Question

A particular atom has the 4th shell as its valence shell. If the difference between the number of electrons between K and N shell and L and M shell is zero, find the electronic configuration of its stable ion.

• A. 2, 8, 8, 2
• B. 2, 8, 18, 8
• C. 2, 8, 8
• D. 2, 8

Answer 1

The correct option is C 2, 8, 8

Number of electrons in shell is given by $2n^2$

where $n$= energy level

Maximum number of electrons in $K$ shell

$⟶2n^2=2\times (1{)}^2=2$ $[n=1]$

Maximum number of electrons in $L$ shell

$⟶2n^2=2\times (2{)}^2=2\times 4=8$ $[n=2]$

Maximum number of electrons in $M$ shell

$⟶2n^2=2\times (3{)}^2=2\times 9=18$ $[n=3]$

Maximum number of electrons in $N$ shell

$⟶2n^2=2\times (4{)}^2=2\times 16=32$ $[n=4]$

Difference between $K$ and $N$ shell is zero. $K$ can accommodate maximum $2$ electron. Therefore, for difference to be zero $N$ should also have $2$ electrons.

$\therefore K=2$ electrons

$N=2$ electrons

Difference between $L$ and $M$ is also zero. Maximum electron in $L=8$. So, for difference to be zero $M$ should also have $8$ electron in it.

$\therefore L=8e^{-}$

$M=8e^{-}$

Electrons per shell in

$K$ $L$ $M$ $N$ [$N$ is the $4^{th}$ valence shell]

$2$ $8$ $8$ $2$

Stable ion is formed by loss of $2$ electrons from $N$ shell.

$\therefore$ Electronic configuration will be $2,8,8$ .