Problem 82 - A Complete Resource Book in Chemistry for JEE Main

Question

The orbital angular momentum of an electron in \(2 \mathrm{~s}\) orbital is (a) \(+\frac{1}{2} \cdot \frac{\mathrm{h}}{2 \pi}\) (b) zero (c) \(\frac{\mathrm{h}}{2 \pi}\) (d) \(\sqrt{2} \frac{\mathrm{h}}{2 \pi}\)

Step-by-Step Solution

Answer

The orbital angular momentum of an electron in a 2s orbital is zero.

Step 1: Understanding electron orbitals

Electrons in atoms occupy orbitals which are defined by quantum numbers. The orbital angular momentum depends on the azimuthal quantum number \(l\). For s orbitals, like the \(2s\) orbital, \(l = 0\).

Step 2: Applying the formula for orbital angular momentum

The orbital angular momentum \(L\) is given by the formula \(L = \sqrt{l(l+1)} \frac{h}{2\pi}\). For the \(2s\) orbital, \(l = 0\), so the expression becomes \(L = \sqrt{0(0+1)} \frac{h}{2\pi} = \sqrt{0} \frac{h}{2\pi} = 0\).

Step 3: Conclusion based on calculation

Since the calculation shows that the orbital angular momentum \(L = 0\), the correct option is (b) zero.

Key Concepts

Quantum Numbers

In the fascinating world of quantum mechanics, electrons in an atom are not simply particles circling around the nucleus. Rather, they exist in specific energy levels and locations described by quantum numbers.
Quantum numbers are like an electron's address within an atom, detailing its energy, position, and spin.

• **Principal Quantum Number (\(n\))**: This number indicates the energy level of an electron and its distance from the nucleus. Higher values mean more energy and a greater average distance from the nucleus.
• **Azimuthal Quantum Number (\(l\))**: This number tells us the shape of the orbital and is integral to determining the angular momentum of an electron.
• **Magnetic Quantum Number (\(m_l\))**: This number reveals the orientation of the orbital in space.
• **Spin Quantum Number (\(m_s\))**: This defines the electron's spin direction within an orbital.

In our exercise involving a \(2s\) orbital, we're particularly interested in how the azimuthal quantum number contributes to the angular momentum.

Azimuthal Quantum Number

The azimuthal quantum number, denoted as \(l\), is a pivotal part of understanding the orbital's shape and the electron's angular momentum.
It can take any integer value from 0 to \(n-1\), where \(n\) is the principal quantum number.
Importantly, the value of \(l\) provides insight into the type of orbital:

• When \(l = 0\), the orbital is an **s orbital**, spherical in shape.
• When \(l = 1\), it's a **p orbital**, dumbbell-shaped.
• When \(l = 2\), the orbital becomes a **d orbital**, often looking like a clover.
• For \(l = 3\), you have an **f orbital**, more complex in shape.

For the \(2s\) orbital under the spotlight in our problem, \(l = 0\). That means the azimuthal quantum number tells us it has a spherical shape and, crucially, that its orbital angular momentum is zero.

s Orbitals

The simplest type of atomic orbitals, **s orbitals** are spherical in shape and can be found in every energy level.
The spherical symmetry of s orbitals allows the electron to be spread out uniformly around the nucleus.
This makes visualizing the electron's probable location a bit like imagining a cloud surrounding the nucleus.
S orbitals hold some special characteristics:

• **Only one per energy level**: Each principal energy level \(n\) has only one s orbital. For example, there is a \(1s\), \(2s\), \(3s\), and so on.
• **Maximum of two electrons**: Due to the Pauli exclusion principle, each s orbital can hold a maximum of two electrons, each with opposite spins.
• **No angular nodes**: As indicated by \(l = 0\), s orbitals lack angular nodes. This means they have no nodal planes bisecting the electron cloud.

In the context of the original exercise, the \(2s\) orbital specifies that \(l = 0\). As a result, the orbital angular momentum is zero, which confirms the correctness of option **(b) zero** for this problem.