Question

Choose the correct statement.

• A. Ionic compounds of \(Sc^{3+}\) and \(Cu^{+}\) are coloured because of \(d-d\) electronic transitions
• B. The order in which the paramagnetic nature of the 4 cations \(Cr^{2+}, Mn^{2+}, V^{2+}\) and \(Fe^{2+}\) varies is \(V^{2+}<Cr^{2+} = Mn^{2+}<Fe^{2+}\)
• C. As the oxidation number of the transition element increases, the ionic nature decreases and the oxides show acidic nature predominantly
• D. The metal cobalt has the electronic configuration \([Ar]3d^5\) in the \(+3\) oxidation state

Hint:

Transition metal oxides in lower oxidation states are generally basic, while those in higher oxidation states are generally acidic due to increased covalent character.

Answer 1

The Correct Option is C

Step 1: Analyse statement (A).
For \(d-d\) transition, partially filled \(d\)-orbitals are required.
The electronic configuration of \(Sc\) is:
\[ Sc = [Ar]3d^14s^2 \] So,
\[ Sc^{3+} = [Ar]3d^0 \] Similarly, copper has configuration:
\[ Cu = [Ar]3d^{10}4s^1 \] So,
\[ Cu^+ = [Ar]3d^{10} \] Since \(Sc^{3+}\) has \(d^0\) configuration and \(Cu^+\) has \(d^{10}\) configuration, \(d-d\) transition is not possible in either case.
Therefore, statement (A) is incorrect.

Step 2: Understand paramagnetism for statement (B).
Paramagnetic nature depends on the number of unpaired electrons.
Higher the number of unpaired electrons, greater will be the paramagnetic character.
For transition metal ions, electrons are first removed from \(4s\)-orbital and then from \(3d\)-orbital.

Step 3: Calculate unpaired electrons in the given ions.
For \(V^{2+}\):
\[ V = [Ar]3d^34s^2 \] \[ V^{2+} = [Ar]3d^3 \] So, it has \(3\) unpaired electrons.
For \(Cr^{2+}\):
\[ Cr = [Ar]3d^54s^1 \] \[ Cr^{2+} = [Ar]3d^4 \] So, it has \(4\) unpaired electrons.
For \(Mn^{2+}\):
\[ Mn = [Ar]3d^54s^2 \] \[ Mn^{2+} = [Ar]3d^5 \] So, it has \(5\) unpaired electrons.
For \(Fe^{2+}\):
\[ Fe = [Ar]3d^64s^2 \] \[ Fe^{2+} = [Ar]3d^6 \] So, it has \(4\) unpaired electrons.
Thus, the correct order of paramagnetic nature is:
\[ V^{2+}<Cr^{2+} = Fe^{2+}<Mn^{2+} \] Hence, statement (B) is incorrect.

Step 4: Analyse statement (C).
As the oxidation number of a transition element increases, the polarising power of the metal ion increases.
Due to this, the covalent character of the compound increases and ionic character decreases.
Also, oxides of transition metals in lower oxidation states are generally basic, while oxides in higher oxidation states are generally acidic.
For example:
\[ MnO \text{ is basic, whereas } Mn_2O_7 \text{ is acidic.} \] Therefore, statement (C) is correct.

Step 5: Analyse statement (D).
Cobalt has atomic number \(27\).
Its electronic configuration is:
\[ Co = [Ar]3d^74s^2 \] In \(+3\) oxidation state, two electrons are removed from \(4s\)-orbital and one electron is removed from \(3d\)-orbital.
So,
\[ Co^{3+} = [Ar]3d^6 \] It is not \([Ar]3d^5\).
Therefore, statement (D) is incorrect.

Step 6: Final selection of correct statement.
Only statement (C) is correct because higher oxidation states of transition metals show lower ionic character and their oxides are predominantly acidic.
Final Answer:
The correct statement is:
\[ \boxed{\text{(C) As the oxidation number of the transition element increases, the ionic nature decreases and the oxides show acidic nature predominantly.}} \]