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For principal quantum number n = 4 the total number of orbitals having l = 3

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answer is B.

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Detailed Solution

$n = 4 \to 1 s^{2}, 2 s^{2}, 2 p^{6}, 3 s^{2}, 3 p^{6}, 3 d^{10}, 4 s^{2}, 4 p^{6}, 4 d^{10}, 4 f^{14}$
So I = (n-1) = 4-1 = 3 which is f subshell contain 7 orbital.

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