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Chapter 7: Problem 49

Which of the following orbitals do not exist: \(1 p, 2 s\) \(2 d, 3 p, 3 d, 3 f, 4 g ?\)

Short Answer

Expert verified
The orbitals \(1 p\), \(2 d\), and \(4 g\) do not exist.

Step by step solution

01

Analyzing the Orbitals

Start from the first orbital: \(1 p\). Here, n=1, and since p corresponds to l=1, it violates the rule that l must be less than n, so \(1 p\) doesn't exist.
02

Move to the Next Orbital

Move to the next orbital \(2 s\). Here, n=2 and s corresponds to l=0, so it doesn't violate the rule. Hence \(2 s\) exists.
03

Continue Analysis

Next, analyze \(2 d\). Here, n=2 and d corresponds to l=2. This also violates the rule, so \(2 d\) doesn't exist.
04

Continue to the Following Orbitals

Continue with \(3 p\), \(3 d\), \(3 f\). The n values are 3, and none of these orbitals violate the rule, so all of these exist.
05

Analyzing the Last Orbital

Finally, analyze \(4 g\). Here, n=4 and g corresponds to l=4. This violates the rule, so \(4 g\) doesn't exist

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Key Concepts

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

Quantum Numbers
Quantum numbers are fundamental to understanding the arrangement of electrons within an atom's electron cloud. They describe the position and energy of electrons as per the quantum mechanics principles.

The four quantum numbers are:
  • Principal quantum number () - It indicates the main energy level or shell of an electron and its distance from the nucleus. It can be any positive integer value.
  • Azimuthal quantum number () - Also known as the orbital angular momentum quantum number, it determines the shape of the orbital and is denoted by the letter . It ranges from 0 to ( - 1), where is the principal quantum number.
  • Magnetic quantum number () - This number gives the orientation of the orbital in space and can range from to positive , including zero.
  • Spin quantum number () - Representing the direction of the electron's spin, it can only be +1/2 or -1/2.
These quantum numbers are vital in determining whether a given orbital like is physically possible. For instance, as per the rules deduced from quantum numbers, does not exist because for the first shell ( = 1), there is only an ( = 0) sublevel available.
Electron Configuration
Electron configuration describes the distribution of electrons of an atom or molecule in atomic or molecular orbits. It provides a detailed account of each electron in an atom, using quantum numbers.

Electrons are arranged according to the Pauli Exclusion Principle, which states that no two electrons can have the same set of four quantum numbers, hence occupying their unique position in an atom. Here's a basic outline of the rules for electron configurations:
  • Aufbau Principle: Electrons are added one by one to the lowest energy orbitals until all of them are placed.
  • Hund's Rule: For orbitals of the same energy, electrons fill unoccupied orbitals first before pairing up.
  • Pauli's Exclusion Principle: Each electron in an atom has a unique set of quantum numbers.
With these guidelines, the question of whether an orbital exists gains clarity. For example, exists, since a orbital can be found within the second shell as per the Aufbau principle.
Orbital Shapes
The shape of atomic orbitals is a visual representation of where an electron is likely to be found around the nucleus. These shapes are pivotal to understanding chemical bonding and the arrangement of electrons.

The azimuthal quantum number () is directly linked to the shape of the orbital:
  • orbitals (l = 0) are spherical.
  • orbitals (l = 1) have a dumbbell shape.
  • orbitals (l = 2) are clover-shaped.
  • orbitals (l = 3) are complex, with more lobes.
Higher values correspond to even more complex shapes. The existence of an orbital is contingent on the correct pairing of the principal quantum number () and azimuthal quantum number (). Therefore, a orbital isn't feasible since must always be less than , which rules out the possibility of a level ( = 4) in the second ( = 2) energy level.

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Most popular questions from this chapter

How does de Broglie's hypothesis account for the fact that the energies of the electron in a hydrogen atom are quantized?

Photodissociation of water \(\mathrm{H}_{2} \mathrm{O}(l)+h \nu \longrightarrow \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g)\) has been suggested as a source of hydrogen. The \(\Delta H_{\mathrm{xn}}^{\circ}\) for the reaction, calculated from thermochemical data, is \(285.8 \mathrm{~kJ}\) per mole of water decomposed. Calculate the maximum wavelength (in \(\mathrm{nm}\) ) that would provide the necessary energy. In principle, is it feasible to use sunlight as a source of energy for this process?

The UV light that is responsible for tanning the skin falls in the 320 - to 400 -nm region. Calculate the total energy (in joules) absorbed by a person exposed to this radiation for \(2.0 \mathrm{~h}\), given that there are \(2.0 \times 10^{16}\) photons hitting Earth's surface per square centimeter per second over a 80-nm (320 nm to \(400 \mathrm{nm}\) ) range and that the exposed body area is \(0.45 \mathrm{~m}^{2}\). Assume that only half of the radiation is absorbed and the other half is reflected by the body. (Hint: Use an average wavelength of \(360 \mathrm{nm}\) in calculating the energy of a photon.

Protons can be accelerated to speeds near that of light in particle accelerators. Estimate the wavelength (in \(\mathrm{nm}\) ) of such a proton moving at \(2.90 \times\) \(10^{8} \mathrm{~m} / \mathrm{s} .\) (Mass of a proton \(\left.=1.673 \times 10^{-27} \mathrm{~kg} .\right)\)

All molecules undergo vibrational motions. Quantum mechanical treatment shows that the vibrational energy, \(E_{\mathrm{vib}},\) of a diatomic molecule like \(\mathrm{HCl}\) is given by $$ E_{\mathrm{vib}}=\left(n+\frac{1}{2}\right) h \nu $$ where \(n\) is a quantum number given by \(n=0,1,2,\) \(3, \ldots\) and \(\nu\) is the fundamental frequency of vibration. (a) Sketch the first three vibrational energy levels for \(\mathrm{HCl}\). (b) Calculate the energy required to excite a HCl molecule from the ground level to the first excited level. The fundamental frequency of vibration for \(\mathrm{HCl}\) is \(8.66 \times 10^{13} \mathrm{~s}^{-1} .\) (c) The fact that the lowest vibrational energy in the ground level is not zero but equal to \(\frac{1}{2} h v\) means that molecules will vibrate at all temperatures, including the absolute zero. Use the Heisenberg uncertainty principle to justify this prediction. (Hint: Consider a nonvibrating molecule and predict the uncertainty in the momentum and hence the uncertainty in the position.)
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