Which of the following ions has the maximum magnetic moment?(a) Mn (b) Fe(c) V (d) Cr
Answers
Answer:
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Answer:
As we learnt in
Calculation of spin only magnetic moment -
\mu =\sqrt{n(n+2)}
\mu =magnetic\:moment\:in(BM)
where n = no. of unpaired electron
n=1......5(no.\:of\:unpaired\:e^{-})
- wherein
V^{2+}\rightarrow3d^{3}=\sqrt{3(3+2)}
=\sqrt{15}
=3.87\:\:\:BM
3+.9\rightarrow 3.9(BM)
Mn^{2+}\Rightarrow 3d^{5},\:\:\mu = \sqrt{5\left ( 5+2 \right )}= \sqrt{35}
Fe^{2+}\Rightarrow 3d^{6}, \:\:\mu = \sqrt{4\left ( 4+2 \right )}= \sqrt{24}
Ti^{2+}\Rightarrow 3d^{2},\:\:\mu = \sqrt{2\left ( 2+2 \right )}= \sqrt{8}
Cr^{2+} \Rightarrow 3d^{4}, \:\: \mu = \sqrt{4\left ( 4+2 \right )}= \sqrt{24}
Option 1)
Mn^{2+}\;
This option is correct.
Option 2)
\; Fe^{2+}\;
This option is incorrect.
Option 3)
\; Ti^{2+}\;
This option is incorrect.
Option 4)
\; Cr^{2+}