(A) The electronic configuration of $Sr$ $(Z=38)$ is $1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 4s^2, 3d^{10}, 4p^6, 5s^2$. Electrons with $l=0$ (s-orbitals) are:

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(A) The electronic configuration of $Sr$ $(Z=38)$ is $1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 4s^2, 3d^{10}, 4p^6, 5s^2$. Electrons with $l=0$ (s-orbitals) are: $1s^2, 2s^2, 3s^2, 4s^2, 5s^2$. Total electrons $x = 2+2+2+2+2 = 10$. Electrons with $l=2$ (d-orbitals) are: $3d^{10}$. Total electrons $y = 10$. Therefore, $(x-y) = 10 - 10 = 0$.

Class 11 Chemistry Structure of Atom Quantum number, Electronic configuration and Shape of orbitals

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In $Sr$ $(Z=38)$, the number of electrons with $l=0$ is $x$, and the number of electrons with $l=2$ is $y$. The value of $(x-y)$ is equal to:
($l=$ Azimuthal quantum number)

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$0$
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$8$
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$-2$
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$2$

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In $Sr$ $(Z=38)$, the number of electrons with $l=0$ is $x$, and the number of electrons with $l=2$ is $y$. The value of $(x-y)$ is equal to: ($l=$ Azimuthal quantum number)

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In $Sr$ $(Z=38)$, the number of electrons with $l=0$ is $x$, and the number of electrons with $l=2$ is $y$. The value of $(x-y)$ is equal to:
($l=$ Azimuthal quantum number)