Question

The number of electrons present in the valence shell of an atom with atomic number $38$ is:
A.$2$
B.$10$
C.$1$
D.$8$

Answer 1

Valence shell: The shell which is at the last of the electronic configuration, is known as valence shell and the number of electrons present in the valence shell is known as valence electrons and they decide the valency of an atom.

Complete step by step answer:
First of all let us talk about orbits, orbitals.
Orbits: These are the spaces in the atom where electrons revolve. The orbits are as $1,2,3,...$
Orbitals: The space where electrons are likely to be found. The orbitals are as $s,p,d$ and $f$ orbitals.
For $s -$orbitals maximum number of electrons it can have is two.
For $p -$orbitals maximum number of electrons it can have is six
For $d -$ orbitals maximum number of electrons it can have is ten.
For $f -$ orbitals maximum number of electrons it can have is fourteen.
Now the electronic configuration will be as: The number of electrons is $38$. Two electrons will be in $1s$ so it will become $1{s^2}$. Then the next two electrons will go into $2s$ and now the configuration will be: $1{s^2},2{s^2}$. Now we are left with $34$ electrons. The next six electrons will go into $2p$orbitals. Now the configuration is as: $1{s^2},2{s^2}2{p^6}$. Now the left electrons are $28$. Similarly next eight electrons will go into $3s$ and $3p$ orbitals. Now the configuration is as: $1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}$. Now the total electrons left are $20$. Now the next two electrons will go into $4s$. And configuration will be as: $1{s^2},2{s^2}2{p^6},3{s^2}3{p^6},4{s^2}$. And then ten electrons will go into $3d$ orbitals. So the configuration of the element will become as $1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}3{d^{10}},4{s^2}$. And the left electrons are eight. Now the next six electrons will go into $4p$ and the configuration will become $1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}3{d^{10}},4{s^2}4{p^6}$. Now finally we are left with two electrons and they will go to $5s$. Now the electronic configuration of the element having atomic number $38$ is as: $1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}3{d^{10}},4{s^2}4{p^6},5{s^2}$.
The number of valence electrons in the element is two.

So option A is correct.

Note:
For a given orbital maximum number of electrons it can hold is determined as $2(2l + 1)$ where $l$ is azimuthal quantum number. For $s$ the value of $l$ is zero, for $p$ the value of $l$ is one and so on. So the maximum number of electrons in $s$ orbitals is two.