If the sum of (n + l) is 6,how many electrons | vedclass

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(B) For a given energy level where $(n + l) = 6$,the possible values for $(n, l)$ are:$1$. If $n = 6, l = 0$ ($6s$ orbital)$2$. If $n = 5, l = 1$ ($5p

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$9$

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(B) For a given energy level where $(n + l) = 6$,the possible values for $(n, l)$ are:$1$. If $n = 6, l = 0$ ($6s$ orbital)$2$. If $n = 5, l = 1$ ($5p$ orbital)$3$. If $n = 4, l = 2$ ($4d$ orbital)$4$. If $n = 3, l = 3$ ($3f$ orbital - not possible as $l Thus,the possible orbitals are $6s$,$5p$,and $4d$.Total number of orbitals = $1$ $(6s)$ + $3$ $(5p)$ + $5$ $(4d)$ = $9$ orbitals.Each orbital can hold a maximum of $2$ electrons,but only one electron in each orbital has a clockwise spin (usually represented as $m_s = +1/2$).Therefore,the number of electrons with clockwise spin = $9$.

If the sum of $(n + l)$ is $6$,how many electrons would have clockwise spin associated with this energy level?

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