Question

which of the following sets of quantum numbers describe valid orbitals? check all that apply. n = 1, l = 0, m = 0 n = 2, l = 1, m = 3 n = 2, l = 2, m = 2 n = 3, l = 0, m = 0 n = 5, l = 4, m = -3 n = 4, l = -2, m = 2

Explanation:

Step1: Recall quantum - number rules

The principal quantum number $n$ can take positive integer values ($n = 1,2,3,\cdots$). The angular - momentum quantum number $l$ can take values from $0$ to $n - 1$. The magnetic quantum number $m$ can take values from $-l$ to $+l$.

Step2: Analyze $n = 1,l = 0,m = 0$

$n=1$ is a positive integer. Since $n = 1$, $l$ can only be $0$ (because $l$ ranges from $0$ to $n - 1$), and when $l = 0$, $m$ can only be $0$ (since $m$ ranges from $-l$ to $+l$). This set is valid.

Step3: Analyze $n = 2,l = 1,m = 3$

$n = 2$ is a positive integer. When $n = 2$, $l$ can be $0$ or $1$. Here $l = 1$. But when $l = 1$, $m$ can take values $- 1,0,1$, not $3$. So this set is invalid.

Step4: Analyze $n = 2,l = 2,m = 2$

$n = 2$ is a positive integer. But when $n = 2$, $l$ can only be $0$ or $1$ (because $l$ ranges from $0$ to $n - 1$), not $2$. So this set is invalid.

Step5: Analyze $n = 3,l = 0,m = 0$

$n = 3$ is a positive integer. When $n = 3$, $l$ can be $0,1,2$. Here $l = 0$, and when $l = 0$, $m$ can only be $0$. This set is valid.

Step6: Analyze $n = 5,l = 4,m=-3$

$n = 5$ is a positive integer. When $n = 5$, $l$ can be $0,1,2,3,4$. Here $l = 4$, and when $l = 4$, $m$ can range from $-4$ to $4$. Since $m=-3$ is in this range, this set is valid.

Step7: Analyze $n = 4,l=-2,m = 2$

$n = 4$ is a positive integer. But $l$ cannot be negative. $l$ ranges from $0$ to $n - 1$, so $l$ should be a non - negative integer. This set is invalid.

Answer:

$n = 1,l = 0,m = 0$; $n = 3,l = 0,m = 0$; $n = 5,l = 4,m=-3$