What is the number of unpaired electrons pres | vedclass

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

Question

(D) The atomic number of $Sc$ is $21$. The electronic configuration of $Sc$ is $[Ar] \ 3d^1 \ 4s^2$. Thus,$Sc^{3+}$ is $[Ar]$,which has $0$ unpaired e

Vedclass · Accepted Answer

$0, 0$

Vedclass · Answer

(D) The atomic number of $Sc$ is $21$. The electronic configuration of $Sc$ is $[Ar] \ 3d^1 \ 4s^2$. Thus,$Sc^{3+}$ is $[Ar]$,which has $0$ unpaired electrons.The atomic number of $Zn$ is $30$. The electronic configuration of $Zn$ is $[Ar] \ 3d^{10} \ 4s^2$. Thus,$Zn^{2+}$ is $[Ar] \ 3d^{10}$,which has $0$ unpaired electrons.Therefore,the number of unpaired electrons in $Sc^{3+}$ and $Zn^{2+}$ are $0$ and $0$ respectively.

What is the number of unpaired electrons present in $Sc^{3+}$ and $Zn^{2+}$ ions respectively?

Similar Questions

Elements up to atomic number $103$ have been synthesized and studied. If a newly discovered element is found to have an atomic number $106,$ its electronic configuration will be

Among the following metals,the most dense is

Which of the following ions exhibits a $3d^0 4s^0$ electronic configuration?

The increasing order of magnetism of $(I)$ $MnSO_4 \cdot 4H_2O$,$(II)$ $FeSO_4 \cdot 7H_2O$,$(III)$ $NiSO_4 \cdot 6H_2O$,and $(IV)$ $CuSO_4 \cdot 5H_2O$ is:

Which of the following electronic configurations represents a transition element?

Vedclass Test Series

Exam Paper Generator

Online Exam Module