Question

The electrons identified by quantum numbers $n$ and $l$ can be placed in the order of increasing energy as:

1. $n=4,l=1$
2. $n=4,l=0$
3. $n=3,l=2$
4. $n=3,l=1$

• A. $3>4<2<1$
• B. $4<2<3<1$
• C. $2<4<1\le 3$
• D. $1<3\le 2<4$

Answer 1

Answer: (B)

$4<2<3<1$

The electrons are filled in the increasing order of $(n+l)$ value.
1.$4+1=5$
2.$4+0=4$
3.$3+2=5$
4.$3+1=4$

If $(n+l)$ value is same for two different $n$ and $l$ value, then electron first fills in the orbital in which the n value is less.

The correct order is $4<2<3<1$.

Hence, option $B$ is correct.