Question

type in the maximum number of electrons that can be present in each shell or subshell below.
$n = 5$ shell:
$n = 2$ shell:
$n = 2, l = 0$ subshell:
$n = 2, l = 1$ subshell:
$3d$ subshell:
$2s$ subshell:
$5f$ subshell:
done

Explanation:

Step1: Recall electron - shell formula

The maximum number of electrons in a shell with principal quantum number $n$ is given by $2n^{2}$.

Step2: Calculate for $n = 5$ shell

Substitute $n = 5$ into $2n^{2}$: $2\times5^{2}=2\times25 = 50$.

Step3: Calculate for $n = 2$ shell

Substitute $n = 2$ into $2n^{2}$: $2\times2^{2}=2\times4 = 8$.

Step4: Recall sub - shell electron capacity

For a sub - shell with angular momentum quantum number $l$, the number of orbitals is $2l + 1$, and each orbital can hold 2 electrons. So the number of electrons in a sub - shell is $2(2l + 1)$.

Step5: Calculate for $n = 2$, $l = 0$ sub - shell

Substitute $l = 0$ into $2(2l + 1)$: $2(2\times0 + 1)=2$.

Step6: Calculate for $n = 2$, $l = 1$ sub - shell

Substitute $l = 1$ into $2(2l + 1)$: $2(2\times1+ 1)=2\times3 = 6$.

Step7: Identify $l$ for $d$ sub - shell

For $d$ sub - shell, $l = 2$. Substitute $l = 2$ into $2(2l + 1)$: $2(2\times2 + 1)=2\times5 = 10$.

Step8: Identify $l$ for $s$ sub - shell

For $s$ sub - shell, $l = 0$. Substitute $l = 0$ into $2(2l + 1)$: $2(2\times0 + 1)=2$.

Step9: Identify $l$ for $f$ sub - shell

For $f$ sub - shell, $l = 3$. Substitute $l = 3$ into $2(2l + 1)$: $2(2\times3+ 1)=2\times7 = 14$.

Answer:

$n = 5$ shell: 50
$n = 2$ shell: 8
$n = 2$, $l = 0$ sub - shell: 2
$n = 2$, $l = 1$ sub - shell: 6
$3d$ sub - shell: 10
$2s$ sub - shell: 2
$5f$ sub - shell: 14