Question

which of the following is an acceptable set of quantum numbers to describe the 2p orbital for the hydrogen atom? a. n = 1, l = 1, m_l = 0 b. n = 2, l = 2, m_l = +1 c. n = 2, l = 0, m_l = 0 d. n = 2, l = 1, m_l = -1 e. n = 1, l = 2, m_l = -1

Explanation:

Step1: Recall quantum - number rules

The principal quantum number $n$ can take values $n = 1,2,3,\cdots$. The angular - momentum quantum number $l$ can take values $l=0,1,\cdots,n - 1$. The magnetic quantum number $m_l$ can take values $m_l=-l,-l + 1,\cdots,0,\cdots,l-1,l$. For a 2p orbital, $n = 2$ and $l=1$.

Step2: Check option A

For option A, $n = 1$, but for $n = 1$, $l$ can only be 0, not 1. So option A is incorrect.

Step3: Check option B

For option B, $n = 2$, but for $n = 2$, $l$ can be 0 or 1, not 2. So option B is incorrect.

Step4: Check option C

For option C, $n = 2$, $l = 0$ represents a 2s orbital, not a 2p orbital. Since for 2p, $l = 1$. So option C is incorrect.

Step5: Check option D

For option D, $n = 2$, $l = 1$ (which is correct for a 2p orbital), and $m_l=-1$ is a valid value for $l = 1$ ($m_l$ for $l = 1$ can be - 1,0,1). So option D is correct.

Step6: Check option E

For option E, $n = 1$, and $l$ cannot be 2 when $n = 1$ since $l$ ranges from 0 to $n - 1$. So option E is incorrect.

Answer:

D. $n = 2$, $l = 1$, $m_l=-1$