Easy
The electronic configuration of the valence shell of $Cu$ is $3d^{10}4s^{1}$ and not $3d^{9}4s^{2}$. How is this configuration explained?
- ADue to the stability of completely filled $d$-orbitals.
- BDue to the stability of half-filled $d$-orbitals.
- CDue to the exchange energy.
- DDue to the lower energy of $3d$ orbitals compared to $4s$.
Explore More
Similar Questions
The electrons identified by quantum numbers $n$ and $l$ :
$A. n=4, l=1$ $B. n=4, l=0$
$C. n=3, l=2$ $D. n=3, l=1$
can be placed in order of increasing energy as :
$A. n=4, l=1$ $B. n=4, l=0$
$C. n=3, l=2$ $D. n=3, l=1$
can be placed in order of increasing energy as :
MediumAIEEE 2012
View Solution"No two electrons in an atom can have the same set of all four quantum numbers." This principle is called:
Which of the following orbitals has the least energy?
Easy
View SolutionThe atomic number of the element (in ground state) having the maximum number of unpaired $3p$ electrons is:
Medium
View SolutionGive the symbol of the element with the lowest atomic number that has three $2p$ electrons.
Easy
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo