# Question 3ccc1

Jul 9, 2017

Because there are $16$ orbitals on the fourth energy level.

#### Explanation:

For starters, you should keep in mind that the maximum number of electrons that can be added on the fourth energy level is equal to $32$, not the maximum number of electrons that an atom can hold up and including the fourth energy level.

Now, the number of subshells present on the fourth energy level, which is described by $n = 4$, is given by the number of values the angular momentum quantum number can take.

You know that

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

In your case, the angular momentum quantum number can take four possible values

$l = \left\{0 , 1 , 2 , 3\right\}$

Now, the number of orbitals present in each subshell is given by the magnetic quantum number, ${m}_{l}$, which depends on the value of the angular momentum quantum number.

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

This means that you have

• $l = 0 \implies {m}_{l} = \left\{0\right\}$
• $l = 1 \implies {m}_{l} = \left\{- 1 , 0 , 1\right\}$
• $l = 2 \implies {m}_{l} = \left\{- 2 , - 1 , 0 , 1 , 2\right\}$
• $l = 3 \implies {m}_{l} = \left\{- 3 , - 2 , - 1 , 0 , 1 , 2 , 3\right\}$

The orbitals look like this--the single orbital present in the $s$ subshell is at the top and the $7$ orbitals present in the $f$ subshell are at the bottom.

If you add up the number of orbitals present in each subshell, you will end up with

overbrace("1 orbital")^(color(blue)("in the s subshell")) + overbrace("3 orbitals")^(color(blue)("in the p subshell")) + overbrace("5 orbitals")^(color(blue)("in the d subshell")) + overbrace("7 orbitals")^(color(blue)("in the f subshell")) = "16 orbitals"

Now, you should that each orbital can hold a maximum of $2$ electrons of opposite spins, i.e. one having spin-up and the other having spin-down $\to$ think Pauli's Exclusion Principle here.

This means that the maximum number of electrons that the fourth energy level can hold is equal to

16 color(red)(cancel(color(black)("orbitals"))) * "2 e"^(-)/(1color(red)(cancel(color(black)("orbital")))) = "32 e"^(-)#

The equation that gives you the maximum number of electrons that can be added on a given energy level $n$ looks like this

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\text{max no. of electrons} = 2 \cdot {n}^{2}}}}$

In your case, $n = 4$ and

${\text{max no. of electrons" = 2 * 4^2 = "32 e}}^{-}$

Notice that ${n}^{2}$ gives the number of orbitals that are present on a given energy level $n$.