How many sets of four quantum numbers are possible for electrons present in ${H e}^{2 -}$ anion?

2

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4

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5

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7

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Solution

The correct option is B $4$
For a multi-electron system, each electron in the system has a unique set of four quantum numbers. Therefore, the system $H e^{2 -}$ has 4 electrons and has thus four unique sets of four quantum numbers.

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F e

n = 3, l = 2, m = + 2

n = 3, ℓ = 2, m = + 2

(i) n = 4, m_{s} = - \frac{1}{2}

(i i) n = 3, l = 0

(i i i) n = 2, l = 1

(i v) n = 2, l = 0

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