The number of 4f electrons in the ground state electronic configuration of Gd$$^{2+}$$ is _________ [Atomic number of Gd = 64]

We have that the atomic number of gadolinium is $$64$$. In the ground state, a neutral gadolinium atom possesses $$64$$ electrons.

To write the electronic configuration, we first recall the sequence in which orbitals are filled (Aufbau principle): $$1s,\, 2s,\, 2p,\, 3s$$, $$3p,\, 4s,\, 3d,\, 4p$$, $$5s,\, 4d,\, 5p,\, 6s$$, $$4f,\, 5d,\, 6p\dots$$

Filling up to $$64$$ electrons according to this order gives the ground-state configuration of neutral Gd as $$[Xe]\,4f^7\,5d^1\,6s^2.$$ Here $$[Xe]$$ represents the complete configuration of xenon (atomic number $$54$$). Thus the distribution beyond xenon is $$4f^7 5d^1 6s^2$$.

Now we need the configuration of the divalent ion $$\text{Gd}^{2+}$$. To obtain a $$+2$$ charge, we remove two electrons from the neutral atom. According to the rule for removal of electrons, electrons are taken first from the subshell having the largest principal quantum number $$n$$, and if two subshells have the same $$n$$, then from the one with the larger azimuthal quantum number $$\ell$$.

Comparing the outer subshells: $$$6s \;(n=6,\ \ell=0), \qquad 5d \;(n=5,\ \ell=2), \qquad 4f \;(n=4,\ \ell=3).$$$ Clearly, $$n=6$$ is the highest, so the two electrons are removed from the $$6s$$ subshell.

Therefore, after ejecting those two $$6s$$ electrons, the configuration for $$\text{Gd}^{2+}$$ is $$[Xe]\,4f^7\,5d^1.$$

In this configuration, the $$4f$$ subshell still contains $$7$$ electrons, while the $$5d$$ subshell contains $$1$$ electron.

Hence, the number of $$4f$$ electrons present in the ground-state electronic configuration of $$\text{Gd}^{2+}$$ is $$7.$$

So, the answer is $$7$$.