Question

Reset ♡ 22 Which of the given species has highest spin only magnetic moment? Only one correct answer А. Ti3+ В. Mn2+ C. Ni2+ D. Sc3+​

Answer 1

Answer: Option B. Mn2+

Explanation:

Mn2+ has the highest spin only magnetic moment because of the maximum number of unpaired electrons in its d-orbital.

The formula for calculating the magnetic moment if given by

μ=  $\sqrt{4s(s+1)}$

​ where s = Spin magnetic moment.

μ=  $\sqrt{n(n+2)}$

​ where n= Number of unpaired electrons.

Both formulas will give the same answer and you can check it my putting s = $\frac{n}{2}$.

Hope this clears your doubt. :)

Answer 2

Among the given species, the species that has the highest spin only magnetic moment is $\[M{n^{2 + }}\]$. Thus, the correct option is B.

Explanation:

• Spin only magnetic moment can be given by the formula$\[{\mu _s} = \sqrt {n(n + 1)} \]$, where, 'n' is the number of unpaired electrons.

• The electronic configuration of titanium is

         $\[T{i^{3 + }}:[Ar]\,4{s^0}\,3{d^1}\]$   $n=1$

        Therefore, its spin only magnetic moment will be

        $\[{\mu _s} = \sqrt {1(1 + 1)} \]$

        $\[{\mu _s} = \sqrt 3 = 1.732\]$

• The electronic configuration of manganese is

        $\[M{n^{2 + }}:[Ar]\,4{s^0}\,3{d^5}\]$       $n=5$

        Therefore, its spin only magnetic moment will be

         $\[{\mu _s} = \sqrt {5(5 + 1)} \]$

         $\[{\mu _s} = \sqrt {30} = 5.47\]$

• The electronic configuration of nickel is

        $\[N{i^{2 + }}:[Ar]\,4{s^0}\,3{d^8}\]$       $n=2$

        Therefore, its spin only magnetic moment will be

         $\[{\mu _s} = \sqrt {2(2 + 1)} \]$

         $\[{\mu _s} = \sqrt 6 = 2.44\]$

• The electronic configuration of scandium is

        $\[S{c^{3 + }}:[Ar]\,4{s^0}\,3{d^0}\]$        $n=0$

        It has no unpaired electrons.

        Therefore, its spin-only magnetic moment will be zero.

• Therefore, The highest spin only magnetic moment is of $\[M{n^{2 + }}\]$.