Angular Solutions to the Schrodinger Equation (7.2.3) Flashcards
• The solutions to Schrödinger’s equation have both a radial and an angular component.
• The solutions to Schrödinger’s equation have both a radial and an angular component.
• The angular solutions describe the shape of the electron orbital; the p orbitals
have directionality and one nodal plane.
• The angular solutions describe the shape of the electron orbital; the p orbitals
have directionality and one nodal plane.
The solutions to Schrödinger’s equation have both
a radial and angular component.
The solutions for the s orbitals were derived by
holding the angular part of the equation constant
and provide information only about the probability of
the presence of an electron at a given distance from
the nucleus; s orbitals are spherical in shape.
The solutions to Schrödinger’s equation have both
a radial and angular component.
The solutions for the s orbitals were derived by
holding the angular part of the equation constant
and provide information only about the probability of
the presence of an electron at a given distance from
the nucleus; s orbitals are spherical in shape.
The angular solutions to Schrödinger’s equation
describe the shape of the electron orbital.
The example shows the equation for one of three p
orbitals. When theta (θ) is 90°, cos θ = 0 and the
probability of finding the electron is zero. Thus the
electron is forbidden in the x-y plane (for the pz
orbital). This is the single angular node
characteristic of the p orbitals. Similar solutions are
obtained for the other two p orbitals.
The p orbitals have directionality, and the probability
function describes a probability “cloud” of a specific
shape in three-dimensional space.
The d orbitals and f orbitals have two and three
angular nodes respectively, giving them different
shapes than the p orbitals.
The angular solutions to Schrödinger’s equation
describe the shape of the electron orbital.
The example shows the equation for one of three p
orbitals. When theta (θ) is 90°, cos θ = 0 and the
probability of finding the electron is zero. Thus the
electron is forbidden in the x-y plane (for the pz
orbital). This is the single angular node
characteristic of the p orbitals. Similar solutions are
obtained for the other two p orbitals.
The p orbitals have directionality, and the probability
function describes a probability “cloud” of a specific
shape in three-dimensional space.
The d orbitals and f orbitals have two and three
angular nodes respectively, giving them different
shapes than the p orbitals.
How many angular nodes does a p orbital have?
1 (D)
The p orbitals have 1 angular node.
Suppose that the radial part of a wave function was equal to 1. What effect would this have on the orbital?
The probability of finding an electron would remain the same as the distance from the nucleus increased. (B)
The radial part of the wave function is equal to 1; therefore, it does not depend on r, the distance from the nucleus. This means that the probability of finding an electron in the orbital will remain the same as the distance from the nucleus increases. In other words, the probability of finding the electron 1 angstrom from the nucleus will be the same as finding it 100 angstroms or even 100 meters from the nucleus. (Naturally, this type of orbital does not actually exist.)
The angular wave function for a 2px orbital is shown below.
(3/4π)1/2 (sin θ) (cos φ)
Where will this orbital have a node?
where sine θ or cosine φ is equal to 0 (C)
A node will occur in the orbital wherever the probability of finding an electron is equal to zero. This occurs where the wave function of the orbital is equal to zero. When sine θ equals 0, the entire wave function will equal zero, as shown below.
(3/4π)1/2 (sin θ) (cos φ) = (3/4π)1/2 (0) (cos φ) = 0
Likewise, the wave function will also equal zero when cosine φ equals 0.
(3/4π)1/2 (sin θ) (cos φ) = (3/4π)1/2 (sin q) (0) = 0
Therefore, a 2px orbital will have a node wherever sine θ or cosine φ is equal to zero.
Which of the following orbitals has three spherical nodes, no planar nodes, and lacks directionality?
4s (B)
A 4s orbital, like all s orbitals, has spherical symmetry (which means that it lacks directionality) and no planar nodes. It also has three spherical nodes as shown in the diagram below. (The darker areas represent nodes.)
For which of the following sets of spherical coordinates (r, θ, φ) would the probability of finding an electron in a 2pz orbital be zero? (Assume that θ is the angle between the point and the z-axis and that φ is the angle between the x-axis and a projection of the point onto the xy-plane.)
(0.2 angstrom, 90°, 5°) (A)
The θ-value of this point is 90°, which means that the angle between the point and the z-axis is 90°. As a result, the point will lie in the xy-plane. Because a 2pz orbital has an angular node in the xy-plane, the probability of finding an electron anywhere in this plane will be equal to zero. As a result the probability of finding an electron at a point with the coordinates (0.2 angstrom, 90°, 5°) will be equal to zero. (For a 2pz orbital, any point with a θ-value of 90° will be in the orbital’s angular node.)
Which of the following is the best description of the radial portion of a 2pz orbital?
It has no radial nodes; in any given direction, the probability of finding an electron decreases exponentially as the distance from the nucleus increases. (C)
Like a 1s orbital, the radial part of a 2p orbital does not have any nodes. Also like a 1s orbital (or any other orbital), the probability of finding an electron decreases as the distance from the nucleus increases. This decrease depends only on the value of the radial part of the wave function, not on the angular part.
What is the probability of finding a particular electron above the yz-plane in a 2px orbital?
50% (C)
A 2px orbital consists of two regions where the probability of finding an electron is relatively high. One of these regions is above the yz-plane, and the other is below it. Each region is exactly equal in shape and size to the other. Therefore, the electron is equally likely to be found in either region, and the probability that the electron will be above the yz-plane is 50%.
Which of the following statements does not describe a 2py orbital?
It has a node along the y-axis. (A)
A 2py orbital does not have a node along the y-axis. In fact, the probability of finding an electron along the y-axis is relatively high. Because a 2py orbital has a node in the xz-plane, it does have nodes along both the x- and z-axes. The probability of finding an electron anywhere along these axes is zero.
The angular wave function for a 2pz orbital depends on which of the following values?
θ, the angle between a point and the z-axis (B)
The angular part of the wave function of a 2pz orbital depends θ, the angle between a point and the z-axis. When θ is equal to 0°, for instance, the probability of finding an electron is quite high; on the other hand, when θ is equal to 90°, the probability of finding an electron is zero. If the wave function did not depend on θ, the probability of finding an electron for any value of θ would be the same.
How is an angular node different from a radial node?
An angular node has only two dimensions, but a radial node has three. (A)
An angular node is represented by a plane, which only has two dimensions. A radial node, however, is represented by a hollow sphere. Therefore, this type of node has three dimensions.
