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Chapter 7: Problem 160

In the ground state of cadmium, Cd, a. how many electrons have \(\ell=2\) as one of their quantum numbers? b. how many electrons have \(n=4\) as one of their quantum numbers? c. how many electrons have \(m_{\ell}=-1\) as one of their quantum numbers? d. how many electrons hav \(m_{s}=-\frac{1}{2}\) as one of their quantum numbers?

Short Answer

Expert verified
a. 18 electrons have \(\ell=2\) as one of their quantum numbers. b. 18 electrons have \(n=4\) as one of their quantum numbers. c. 4 electrons have \(m_\ell=-1\) as one of their quantum numbers. d. 24 electrons have \(m_s=-\frac{1}{2}\) as one of their quantum numbers.

Step by step solution

01

1. Determine electron configuration of Cadmium

Using Aufbau's principle, the electron configuration of Cd can be represented as follows: \[1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10}\] This represents the distribution of electrons in various orbitals for the ground state of cadmium. Now we will analyze each part of the exercise: a. Finding the number of electrons with \(\ell=2\)
02

2. Identify \(\ell=2\) orbitals in the electron configuration

In the electron configuration, orbitals with \(\ell=2\) are denoted by p orbitals. We need to find all the p orbitals in the electron configuration.
03

3. Count the electrons in \(\ell=2\) orbitals

In cadmium's ground state electron configuration, 2p, 3p and 4p are the orbitals with \(\ell=2\). We will count the electrons in these orbitals: \(2p^6 \Rightarrow 6\) \: 3p^6 \Rightarrow 6\) \: 4p^6 \Rightarrow 6\) So, there are a total of \(6+6+6=18\) electrons with \(\ell=2\) as one of their quantum numbers. b. Finding the number of electrons with \(n=4\)
04

4. Identify \(n=4\) orbitals in the electron configuration

In the electron configuration, the orbitals with the principal quantum number \(n=4\) are 4s, 4p, and 4d.
05

5. Count the electrons in \(n=4\) orbitals

We will count the electrons in these orbitals: \(4s^2 \Rightarrow 2\) \: 4p^6 \Rightarrow 6\) \: 4d^{10} \Rightarrow 10\) So, there are a total of \(2+6+10 = 18\) electrons with \(n=4\) as one of their quantum numbers. c. Finding the number of electrons with \(m_\ell=-1\)
06

6. Identify orbitals with corresponding \(m_\ell\) values

In the electron configuration, we identify the orbitals and corresponding \(m_\ell\) values as follows: \[\begin{array}{c|c} Orbital & \: \: m_{\ell} \: \: \\ \hline s & 0 \\ p & -1,0,1 \\ d & -2,-1,0,1,2 \\ f & -3,-2,-1,0,1,2,3 \end{array}\] Using this table, we identify orbitals with \(m_\ell=-1\) as 2p (\(2nd\: in\: p\)), 3p (\(2nd\: in\: p\)), 4p (\(2nd\: in\: p\)), and 3d (\(2nd\: in\: d\)).
07

7. Count the electrons in orbitals with \(m_\ell=-1\)

Now we will count the electrons in these orbitals with \(m_\ell=-1\): \[\begin{array}{c|c} Orbital & Number\: of\: electrons \\ \hline 2p & 1 \\ 3p & 1 \\ 4p & 1 \\ 3d & 1 \end{array}\] So, there are a total of \(1+1+1+1=4\) electrons with \(m_\ell=-1\) as one of their quantum numbers. d. Finding the number of electrons with \(m_s=-\frac{1}{2}\)
08

8. Calculate the electrons with \(m_s=-\frac{1}{2}\)

All electrons have either \(m_s=\frac{1}{2}\) or \(m_s=-\frac{1}{2}\). In each orbital, half of the electrons will have \(m_s=-\frac{1}{2}\). Cadmium has a total of 48 electrons. Therefore, half of these electrons will have \(m_s = -\frac{1}{2}\). So, there are \(48 / 2 = 24\) electrons with \(m_s=-\frac{1}{2}\) as one of their quantum numbers. In conclusion: a. 18 electrons have \(\ell=2\) as one of their quantum numbers. b. 18 electrons have \(n=4\) as one of their quantum numbers. c. 4 electrons have \(m_\ell=-1\) as one of their quantum numbers. d. 24 electrons have \(m_s=-\frac{1}{2}\) as one of their quantum numbers.

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Most popular questions from this chapter

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