Answer 1

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).

∴ 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.