Why do 2d, 1d, and 3f orbitals not exist?

Sep 12, 2017

Each energy level listed does not contain the given sublevels in the ground state.

Explanation:

In the ground state for each energy level:

In the 2nd energy level, electrons are located only in the s and p sublevels, so there are no d orbitals.

In the 1st energy level, electrons occupy only in the s sublevel, so there is no d sublevel.

In the 3rd energy level, electrons occupy only the s, p, and d sublevels, so there is no f sublevel.

Sep 12, 2017

Because those angular momenta are too high for the given quantum levels.

Recall that the first two quantum numbers are:

• $n = 1 , 2 , 3 , . . .$

• $l = 0 , 1 , 2 , . . . , n - 1$ $\leftrightarrow$ $s , p , d , f , g , h , i , k , . . .$

where $n$ is the principal quantum number indicating the energy level, and $l$ is the angular momentum quantum number indicating the shape of the orbital.

Since $l$ can be no greater than $n - 1$ (i.e. ${l}_{\max} = n - 1$), it follows that the maximum $l$ for each energy level is:

$n = 1 \implies {l}_{\max} = 0 \implies s$

$n = 2 \implies {l}_{\max} = 1 \implies p$

$n = 3 \implies {l}_{\max} = 2 \implies d$

$n = 4 \implies {l}_{\max} = 3 \implies f$

$\vdots \text{ "" "" "" "" "" "" } \vdots$

As a result, the highest angular momentum orbitals we have are $1 s$, $2 p$, $3 d$, $4 f$, etc. We cannot have $1 p$, $2 d$, $3 f$, $4 g$, etc.

In other words, we have only:

$1 s$

$2 s , 2 p$

$3 s , 3 p , 3 d$

$4 s , 4 p , 4 d , 4 f$

$\vdots \text{ "" "" "" } \ddots$