Given below are the quantum numbers for 4 electrons.
A. $$n=3, l=2, m_l=1, m_s=+\frac{1}{2}$$
B. $$n=4, l=1, m_l=0, m_s=+\frac{1}{2}$$
C. $$n=4, l=2, m_l=-2, m_s=-\frac{1}{2}$$
D. $$n=3, l=1, m_l=-1, m_s=+\frac{1}{2}$$
The correct order of increasing energy is

We need to arrange the four electrons in order of increasing energy using the (n+l) rule. When two orbitals have the same (n+l) value, the one with the lower n has lower energy.

Electron A: $$n=3, l=2$$ — this is a 3d electron. We have $$n+l = 3+2 = 5$$.

Electron B: $$n=4, l=1$$ — this is a 4p electron. We have $$n+l = 4+1 = 5$$.

Electron C: $$n=4, l=2$$ — this is a 4d electron. We have $$n+l = 4+2 = 6$$.

Electron D: $$n=3, l=1$$ — this is a 3p electron. We have $$n+l = 3+1 = 4$$.

Now arranging by increasing (n+l): D has $$n+l = 4$$ (lowest), then A and B both have $$n+l = 5$$. Among A (n=3) and B (n=4), A has a lower n, so A has lower energy than B. Finally, C has $$n+l = 6$$ (highest).

The correct order of increasing energy is: $$D < A < B < C$$.

Hence, the correct answer is Option B.