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How many possible values for \(l\) and \(m_{l}\) are there when (a) \(n=3\); (b) \(n=5\) ?

Short Answer

Expert verified
When (a) \(n = 3\), there are 3 possible values for \(l\) and 9 possible values for \(m_l\). When (b) \(n = 5\), there are 5 possible values for \(l\) and 25 possible values for \(m_l\).

Step by step solution

01

Find the possible values of \(l\) for \(n=3\)

According to rule 1, the possible values of \(l\) are \(0, 1, 2\).
02

Find the possible values of \(m_l\) for each value of \(l\)

For \(l=0\), \(m_l = 0\). For \(l=1\), we have \(m_l = -1, 0, 1\). For \(l=2\), we have \(m_l = -2, -1, 0, 1, 2\).
03

Count the total number of possible values for \(l\) and \(m_l\)

We have 3 possible values for \(l\) and 1 + 3 + 5 = 9 possible values for \(m_l\). (b) \(n=5\)
04

Find the possible values of \(l\) for \(n=5\)

According to rule 1, the possible values of \(l\) are \(0, 1, 2, 3, 4\).
05

Find the possible values of \(m_l\) for each value of \(l\)

For \(l=0\), \(m_l = 0\). For \(l=1\), we have \(m_l = -1, 0, 1\). For \(l=2\), we have \(m_l = -2, -1, 0, 1, 2\). For \(l=3\), we have \(m_l = -3, -2, -1, 0, 1, 2, 3\). For \(l=4\), we have \(m_l = -4, -3, -2, -1, 0, 1, 2, 3, 4\).
06

Count the total number of possible values for \(l\) and \(m_l\)

We have 5 possible values for \(l\) and 1 + 3 + 5 + 7 + 9 = 25 possible values for \(m_l\).

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Key Concepts

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

Atomic Orbitals
Atomic orbitals are specific areas within an atom where there is a high probability of finding electrons. According to quantum mechanics, electrons are not seen as particles at fixed locations but as wave-like distributions with uncertain positions. This view portrays electrons as existing in clouds or orbitals around the nucleus. Each orbital is associated with a particular energy level and shape, which are determined by quantum numbers.

One such quantum number is the principal quantum number, denoted as \(n\), which indicates the main energy level or shell of an electron and its distance from the nucleus. The value of \(n\) is always a positive integer (\(1, 2, 3, \dots\)). The exercise given considers \(n=3\) and \(n=5\), pointing to the third and fifth energy levels where electrons may reside.

The angular momentum quantum number, represented as \(l\), gives the shape of the orbital. For any given value of \(n\), \(l\) ranges from \(0\) to \(n-1\). Therefore, for \(n=3\), the possible values for \(l\) would be \(0, 1, 2\), correlating to s, p, and d orbitals respectively. Similarly, for \(n=5\), \(l\) can be \(0, 1, 2, 3, 4\), signifying s, p, d, f, and g orbitals.
Quantum Mechanics in Chemistry
Quantum mechanics is fundamental in explaining the behavior of electrons in atoms, forming the basis for modern chemistry. The principles of quantum mechanics guide us in understanding the electronic structure of atoms, which is pivotal for predicting chemical properties and reactivity.

Quantum mechanics employs quantum numbers, which include the principal quantum number \(n\), angular momentum quantum number \(l\), magnetic quantum number \(m_{l}\), and spin quantum number \(s\). These quantum numbers are integral in defining the state of an electron in an atom, where the magnetic quantum number \(m_{l}\) describes the orientation of the orbital in space and can take on values from \(-l\) to \(+l\), including zero. For example, when \(l=1\), \(m_{l}\) can have values \(-1, 0, +1\).

This intricate quantum world governs not just the micro-environment within atoms but also informs the macroscopic properties of materials and reactions in chemistry. This demonstrates a profound relationship between the observed chemical behavior and the complex, often invisible world detailed by quantum numbers and their associated electrons.
Electron Configuration
Electron configuration describes how electrons are distributed in an atom's orbitals. The arrangement of electrons is grounded on the principles of quantum mechanics and follows a set of rules such as the Aufbau principle, Hund's rule, and the Pauli Exclusion Principle to minimize the energy and maintain stability.

The Aufbau principle suggests that electrons fill orbitals starting from the lowest energy level, progressing to higher levels. Hund's rule states that electrons will fill degenerate orbitals (orbitals of the same energy level) singly before pairing up, and according to the Pauli exclusion principle, no two electrons in an atom can have the same set of four quantum numbers.

For instance, when we consider the principal quantum number \(n=3\), there are three sublevels as per the values of \(l\): s (\(l=0\)), p (\(l=1\)), and d (\(l=2\)). Accordingly, the sublevels will have 1, 3, and 5 orbitals each, given by the possible values of \(m_{l}\). This pattern allows chemists to determine the possible electron configurations and predict the chemical characteristics of elements. As students progress through chemistry, they learn to use these rules to write out the electron configuration for elements, which in turn is crucial for understanding the periodic table and chemical bonding.

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Most popular questions from this chapter

(a) What is the frequency of radiation whose wavelength is \(10.0 \AA\) ? (b) What is the wavelength of radiation that has a frequency of \(7.6 \times 10^{10} \mathrm{~s}^{-1} ?\) (c) Would the radiations in part (a) or part (b) be detected by an X-ray detector? (d) What distance does electromagnetic radiation travel in \(25.5 \mathrm{fs}\) ?

The series of emission lines of the hydrogen atom for which \(n_{f}=3\) is called the Paschen series. (a) Determine the region of the electromagnetic spectrum in which the lines of the Paschen series are observed. (b) Calculate the wavelengths of the first three lines in the Paschen series - those for which \(n_{i}=4,5\), and 6 .

(a) The average distance from the nucleus of a \(3 s\) electron in a chlorine atom is smaller than that for a \(3 p\) electron. In light of this fact, which orbital is higher in energy? (b) Would you expect it to require more or less energy to remove a \(3 s\) electron from the chlorine atom, as compared with a \(2 p\) electron? Explain.

The "magic numbers" in the periodic table are the atomic numbers of elements with high stability (the noble gases): \(2,10,18,36,54\), and \(86 .\) In terms of allowed values of orbitals and spin quantum numbers, explain why these electron arrangements correspond to special stability.

Identify the specific element that corresponds to each of the following electron configurations: (a) \(1 s^{2} 2 s^{2}\), (b) \(1 s^{2} 2 s^{2} 2 p^{4}\), (c) \([\mathrm{Ar}] 4 s^{1} 3 d^{5}\), (d) \([\mathrm{Kr}] 5 s^{2} 4 d^{10} 5 p^{4}\), (e) \(1 s\) ?

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