The following sets of quantum numbers represent four electrons in an atom. (i) $$n = 4, l = 1$$ (ii) $$n = 4, l = 0$$ (iii) $$n = 3, l = 2$$ (iv) $$n = 3, l = 1$$. The sequence representing increasing order of energy, is

For multi-electron atoms the energy of a subshell is decided by the $$\,(n+l)\,$ rule (also called the Madelung rule):

1. Subshells with lower $$n+l$$ have lower energy.
2. If two subshells have the same $$n+l$$, the subshell with lower $$n$$ is lower in energy.

List each set of quantum numbers, identify the subshell and calculate $$n+l$$:

• (i) $$n = 4,\; l = 1 \;\;(4p)$$ → $$n+l = 4+1 = 5$$

• (ii) $$n = 4,\; l = 0 \;\;(4s)$$ → $$n+l = 4+0 = 4$$

• (iii) $$n = 3,\; l = 2 \;\;(3d)$$ → $$n+l = 3+2 = 5$$

• (iv) $$n = 3,\; l = 1 \;\;(3p)$$ → $$n+l = 3+1 = 4$$

Arrange by increasing $$n+l$$ first:

• $$n+l = 4$$: 4s (ii) and 3p (iv)
• $$n+l = 5$$: 3d (iii) and 4p (i)

For ties, compare $$n$$:

• Between 3p (iv) and 4s (ii): lower $$n$$ (3) means 3p is lower in energy than 4s.

• Between 3d (iii) and 4p (i): lower $$n$$ (3) means 3d is lower in energy than 4p.

Hence the overall increasing energy order is

$$3p \lt 4s \lt 3d \lt 4p$$

Writing this sequence with the given Roman numerals:

$$(iv) \lt (ii) \lt (iii) \lt (i)$$

Therefore, the correct choice is:
Option B which is: (iv) < (ii) < (iii) < (i)