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In its ground state, carbon's \(2 p\) electrons interact to pro. duce \(j_{T}=0 .\) Given Hund's rule. what does this say about the total orbital angular momentum of these electrons?

Short Answer

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The total orbital angular momentum of carbon's 2p electrons in their ground state, according to Hund's rule, is 0.

Step by step solution

01

Understanding Hund's Rule

According to Hund's Rule, for any set of orbitals (in this case, the 2p orbitals of carbon), electrons will fill each of these orbitals singly (with their spin pointing in the same direction) before they start pairing up in the same orbital. The total spin angular momentum (\(j_{T}\)) is the sum of the individual spins of the electrons.
02

Applying Hund's Rule to Carbon

The 2p orbitals of a carbon atom in its ground state contain two electrons in each of the three orbitals, for a total of six electrons. Given Hund's Rule, each of these electrons will fill the orbitals singly before pairing up, meaning their spins will be aligned. When two electrons occupy the same orbital, their spin angular momenta cancel out, resulting in a total spin angular momentum of 0.
03

Determining Total Orbital Angular Momentum

Given that the total spin angular momentum (\(j_{T}\)) is 0, this implies that the total orbital angular momentum must also be 0, as these two quantities are related by the vector sum of the individual orbital and spin angular momenta of all electrons within the atom. Thus, the total orbital angular momentum of carbon's 2p electrons in their ground state must be 0.

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Key Concepts

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

Orbital Angular Momentum
Orbital angular momentum is a fundamental concept in quantum mechanics that helps us understand how electrons move within an atom. It arises due to the electron's motion around the nucleus and is quantized, meaning it can only take on certain discrete values. The orbital angular momentum of an electron in an atom is given by the formula:
  • \[ L = ext{n(n-1)} \ ext{where} \ L ext{ is the orbital angular momentum, and n is the principal quantum number.}\]
For carbon's 2p electrons, the principal quantum number \(n = 2\) and the azimuthal quantum number \(l = 1\), which means the electrons have angular momentum quantized as \(l(l+1)\). In this particular state, the total orbital angular momentum is represented by the sum of the angular momenta vectors of the individual electrons. In carbon's ground state, however, all the individual electron angular momenta cancel each other out due to the symmetrical filling of the orbitals, leading to a total orbital angular momentum of 0.
This cancellation aligns with Hund's rule and contributes to our understanding of atomic stability and electron interactions.
Spin Angular Momentum
Spin angular momentum refers to the intrinsic form of angular momentum carried by particles like electrons. Each electron has a spin, which is an intrinsic property represented by a quantum number, \(s = \frac{1}{2}\), and its spin angular momentum is given by:
  • \[ S = \sqrt{s(s+1)} \hbar \ ext{where} \ \hbar ext{ is the reduced Planck's constant.}\]
This property of electrons is quantized and can have one of two possible orientations: "up" or "down." These correspond to the spin quantum numbers \(m_s = +\frac{1}{2}\) or \(m_s = -\frac{1}{2}\).
When applying Hund's rule, carbon's 2p electrons initially fill each of the three 2p orbitals singly with parallel spins to maximize the total spin angular momentum. In the ground state, each electron pairs up with another having opposite spin within the same orbital, resulting in their net spin angular momentum canceling out, producing a total spin angular momentum of 0.
Carbon 2p Electrons
Carbon is an element with atomic number 6, and its electron configuration includes electrons in the 2p orbitals. These orbitals have a characteristic "dumbbell" shape and can hold up to six electrons distributed across three orbitals.
In its ground state, carbon's 2p orbital configuration is explained by Hund's rule, which dictates that electrons preferentially occupy separate orbitals with aligned spins before pairing. Specifically, carbon's 2p orbitals will have two singly occupied orbitals and one empty orbital when considering just the two electrons in the 2p orbitals. These electrons maintain parallel spins to maximize spin angular momentum before any pairing occurs.
Finally, when the 2p electrons fully pair, they occupy the orbitals in a way that any spin or orbital angular momentum present cancels out completely, resulting in the characteristic zero net angular momentum commonly observed in the ground state of many atoms, including carbon.

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Most popular questions from this chapter

Compare and contrast the angular momentum and magnetic moment related to or bital mot ion with those that ane intrinsic.

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The general form for symmetric and antisymmetric wave functions is \(\psi_{n}\left(x_{1}\right) \psi_{n}\left(x_{2}\right) \pm \psi_{n} \cdot\left(x_{1}\right) \psi_{n}\left(x_{2}\right)\) but it is not nornalized. (a) In applying quantum mechanics, we usually deal with quantum states that are "orthonormal." That is, if we integrate over all space the square of any individual-particle function, such as \(\psi_{n}^{*}(x) \psi_{n}(x)\) or \(\psi_{i}^{\circ}(x) \psi_{n}(x),\) we get \(1,\) but for the product of different individual-particle functions, such as \(\psi_{n}^{\circ}(x) \psi_{u}(x)\), we get 0 . This happens to be true for all the systems in which we have obtained or tabulated sets of wave functions (e.g., the particle in a box, the harmonic oscillator, and the hydrogen atom). Assuming that this holds, what multiplicative constant would normalize the symmetric and antisymmetric functions? (b) What value \(A\) gives the vector \(\mathbf{V}=A(\hat{\mathbf{x}} \pm \hat{\mathbf{y}})\) unit length? (c) Discuss the relationship between your answers in (a) and (b).

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