A) 0
B) 1.73BM
C) 2.83BM
D) 4.9BM
Correct Answer: D
Solution :
The spin only magnetic moment of \[1.27f\] \[\text{1}.0\text{1}\times \text{1}{{0}^{\text{5}}}\text{ N}/{{\text{m}}^{\text{2}}}\] \[\text{9}.\text{13}\times \text{1}{{0}^{\text{4}}}\text{ N}/{{\text{m}}^{\text{2}}}\] \[\text{9}.\text{13}\times \text{1}{{0}^{\text{3}}}\text{N}/{{\text{m}}^{\text{2}}}\] \[\text{18}.\text{26 N}/{{\text{m}}^{\text{2}}}\] (number of unpaired electrons) Magnetic moment \[\text{2}.\text{25}\times \text{1}{{0}^{\text{3}}}\text{min}\] \[\text{3}.\text{97}\times \text{1}{{0}^{\text{3}}}\text{min}\] \[9.13\times {{10}^{3}}N/{{m}^{2}}\]
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