Question

what is the $m_s$ value (magnetic spin) for the electronic configuration diagram for aluminum? 1. sulfur, s, 16 2. argon, ar, 18 3. aluminum, al, 13 4. sodium, na, 11

Explanation:

Step1: Determine the electron configuration of aluminum

Aluminum (Al) has an atomic number of 13. Its electron configuration is \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{1}\).

Step2: Analyze the spin of the out - most electron

The out - most electron is in the \(3p\) orbital. According to the rules of electron spin in orbital diagrams, when there is an unpaired electron (as in the \(3p\) orbital of Al), the spin quantum number \(m_s\) can be either \(+\frac{1}{2}\) or \(-\frac{1}{2}\). In the given orbital diagram for aluminum, the unpaired electron is shown with an upward arrow, which corresponds to \(m_s =+\frac{1}{2}\)

Answer:

\(+\frac{1}{2}\)