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Question: Answered & Verified by Expert
Magnetic moment of $\mathrm{Cr}^{2+}$ is nearest to
Chemistryd and f Block ElementsAIIMSAIIMS 2012
Options:
  • A $\mathrm{Fe}^{2+}$
  • B $\mathrm{Mn}^{2+}$
  • C $\mathrm{Co}^{2+}$
  • D $\mathrm{Ni}^{2+}$
Solution:
1205 Upvotes Verified Answer
The correct answer is: $\mathrm{Fe}^{2+}$
$\mathrm{Cr}^{2+}=3 d^4$, No. of unpaired electrons $(n)=4$
Magnetic moment $=\sqrt{(+2)} \mathrm{BM}$
$=\sqrt{4(4+2)}=\sqrt{24}=4.89 \mathrm{BM}$
$\mathrm{Fe}^{2+}=3 d^6$, No. of unpaired electrons $(n)=4$
Magnetic moment $=\sqrt{4(4+2)} \mathrm{BM}$
$=\sqrt{24}=4.89 \mathrm{BM}$
$\mathrm{Mn}^{2+}=3 d^5$, No. of unpaired electrons $(n)=5$
Magnetic moment $=\sqrt{5(5+2)} \mathrm{BM}$
$=\sqrt{35}=5.91 \mathrm{BM}$
$\mathrm{Co}^{2+}=3 d^7$, No. of unpaired electrons $(n)=3$
Magnetic moment $=\sqrt{3(3+2)} \mathrm{BM}$
$=\sqrt{15}=3.87 \mathrm{BM}$
$\mathrm{Ni}^{2+}=3 d^8$, No. of unpaired electrons $(n)=2$
Magnetic moment $=\sqrt{2(2+2)} \mathrm{BM}$
$=\sqrt{\circledast}=2.82 \mathrm{BM}$

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