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\(f\) orbitals be able to hold?

An orbital is determined by three quantum numbers: principal quantum number \(n\), angular momentum quantum number \(\ell\), and magnetic quantum number \(m_\ell\). However, it can hold electrons with two different spin magnetic quantum numbers (\(m_s\)) under normal circumstances. In this exercise, we are assuming \(m_s\) can have three values instead of two. So each orbital can now hold 3 electrons instead of 2. Answer: An orbital can hold 3 electrons.

In the periodic table, elements are arranged by increasing atomic number. The elements in a given period have the same principal quantum number \(n\). For the first period (n=1), there is only one possible value for \(\ell\), which is \(\ell=0\). Therefore, we have only one orbital (called 1s), and according to the answer in part a, it can hold 3 electrons. This means there would be 3 elements in the revised first period. For the second period (n=2), there are two possible values for \(\ell\), which are \(\ell=0\) and \(\ell=1\). This gives us a total of 2 orbitals for \(\ell=0\) (2s orbitals) and 6 orbitals for \(\ell=1\) (2p orbitals). Each of these orbitals can hold 3 electrons. Therefore, there will be 8 orbitals x 3 electrons per orbital = 24 elements in the second period. Answer: The first period would contain 3 elements, and the second period would contain 24 elements.

The first transition metal series contains elements with partially filled d orbitals (3d orbitals, since n=3 and \(\ell=2\)) in their ground state. There are 5 possible values for \(m_\ell\) in the d orbitals, so there are 5 orbitals with 3 electrons per orbital. Thus, there will be 5 orbitals x 3 electrons per orbital = 15 elements contained in the first transition metal series. Answer: The first transition metal series would contain 15 elements.

For the 4f orbitals, we have n=4 and \(\ell=3\). There are 7 possible values for \(m_\ell\) in f orbitals, so there are 7 orbitals in the 4f subshell. Each of these orbitals can hold 3 electrons. Therefore, the set of 4f orbitals can hold 7 orbitals x 3 electrons per orbital = 21 electrons. Answer: The set of 4f orbitals can hold 21 electrons.