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) \(\mathrm{Ti}^{\prime 4}, \mathrm{~V}^{3+}, \mathrm{Cr}^{2+}, \mathrm{Mn}^{3+}\) (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) is the correct choice as all ions have a 3d\(^2\) configuration.

Step 1: Determine the Electronic Configuration of Each Shell

To solve this problem, we first need to find the general electronic configuration for each neutral atom. The electron configuration based on atomic numbers is as follows: - Ti (22 electrons): [Ar] 3d\(^2\) 4s\(^2\) - V (23 electrons): [Ar] 3d\(^3\) 4s\(^2\)- Cr (24 electrons): [Ar] 3d\(^5\) 4s\(^1\) (due to half-filled stability)- Mn (25 electrons): [Ar] 3d\(^5\) 4s\(^2\)

Step 2: Calculate Configurations for Each Ion in Option (a)

For option (a): Ti\(^{3+}\) would lose three electrons (4s\(^2\) and 3d\(^1\)), resulting in 3d\(^1\). V\(^{2+}\) would lose two electrons (4s\(^2\)), resulting in 3d\(^3\). Cr\(^{3+}\) would lose three electrons (4s\(^1\) and 3d\(^2\)), resulting in 3d\(^3\). Mn\(^{4+}\) would lose four electrons (4s\(^2\) and 3d\(^3\)), resulting in 3d\(^3\). None have d\(^2\).

Step 3: Calculate Configurations for Each Ion in Option (b)

For option (b): Ti\(^{+}\) would lose one electron (4s\(^1\)), making it 3d\(^2\). V\(^{4+}\) loses four electrons (4s\(^2\) and 3d\(^2\)), making it 3d\(^1\). Cr\(^{6+}\) loses six electrons (4s\(^1\) and 3d\(^5\)), resulting in 3d\(^0\). Mn\(^{7+}\) loses seven electrons (4s\(^2\) and 3d\(^5\)), making it 3d\(^0\). Not all have d\(^2\).

Step 4: Calculate Configurations for Each Ion in Option (c)

For option (c): Ti\(^{4+}\) loses four electrons, resulting in 3d\(^0\). V\(^{3+}\) loses three electrons (4s\(^2\) and 3d\(^1\)), making it 3d\(^2\). Cr\(^{2+}\) loses two electrons (4s\(^1\) and 3d\(^1\)), making it 3d\(^4\). Mn\(^{3+}\) loses three electrons (4s\(^2\) and 3d\(^1\)), making it 3d\(^4\). Not all have d\(^2\).

Step 5: Calculate Configurations for Each Ion in Option (d)

For option (d): Ti\(^{2+}\) loses two electrons (4s\(^2\)), leaving it 3d\(^2\). V\(^{3+}\) loses three electrons (4s\(^2\) and 3d\(^1\)), leaving it 3d\(^2\). Cr\(^{4+}\) loses four electrons (4s\(^1\) and 3d\(^3\)), leaving it 3d\(^2\). Mn\(^{5+}\) loses five electrons (4s\(^2\) and 3d\(^3\)), leaving it 3d\(^2\). Hence, all ions here have a d\(^2\) configuration.

Key Concepts

Electronic Configuration

Electronic configuration refers to the distribution of electrons in an atom or ion. It describes how electrons are organized around the nucleus in different energy levels, subshells, and orbitals. For transition metals, which include elements found in the d-block of the periodic table, the electron filling follows a specific order.

Typically, electrons fill the 3d orbitals after the 4s orbital, but there are exceptions, especially in the case of ions.

• The Aufbau principle guides electron configuration as electrons populate the lowest energy orbital available.
• For example, the standard electronic configuration for titanium (Ti) is \[ \text{[Ar]} \ 3d^2 \ 4s^2 \].
• When forming transition metal ions, electrons are usually removed from the s orbital before the d orbital.

Understanding electronic configuration is crucial to determining many properties of transition metal ions, such as magnetic behavior and chemical reactivity.

Ionization States

Ionization states refer to the charge of an ion as a result of losing or gaining electrons. Transition metals can exhibit a wide variety of ionization states due to their ability to lose different numbers of electrons.

When an atom is ionized, it loses electrons from its outermost orbitals. For transition metals, this typically means removing electrons from the 4s orbital first, even though they are filled after the 3d in neutral atoms.

• Higher ionization states occur when more electrons are removed. For instance, Ti can lose two electrons to become Ti\(^{2+}\).
• Each lost electron adds a positive charge to the ion, and the electronic configuration changes accordingly.
• Due to varying ionization states, transition metals can form a wide array of compounds.

This flexibility in ionization states allows transition metals to participate in diverse chemical reactions, making them important in many industrial and biochemical processes.

Atomic Number

The atomic number of an element is the count of protons in its nucleus, which is also the number of electrons in a neutral atom. It determines the element's position in the periodic table.

For transition metals, knowing the atomic number is essential to understanding their electronic configuration and behavior.

• For example, titanium (Ti) has an atomic number of 22, meaning it has 22 protons and, in its neutral state, 22 electrons.
• The atomic number also influences the electron configuration, as it dictates the number of electrons to be organized into orbitals.
• Understanding the atomic number helps in predicting the types of chemical bonds it can form based on electron distribution.

Recognizing the link between atomic number and elemental properties is foundational when studying the periodic table and chemical behaviors.

3d Orbital

The 3d orbital is part of the electron configuration for transition metals, characterized by its ability to hold up to ten electrons. The 3d orbitals are crucial in defining the chemistry of transition metals as they often participate directly in bonding.

In transition metals, the 3d and 4s orbitals are very close in energy levels. This leads to unique filling orders and electronic configurations.

• The distribution of electrons in the 3d orbital strongly influences a metal's magnetic and chemical properties.
• Observing electrons in these orbitals can help predict the stability of various ionization states and how a metal might interact chemically with other elements.
• Analyzing the 3d orbital configuration is key to understanding the catalytic activities of many transition metals.

The peculiarities of the 3d orbital demonstrate why transition metals have such varied and interesting chemistry compared to elements in other parts of the periodic table.