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The number of electrons with magnetic quantum number,$m_l=0$ in the elements with atomic numbers $Z=24$ and $Z=29$ are respectively

The atomic number of an element is $35$. What is the total number of electrons present in all the $p-$ orbitals of the ground state atom of that element?

In the ground state of atomic $Fe \,(Z = 26)$,the spin-only magnetic moment is ...... $\times 10^{-1} \,BM$. (Round off to the nearest integer).
$[$Given $: \sqrt{3} = 1.73, \sqrt{2} = 1.41]$

The correct set of four quantum numbers for the valence electron of rubidium $(Z = 37)$ is:

The orbital angular momentum of an electron in $2s$ orbitals is

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