Chapter 5.2, Problem 11LC

Interpretation Introduction

Interpretation: To determine the energy of sublevel in decreasing order.

Concept Introduction: The placement of electrons inside orbitals around a nuclear nucleus is known as electronic configuration, also known as electronic structure as well as electron configuration.

Answer to Problem 11LC

The descending order of energy will be 3d > 4s >3p >3s>2p.

Explanation of Solution

For the highest value of $\text{n}+\text{l}$ , the energy of the sublevel will be high. If $\left(\text{n+l}\right)$ value will be same, then that orbital will have higher energy which possesses a higher principal quantum number (n) value.

The $\left(\text{n+l}\right)$ value for $\text{2p}$ sublevel: n = 2 and l = 1, by putting these value $\text{(2+1)=3}$ .

The $\left(\text{n+l}\right)$ value for $\text{4s}$ sublevel: n = 4 and l = 0, by putting these value $\text{(4+0)}\text{=}\text{4}$ .

The $\left(\text{n+l}\right)$ value for $\text{3s}$ sublevel: n = 3 and l = 0, by putting these value $\text{(3+0)}\text{=}\text{3}$ .

The $\left(\text{n+l}\right)$ value for $\text{3d}$ sublevel: n = 3 and l = 2, by putting these values $\text{(3+2)=5}$ .

The $\left(\text{n+l}\right)$ value for $\text{3p}$ sublevel: n = 3 and l = 1, by putting these values $\text{(3+1)=4}$ .

So, the descending order of energy will be 3d > 4s >3p >3s>2p.