Number of electrons present in 4f orbital of Ho$$^{3+}$$ ion is _________ (Given Atomic No. of Ho = 67)

We have Atomic Number $$Z = 67$$ for holmium (Ho). For a neutral atom, the number of electrons equals the atomic number, so a Ho atom contains $$67$$ electrons.

First, we recall the order in which subshells are filled, which is dictated by the $$(n+\ell)$$ rule. The sequence immediately after the xenon core $$[Xe]$$ (which accounts for $$54$$ electrons) is: $$6s \rightarrow 4f \rightarrow 5d$$.

Subtracting the core electrons from the total gives the electrons still to be placed: $$67 - 54 = 13$$ electrons.

By the filling order:

• The $$6s$$ subshell can accommodate $$2$$ electrons, so we put $$2$$ there, leaving $$13 - 2 = 11$$ electrons.
• These remaining $$11$$ electrons now enter the $$4f$$ subshell (which can hold up to $$14$$), giving $$4f^{11}$$.

So the ground-state electron configuration of neutral holmium is

$$[Xe]\;4f^{11}\,6s^{2}.$$

Now we consider the triply-charged ion $$\text{Ho}^{3+}$$. When positive ions are formed, electrons are removed first from the subshell having the highest principal quantum number $$n$$. Here, $$n=6$$ for the $$6s$$ electrons and $$n=4$$ for the $$4f$$ electrons, so:

• Remove both $$6s$$ electrons: $$6s^{2} \longrightarrow 6s^{0}$$ (this uses up two charges).
• One more electron must be removed; the next highest-energy electrons come from the $$4f$$ subshell. Thus $$4f^{11} \longrightarrow 4f^{10}$$.

Therefore, the electron configuration of $$\text{Ho}^{3+}$$ is

$$[Xe]\;4f^{10}.$$

The number of electrons present in the $$4f$$ orbital of $$\text{Ho}^{3+}$$ is simply the exponent on $$4f$$, namely $$10$$.

So, the answer is $$10$$.