Question

The total number of electrons present in all the ‘s’ orbitals, all the ‘p’ orbitals, and all the' orbitals of the cesium ion are respectively
A) 11, 26, 10
B) 10, 24, 20
C) 8, 22, 24
D) 12, 20, 22

Answer 1

The atomic number of a neutral atom indicates the number of electrons present in it. In the case of ions, the numbers of electrons present depend on the charge carried by ions. The electronic configuration of an element gives information about the numbers of electrons present in the orbitals.

Complete solution:
Cesium is the s-block element present in the first Group and 6th period in the periodic table. The atomic number of the cesium is 55.
The condensed form of the electronic configuration of the cesium is as follows:
$\left[ {Xe} \right]\,6{s^1}$
The expanded form of the electronic configuration of the cesium is as follows:
$1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^2}3{d^{10}}4{p^6}5{s^2}4{d^{10}}5{p^6}6{s^1}$
Here, 55 electrons are present in the s, p, and d orbitals of the cesium atom.
As we know that cesium is s-block alkali metal its common oxidation state is +1 that is it loses one electron from the outermost 6s orbital and form the cation having a +1 charge and attain the stable electronic configuration of xenon.
Now, the total numbers of electrons in the cesium cation are 54.
The electronic configuration of cesium cation $C{s^ + }$ is as follows:
$1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^2}3{d^{10}}4{p^6}5{s^2}4{d^{10}}5{p^6}$
Here, the total numbers of ‘s’ orbital electrons are 10, the total number of ‘p’ orbital electrons are 24, and the total number of ‘d’ orbital electrons are 20.
In option 1, the total number of electrons present are 55 which is incorrect for cesium cation, therefore option 1)11, 26, 10 is incorrect.
In option 2, the total electrons count is 54, and their distribution in s, p, and d orbitals is also the same as we have calculated above.
In option 3, the total electron count is 54 but electron distribution in orbitals is not correct therefore option 3)8, 22, 24 is incorrect.
Also, In option 4 total electron count is 54 but electron distribution in orbitals is not correct therefore option 4)12, 20, 22is incorrect.

Hence, the option (B) is the correct answer to the question.

Note: Cesium is an s-block element present in the 6th period having atomic number 55. The most common oxidation state of the cesium is +1. The electronic configuration of the cesium is$\left[ {Xe} \right]\,6{s^1}$.