Arrange Atomic Orbitals in Order of Increasing Energy

To determine the order of increasing energy of atomic orbitals in a multi-electron atom, we use the Bohr-Bury Rule (also known as the $(n + l)$ rule):

1. An orbital with a lower value of $(n + l)$ has lower energy.
2. If two orbitals have the same value of $(n + l)$, the orbital with the lower principal quantum number $n$ has lower energy.

Let us calculate the $(n + l)$ value for each given orbital:

• A. $n = 3, l = 2, m = +1 \implies 3d\text{ orbital}$ $n + l = 3 + 2 = 5$

• B. $n = 4, l = 0, m = 0 \implies 4s\text{ orbital}$ $n + l = 4 + 0 = 4$

• C. $n = 6, l = 1, m = 0 \implies 6p\text{ orbital}$ $n + l = 6 + 1 = 7$

• D. $n = 5, l = 1, m = +1 \implies 5p\text{ orbital}$ $n + l = 5 + 1 = 6$

• E. $n = 2, l = 1, m = +1 \implies 2p\text{ orbital}$ $n + l = 2 + 1 = 3$

Arranging these $(n + l)$ values in ascending order gives: $3 (\text{E}) < 4 (\text{B}) < 5 (\text{A}) < 6 (\text{D}) < 7 (\text{C})$

Therefore, the correct order of increasing energy of the given atomic orbitals is: $\text{E} < \text{B} < \text{A} < \text{D} < \text{C}$

This matches Option D.