Question

Calculate the minimum and maximum number of electrons which may have magnetic quantum number, $m = +1$ and spin quantum number, $s = -\frac{1}{2}$ in chromium (Cr) :

• A. 0, 1
• B. 1, 2
• C. 2, 3

Answer 1

Answer: (C)

2, 3

Answer 2

Chromium (Cr) has an atomic number of 24, and its electronic configuration is $1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 3d^5$. We need to find the number of electrons with $m=+1$ and $s=-1/2$.

1. p-subshells ($2p^6$ and $3p^6$): For p-subshells, $l=1$, so $m$ can be $-1, 0, +1$. Each orbital can hold 2 electrons with opposite spins. In a filled p-subshell (like $2p^6$ and $3p^6$), the orbital with $m=+1$ contains two electrons: one with $s=+1/2$ and one with $s=-1/2$. Thus, from each of the $2p^6$ and $3p^6$ subshells, there is 1 electron with $m=+1$ and $s=-1/2$. Total from p-subshells = $1 + 1 = 2$ electrons.

2. d-subshell ($3d^5$): For d-subshells, $l=2$, so $m$ can be $-2, -1, 0, +1, +2$. The orbital with $m=+1$ can hold a maximum of 2 electrons.

   • Minimum number of electrons with $m=+1$ and $s=-1/2$: According to Hund's rule, for the $3d^5$ configuration, the 5 electrons will occupy the 5 degenerate orbitals singly with parallel spins. If all spins are $+1/2$, then the electron in the $m=+1$ orbital has $s=+1/2$. In this case, there are 0 electrons with $m=+1$ and $s=-1/2$ from the $3d^5$ subshell.
   • Maximum number of electrons with $m=+1$ and $s=-1/2$: To maximize the count, the $m=+1$ orbital can be doubly occupied. This would mean one electron with $s=+1/2$ and one electron with $s=-1/2$. Thus, there is 1 electron with $m=+1$ and $s=-1/2$ from the $3d^5$ subshell.

Total Minimum: Electrons from p-subshells + Minimum from d-subshell $= 2 + 0 = 2$ electrons.

Total Maximum: Electrons from p-subshells + Maximum from d-subshell $= 2 + 1 = 3$ electrons.

Therefore, the minimum number of electrons is 2, and the maximum number is 3.