Match List I with List II

Choose the correct answers from the options given below:

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

The correct answer is option 2. A-I, III; B-I, III; C-II, IV; D-I, III.

Let us delve into the details of how we match the ions in List I with their corresponding electronic configurations and magnetic properties in List II.

To match the ions with the correct options, we first need to determine the electronic configuration of each ion after losing the appropriate number of electrons. This will also help us determine if the ion is paramagnetic (unpaired electrons) or diamagnetic (all electrons are paired).

A. \(Cr^{3+}\)

Atomic number of Chromium (Cr): 24

Ground state electronic configuration of Cr: \( [Ar] 3d^5 4s^1 \)

\(Cr^{3+}\) ion: The ion is formed by removing three electrons from the atom.

\(Cr^{3+}\) configuration: Remove two electrons from the 4s and one electron from the 3d, leaving \( [Ar] 3d^3 \).

Configuration: \( d^3 \)

Magnetic property: Since the \( d^3 \) configuration has three unpaired electrons, \(Cr^{3+}\) is paramagnetic.

B. \(Mn^{4+}\)

Atomic number of Manganese (Mn): 25

Ground state electronic configuration of Mn: \( [Ar] 3d^5 4s^2 \)

\(Mn^{4+}\) ion: The ion is formed by removing four electrons from the atom.

\(Mn^{4+}\) configuration: Remove two electrons from the 4s and two from the 3d, leaving \( [Ar] 3d^3 \).

Configuration: \( d^3 \)

Magnetic property: The \( d^3 \) configuration means that there are three unpaired electrons, so \(Mn^{4+}\) is paramagnetic.

C. \(Cu^{+}\)

Atomic number of Copper (Cu): 29

Ground state electronic configuration of Cu: \( [Ar] 3d^{10} 4s^1 \)

\(Cu^{+}\) ion: The ion is formed by removing one electron.

\(Cu^{+}\) configuration: Remove the single 4s electron, leaving \( [Ar] 3d^{10} \).

Configuration: \( d^{10} \)

Magnetic property: The \( d^{10} \) configuration has all electrons paired, making \(Cu^{+}\) diamagnetic.

D. \(V^{2+}\)

Atomic number of Vanadium (V): 23

Ground state electronic configuration of V: \( [Ar] 3d^3 4s^2 \)

\(V^{2+}\) ion: The ion is formed by removing two electrons.

\(V^{2+}\) configuration: Remove the two 4s electrons, leaving \( [Ar] 3d^3 \).

Configuration: \( d^3 \)

Magnetic property: With three unpaired electrons in the \( d^3 \) configuration, \(V^{2+}\) is paramagnetic.

Now that we have the configurations and magnetic properties, we can match each ion with the correct options:

A. \(Cr^{3+}\):

Configuration: \( d^3 \) → Matches with I.

Magnetic Property: Paramagnetic → Matches with III.

B. \(Mn^{4+}\):

Configuration: \( d^3 \) → Matches with I.

Magnetic Property: Paramagnetic → Matches with III.

C. \(Cu^{+}\):

Configuration: \( d^{10} \) → Matches with II.

Magnetic Property: Diamagnetic → Matches with IV.

D. \(V^{2+}\):

Configuration: \( d^3 \) → Matches with I.

Magnetic Property: Paramagnetic → Matches with III.

Thus the correct answer is option 2. A-I, III; B-I, III; C-II, IV; D-I, III