The number of s-electrons present in an ion with 55 protons in it… — Solution | JEE Chemistry

MCQEasyJEE 2023Electronic Configuration

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

Answer

Correct answer:D

Standard Method

Given: The ion has $55$ protons and is in the unipositive state.

Find: The number of $s$ -electrons present in that ion.

The element with $55$ protons is cesium, so $Z = 55$ and the element is Cs.

The electronic configuration of neutral cesium is

\text{Cs} : [\text{Xe}]\,6s^1

For the unipositive ion, it loses one electron:

\text{Cs}^+ : [\text{Xe}]

Now count the electrons present in all occupied $s$ -subshells:

1s^2,\; 2s^2,\; 3s^2,\; 4s^2,\; 5s^2

These contribute

2 \times 5 = 10

Therefore, the number of $s$ -electrons is $10$ . The correct option is D.

Counting the $6s^1$ electron of neutral Cs even after forming $\text{Cs}^+$ is incorrect because the unipositive ion has already lost that electron. First write the ion configuration, then count the occupied $s$ -electrons.
Using the atomic number directly as the number of $s$ -electrons is wrong because $55$ gives the total number of protons, not the number of electrons in $s$ -subshells. Write the electronic configuration and count only $1s, 2s, 3s, 4s,$ and $5s$ .

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