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Chapter 6: Problem 46

How many nodal surfaces through the nucleus (planar nodes) are associated with each of the following atomic orbitals? (a) \(4 f\) (b) \(2 p\) (c) \(6 s\)

Short Answer

Expert verified
(a) 3, (b) 1, (c) 0.

Step by step solution

01

Understand the Concept

Nodal surfaces are regions in an atomic orbital where the probability of finding an electron is zero. These occur due to wave interference, causing nodes. Planar nodes, or nodal planes, relate to the angular part of wave functions.
02

Identify Nodal Surface Formula

The number of nodal surfaces, or planar nodes, in an orbital can be determined by the formula: \[ \text{Number of planar nodes} = \ell \]where \(\ell\) is the azimuthal quantum number (orbital angular momentum quantum number).
03

Determine Azimuthal Quantum Number (a)

For a 4f orbital, the azimuthal quantum number, \(\ell\), is 3 (\(f\) corresponds to \(\ell = 3\)). \[ \text{Planar nodes} = \ell = 3 \]
04

Determine Azimuthal Quantum Number (b)

For a 2p orbital, the azimuthal quantum number, \(\ell\), is 1 (\(p\) corresponds to \(\ell = 1\)). \[ \text{Planar nodes} = \ell = 1 \]
05

Determine Azimuthal Quantum Number (c)

For a 6s orbital, the azimuthal quantum number, \(\ell\), is 0 (\(s\) corresponds to \(\ell = 0\)). \[ \text{Planar nodes} = \ell = 0 \]
06

Conclusion

The number of nodal surfaces through the nucleus for each orbital is: (a) 3 for 4f, (b) 1 for 2p, and (c) 0 for 6s.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Nodal Surfaces
Nodal surfaces are fascinating features within atomic orbitals where the probability of finding an electron is precisely zero. These regions are crucial for understanding the shape and behavior of orbitals.
Simply put, nodal surfaces arise due to wave-like behavior of electrons that create interference patterns. When these patterns cancel out completely in certain areas, we get nodal surfaces.

There are different types of nodes, but nodal surfaces specifically refer to the areas or regions, like planes or spherical regions, where no electrons can be found.
  • Radial nodes occur when there's a change in the orbital's distance from the nucleus.
  • Planar nodes, linked to angular momentum, happen in plane-like regions through the nucleus.
Understanding nodal surfaces helps us to predict many chemical properties and reactions.
Azimuthal Quantum Number
The azimuthal quantum number, often symbolized as \(\ell\), is an essential quantum number in atomic physics. It tells us about the shape of an electron's orbital and its angular momentum around the nucleus.
This quantum number is intimately connected to the idea of nodal surfaces, specifically planar nodes. In terms of formulas, it's essential for determining the number of planar nodes.

The azimuthal quantum number takes on different integer values, each corresponding to different orbital types, like:
  • \(\ell = 0 \) for s-orbitals.
  • \(\ell = 1 \) for p-orbitals.
  • \(\ell = 2 \) for d-orbitals.
  • \(\ell = 3 \) for f-orbitals.
This number not only influences the number of planar nodes but is also pivotal in defining an atom's electron configuration and its chemical properties.
Planar Nodes
Planar nodes occur in atomic orbitals due to the wave nature of electrons and can be thought of as flat, two-dimensional areas within an orbital. At these nodes, the probability of finding an electron drops to zero. They are intrinsically linked to the azimuthal quantum number.
The number of planar nodes in an orbital is determined by the \(\ell\) value, making them synonymous with nodal planes.
  • For example, a 2p orbital (\(\ell = 1\)) has one planar node.
  • A 4f orbital (\(\ell = 3\)) features three planar nodes.
  • No planar nodes for s-orbitals (\(\ell = 0\)), such as in a 6s orbital.
Understanding planar nodes not only enhances our grasp of atomic structure but also explains how electrons are distributed and how these distributions affect chemical reactivity and bonding.

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Most popular questions from this chapter

An energy of \(3.3 \times 10^{-19} \mathrm{J} /\) atom is required to cause a cesium atom on a metal surface to lose an electron. Calculate the longest possible wavelength of light that can ionize a cesium atom. In what region of the electromagnetic spectrum is this radiation found?

A beam of electrons \(\left(m=9.11 \times 10^{-31} \mathrm{kg} / \text { electron }\right)\) has an average speed of \(1.3 \times 10^{8} \mathrm{m} / \mathrm{s}\). What is the wavelength of electrons having this average speed?

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If sufficient energy is absorbed by an atom, an electron can be lost by the atom and a positive ion formed. The amount of energy required is called the ionization energy. In the H atom, the ionization energy is that required to change the electron from \(n=1\) to \(n=\) infinity. Calculate the ionization energy for the He \(^{+}\) ion. Is the ionization energy of the He \(^{+}\) more or less than that of H? (Bohr's theory applies to He \(^{+}\) because it, like the H atom, has a single electron. The electron energy, however, is now given by \(E=-Z^{2} R h c / n^{2},\) where \(Z\) is the atomic number of helium.)

Place the following types of radiation in order of increasing energy per photon: (a) radiation within a microwave oven (b) your favorite radio station (c) gamma rays from a nuclear reaction (d) red light from a neon sign (e) ultraviolet radiation from a sun lamp
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