(B) The electronic configuration of $Cr$ $(Z = 24)$ is $1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 3d^5$.For azimuthal quantum number $l = 1$,the electrons are in

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(B) The electronic configuration of $Cr$ $(Z = 24)$ is $1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 3d^5$.For azimuthal quantum number $l = 1$,the electrons are in $p$-orbitals ($2p^6$ and $3p^6$). Total electrons = $6 + 6 = 12$.For azimuthal quantum number $l = 2$,the electrons are in $d$-orbitals $(3d^5)$. Total electrons = $5$.Therefore,the number of electrons for $l = 1$ and $l = 2$ are $12$ and $5$ respectively.

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

Medium

Considering the ground state of $Cr$ atom $(Z = 24)$,the number of electrons associated with azimuthal quantum numbers $l = 1$ and $l = 2$ are,respectively:

A
$16$ and $5$
B
$12$ and $5$
C
$16$ and $4$
D
$12$ and $4$

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Structure of Atom

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Quantum number, Electronic configuration and Shape of orbitals

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List-$I$ (Quantum number)	List-$II$ (Orbital)
$a. n=2, \ell=1$	$i. 2s$
$b. n=3, \ell=2$	$ii. 3s$
$c. n=3, \ell=0$	$iii. 2p$
$d. n=2, \ell=0$	$iv. 3d$

Considering the ground state of $Cr$ atom $(Z = 24)$,the number of electrons associated with azimuthal quantum numbers $l = 1$ and $l = 2$ are,respectively:

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Match List-$I$ with List-$II$ :List-$I$ (Quantum number)List-$II$ (Orbital)$a. n=2, \ell=1$$i. 2s$$b. n=3, \ell=2$$ii. 3s$$c. n=3, \ell=0$$iii. 2p$$d. n=2, \ell=0$$iv. 3d$Choose the correct answer from the options given below :

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Match List-$I$ with List-$II$ :
List-$I$ (Quantum number) List-$II$ (Orbital)
$a. n=2, \ell=1$ $i. 2s$
$b. n=3, \ell=2$ $ii. 3s$
$c. n=3, \ell=0$ $iii. 2p$
$d. n=2, \ell=0$ $iv. 3d$

Choose the correct answer from the options given below :