Ahydrogen atom orbital has n=5 and ml=−2 .what are the possible values of l for this orbital?

Answer : There are three possible values of 'l' which are 2, 3 and 4.

Solution : Given,

n = 5

There are 4 quantum numbers :

        values are 1, 2, 3, 4, .........

        values are 0 to (n-1)

        values are +l to -l

        values are to

when n = 5 then the value of l are,

l = 0, 1, 2, 3, 4

At l = 0,  

At l = 1,  

At l = 2,  

At l = 3,  

At l = 4,  

So, when then the possible values of l are,

l = 2, 3, 4

 H-N-H bond angles : NH₂⁻ <  NH₃ <  NH₄⁺

Explanation:

The question tell us that the central nitrogen has four pairs of electron domains. Four pairs of electron domains predict a tetrahedral geometry but the molecular geometry will be affected by the presence of lone pair of electron and they are the ones that require more space because they have greater repulsion.

In NH₄⁺ all the 4 pairs are used in bonding with H, therefore the molecular geometry will be tetrahedral with H-N-H bond angles of 109 º.

In NH₃ we have one lone pair of electron which push the H-N-H angles less because of repulsion.

In NH₂⁻, there are two pairs of lone pairs of electrons which push the H-N-H even more.

So the order of increasing H-N-H  bond angle is  NH₂⁻ <  NH₃ <  NH₄⁺