Cr^{2+} has a magnetic moment same as that of | vedclass

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

Question

(A) The magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$, where $n$ is the number of unpaired electrons.For $Cr^{2+}

Vedclass · Accepted Answer

$Fe^{2+}$

Vedclass · Answer

(A) The magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$, where $n$ is the number of unpaired electrons.For $Cr^{2+}$ $(Z=24)$: Electronic configuration is $[Ar] 3d^4$. Number of unpaired electrons $(n)$ = $4$.For $Fe^{2+}$ $(Z=26)$: Electronic configuration is $[Ar] 3d^6$. Number of unpaired electrons $(n)$ = $4$.For $Cu^{2+}$ $(Z=29)$: Electronic configuration is $[Ar] 3d^9$. Number of unpaired electrons $(n)$ = $1$.For $Co^{2+}$ $(Z=27)$: Electronic configuration is $[Ar] 3d^7$. Number of unpaired electrons $(n)$ = $3$.For $Cu^{+}$ $(Z=29)$: Electronic configuration is $[Ar] 3d^{10}$. Number of unpaired electrons $(n)$ = $0$.Since $Cr^{2+}$ and $Fe^{2+}$ both have $4$ unpaired electrons, they have the same magnetic moment.

$Cr^{2+}$ has a magnetic moment same as that of

Similar Questions

The set having ions which are coloured and paramagnetic both is :

Which of the following is a diamagnetic transition metal ion?

If a titanium compound exhibits a magnetic moment of $1.73 \, B.M.$,what is the oxidation state of $Ti$ in the compound?

Write the ground state electronic configurations of $Sc, Ti, V, Cr, Mn, Fe, Co, Ni, Cu$.

In which of the following transition metal ions is $d-d$ transition possible?

Vedclass Test Series

Exam Paper Generator

Online Exam Module