Practicing Success

Target Exam

CUET

Subject

Chemistry

Chapter

Inorganic: Coordination Compounds

Question:

Which of the following is diamagnetic ion?

Options:

Sc3+

Cr3+

Ti3+

V2+

Correct Answer:

Sc3+

Explanation:

The correct answer is option 1. \(Sc^{2+}\).

Let us delve deeper into why Scandium ion in the +3 oxidation state (Sc3+) is diamagnetic, while the others are not. Diamagnetism occurs when all electrons in an ion or molecule are paired. When all electrons are paired, their magnetic moments cancel each other out, resulting in no net magnetic moment. This property is typical for ions or molecules with completely filled electron shells or subshells.

1. Scandium ion \((Sc^{3+})\): Scandium \((Sc)\) in its ground state \(([Ar] 3d^1 4s^2)\) has one unpaired electron in the \(3d\) orbital. When Scandium loses 3 electrons

2. Chromium ion \((Cr^{3+})\): Chromium \((Cr)\) in its ground state \(([Ar] 3d^5 4s^1)\) has 5 unpaired electrons in the \(3d\) orbital. When Chromium loses 3 electrons to form \(Cr^{3+}\) \(([Ar] 3d^3)\), there are still 3 unpaired electrons remainin\(3d\) orbital. When Titanium loses 3 electrons to form \(Ti^{3+}\) \(([Ar] 3d^1)\), there is 1 unpaired electron remaining.

3. Titanium ion \((Ti^{3+})\): Titanium \((Ti)\) in its ground state ([Ar] 3d^2 4s^2) has 2 unpaired electrons in the 3d orbital. When Titanium loses 3 electrons to form \(Ti^{3+}\) \(([Ar] 3d^1)\), there is 1 unpaired electron remaining. \(Ti^{3+}\) has 1 unpaired electron, so it is paramagnetic.

4. Vanadium ion \((V^{2+})\): Vanadium (V) in its ground state \(([Ar] 3d^3 4s^2)\) has 3 unpaired electrons in the \(3d\) orbital. When Vanadium loses 2 electrons to form \(V^{2+}\) \(([Ar] 3d^3)\), there are still 3 unpaired electrons remaining. \(V^{2+}\) has 3 unpaired electrons, so it is paramagnetic.

Diamagnetic Ion: Scandium ion in the +3 oxidation state \((Sc^{3+})\) is diamagnetic because it has a completely filled 3d orbital with all electrons paired \(([Ar] 3d^0)\).

Paramagnetic Ions: Chromium ion \((Cr^{3+})\), Titanium ion \((Ti^{3+})\), and Vanadium ion \((V^{2+})\) all have unpaired electrons in their 3d orbitals, making them paramagnetic. This distinction arises from the electronic configurations of these ions after losing electrons to achieve their respective oxidation states.