Total number of electrons present in a calcium atom having $(n + ℓ) = 4$ is

9

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6

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8

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10

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Solution

The correct option is C $8$
Answer (3)
Ca( Z = $20$ ) : $1 s^{2}, 2 s^{2}, 2 p^{6}, 3 s^{2}, 3 p^{6}, 4 s^{2},$
For, $3 p$ and $4 s$ orbitals
$(n + ℓ) = 4$
Total number of electrons $= 6 + 2 = 8$ .

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