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Which of the following represent impossible combinations of \(n\) and \(l:\) (a) \(1 p\), (b) \(4 s\), (c) \(5 f\), (d) \(2 d ?\)

Short Answer

Expert verified
The impossible combinations of \(n\) and \(l\) are (a) 1p and (d) 2d.

Step by step solution

01

Combination (a) - 1p#

In this case, \(n = 1\) and \(l\) is represented by the letter 'p', which corresponds to the value \(l = 1\). Since the only possible value for \(l\) when \(n = 1\) is 0, this combination is impossible.
02

Combination (b) - 4s#

In this case, \(n = 4\) and \(l\) is represented by the letter 's', which corresponds to the value \(l = 0\). Since 0 is a possible value for \(l\) when \(n = 4\), this combination is valid.
03

Combination (c) - 5f#

In this case, \(n = 5\) and \(l\) is represented by the letter 'f', which corresponds to the value \(l = 3\). The value of \(l\) can range from 0 to 4 when \(n = 5\), so this combination is valid.
04

Combination (d) - 2d#

In this case, \(n = 2\) and \(l\) is represented by the letter 'd', which corresponds to the value \(l = 2\). Since the possible value for \(l\) when \(n = 2\) is 0 or 1, this combination is impossible. In conclusion, the impossible combinations of \(n\) and \(l\) are (a) 1p and (d) 2d.

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Key Concepts

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

n (principal quantum number)
The principal quantum number, denoted by \( n \), is a fundamental component in the description of electron configurations within an atom. This number is always a positive integer, starting from 1, and provides information about the size and energy level of an electron's orbital. As the value of \( n \) increases, the electron is found further from the nucleus, and the energy of the electron also increases.

Key characteristics of the principal quantum number include:
  • Determines the shell, or main energy level, in which an electron resides.
  • Influences the number of possible orbitals within that energy level.
  • Directly correlates with the average distance of the electron from the nucleus—higher \( n \) values indicate further distance.
Understanding \( n \)'s role is essential as it forms the backbone of the quantum mechanical model of the atom, establishing the baseline for other quantum numbers and electron behaviors.
l (azimuthal quantum number)
The azimuthal quantum number, symbolized by \( l \), provides additional detail about the shape of an electron's orbital beyond the principal quantum number \( n \). For any given principal quantum number \( n \), the value of \( l \) can range from 0 to \( n-1 \). Each integer value of \( l \) corresponds to a different orbital shape:
  • \( l = 0 \): s orbital (spherical)
  • \( l = 1 \): p orbital (dumbbell-shaped)
  • \( l = 2 \): d orbital (clover-shaped)
  • \( l = 3 \): f orbital (complex shapes)
The azimuthal quantum number not only determines the shape of the orbitals but also contributes to the angular momentum of electrons within the orbitals. It is key to identifying the subshells within each principal energy level, showing the diversity of electron organization within atoms. For example, when \( n = 2 \), valid \( l \) values are 0 and 1, but not 2—hence, 2d is not possible, as described in the exercise.
Electron orbitals
Electron orbitals are regions within an atom where there is a high probability of finding electrons. These are defined by specific quantum numbers: the principal quantum number \( n \), the azimuthal quantum number \( l \), and others. Each orbital type has a distinctive shape and capacity:
  • s orbitals: Spherical and can hold up to 2 electrons.
  • p orbitals: Dumbbell-shaped and can hold up to 6 electrons, across three p orbitals.
  • d orbitals: Cloverleaf shapes and can accommodate up to 10 electrons, distributed across five d orbitals.
  • f orbitals: More complex shapes with a capacity of 14 electrons, spanning seven f orbitals.
These configurations are modified by the number of protons in the nucleus and result in the unique chemical properties of each element. An understanding of electron orbitals is critical to explain chemical bonding, molecular geometry, and the behavior of elements in different states.

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Most popular questions from this chapter

In the transition metals (the \(d\) -block), the electron configuration of cations is different than what you might expect. Instead of the \(\mathrm{d}\) electrons being lost first, \(s\) electrons are lost first. For example, the electron configuration of iron, \(\mathrm{Fe}\), is \([\mathrm{Ar}] 4 \mathrm{~s}^{2} 3 d^{6} ;\) but the electron configuration of \(\mathrm{Fe}^{2+}\) is \([\mathrm{Ar}] 3 d^{6} ;\) the \(4 \mathrm{~s}\) electrons are eliminated to make the cation. Write out the electron configurations of (a) \(\mathrm{Zn}^{2+}\) (b) \(\mathrm{Pt}^{2+}\) (c) \(\mathrm{Cr}^{3+}\) (d) \(\mathrm{Ti}^{4+}\).

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If you put 120 volts of electricity through a pickle, the pickle will smoke and start glowing an orange-yellow color. The light is emitted because the sodium ions in the pickle become excited; their return to the ground state results in light emission (see Figure \(6.13 \mathrm{~b}\) and Sample Exercise 6.3). (a) The wavelength of this emitted light is \(589 \mathrm{~nm}\). Calculate its frequency. (b) What is the energy of \(0.10\) mole of these photons? (c) Calculate the energy gap between the excited and ground states for the sodium ion. (d) If you soaked the pickle for a long time in a different salt solution, such as strontium chloride, would you still observe \(589 \mathrm{~nm}\) light emission? Why or why not?

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