Arrange the following in the decreasing order of their magnetic moment values:

A. Sc

B. V

C. Mn

D. Fe

E. Ni

Choose the correct answer from the option given below:

C > D > B > E > A

The correct answer is option 4. C > D > B > E > A.

The magnetic moment of an atom is primarily determined by the presence of unpaired electrons. Unpaired electrons possess intrinsic magnetic moments and contribute to the overall magnetic behavior of the atom. The more unpaired electrons an atom has, the higher its magnetic moment.

Let us break down the electronic configurations of the given elements:

A. Sc (Scandium): [Ar] 3d¹ 4s² - Scandium has one unpaired electron in the 3d orbital.

Thus, magnetic moment,

\(\mu = \sqrt{n(n + 2)}\)

or, \(\mu = \sqrt{1(1 + 2)} = \sqrt{3}\)

B. V (Vanadium): [Ar] 3d³ 4s² - Vanadium has three unpaired electrons in the 3d orbital.

Thus, magnetic moment,

\(\mu = \sqrt{n(n + 2)}\)

or, \(\mu = \sqrt{3(3 + 2)} = \sqrt{15}\)

C. Mn (Manganese): [Ar] 3d⁵ 4s² - Manganese has five unpaired electrons in the 3d orbital.

Thus, magnetic moment,

\(\mu = \sqrt{n(n + 2)}\)

or, \(\mu = \sqrt{5(5 + 2)} = \sqrt{35}\)

D. Fe (Iron): [Ar] 3d⁶ 4s² - Iron has four unpaired electrons in the 3d orbital.

Thus, magnetic moment,

\(\mu = \sqrt{n(n + 2)}\)

or, \(\mu = \sqrt{4(4 + 2)} = \sqrt{24}\)

E. Ni (Nickel): [Ar] 3d⁸ 4s² - Nickel has two unpaired electrons in the 3d orbital.

Thus, magnetic moment,

\(\mu = \sqrt{n(n + 2)}\)

or, \(\mu = \sqrt{2(2 + 2)} = \sqrt{8}\)

So, the decreasing order of magnetic moment values is C > D > B > E > A, as indicated in option 4.