Since there have the right to be <-ℓ, ℓ> orientations and since the orbital type f has ℓ = 3, we should have actually 7 possible orientations through 2 spins, so $7 imes 2 = 14$, therefore I believed the exactly answer to be D (14).

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However, I obtained it wrong and the exactly answer is marked as C (2). Is it an error in the test, or am I absent something?

You"re correct that there space seven possible spatial orientations for an f-type orbital, and also hence seven feasible orbitals in one f-type sub-shell. However, the inquiry specifically asks because that the maximum variety of electrons in one such orbital, and

*any*single atomic orbital, nevertheless of the sub-shell form specified by $l$, deserve to only organize two electrons. This is through virtue that the Pauli exclusion principle. Fourteen would certainly be the maximum variety of electrons across an entire f-type sub-shell, but the question just asks about one orbital.

The question specifically ask that no.of electron an orbit of f subshell can hold.... Together we recognize that f subshell save 7 orbital and each orbital have the right to hold best 2 electons so exactly answer would be 2..we deserve to simply understand this by taking the real life example imagine that there is house named f which consists of 7 rooms so similarly in this case house is a subshell and each room is an orbital... Give thanks to you,

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