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Given this set of quantum numbers for a multi-electron atom: 2, 0, 0, 1/2 and 2, 0, 0, -1/2. What is the next higher allowed set of n and l quantum numbers for this atom in its ground state?

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a
n = 2, l = 0
b
 n = 2, l = 1
c
n = 3, l = 0
d
 n = 3, l = 1 

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detailed solution

Correct option is B

When given a set of quantum numbers, the 2s orbital of a multi-electron atom is represented as n=2,l=0.

For this atom in the ground state, the next highest permitted set of "n" and "l" quantum numbers is "n=2,l=1." This is equivalent to the 2p orbital.

Recall that the orbital with a greater sum (n+l) value has a higher energy.

(n+l)=(2+0)=2 for the 2s orbital.

Hence, the correct option is B

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