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):

1. A

0, 1

2. B

1,2

3. C

4,6

4. D

2,3

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Solution:

Out of 6 electrons in 2p and 3p must have one electron with m = + 1 and s = -$\frac{1}{2}$ but in 3d-subshell an orbital having m = +1 may have spin quantum no.
Therefore, minimum an maximum possible values are 2 and 3 respectively.

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