Question

Quantum numbers of the electrons are given
$\left(i\right)n=4,m_s=-\frac{1}{2}$

$\left(ii\right)n=3,l=0$

$\left(iii\right)n=2,l=1$

$\left(iv\right)n=2,l=0$
How many electrons in an atom may have these quantum numbers?

• A. $16,2,6,2$
• B. $32,18,8,8$
  
• C. $16,6,8,6$
• D. $16,2,6,4$

Answer 1

The correct option is A $16,2,6,2$

Option A is correct

Solution:

Total number of electrons in n shell =$2n^2$
total number of electrons in a orbitals = $2$
(a) $n=4,m_s=-1/2$
total number of electrons in n = $4shell=2(4{)}^2=2\times 16=32$
but we know, there are two spin quantum number positive spin and negative spin .
hence, half of electrons are positive spin and half of electrons are negative spin .
so, $16$ electrons have $m_s/=-1/2$
(b) n = 3 , l = 0
it is $3s$ orbital . so, number of electrons in $3s$ orbital = $2$

(C) n=2, l=1

$2p=6$

as $p$ shell have $3$ orbital and $1$ orbital can accommodate $2$ electrons.

Option A is correct answer