Consider the ground state of Cr (Z = 24). The | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The electronic configuration of $Cr$ $(Z = 24)$ is $1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 3d^5, 4s^1$.For $l = 1$ (which corresponds to $p$-orbitals),the

Vedclass · Accepted Answer

$12$ and $5$

Vedclass · Answer

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

Consider the ground state of $Cr$ $(Z = 24)$. The numbers of electrons with the azimuthal quantum numbers,$l = 1$ and $l = 2$ are,respectively:

Similar Questions

Which statement is not true regarding the $2s$ orbital?

Explain $d$ orbital.

List the quantum numbers ($l$ and $m_{l}$) for an electron in a $3d$ orbital.

Which electronic configuration must represent an atom in an excited state?

Which of the following orbitals has/have zero probability of finding the electron in the $xy$ plane?

Vedclass Test Series

Exam Paper Generator

Online Exam Module