# Question add1d

Sep 6, 2017

$9$

#### Explanation:

The number of orbitals present in an energy level denoted by the principal quantum number $n$ is given by

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{\text{no. of orbitals} = {n}^{2}}}}$

$n = 3$

and so

$\text{no. of orbitals} = {3}^{2} = 9$

To verify this, use the fact that the angular momentum quantum number, $l$, which denotes the energy subshell in which an electron is located inside an atom, can take the possible values

$l = \left\{0 , 1 , 2 , \ldots , n - 1\right\}$

In your case, the third energy level has a total of $3$ subshells

$l = \left\{0 , 1 , 2\right\}$

Now, each subshell can hold a specific number of orbitals as given by the possible values of the magnetic quantum number, ${m}_{l}$

${m}_{l} = \left\{- l , - \left(l - 1\right) , \ldots , - 1 , 0 , 1 , \ldots , \left(l - 1\right) , l\right\}$

This means that you have

• $l = 0 \implies {m}_{l} = \left\{0\right\} \to$ the $s$ subshell contains $1$ orbital
• $l = 1 \implies {m}_{l} = \left\{- 1 , 0 , 1\right\} \to$ the $p$ subshell contains $3$ orbitals
• $l = 2 \implies {m}_{l} = \left\{- 2 , - 1 , 0 , 1 , 2\right\} \to$ the $d$ subshell contains $5$ orbitals

You can thus say that the third energy level contains a total of

overbrace("1 orbital")^(color(blue)("in the 3s subshell")) + overbrace("3 orbitals")^(color(blue)("in the 3p subshell")) + overbrace("5 orbitals")^(color(blue)("in the 3d subshell")) = overbrace("9 orbitals")^(color(blue)("on the third energy level"))#