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

When the azimuthal quantum number has a value of $l = 0$,the shape of the orbital is

The order of increasing energies of the orbitals follows:

For a $d$ electron,the orbital angular momentum is

Which of the following represents an excited state of an atom?

Which of the following electronic configurations has the maximum energy?

Vedclass Test Series

Exam Paper Generator

Online Exam Module