In the subshell I =3, (a) what is the greatest (most positive) ${\mathbf{m}}_{\mathbf{1}}$value, (b) how many states are available with the greatest ${\mathbf{m}}_{\mathbf{1}}$ value, and (c) what is the total number of states available in the subshell?

The quantum number of the subshell is I=3.

Using the concept of the possible value of magnetic quantum number for every value of the orbital quantum number. The maximum value of m is + I . Thus, every electron has two spin orientations and thus there are two states.

Now, for every value of I, there are ( 2I +1) magnetic angular quantum number values. Thus, considering the case of two spins for every electron, have 2(2I+1) electron states.

Formula:

The total possible electron states for every value is,

${\mathrm{N}}_{\mathrm{I}}=2\left(2\mathrm{I}+1\right)$ ….. (1)

Using the concept, the greatest ${\mathrm{m}}_1$value for I = 3 is as follow.

$$\begin{gathered} {\mathrm{m}}_1=+\mathrm{I} \\ =3 \end{gathered}$$

Hence, the greatest value is 3.

Using the concept of spins orientation, the states available with the greatest ${\mathrm{m}}_{\mathcal{l}}\left(=3\right)$value is given by:

${\mathrm{m}}_{\mathrm{s}}=\pm \frac{1}{2}$

Hence, there are two states available with that value.

Using the given value I = 3 in equation (1), the total number of states available in the subshell is as given by,

$$\begin{gathered} {\mathrm{N}}_{\mathrm{I}}=2\left(2\times 3+1\right) \\ =14 \end{gathered}$$

Hence, there are 14 electron states in the subshell.