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  • Match the columns: List 1 <td style="text

Match the columns:

List 1 List 2
i) \mathrm{Cr^{3+}} A) 0
ii) \mathrm{Cu^{+}} B) 3.87 BM
iii) \mathrm{Fe^{3+}} C) 4.90 BM
iv) \mathrm{Fe^{2+}} D) 5.92 BM

 

Option: 1

I-A, ii-B, iii-C, iv-D 


Option: 2

I-D, ii-C, iii-B, iv-A


Option: 3

I-B, ii-A, iii-D, iv- C


Option: 4

I-B, ii-A, iii-C, iv-D


Answers (1)

best_answer

The number of unpaired electrons in \mathrm{Cr}^{3+}, \mathrm{Cu}^{+}, \mathrm{Mn}^{2+} \text{and} \; \mathrm{Fe}^{2+} is 3,0,5 and 4. Therefore the corresponding spin only magnetic moment is 3.87 BM, Zero, 5.92 BM and 4.90 BM respectively. Formula for magnetic moment is \mathrm{\sqrt{n(n+2)}} where n is number of unpaired electrons.

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