Which electron is associated with the least e | vedclass

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

Question

(B) The energy of an electron in an orbital is determined by the $(n+l)$ rule (also known as the $Aufbau$ principle).$I$. The energy of an orbital inc

Vedclass · Accepted Answer

$n = 4, l = 0, m = 0, s = +1/2$

Vedclass · Answer

(B) The energy of an electron in an orbital is determined by the $(n+l)$ rule (also known as the $Aufbau$ principle).$I$. The energy of an orbital increases as the sum of the principal quantum number $(n)$ and the azimuthal quantum number $(l)$ increases.$II$. If two orbitals have the same $(n+l)$ value,the one with the lower $n$ value has lower energy.$III$. Calculating $(n+l)$ for each option:$(A)$ $n=3, l=2 \implies n+l = 3+2 = 5$$(B)$ $n=4, l=0 \implies n+l = 4+0 = 4$$(C)$ $n=4, l=1 \implies n+l = 4+1 = 5$$(D)$ $n=5, l=0 \implies n+l = 5+0 = 5$Since option $(B)$ has the lowest $(n+l)$ value of $4$,it corresponds to the electron with the least energy.

Which electron is associated with the least energy?

Similar Questions

What is the number of $d$-electrons present in $Fe^{2+}$ ($Fe$ atomic number = $26$)?

Given,
$(A)$ $n=5, m_l=-1$
$(B)$ $n=3, l=2, m_l=-1, m_s=+\frac{1}{2}$
The maximum number of electron$(s)$ in an atom that can have the quantum numbers as given in $(A)$ and $(B)$ respectively are:

The number of nodal planes in a $p_x$ orbital is:

For hydrogen-like atoms,the average distance of an electron from the nucleus for the orbitals belonging to the $4^{th}$ orbit is minimum for:

$(i)$ An atomic orbital has $n=3$. What are the possible values of $l$ and $m_l$?
$(ii)$ List the quantum numbers ($m_l$ and $l$) of electrons for $3d$ orbital.
$(iii)$ Which of the following orbitals are possible? $1p$,$2s$,$2p$,and $3f$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module