Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

Magnetic moment of $Fe^{a+}(z=26)$ is $\sqrt{24}$BM.Hence number of unpaired electrons and value of a respectively are :


Can you answer this question?

1 Answer

0 votes
Magnetic moment $\sqrt{n(n+2)}$BM
$\Rightarrow n(n+2)=24$
$\Rightarrow n=4$
Thus, $Fe^{2+}$ has four unpaired electrons,
i.e., It is $Fe^{2+}$ or $[Ar]3d^6$
Hence (a) is the correct answer.
answered Jan 10, 2014 by sreemathi.v
edited Mar 19, 2014 by mosymeow_1

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App