In an element with atomic number (Z) = 25,the | vedclass

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(D) The electronic configuration of an element with atomic number $(Z) = 25$ is $1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 4s^2, 3d^5$.For $(n+l) = 3$:- $2p$ orbi

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$16$

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(D) The electronic configuration of an element with atomic number $(Z) = 25$ is $1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 4s^2, 3d^5$.For $(n+l) = 3$:- $2p$ orbital $(n=2, l=1)$: $(n+l) = 2+1 = 3$. Number of electrons = $6$.- $3s$ orbital $(n=3, l=0)$: $(n+l) = 3+0 = 3$. Number of electrons = $2$.So,$x = 6 + 2 = 8$.For $(n+l) = 4$:- $3p$ orbital $(n=3, l=1)$: $(n+l) = 3+1 = 4$. Number of electrons = $6$.- $4s$ orbital $(n=4, l=0)$: $(n+l) = 4+0 = 4$. Number of electrons = $2$.So,$y = 6 + 2 = 8$.The value of $(x+y) = 8 + 8 = 16$.

In an element with atomic number $(Z) = 25$,the number of electrons with $(n+l)$ value equal to $3$ and $4$ are $x$ and $y$ respectively. The value of $(x+y)$ is

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