Question

Arrange the given metal ions in increasing order of number of unpaired electrons in the low spin complexes formed by \( \text{Mn}^{3+}, \text{Cr}^{3+}, \text{Fe}^{3+}, \text{Co}^{3+} \)

• A. \( \text{Co}^{3+}<\text{Fe}^{3+}<\text{Mn}^{3+}<\text{Cr}^{3+} \)
• B. \( \text{Co}^{3+}<\text{Mn}^{3+}<\text{Fe}^{3+}<\text{Cr}^{3+} \)
• C. \( \text{Cr}^{3+}<\text{Mn}^{3+}<\text{Co}^{3+}<\text{Fe}^{3+} \)
• D. \( \text{Cr}^{3+}<\text{Mn}^{3+}<\text{Co}^{3+}<\text{Fe}^{3+} \)

Hint:

For low spin octahedral complexes, always pair electrons in \(t_{2g}\) orbitals first. Higher oxidation states like \( \text{Co}^{3+} \) strongly favor low spin configuration.

Answer 1

The Correct Option is A

Step 1: Electronic Configurations of the Metal Ions 

The low-spin configuration occurs when a strong field ligand is present, which forces electrons to pair up as much as possible in the lower energy orbitals. - **Mn\(^{3+}\)** (Manganese ion): Manganese has an atomic number of 25, and its electronic configuration is: \[ \text{Mn:} \, [Ar] \, 3d^5 4s^2 \] For Mn\(^{3+}\), we remove three electrons: \[ \text{Mn}^{3+}: \, [Ar] \, 3d^4 \] This results in 4 unpaired electrons in the \( 3d \) orbitals. - **Cr\(^{3+}\)** (Chromium ion): Chromium has an atomic number of 24, and its electronic configuration is: \[ \text{Cr:} \, [Ar] \, 3d^5 4s^1 \] For Cr\(^{3+}\), we remove three electrons: \[ \text{Cr}^{3+}: \, [Ar] \, 3d^3 \] This results in 3 unpaired electrons in the \( 3d \) orbitals. - **Fe\(^{3+}\)** (Iron ion): Iron has an atomic number of 26, and its electronic configuration is: \[ \text{Fe:} \, [Ar] \, 3d^6 4s^2 \] For Fe\(^{3+}\), we remove three electrons: \[ \text{Fe}^{3+}: \, [Ar] \, 3d^5 \] This results in 5 unpaired electrons in the \( 3d \) orbitals. - **Co\(^{3+}\)** (Cobalt ion): Cobalt has an atomic number of 27, and its electronic configuration is: \[ \text{Co:} \, [Ar] \, 3d^7 4s^2 \] For Co\(^{3+}\), we remove three electrons: \[ \text{Co}^{3+}: \, [Ar] \, 3d^6 \] This results in 4 unpaired electrons in the \( 3d \) orbitals.

Step 2: Order of Unpaired Electrons

Now that we know the number of unpaired electrons for each ion, we can arrange them in increasing order: - Cr\(^{3+}\) has 3 unpaired electrons. - Mn\(^{3+}\) has 4 unpaired electrons. - Co\(^{3+}\) has 4 unpaired electrons. - Fe\(^{3+}\) has 5 unpaired electrons. Therefore, the increasing order of the number of unpaired electrons is: \[ \text{Cr}^{3+} < \text{Mn}^{3+} = \text{Co}^{3+} < \text{Fe}^{3+} \]

Step 3: Final Answer

The correct answer is: \[ \boxed{\text{Cr}^{3+} < \text{Mn}^{3+} = \text{Co}^{3+} < \text{Fe}^{3+}} \]