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# The number of electrons with the azimuthal quantum number $l=1$ 1nd $l=2$ in ground state of $_{24}Cr$ is respectively :

$\begin{array}{1 1}(a)\;16\;and\;5\\(b)\;12\;and\;5\\(c)\;16\;and\;4\\(d)\;12\;and\;4\end{array}$

$1s^2,2s^22p^6\Rightarrow l=1$
$3s^23p^63d^5,4s^1\Rightarrow l=1,l=2$
Hence (b) is the correct answer.