Five valence electrons of P^{15} are labelled | vedclass

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(B) The electronic configuration of $P$ $(Z=15)$ is $1s^2 2s^2 2p^6 3s^2 3p^3$.The five valence electrons are in $3s^2$ and $3p^3$.For $3s$ orbital: $

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$AB$

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(B) The electronic configuration of $P$ $(Z=15)$ is $1s^2 2s^2 2p^6 3s^2 3p^3$.The five valence electrons are in $3s^2$ and $3p^3$.For $3s$ orbital: $n=3, l=0, m=0$. Both electrons $A$ and $B$ have the same $(n, l, m)$ values.For $3p$ orbitals: $n=3, l=1$. The three electrons $X, Y, Z$ are in different $m$ values ($m=-1, 0, +1$ respectively).Since $B$ has spin $+\frac{1}{2}$,$A$ must have spin $-\frac{1}{2}$.Since $Z$ has spin $+\frac{1}{2}$,and by Hund's rule,$X, Y, Z$ all have spin $+\frac{1}{2}$.Electrons with the same $(n, l, m)$ are those in the same orbital.$A$ and $B$ are in the same $3s$ orbital,so they share $(n, l, m)$.$X, Y, Z$ are in different $3p$ orbitals,so they do not share $(n, l, m)$ with each other.Therefore,the group of electrons with the same $(n, l, m)$ is $AB$.

Five valence electrons of $P^{15}$ are labelled as $\boxed{AB}_{3s} \,\boxed{X|Y|Z}_{3p}$. If the spin quantum numbers of $B$ and $Z$ are $+\frac{1}{2}$,the group of electrons with three of the quantum numbers $(n, l, m)$ the same are:

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