Question

Which of the following sets of 3 quantum numbers is/are possible?

`a)` `2,1,1`
`b)` `3,2,-2`
`c)` `2,0,0`
`d)` `2,0,-1`
`e)` `1,1,-1`

Answer 1

This is a situation where you'll just have to remember the rules... The following rules are relevant:

• The principal quantum number `n` is always one more than the maximum angular momentum `l`. That is, `l_(max) = n-1`.
• The set of valid `m_l` must be in the range `{0, pm1, pm2, . . . , pm l}`. Thus, if for example, `l = 2`, `m_l` cannot be `-3` because `|-3| > 2`.

These are all that are necessary to determine the impossible combinations. I will identify the possible quantum number combinations, and I will leave you to determine why the remaining combinations are incorrect.

Possible:

• `a)`: `(n,l,m_l) = (2,1,1)`. This designates one of the `2p` orbitals.
• `b)` `(n,l,m_l) = (3,2,-2)`. This designates one of the `3d` orbitals.
• `c)` `(n,l,m_l) = (2,0,0)`. This designates the one `2s` orbital.

Impossible: the rest, because they violate the above rules. Which one violates which rule? What is a valid correction to the combination?

Here is one example; `d)` has `l = 0`, but `m_l` is not in the set `{0}` since `|-1| > 0`. Therefore, `a)` is impossible. One can fix this by letting `m_l = 0`, since `0` is in the set `{0}`. If we have `(n,l,m_l) = (1,0,0)`, we have a `1s` orbital.