e. #LiH*AlH_3#. Lithium aluminum hydride can still potentially deliver up to 4 equiv of hydride.
For #NO#, 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.
