Beryllium's fourth electron will have the fou | vedclass

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

Question

(C) The atomic number of Beryllium $(Be)$ is $4$. Its electronic configuration is $1s^2 2s^2$. The fourth electron enters the $2s$ orbital. For the $2

Vedclass · Accepted Answer

$n=2, l=0, m=0, s=-1/2$

Vedclass · Answer

(C) The atomic number of Beryllium $(Be)$ is $4$. Its electronic configuration is $1s^2 2s^2$. The fourth electron enters the $2s$ orbital. For the $2s$ orbital,the principal quantum number $n=2$ and the azimuthal quantum number $l=0$. Since $l=0$,the magnetic quantum number $m=0$. According to the Pauli exclusion principle,the second electron in the $2s$ orbital must have an opposite spin to the first,so $s=-1/2$. Thus,the four quantum numbers for the fourth electron are $n=2, l=0, m=0, s=-1/2$.

Beryllium's fourth electron will have the four quantum numbers :

Similar Questions

The orbital angular momentum of an electron in $3s$ orbital is $\frac{xh}{2\pi}$. The value of $x$ is . . . . . . .

The number of unpaired electrons in cobalt atom is (atomic number of $Co = 27$)

What is a nodal surface?

"No two electrons in an atom can have the identical set of four quantum numbers." The statement is known as

For the energy levels in an atom,which one of the following statements is correct?

Vedclass Test Series

Exam Paper Generator

Online Exam Module