Problem 44 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

Which of the following satement(s) is (are) correct? [1998 - 2 Marks] (a) The electronic configuration of \(\mathrm{Cr}\) is \([\mathrm{Ar}] 3 \mathrm{~d}^{5} 4 \mathrm{~s}^{\mathrm{I}}\). (Atomic Number of \(\mathrm{Cr}=24)\) (b) The magnetic quantum number may have a negative value. (c) In silver atom, 23 electrons have a spin of one type and 24 of the opposite type. (Atomic Number of Ag \(=47\) ) (d) The oxidation state of nitrogen in \(\mathrm{HN}_{3}\) is \(-3\).

Step-by-Step Solution

Answer

Statements (a), (b), and (c) are correct.

Step 1: Review Option (a)

Consider the atomic number of chromium (Cr), which is 24. The electronic configuration can be determined by filling electrons in the following order: 1s², 2s², 2p⁶, 3s², 3p⁶, 3d⁵, 4s¹. Chromium is known for an unusual electron configuration due to its stability with half-filled d-orbitals. Thus, the configuration [Ar] 3d⁵ 4s¹ is indeed correct.

Step 2: Examine Option (b)

The magnetic quantum number, denoted as m, can take values from -l to +l, where l is the azimuthal quantum number. This means m can indeed have both positive and negative values. Therefore, option (b) is correct.

Step 3: Analyze Option (c)

For silver (Ag), with an atomic number of 47, the electronic configuration is [Kr] 4d¹⁰ 5s¹. The innermost 1s² contains 2 electrons with opposite spins, continuing up to 5s¹ where an additional electron adds a spin. In total, 24 electrons will have one spin type and 23 will have the opposite, due to unpaired electrons in the 5s orbital which have a single spin direction. Hence, this statement is correct.

Step 4: Calculate the Oxidation State in (d)

In the compound HN₃, calculate the oxidation state of nitrogen. Hydrogen is +1. Assume x is the oxidation state of each nitrogen: +1 + 3x = 0. Solving for x gives x = -1/3 per nitrogen atom, not -3 overall. Therefore, statement (d) is incorrect.

Key Concepts

Quantum Numbers

Quantum numbers are essential for describing the unique quantum state of an electron in an atom. There are four types of quantum numbers:

• Principal Quantum Number (\( n \)): Describes the main energy level or shell of an electron. Higher values of \( n \) indicate electrons farther from the nucleus.
• Azimuthal or Angular Momentum Quantum Number (\( l \)): Determines the shape of the electron's orbital and ranges from 0 to \( n-1 \). Commonly corresponds to s, p, d, and f orbitals.
• Magnetic Quantum Number (\( m \)): Dictates the orientation of the orbital in space. It ranges from \( -l \) to \( +l \), and thus can have negative, positive, or zero values.
• Spin Quantum Number (\( s \)): Describes the spin of the electron, with possible values of +1/2 or -1/2.

Each electron in an atom has a specific set of these quantum numbers, ensuring no two electrons share the exact same state, as per Pauli's Exclusion Principle.

Chromium Electron Configuration

Chromium presents a fascinating case in electron configurations due to its preference for a half-filled d orbital. Normally, electrons are added sequentially to sub-shells based on the Aufbau principle. However, chromium displays an exception with the configuration \([\text{Ar}] 3d^5 4s^1\) rather than the expected \([\text{Ar}] 3d^4 4s^2\).

This can be explained by its tendency to achieve a lower-energy, more stable state. The half-filled \(3d^5\) orbital provides enhanced stability because of symmetric electron distribution and reduced electron-electron repulsion.

Having this unique configuration does not violate any chemical laws; it often reflects on the resourceful nature of the atom to reach an energetically favorable state.

Oxidation State of Nitrogen

Determining the oxidation state of an element within a compound helps in understanding its electron loss or gain. For hydrogen azide (\( ext{HN}_{3}\)), the formula expresses the complex nature of nitrogen's oxidation states. Here’s how it breaks down:

• Hydrogen (H) generally has an oxidation state of +1.
• Assume \(x\) is the oxidation state of nitrogen.

To balance the formula for a neutral compound:\[+1 + 3x = 0 \]Solving for \(x\) gives \(x = -1/3\).

This fractional oxidation state indicates a shared distribution of electrons among nitrogen atoms in HN₃. Although uncommon, such partial states occur in molecules with resonance or unconventional bonding.

Magnetic Quantum Number

The magnetic quantum number (\( m \)) is a critical component of quantum mechanics and addresses the spatial orientation of electron orbitals. It works in conjunction with the azimuthal quantum number (\( l \)), giving index values ranging from \( -l \) to \( +l \).

For example, if \( l = 2 \), representing a d-subshell, \( m \) can assume values -2, -1, 0, +1, and +2. This range allows for five different orientations, corresponding to the five d-orbitals (\(d_{xy}, d_{xz}, d_{yz}, d_{x^2-y^2}, d_{z^2}\)).

Negative values for the magnetic quantum number provide options for electronic transitions and bonding orientations in complex atomic structures. This flexibility is crucial for understanding phenomena such as electron paramagnetism and NMR spectroscopy.