Two stable ions of the first transition series have the largest number of unpaired electrons and have the highest magnetic moment are:
Answer & explanation
Correct answer: option 3
The correct answer is option 3. \(Fe^{3+}\) and \(Mn^{2+}\).
The electronic configuration of \(Fe^{3+}\) is \([Ar]3d^5\), which means that it has 5 unpaired electrons. The electronic configuration of \(Mn^{2+}\) is \([Ar]3d^5\), which means that it has 5 unpaired electrons.
The magnetic moment of an ion is directly proportional to the number of unpaired electrons. Therefore, \(Fe^{3+}\) and \(Mn^{2+}\) have the highest magnetic moment among the stable ions of the first transition series. Their magnetic moment will be same as they have same of unpaired electrons, i.e., \(n=5\). SO, the magnetic moment will be
\(\mu = \sqrt{5(5+2)} = \sqrt{35} = 5.92 BM \)
The other answer choices are incorrect. \(Ni^{2+}\) and \(Co^{2+}\) have 2 and 3 unpaired electron, respectively. \(Cr^{2+}\) and \(Fe^{2+}\) have 3 and 4 unpaired electrons, respectively. \(Ti^{4+}\) has no unpaired electrons.