Q.(i) How many orbitals are possible for n = 4?

The total number of orbitals for a given n is n^2, giving 16 orbitals for n = 4; nitrogen's 7 electrons fill as 1s2 2s2 2p3, with Hund's rule placing one electron in each of the three 2p orbitals before any pairing.

(This question offered an OR alternative on Pauling's method for ionic radius; the primary parts (i) and (ii) below are answered.)

(i) Number of orbitals for n = 4:

For a given principal quantum number n, the azimuthal quantum number l can take values 0, 1, 2, ..., (n-1), and for each value of l there are (2l+1) orbitals (values of the magnetic quantum number m).

For n = 4: l = 0 (s), 1 (p), 2 (d), 3 (f)

• l = 0 (4s): 1 orbital
• l = 1 (4p): 3 orbitals
• l = 2 (4d): 5 orbitals
• l = 3 (4f): 7 orbitals Total = 1 + 3 + 5 + 7 = 16 orbitals

This matches the general formula: total orbitals for a shell = n^2 = 4^2 = 16.

(ii) Electronic configuration and orbital diagram of nitrogen (Z = 7):

Nitrogen has 7 electrons, filled according to the Aufbau principle: 1s2 2s2 2p3.

Orbital diagram (each box is one orbital, arrows are electrons):

1s: paired (up-down)

2s: paired (up-down) …