Problem 140 - A Complete Resource Book in Chemistry for JEE Main

Question

Match the following Column-I (a) \([\mathrm{Ar}] 3 \mathrm{~d}^{\mathrm{s}} 4 \mathrm{~s}^{2}\) (b) \([\mathrm{Ar}] 3 \mathrm{~d}^{10}\) (c) \([\mathrm{Ar}] 3 \mathrm{~d}^{1}\) (d) \([\mathrm{Ar}] 3 \mathrm{~d}^{9}\) Column-II (p) \(\mathrm{Cu}^{2+}\) (q) \(\mathrm{Zn}^{2+}\) (r) \(\mathrm{T}_{1}^{+3}\) (s) \(\mathrm{Cu}^{+}\) (t) Ni

Step-by-Step Solution

Answer

(a)-(t), (b)-(q), (c)-(r), (d)-(s).

Step 1: Understanding Electron Configurations

First, we need to understand the given electron configurations and match them with possible ions or elements. The configuration \( [\mathrm{Ar}] 3 \mathrm{d}^{\mathrm{s}} 4 \mathrm{s}^{2} \) indicates that the element is in the first row of transition metals, specifically involving the 3d subshell as it's filled after 4s. The correct match depends on identifying the electron count for specific ions.

Step 2: Analyze Configuration (a)

The electron configuration in (a) \( [\mathrm{Ar}] 3 \mathrm{d}^{\mathrm{s}} 4 \mathrm{s}^{2} \) corresponds to nickel (Ni) with \( 3d^{8} \) electrons and \( 4s^{2} \) for a neutral atom. Therefore, the correct match is (t) Ni.

Step 3: Analyze Configuration (b)

(b) \( [\mathrm{Ar}] 3 \mathrm{d}^{10} \) corresponds to a fully filled d subshell. This configuration matches \( \mathrm{Zn}^{2+} \), as zinc loses two electrons from the \( 4s^{2} \) to form the ion. Therefore, the correct match for (b) is (q) \( \mathrm{Zn}^{2+} \).

Step 4: Analyze Configuration (c)

(c) \[ [\mathrm{Ar}] 3 \mathrm{d}^{1} \] corresponds to only one electron in the 3d subshell. This would be the configuration for an ion formed by removal of electrons to leave a single d electron. The match is likely to be (r) \( \mathrm{T}_{1}^{+3} \).

Step 5: Analyze Configuration (d)

(d) \[ [\mathrm{Ar}] 3 \mathrm{d}^{9} \] indicates a situation most fitting of copper's electronic reduction. For \( \mathrm{Cu}^+ \), the electronic configuration would reduce by losing one electron from \( 4s^{1} \) to remain most stable, hence it matches (s) \( \mathrm{Cu}^{+} \).

Key Concepts

Transition Metals

Transition metals are elements found in the center of the periodic table, specifically in groups 3 to 12. These metals have a unique ability to form a wide variety of oxidation states due to their partially filled d sub-shells. Generally, the electron configuration of transition metals can be intriguing as they initially fill the 4s orbital before starting to fill the 3d sub-shell. This can sometimes seem counterintuitive to students.

These metals include elements such as iron, copper, and nickel, and they are known for their remarkable properties like malleability, corrosion resistance, and the ability to conduct electricity. These elements play significant roles in many industrial applications, such as in the fabrication of alloys and in catalysis for chemical reactions. Understanding how their electron configurations influence their properties and behaviors underlies much of inorganic chemistry.

Matching Electron Configurations

Electron configurations can tell us a lot about the chemical properties and behavior of an element or ion. Matching these configurations helps us to identify specific substances or ions. Upon examining an electron configuration, we should first identify the number of electrons, then relate that to the position in the periodic table.

For instance, the configuration \[\mathrm{Ar}] 3 \mathrm{d}^{10} \] is fully filled, suggesting that it belongs to an ionized form of zinc, such as \mathrm{Zn}^{2+}\, which loses its \[4s^2\] electrons. \[\mathrm{Ar}] 3 \mathrm{d}^{9}\] is associated with \mathrm{Cu}^+\, indicating a subtraction of one electron from its copper parent atom. These exercises train students to understand how electronic structures change when atoms gain or lose electrons.

Ionic Charges

Ionic charges occur due to electrons being transferred between atoms, leading to an imbalance in protons and electrons. This results in an atom becoming positively or negatively charged. The number of electrons lost or gained determines the magnitude of the charge on the ion.

For transition metals, the ionic charge can vary since they can lose different numbers of electrons from their s and d orbitals. For example, zinc typically forms \mathrm{Zn}^{2+}\ by losing its two \[4s\] electrons. Understanding these charges is crucial for predicting how ions will interact in chemical reactions, as they determine the reactivity and the types of bonds an ion will form.

Electron Removal in Ions

When forming ions, the removal of electrons involves subtracting from the outermost sub-shell first. In transition metals, this typically involves the 4s electrons being removed prior to any 3d electrons, despite 4s being filled before the 3d sub-shell.

This concept is particularly key for forming transition metal cations. For instance, when forming \(\mathrm{Cu}^{+}\), one electron is removed from the \[4s\] rather than the \[3d\]. This is because electrons in the 3d sub-shell remain more stable and are less likely to be removed in the initial ionization stages when compared to 4s electrons. The proper understanding of electron removal is significant when predicting and explaining the stability and formation of specific ions in both physical and chemical properties.