Match List I with List II

Choose the correct answer from the options given below:

A-III, B-I, C-IV, D-II

The correct answer is option 3. A-III, B-I, C-IV, D-II.

To match the transition metal ions in List I with their corresponding values of magnetic moment (in Bohr magneton, BM) in List II let us first find out the values of each ion:

We know,

Magnetic moment, \(\mu = \sqrt{n(n + 2)}BM\), where n = number of unpaired electrons

A. \(Fe^{2+}\) (Iron(II) ion)

Iron(II) ion (\(Fe^{2+}\)) has a \(3d^6\) configuration  with four unpaired electrons \((n)\) in its d-orbitals. So, the magnetic moment will be:

\(\mu = \sqrt{4(4 + 2)}BM\)

or, \(\mu = \sqrt{24}BM\)

or, \(\mu = 4.90 BM\)

B. \(Ni^{2+}\) (Nickel(II) ion)

Nickel(II) ion (\(Ni^{2+}\)) has a \(3d^8\) configuration  with two unpaired electrons \((n)\) in its d-orbitals. So, the magnetic moment will be:

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

or, \(\mu = \sqrt{8}BM\)

or, \(\mu = 2.84 BM\)

C. \(Mn^{2+}\) (Manganese(II) ion)

Manganese(II) ion (\(Mn^{2+}\)) has a \(3d^5\) configuration  with five unpaired electrons \((n)\) in its d-orbitals. So, the magnetic moment will be:

\(\mu = \sqrt{5(5 + 2)}BM\)

or, \(\mu = \sqrt{35}BM\)

or, \(\mu = 5.92 BM\)

D. \(Zn^{2+}\) (Zinc(II) ion)

Zinc(II) ion (\(Zn^{2+}\)) has a \(3d^{10}\) configuration  with zero unpaired electrons \((n)\) in its d-orbitals. So, the magnetic moment will be:

\(\mu = 0 BM\)

So, the correct match is option 3. A-III, B-I, C-IV, D-II