Problem 61 - A Complete Resource Book in Chemistry for JEE Main

Question

Among the following series of transition metal ions, the one where all metal ion have \(3 \mathrm{~d}^{2}\) electronic configuration is (a) \(\mathrm{Ti}^{3+}, \mathrm{V}^{2+}, \mathrm{Cr}^{3+}, \mathrm{Mn}^{4+}\) (b) \(\mathrm{Ti}^{+}, \mathrm{V}^{4+}, \mathrm{Cr}^{6+}, \mathrm{Mn}^{7+}\) (c) \(\mathrm{Ti}^{4+}, \mathrm{V}^{3+}, \mathrm{Cr}^{2+}, \mathrm{Mn}^{3+}\) (d) \(\mathrm{Ti}^{2+}, \mathrm{V}^{3+}, \mathrm{Cr}^{4+}, \mathrm{Mn}^{5+}\) (At. wt \(\mathrm{Ti}=22, \mathrm{~V}=23, \mathrm{Cr}=24, \mathrm{Mn}=25\) )

Step-by-Step Solution

Answer

Option (d) \(\mathrm{Ti}^{2+}, \mathrm{V}^{3+}, \mathrm{Cr}^{4+}, \mathrm{Mn}^{5+}\) is the correct series.

Step 1: Understanding the Problem

We need to determine which series of metal ions has the electronic configuration of \(3d^2\). Each metal ion will lose electrons corresponding to its oxidation state. Transition metals typically lose 4s electrons before 3d electrons. We'll calculate the electronic configuration of each ion.

Step 2: Determine Electron Configuration of Ti$^{3+}$

Titanium (Ti) has an atomic number of 22. Its neutral atom configuration is [Ar] 3d$^2$ 4s$^2$. Ti$^{3+}$ loses three electrons, so it loses two 4s electrons and one 3d electron, resulting in a configuration of [Ar] 3d$^1$.

Step 3: Determine Electron Configuration of V$^{2+}$

Vanadium (V) has an atomic number of 23. Its neutral atom configuration is [Ar] 3d$^3$ 4s$^2$. V$^{2+}$ loses two, so it loses two 4s electrons first, resulting in a configuration of [Ar] 3d$^3$.

Step 4: Determine Electron Configuration of Cr$^{3+}$

Chromium (Cr) has an atomic number of 24. Its neutral atom configuration is [Ar] 3d$^5$ 4s$^1$. Cr$^{3+}$ loses three electrons: one 4s and two 3d, resulting in a configuration of [Ar] 3d$^3$.

Step 5: Determine Electron Configuration of Mn$^{4+}$

Manganese (Mn) has an atomic number of 25. Its neutral atom configuration is [Ar] 3d$^5$ 4s$^2$. Mn$^{4+}$ loses four electrons, first its two 4s electrons and then two 3d electrons, resulting in a configuration of [Ar] 3d$^3$.

Step 6: Evaluate the Given Options

Based on the configurations: - For (a), only Ti$^{3+}$ doesn't fit 3d$^2$. - For (b), all ions lack the 3d$^2$ configuration. - For (c), every ion again deviates from 3d$^2$. - For (d), we find that each ion has the electronic configuration [Ar] 3d$^2$ after losing the appropriate number of electrons.

Key Concepts

Electronic Configuration

Understanding electronic configurations is vital in predicting how atoms and ions behave chemically. The electronic configuration describes how electrons are distributed among the orbitals around an atom's nucleus. For transition metals, the electrons are filled in the order of energy levels, focusing primarily on the 3d and 4s orbitals. After the 4s orbital fills, electrons will start populating the 3d orbital. For example, a neutral titanium (Ti) atom has an electronic configuration of

• [Ar] 3d$^2$ 4s$^2$

This configuration tells us that titanium has two electrons in the 3d orbital and two in the 4s orbital. Knowing this allows chemists to predict how titanium will engage in bonding or reactions.

When moving to ions, especially in transition metals, it’s crucial to remember that electrons are removed from the 4s orbital before the 3d orbital, even though 4s is filled first. This unique order of losing electrons greatly influences the properties of the transition metals, making them distinct from other elements on the periodic table.

Oxidation States of Metals

Transition metals are especially interesting due to their wide range of oxidation states. An oxidation state is a concept used to describe the degree of oxidation or reduction (gain or loss of electrons) a metal can undergo.

For example, titanium (Ti) can exist in several oxidation states, like Ti$^{2+}$, Ti$^{3+}$, and Ti$^{4+}$. When titanium forms a Ti$^{3+}$ ion, it loses three electrons overall. This electron loss occurs mainly from the 4s orbital first, followed by electrons from the 3d orbital.

• Losing the 4s electrons changes the chemical behavior of the metal significantly.

Oxidation states not only affect the electronic configuration but also the chemical properties, stability, and the reactivity of metals. The ability of transition metals to exist in multiple oxidation states is one of the reasons they are widely used in various chemical processes and catalysis.

Electron Configuration of Ions

The electron configuration of ions differs from that of neutral atoms, especially for transition metals. As ions form, electrons are removed from the atom to result in positively charged ions, or in some cases, added to form negatively charged ions.

• For instance, manganese (Mn) with an atomic number of 25 has a neutral electronic configuration of [Ar] 3d$^5$ 4s$^2$.
• When forming an Mn$^{4+}$ ion, it loses four electrons, two from the 4s orbital, and two more from the 3d orbital, resulting in an electron configuration of [Ar] 3d$^3$.

Knowing the configuration of ions is crucial for understanding their behavior in an electrochemical cell, reaction mechanics, and bonding with other elements. Transition metals' adaptability in losing electrons allows them to participate in a wide range of chemical reactions, making them incredibly valuable in industrial and biochemical applications.