Question

Given set of quantum numbers for a multi-electron atom is:

$n$	$l$	$m$	$s$
$2$	$0$	$0$	$+1/2$
$2$	$0$	$0$	$-1/2$

What is the next higher allowed set of '$n$' and '$l$' quantum numbers for this atom in the ground state?

• A. $n=2,l=0$
• B. $n=2,l=1$
• C. $n=3,l=0$
• D. $n=3,l=1$

Answer 1

The correct option is B $n=2,l=1$

Given a set of quantum numbers, $n=2,l=0$ for a multi-electron atom refers to 2s orbital.

The next higher allowed set of 'n' and 'l' quantum numbers for this atom in the ground state is $n=2,l=1$. This corresponds to 2p orbital.

Note: The orbital with a higher value of the sum $(n+l)$ has higher energy.

For 2s orbital $(n+l)=(2+0)=2$

For 2p orbital $(n+l)=(2+1)=3$

Hence, the correct option is $\text{B}$