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Chapter 7: Problem 158

Answer the following questions assuming that \(m_{s}\) could have three values rather than two and that the rules for \(n, \ell\), and \(m_{\ell}\) are the normal ones. a. How many electrons would an orbital be able to hold? b. How many elements would the first and second periods in the periodic table contain? c. How many elements would be contained in the first transition metal series? d. How many electrons would the set of 4 forbitals be able to hold?

Short Answer

Expert verified
a. An orbital could hold up to 3 electrons. b. The first period would have 3 elements, and the second period would have 12 elements. c. The first transition metal series would have 15 elements. d. The set of 4f orbitals could hold 21 electrons.

Step by step solution

01

Determine the number of electrons in an orbital

Using the given magnetic quantum number values \(m_s=-\frac{1}{2}, 0, \frac{1}{2}\), since an orbital is described by a unique set of quantum numbers (\(n, \ell, m_\ell, m_s\)), we observe that for each orbital, the \(m_s\) value can only change. So, each orbital can accommodate three different \(m_s\) values, meaning an orbital can hold up to 3 electrons. Answer to (a): An orbital could hold up to 3 electrons.
02

Calculate the number of elements in the first and second periods of the periodic table

To determine the number of elements in a period, we will use the given rules for quantum numbers: - The number of the first period is when \(n = 1\). In this case, only the 1s orbital is available, and it can hold 3 electrons (as calculated in step 1). Therefore, there are 3 elements in the first period. - The number of the second period is when \(n = 2\). In this case, we have the 2s and 2p orbitals available. The 2s orbital can hold 3 electrons, and the 2p orbitals can hold 3 orbitals * 3 electrons = 9 electrons. Thus, there are a total of 12 elements in the second period. Answer to (b): The first period would have 3 elements, and the second period would have 12 elements.
03

Calculate the number of elements in the first transition metal series

The first transition metal series corresponds to \(n=3\). In this case, we have the 3d orbitals available. There are 5 orbitals in the 3d subshell, and each orbital can hold 3 electrons each. Therefore, there are a total of 5 orbitals * 3 electrons = 15 electrons, which corresponds to 15 elements in the first transition metal series. Answer to (c): The first transition metal series would have 15 elements.
04

Calculate the number of electrons in the set of 4f orbitals

The 4f orbitals correspond to \(n=4\) and \(\ell=3\). There are a total of 7 orbitals in the 4f subshell (since \(m_\ell\) can take values between \(-\ell\) and \(+\ell\), i.e. -3, -2, -1, 0, 1, 2, and 3). Each of the 7 orbitals can hold up to 3 electrons. Hence, the 4f orbitals can hold a total of 7 orbitals * 3 electrons = 21 electrons. Answer to (d): The set of 4f orbitals could hold 21 electrons.

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Most popular questions from this chapter

It takes \(208.4 \mathrm{~kJ}\) of energy to remove 1 mole of electrons from an atom on the surface of rubidium metal. How much energy does it take to remove a single electron from an atom on the surface of solid rubidium? What is the maximum wavelength of light capable of doing this?

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