The number of electrons with (n+ell) values e | vedclass

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

Question

(C) The atomic number $Z = 24$ corresponds to Chromium $(Cr)$.The electronic configuration of $Cr$ is $1s^2 2s^2 2p^6 3s^2 3p^6 3d^5 4s^1$.For $(n+\el

Vedclass · Accepted Answer

$8, 7, 5$

Vedclass · Answer

(C) The atomic number $Z = 24$ corresponds to Chromium $(Cr)$.The electronic configuration of $Cr$ is $1s^2 2s^2 2p^6 3s^2 3p^6 3d^5 4s^1$.For $(n+\ell) = 3$: The orbitals are $2p$ $(n=2, \ell=1)$ and $3s$ $(n=3, \ell=0)$. Total electrons $= 6 + 2 = 8$.For $(n+\ell) = 4$: The orbitals are $3p$ $(n=3, \ell=1)$ and $4s$ $(n=4, \ell=0)$. Total electrons $= 6 + 1 = 7$.For $(n+\ell) = 5$: The orbital is $3d$ $(n=3, \ell=2)$. Total electrons $= 5$.

The number of electrons with $(n+\ell)$ values equal to $3, 4$ and $5$ in an element with atomic number $(Z) = 24$ are respectively $(n =$ principal quantum number and $\ell =$ azimuthal quantum number).

Similar Questions

What is the difference between the $2s$ orbital of hydrogen and the $2s$ orbital of other multi-electron atoms?

If the value of $(n+\ell)$ is not greater than $3$,which of the following subshells is not possible $-$

Which of the following orbitals has the least energy?

Give the symbol of the element with the lowest atomic number that has three $2p$ electrons.

Which of the following represents the correct order of energy for the orbitals?

Vedclass Test Series

Exam Paper Generator

Online Exam Module