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(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?

Short Answer

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(a) The 3s orbital is lower in energy than the 3p orbital due to its smaller average distance from the nucleus and the shielding effect. (b) It requires less energy to remove a 3s electron from the chlorine atom compared to a 2p electron, because the 3s electron is in a higher energy level and experiences a greater effective nuclear charge.

Step by step solution

01

Part (a): Compare average distances of 3s and 3p electrons in chlorine

We are given that the average distance from the nucleus of a 3s electron in a chlorine atom is smaller than that for a 3p electron. According to the concept of shielding effect, the closer an electron is to the nucleus, the lower its energy level due to increased electrostatic attraction towards the nucleus. Therefore, the 3s orbital is lower in energy than the 3p orbital.
02

Part (b): Compare energy required to remove a 3s electron vs a 2p electron

To determine which electron requires more energy to be removed, we will consider their principal quantum numbers and shielding effect. The 3s electron is in the third energy level while the 2p electron is in the second energy level. Generally, electrons in higher energy levels require less energy to be removed. However, we also need to consider the shielding effect. In the case of chlorine, the 3s electrons shield the 2p electrons from the nucleus. Consequently, the effective nuclear charge experienced by the 2p electrons is less than the 3s electrons. Considering both the principal quantum number and the shielding effect, we come to the conclusion that it requires less energy to remove a 3s electron than a 2p electron from the chlorine atom.

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Key Concepts

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

Shielding Effect
In atomic physics, the shielding effect plays a crucial role in determining the energy levels of electrons. When discussing the shielding effect, it is important to understand how inner electrons can "shield" the outer electrons from the full force of the electrostatic attraction of the nucleus. This concept is key when analyzing the electron energy levels in multi-electron atoms.

You can think of the shielding effect as a barrier created by the electrons between the nucleus and the outer electrons. The more layers of electrons there are, the more effective this shielding becomes, which leads to outer electrons experiencing a reduced nuclear charge. This is often termed as "effective nuclear charge."

Because of this, electrons closer to the nucleus experience a stronger electrostatic attraction and thus are often found to be lower in energy than those which are further away. For instance, in a chlorine atom, a 3s electron experiences more of the nucleus's pull compared to a 3p electron due to less shielding. This is why the 3s orbital is typically lower in energy than the 3p orbital.
Principal Quantum Number
The principal quantum number, denoted as \(n\), is a fundamental concept in quantum mechanics that provides crucial insights into the electron's energy level and its general region around the nucleus. It is a positive integer (1, 2, 3, etc.) that essentially represents the shells of electrons around an atom.

As the principal quantum number increases, the average distance of an electron from the nucleus also increases. Consequently, electrons in higher principal quantum number shells, such as \(n=3\), generally have higher energy than those in lower quantum number shells like \(n=2\). For example, a 3s electron (with principal quantum number 3) in a chlorine atom is at a higher energy level than a 2p electron because of its position in a higher energy shell.

However, within a given principal quantum number, the energy differences also depend on other factors such as the subshell type (s, p, d, or f) and the shielding effect. Hence, thorough consideration of these factors is necessary when analyzing electron energy levels.
Electrostatic Attraction
Electrostatic attraction is a fundamental force acting between opposite charges. In the context of atoms, this force is between the negatively charged electrons and the positively charged protons in the nucleus.

The strength of this attraction plays a crucial role in determining the electron's energy level and its stability within an atom. Electrons that are closer to the nucleus are subjected to a higher electrostatic attraction, which makes them lower in energy and more tightly bound to the atom.

In multi-electron atoms like chlorine, the distribution and arrangement of electrons result in varying degrees of electrostatic attraction for electrons in different orbitals. For example, a 3s electron experiences more electrostatic attraction from the nucleus compared to a 3p electron due to its proximity and less shielding from inner electrons. This explains why the 3s orbital is generally at a lower energy level than the 3p orbital in the chlorine atom.

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Most popular questions from this chapter

Label each of the following statements as true or false. For those that are false, correct the statement. (a) Visible light is a form of electromagnetic radiation. (b) Ultraviolet light has longer wavelengths than visible light. (c) X rays travel faster than microwaves. (d) Electromagnetic radiation and sound waves travel at the same speed.

List the following types of electromagnetic radiation in order of descending wavelength: (a) UV lights used in tanning salons \((300-400 \mathrm{nm}) ;\) (b) radiation from an FM radio station at \(93.1 \mathrm{MHz}\) on the dial; (c) radiation from mobile phones \((450-2100 \mathrm{MHz}) ;\) (d) the yellow light from sodium vapor streetlights; (e) the red light of a light-emitting diode, such as in an appliance's display.

It is possible to convert radiant energy into electrical energy using photovoltaic cells. Assuming equal efficiency of conversion, would infrared or ultraviolet radiation yield more electrical energy on a per-photon basis?

Titanium metal requires light with a maximum wavelength of \(286 \mathrm{nm}\) to emit electrons. (a) What is the minimum energy of the photons necessary to emit electrons from titanium via the photoelectric effect? (b) What is the frequency of this radiation? (c) Is it possible to eject electrons from titanium metal using infrared light? (d) If titanium is irradiated with light of wavelength \(276 \mathrm{nm}\), what is the maximum possible kinetic energy of the emitted electrons?

The familiar phenomenon of a rainbow results from the diffraction of sunlight through raindrops. (a) Does the wavelength of light increase or decrease as we proceed outward from the innermost band of the rainbow? (b) Does the frequency of light increase or decrease as we proceed outward? [Section 6.3]

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