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Chapter 7: Problem 76

Give the maximum number of electrons in an atom that can have these quantum numbers: a. \(n=0, \ell=0, m_{\ell}=0\) b. \(n=2, \ell=1, m_{\ell}=-1, m_{s}=-\frac{1}{2}\) c. \(n=3, m_{s}=+\frac{1}{2}\) d. \(n=2, \ell=2\) e. \(n=1, \ell=0, m_{\ell}=0\)

Short Answer

Expert verified
a. 0 electrons b. 1 electron c. 9 electrons d. 0 electrons e. 2 electrons

Step by step solution

01

a. \(n=0\), \(\ell=0\), \(m_\ell=0\)

The principal quantum number n must be a positive integer (n > 0), but here n is equal to 0. Therefore, this quantum number set is not valid, and there are no electrons associated with these quantum numbers.
02

b. \(n=2\), \(\ell=1\), \(m_\ell=-1\), \(m_s=-\frac{1}{2}\)

These quantum numbers are valid according to the rules. Since the spin quantum number (m_s) is given and uniquely defines an electron, there can be only one electron with these quantum numbers in an atom.
03

c. \(n=3\), \(m_s=+\frac{1}{2}\)

In this case, the angular momentum quantum number (\(\ell\)) and magnetic quantum number (\(m_\ell\)) are not provided. Based on the rules, for a given principal quantum number n, there can be \(n^2\) orbitals and 2 electrons per orbital (one for each spin). For n=3, there are \(3^2 = 9\) orbitals. Since \(m_s\) is specified, each orbital can accommodate only one electron with this spin. Therefore, the maximum number of electrons with these quantum numbers is 9.
04

d. \(n=2\), \(\ell=2\)

The angular momentum quantum number (\(\ell\)) has a maximum value of n-1. In this case, \(\ell\) is equal to n, which is not valid. Thus, there are no electrons associated with these quantum numbers.
05

e. \(n=1\), \(\ell=0\), \(m_\ell=0\)

These quantum numbers are valid according to the rules. Since the magnetic quantum number (m_ell) and the angular momentum quantum number (l) are both given and zero, there is only one orbital associated with these quantum numbers. Each orbital can accommodate 2 electrons (one with spin +1/2 and one with spin -1/2). Therefore, the maximum number of electrons with these quantum numbers is 2.

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Most popular questions from this chapter

Using only the periodic table inside the front cover of the text, write the expected ground-state electron configurations for a. the third element in Group \(5 \mathrm{~A}\). b. element number 116 . c. an element with three unpaired \(5 d\) electrons. d. the halogen with electrons in the \(6 p\) atomic orbitals.

Calculate the velocities of electrons with de Broglie wavelengths of \(1.0 \times 10^{2} \mathrm{~nm}\) and \(1.0 \mathrm{~nm}\), respectively.

The four most abundant elements by mass in the human body are oxygen, carbon, hydrogen, and nitrogen. These four elements make up about \(96 \%\) of the human body. The next four most abundant elements are calcium, phosphorus, magnesium, and potassium. Write the expected ground-state electron configurations for these eight most abundant elements in the human body.

Francium, Fr, is a radioactive element found in some uranium minerals and is formed as a result of the decay of actinium. a. What are the electron configurations of francium and its predicted most common ion? b. It has been estimated that at any one time, there is only one (1.0) ounce of francium on earth. Assuming this is true, what number of francium atoms exist on earth? c. The longest-lived isotope of francium is \({ }^{223} \mathrm{Fr}\). What is the total mass in grams of the neutrons in one atom of this isotope?

The successive ionization energies for an unknown element are \(I_{1}=896 \mathrm{~kJ} / \mathrm{mol}\) \(I_{2}=1752 \mathrm{~kJ} / \mathrm{mol}\) \(I_{3}=14,807 \mathrm{~kJ} / \mathrm{mol}\) \(I_{4}=17,948 \mathrm{~kJ} / \mathrm{mol}\) To which family in the periodic table does the unknown element most likely belong?
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