(i) n = 3 (Given)
For a given value of n, l can have values from 0 to (n – 1).
$\therefore$ For n = 3
l = 0, 1, 2
For a given value of l, ml can have (2l + 1) values.
For l = 0, m = 0 l = 1, m = – 1, 0, 1 l = 2, m = –
2, – 1, 0, 1, 2
$\therefore$ For n = 3
l = 0, 1, 2
${\mathrm{m}}_0$ = 0
${\mathrm{m}}_1$ = – 1, 0, 1 m2 = –2, – 1, 0, 1, 2

(ii) For 3d orbital, l = 2.
For a given value of l, ml can have (2l + 1) values i.e., 5 values.
$\therefore$ For l = 2 m2 = – 2, – 1, 0, 1, 2

(iii) Among the given orbitals only 2s and 2p are possible. 1p and 3f cannot exist.
For p-orbital, l = 1.
For a given value of n, l can have values from zero to (n – 1).
$\therefore$ For l is equal to 1, the minimum value of n is 2.

Similarly,
For f-orbital, l = 4.
For l = 4, the minimum value of n is 5.
Hence, 1p and 3f do not exist.