The spin magnetic moment of cobalt in the compound named Mercuric tetrathiocyanato cobaltate (II) is:

\(\sqrt{15}\)

The correct answer is option 3. \(\sqrt{15}\).

To determine the spin magnetic moment (\( \mu_s \)) of a metal ion, you can use the formula:
\[ \mu_s = \sqrt{n(n + 2)} \]
where \(n\) is the number of unpaired electrons.
For cobalt in the complex \([Co(SCN)_4]^{2-}\) (Mercuric tetrathiocyanato cobaltate (II)), you need to determine the number of unpaired electrons.
The electron configuration of neutral cobalt (\( \text{Co}^{2+} \)) is \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^7\). When cobalt forms a \(2+\) ion, it loses two electrons from the 3d subshell, leaving \(3d^7\).
In the \(3d^7\) configuration, there are three unpaired electrons. Therefore, \(n = 3\).
Now, plug this value into the formula:
\[ \mu_s = \sqrt{3(3 + 2)} \]
\[ \mu_s = \sqrt{15} \]
So, the correct answer is option 3: \(\sqrt{15}\).