Question

A neutral atom of an element has 2K, 8L and 5M electrons. What is the total number of electrons having $n+l=3$ energy level?

• A. 6
• B. 8
• C. 10
• D. 4

Answer 1

Answer: (B)

8

The number of electrons in the shells is given as 2 in K shell, 8 in L shell and 5 in the M shell.

Thus, the atom has in total of 2+8+5 = 15 electrons.

Therefore, the atom is phosphorus.

The electronic configuration will be as follows :

$1s^{2}2s^{2}2p^{6}3s^{2}3p^3$

$n$ is a principal quantum number.

$n$ is the number of principal shell.

$l$ is an azimuthal quantum number.

$l=$ $0$ for s orbital.

$l=$ $1$ for p orbital.

$l=$ $2$ for d orbital.

$l=$ $3$ for f orbital.

$n+l=3$ is possible in 2 cases.

1) If $n=2$ and $l=1$

It is a $2p$ subshell.

Phosphorus has 6 electrons in $2p$ subshell.

2) If $n=3$ and $l=0$

It is a $3s$ subshell.

Phosphorus has 2 electrons in $3s$ subshell.

So, total number of electrons having $n+l=3$ are $6+2=8$

Hence, correct answer is option $\left(B\right)$.