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(C) For a given principal quantum number $n$,the total number of orbitals is given by the formula $n^2$.For $n = 3$,the total number of orbitals is $n

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(C) For a given principal quantum number $n$,the total number of orbitals is given by the formula $n^2$.For $n = 3$,the total number of orbitals is $n^2 = (3)^2 = 9$.Alternatively,for $n = 3$,the possible values of the azimuthal quantum number $l$ are $0, 1, 2$.- For $l = 0$ ($3s$ orbital): $1$ orbital $(m_l = 0)$- For $l = 1$ ($3p$ orbitals): $3$ orbitals $(m_l = -1, 0, +1)$- For $l = 2$ ($3d$ orbitals): $5$ orbitals $(m_l = -2, -1, 0, +1, +2)$Total orbitals = $1 + 3 + 5 = 9$.

What is the total number of orbitals associated with the principal quantum number $n = 3$?

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