For how many electrons in an atom is n = 3, m | vedclass

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(A) For a given principal quantum number $n = 3$,the possible values of the azimuthal quantum number $l$ are $0, 1, 2$.For $l = 0$ ($3s$ orbital),$m_l

Vedclass · Accepted Answer

$4$

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(A) For a given principal quantum number $n = 3$,the possible values of the azimuthal quantum number $l$ are $0, 1, 2$.For $l = 0$ ($3s$ orbital),$m_l = 0$.For $l = 1$ ($3p$ orbital),$m_l = -1, 0, +1$.For $l = 2$ ($3d$ orbital),$m_l = -2, -1, 0, +1, +2$.The condition $m_l = 1$ is satisfied by the $3p$ orbital $(m_l = 1)$ and the $3d$ orbital $(m_l = 1)$.Each orbital can hold a maximum of $2$ electrons with opposite spins.Therefore,the $3p$ orbital with $m_l = 1$ can hold $2$ electrons,and the $3d$ orbital with $m_l = 1$ can hold $2$ electrons.Total number of electrons = $2 + 2 = 4$.

For how many electrons in an atom is $n = 3, m_l = 1$ possible?

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