The electrons identified by quantum numbers $$n$$ and $$l$$: (a) $$n=4, l=1$$ (b) $$n=4, l=0$$ (c) $$n=3, l=2$$ (d) $$n=3, l=1$$ Can be placed in order of increasing energy as:

For one-electron atoms and for multishell electrons in multielectron atoms, the relative energy of an orbital is decided by the $$n+l$$ rule (also called the Aufbau principle).

Rule:
1. Orbitals having the smaller value of $$(n+l)$$ possess lower energy.
2. If two orbitals have the same $$(n+l)$$ value, the orbital with the smaller principal quantum number $$n$$ is lower in energy.

Compute $$(n+l)$$ for each given set of quantum numbers:

(a) $$n=4,\;l=1 \;(\text{4p}) \;\Rightarrow\; n+l = 4+1 = 5$$
(b) $$n=4,\;l=0 \;(\text{4s}) \;\Rightarrow\; n+l = 4+0 = 4$$
(c) $$n=3,\;l=2 \;(\text{3d}) \;\Rightarrow\; n+l = 3+2 = 5$$
(d) $$n=3,\;l=1 \;(\text{3p}) \;\Rightarrow\; n+l = 3+1 = 4$$

Step 1 - Compare $$(n+l)$$ values:
• The smallest value is 4 → orbitals (b) and (d).
• The next value is 5 → orbitals (a) and (c).

Step 2 - Tie-break when $$(n+l)$$ is equal:
For $$(n+l)=4$$: choose the smaller $$n$$.
  • (d) has $$n=3$$, (b) has $$n=4$$, so (d) lies lower than (b).
For $$(n+l)=5$$: again choose the smaller $$n$$.
  • (c) has $$n=3$$, (a) has $$n=4$$, so (c) lies lower than (a).

Overall increasing energy order:
$$\text{(d)} \lt \text{(b)} \lt \text{(c)} \lt \text{(a)}$$

Thus the correct option is:
Option B which is: (d) < (b) < (c) < (a)