To identify the correct set of quantum numbers for the unpaired electron in a chlorine atom, we first need to understand the electronic configuration of chlorine. The electronic configuration of chlorine (atomic number 17) is $1s^2 2s^2 2p^6 3s^2 3p^5$. The outer shell configuration, where the unpaired electron resides, is $3p^5$. This means the unpaired electron is in the 3p orbital.

The quantum numbers designate the following:

• $n$ is the principal quantum number, indicating the energy level.
• $l$ is the azimuthal (or angular momentum) quantum number, indicating the shape of the orbital.
• $m_l$ (or simply $m$ here) is the magnetic quantum number, indicating the orientation of the orbital.

For the 3p orbital, which is where our unpaired electron is:

• $n = 3$ because the electron is in the third energy level.
• $l = 1$ because for p orbitals, $l = 1$.
• $m$ could be -1, 0, or 1 because for $l = 1$, $m_l$ can take values from $-l$ to $+l$.

Given that chlorine's 3p orbital has 5 electrons, they would fill the three p orbitals ($m_l = -1, 0, 1$) as follows: one electron each in the $m_l = -1$ and $m_l = 1$ orbitals, and the three remaining electrons would begin to pair up. Since each p orbital (with a distinct $m_l$ value) can hold two electrons, by the time we are placing the fifth electron, we have already placed two electrons in the $m_l = -1$ and $0$ orbitals (assuming we fill them first and follow Hund’s rule, which states that electrons fill each orbital singly before any orbital gets a second electron). Thus, the unpaired electron would be the one in the $m_l = 1$ orbital because it is the last to receive an electron, making it unpaired.

Therefore, the correct set of quantum numbers for the unpaired electron in a chlorine atom is:

• $n = 3$
• $l = 1$
• $m = 1$

This matches with Option C: $n = 3$, $l = 1$, $m = 1$.