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(A) For a given principal quantum number $n$,the possible values of azimuthal quantum number $l$ range from $0$ to $n-1$.For $n=4$,the possible values

Vedclass · Accepted Answer

$4$

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(A) For a given principal quantum number $n$,the possible values of azimuthal quantum number $l$ range from $0$ to $n-1$.For $n=4$,the possible values of $l$ are $0, 1, 2, 3$,which correspond to the $4s, 4p, 4d,$ and $4f$ subshells,respectively.The magnetic quantum number $m_l$ for a given $l$ ranges from $-l$ to $+l$.- For $4s$ $(l=0)$: $m_l = 0$ ($1$ orbital)- For $4p$ $(l=1)$: $m_l = -1, 0, +1$ ($1$ orbital with $m_l=0$)- For $4d$ $(l=2)$: $m_l = -2, -1, 0, +1, +2$ ($1$ orbital with $m_l=0$)- For $4f$ $(l=3)$: $m_l = -3, -2, -1, 0, +1, +2, +3$ ($1$ orbital with $m_l=0$)Thus,there is exactly one orbital with $m_l=0$ in each subshell.Total number of orbitals with $n=4$ and $m_l=0$ is $1+1+1+1 = 4$.

The maximum number of orbitals which can be identified with $n=4$ and $m_l=0$ is $\qquad$

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