The value of magnetic quantum number of the outermost electron of Zn$$^+$$ ion is _________. (Integer answer)

We begin by recalling the atomic number of zinc. Elemental zinc has $$Z = 30$$, therefore a neutral zinc atom possesses $$30$$ electrons.

The ground-state electronic configuration of a neutral zinc atom, written in the usual energy order $$1s \rightarrow 2s \rightarrow 2p \rightarrow 3s \rightarrow 3p \rightarrow 4s \rightarrow 3d$$, is

$$\text{Zn} : [\text{Ar}]\,3d^{10}\,4s^{2}$$

where $$[\text{Ar}]$$ represents the filled inner core $$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}$$.

Now we form the ion $$\text{Zn}^{+}$$. Making a cation means removing electrons, and in transition-metal atoms the electrons are always lost first from the subshell with the highest principal quantum number $$n$$. Here the $$4s$$ electrons (with $$n = 4$$) are removed before any $$3d$$ electrons (with $$n = 3$$).

Consequently, removing one electron from the neutral zinc atom takes away one of the two $$4s$$ electrons:

$$\text{Zn}^{+} : [\text{Ar}]\,3d^{10}\,4s^{1}$$

The single remaining outermost electron therefore resides in the $$4s$$ orbital.

For an $$s$$ subshell we know the azimuthal (orbital) quantum number is

$$l = 0$$

The magnetic quantum number $$m_l$$ can take all integer values from $$-l$$ to $$+l$$, inclusive. Stating the rule explicitly:

$$ m_l = -l,\, -(l-1),\,\dots,\,0,\,\dots,\,(l-1),\, +l $$

When $$l = 0$$ we have only one possible value, because the range collapses to a single integer:

$$m_l = 0$$

This sole value characterises every $$s$$ orbital, including the $$4s$$ orbital that contains the outermost electron of $$\text{Zn}^{+}$$.

Hence, the correct answer is Option 0.