Question

How many total orbitals in shell $n=4$? What is the relationship between the total number of shell and the quantum number n for that shell?

Answer 1

$n^2$ orbitals in each energy level, and $n$ subshells in each energy level.

I assume you kind of recognize quantum numbers...

1. $n$ is the principal quantum number, the energy level. $n=1,2,\mathrm{3...}$

2. $l$ is the angular momentum quantum number, corresponding to the shape of the orbitals of that kind. $l=0,1,2,3,..,n-1$. That is, $l_{max}=n-1$.

3. $m_l$ is the magnetic quantum number, corresponding to each orbital of that shape. $m_l=(-l,-l+1,...,0,..,l-1,l+1)$. That is, $|m_l|\le l$.

4. $m_s$ is the spin quantum number for electrons. $m_s=\pm \frac{1}{2}$

For $n=4$, the maximum $l$ is therefore $4-1=3$. Of course, there is more than one value of $l$ for one value of $n$.

That means:

$n=4$

$l=0$

$m_1=\left(0\right)$

$l=1$

$m_l=(-1,0,+1)$

$l=2$

$m_l=(-2,-1,0,+1,+2)$\

$l=3=l_{max}$

$m_l=(-3,-2,-1,0,+1,+2,+3)$

and each $m_l$ value corresponds to one orbital. We have $4$ subshells in this case; $s,p,d,f\leftrightarrow 0,1,2,3$ for the value of $l$.

We have an odd number of orbitals per subshell $(2l+1)$, and so:

$2\left(0\right)+1+2\left(1\right)+1+2\left(2\right)+1+2\left(3\right)+1$

$=1+3+5+7$

$=16$ orbitals in the $n=4$ energy level.

If you repeat the process for $n=3$, you would find $l_{max}=2$ and there are $9$ orbitals in $n=3$.

$n=3$

$l=0$

$m_i=\left(0\right)$

$l=1$

$m_l=(-1,0,+1)$

$l=2=l_{max}$

$m_l=(-2,-1,0,+1,+2)$

and each $m_l$ value corresponds to one orbital. We have $3$ subshells in this case; $s,p,d\leftrightarrow 0,1,2$ for the value of $l$.

If you repeat the process for $n=2$, you would find $l_{max}=1$ and there are $4$ orbitals in $n=2$.

$n=2$

$l=0$

$m_l=\left(0\right)$

$l=1=l_{max}$

$m_l=(-1,0,+1)$

and each $m_l$ value corresponds to one orbital. We have $2$ subshells in this case; $s,p\leftrightarrow 0,1$ for the value of $l$.

If you repeat the process for $n=1$, you would find $l_{max}=0$ and there is $1$ orbital in $n=1$.

$n=1$

$l=0=l_{max}$

$m_l=\left(0\right)$

nd each $m_l$ value corresponds to one orbital. We have $1$ subshell in this case; $s\leftrightarrow 0$ for the value of $l$.

Thus, we have $n^2$ orbitals in one energy level, and $n$ subshells in one energy level.