Problem 21
Question
For a \(d\)-electron, the orbital angular momentum is (a) \(\sqrt{6}(h / 2 \pi)\) (b) \(\sqrt{2}(h / 2 \pi)\) (c) \((h / 2 \pi)\) (d) \(2(h / 2 \pi)\)
Step-by-Step Solution
Verified Answer
The orbital angular momentum is \(\sqrt{6}\frac{h}{2\pi} \), so the answer is (a).
1Step 1: Understanding Orbital Angular Momentum
The formula for orbital angular momentum \( L \) is given by \( L = \sqrt{l(l + 1)} \frac{h}{2\pi} \), where \( l \) is the azimuthal quantum number (also known as the angular momentum quantum number) and \( h \) is Planck's constant.
2Step 2: Identify the Quantum Number
For a \(d\)-electron, the azimuthal quantum number \( l \) is 2. This is because \(d\)-orbitals correspond to \( l = 2 \) in the quantum mechanical model of the atom.
3Step 3: Plug into the Formula
Substitute \( l = 2 \) into the formula for \( L \):\[L = \sqrt{2(2 + 1)} \frac{h}{2\pi} = \sqrt{6} \frac{h}{2\pi}\]
4Step 4: Compare with Options
Compare the calculated value \( \sqrt{6} \frac{h}{2\pi} \) with the given options. Option (a) matches the calculated value.
Key Concepts
Azimuthal Quantum Numberd-electronQuantum Mechanical Model
Azimuthal Quantum Number
The azimuthal quantum number, often symbolized as \( l \), is crucial in the quantum mechanical model for determining the shape of an electron's orbital. This number is one of the four quantum numbers used to describe the unique quantum state of an electron. It directly influences the electron's angular momentum.
The azimuthal quantum number can take integer values ranging from 0 to \( n-1 \), where \( n \) is the principal quantum number. Each value of \( l \) corresponds to a specific orbital type:
The azimuthal quantum number can take integer values ranging from 0 to \( n-1 \), where \( n \) is the principal quantum number. Each value of \( l \) corresponds to a specific orbital type:
- \( l = 0 \) corresponds to an s-orbital.
- \( l = 1 \) corresponds to a p-orbital.
- \( l = 2 \) corresponds to a d-orbital.
- \( l = 3 \) corresponds to an f-orbital.
d-electron
A \(d\)-electron is simply an electron that occupies a d-orbital, which is a specific type of atomic orbital characterized by the azimuthal quantum number \( l = 2 \). D-orbitals come into play starting from the third energy level (or shell) in an atom.
These orbitals have some of the most complex shapes known in quantum mechanics, with their unconventional clover-like structures. Because of this complexity, d-electrons are involved in intricate bonding and are significant in defining the chemistry of transition metals.
These orbitals have some of the most complex shapes known in quantum mechanics, with their unconventional clover-like structures. Because of this complexity, d-electrons are involved in intricate bonding and are significant in defining the chemistry of transition metals.
- D-orbitals can hold a total of 10 electrons.
- This makes them important in filling the electron shell of transition metals.
- They give rise to unique properties like variable oxidation states and magnetic behavior.
Quantum Mechanical Model
The quantum mechanical model of the atom is a modern description that explains the nature of electrons within an atom. It moves beyond the classic "solar system" model by introducing probability clouds, known as orbitals, where electrons are likely to be found.
This model interprets electron behavior using four quantum numbers:
This model interprets electron behavior using four quantum numbers:
- The principal quantum number \( n \), indicating the main energy level.
- The azimuthal quantum number \( l \), which specifies the orbital shape.
- The magnetic quantum number \( m_l \), identifying the orientation of the orbital.
- The spin quantum number \( m_s \), representing the electron's spin direction.
Other exercises in this chapter
Problem 20
The electrons, identified by quantum numbers \(n\) and \(l\), (i) \(n=4, l=1\), (ii) \(n=4, l\) \(=0\), (iii) \(n=3, l=2\), and (iv) \(n=3, l=1\) can be placed
View solution Problem 21
The wave number of the first emission line in the Balmer series of \(\mathrm{H}\)-Spectrum is: \((\mathrm{R}=\) Rydberg constant \()\) : (a) \(\frac{5}{36} R\)
View solution Problem 22
The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is \(\left[a_{0}\right.\) is Bohr radius] : (a) \(\frac{h^{2}}{4 \pi^{2} m a_{0}^{
View solution Problem 22
The orbital angular momentum of an electron in \(2 s\) orbital is: (a) \(+\frac{1}{2} \cdot \frac{h}{2 \pi}\) (b) Zero (c) \(\frac{h}{2 \pi}\) (d) \(\sqrt{2} \c
View solution