Problem 41 - A Complete Resource Book in Chemistry for JEE Main

Question

The atomic number of \(\mathrm{Ni}\) and \(\mathrm{Cu}\) are 28 and 29 respectively. The electronic configuration \(1 \mathrm{~s}^{2} 2 \mathrm{~s}^{2} 2 \mathrm{p}^{6} 3 \mathrm{~s}^{2} 3 \mathrm{p}^{6}\) \(3 \mathrm{~d}^{10}\) represents (a) \(\mathrm{Cu}^{+}\) (b) \(\mathrm{Cu}^{2+}\) (c) \(\mathrm{Ni}^{2+}\) (d) \(\mathrm{Ni}\)

Step-by-Step Solution

Answer

The electronic configuration represents (a) \(\text{Cu}^{+}\).

Step 1: Recall Atomic Configurations

For Ni (Nickel), with atomic number 28, its neutral atomic configuration is: \[ 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 3d^8 \, 4s^2 \]For Cu (Copper), with atomic number 29, its neutral atomic configuration is: \[ 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 3d^{10} \, 4s^1 \]

Step 2: Identify the Given Configuration

The problem gives us the electronic configuration \[ 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 3d^{10} \]This configuration lacks electrons in the 4s orbital.

Step 3: Determine Missing Electrons

The given configuration has all orbitals filled up to 3d^10 but has no electrons in higher orbitals like 4s. This is indicative of a copper ion with electrons removed from the outermost shell.

Step 4: Understand Ionization of Copper

In the case of copper, when it loses two electrons, one would typically come from the 4s orbital and one from the 3d orbitals, leading to a configuration:\[ 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 3d^{9} \]At first oxidation state (\(\text{Cu}^{1+}\)), configuration becomes:\[ 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 3d^{10} \]

Step 5: Compare Options

Compare the given electronic configuration with the possible ion configurations:- \(\text{Cu}^{+}\): \[ 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 3d^{10} \]- \(\text{Cu}^{2+}\): \[ 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 3d^{9} \]- \(\text{Ni}^{2+}\): \[ 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 3d^8 \]- \(\text{Ni}\): \[ 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 3d^8 \, 4s^2 \]

Step 6: Conclude the Answer

The given configuration \( 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 3d^{10} \) matches the configuration for \(\text{Cu}^{+}\) completely and therefore the correct answer is (a) \(\text{Cu}^{+}\).

Key Concepts

Atomic Number

The atomic number is a fundamental property of an element that helps us understand its identity and position within the periodic table. It is defined as the number of protons found in the nucleus of an atom and is unique to each element.
This means no two elements can have the same atomic number.
For example, Nickel (Ni) has an atomic number of 28, indicating it has 28 protons, while Copper (Cu) has an atomic number of 29, meaning it has 29 protons.
Understanding atomic numbers is crucial when working with elements in chemistry:

• It helps in determining the element's identity.
• It provides insight into the chemical properties of the element.
• It helps predict how the element will interact with others in a chemical reaction.

Since the atomic number reflects the number of protons, it also equals the number of electrons in a neutral atom, which significantly influences the electronic configuration.

Copper Ion

Copper ions are formed when copper atoms lose electrons.
This process is known as ionization, which results in positively charged ions.
Copper has two common ionic forms:

• **Cuprous Ion (\( ext{Cu}^{+} \)):** This ion is formed when a copper atom loses one electron. Its electronic configuration reflects the loss of one 4s electron, shown as \[ 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 3d^{10} \].
• **Cupric Ion (\( ext{Cu}^{2+} \)):** This ion is created when copper loses two electrons, generally one from the 4s orbital and one from the 3d orbital. The resulting configuration is \[ 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 3d^9 \].

These ions are crucial in various applications, such as ionic compounds and catalysts in chemical reactions. The difference in electron count between these two copper ions affects their chemical behavior and the type of compounds they can form.

Oxidation States

An oxidation state, often referred to as oxidation number, represents the degree of oxidation (loss of electrons) of an atom in a chemical compound.
It's an essential concept for understanding redox reactions where electron transfer occurs.
For elements like copper, which can exist in multiple oxidation states, this concept is vital:

• **Copper's Common Oxidation States:** Copper can exist in oxidation states of +1 (\( ext{Cu}^{+} \)) and +2 (\( ext{Cu}^{2+} \)).
• **Understanding Reactions:** These states make copper versatile in forming different compounds and participate in various chemical processes, especially in catalysis and metallurgical applications.

Transition metals, like copper, often show variable oxidation states due to the closeness in energy between their s and d orbitals, allowing them to lose different numbers of electrons.

Transition Metals

Transition metals are elements found in the central block of the periodic table, specifically in groups 3 through 12.
They are characterized by their ability to form multiple oxidation states and to have partially filled d orbitals.
Key features of transition metals include:

• **Electron Configuration:** Due to their d subshells, they exhibit complex electronic structures including variable oxidation states.
  This versatility is due to their electrons' ability to be distributed among the s and d orbitals.
• **Unique Properties:** Transition metals are known for their conductivity, malleability, and high melting points. They often form colored compounds, making them useful in pigmentation and dyes.
• **Catalytic Ability:** Many transition metals can catalyze chemical reactions, playing critical roles in industrial and biological processes.

With Copper as an example, you can observe the classic features of transition metals through its varying oxidation states and its role as a catalytic agent in processes like electric conductivity and alloy creation.