The number of s-electrons present in an ion with 55 protons in its unipositive state is

Explanation

The ion contains 55 protons.

An element with atomic number

$$Z=55$$

is cesium, $$Cs$$.

The electronic configuration of neutral cesium is

$$1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^6\ 4s^2\ 3d^{10}\ 4p^6\ 5s^2\ 4d^{10}\ 5p^6\ 6s^1$$

or

$$[Xe],6s^1$$

The ion is in the unipositive state, which means it has a charge of $$+1$$.

To form $$Cs^+$$, the outermost electron is removed from the $$6s$$ orbital.

Therefore, the electronic configuration of $$Cs^+$$ becomes

$$1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^6\ 4s^2\ 3d^{10}\ 4p^6\ 5s^2\ 4d^{10}\ 5p^6$$

Now count the electrons present in all the $$s$$ subshells.

From $$1s$$

$$2$$ electrons

From $$2s$$

$$2$$ electrons

From $$3s$$

$$2$$ electrons

From $$4s$$

$$2$$ electrons

From $$5s$$

$$2$$ electrons

The $$6s$$ orbital is empty in $$Cs^+$$.

Hence, the total number of $$s$$ electrons is

$$2+2+2+2+2=10$$

Therefore, the number of $$s$$ electrons present in the unipositive ion is

$$10$$

Hence, the correct answer is

$$10$$