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Magentic mument of an ions of iron is 5.92BM calculate unpaired electrons and show electronic configration

Magentic mument of an ions of iron is 5.92BM calculate unpaired electrons and show electronic configration

Grade:12

2 Answers

Tejas Shripal
12 Points
6 years ago
Magnetic moment = \sqrt{n(n+2)}
5.92 = \sqrt{n(n+2)}
Square on both sides,
35 = n(n+2)
n^{2}+2n -35=0
After solving this you get n=5 , -7.
And since n is an postive integer ,we take 5 into consideration. The number of unpaired electrons is 5.
 
Tejas Shripal
12 Points
6 years ago
And about electronic config, Fe has an electronic config [Ar] 3d6 4s2, Here only 4 electrons are upaired. By the question 5 are unpaired. fe+3 has 5 unpaired e- and its electronic config is 1s2 2s2 2p6 3s2 3p6 3d5.

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