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Chapter 2: Problem 97

Give a possible set of values of the four quantum numbers for all the electrons in a boron atom and a nitrogen atom if each is in the ground state.

Short Answer

Expert verified
For a boron atom in the ground state, the set of quantum numbers for all the electrons is: - Electron 1: \(n = 1\), \(l = 0\), \(m_l = 0\), \(m_s = +\frac{1}{2}\) - Electron 2: \(n = 1\), \(l = 0\), \(m_l = 0\), \(m_s = -\frac{1}{2}\) - Electron 3: \(n = 2\), \(l = 0\), \(m_l = 0\), \(m_s = +\frac{1}{2}\) - Electron 4: \(n = 2\), \(l = 0\), \(m_l = 0\), \(m_s = -\frac{1}{2}\) - Electron 5: \(n = 2\), \(l = 1\), \(m_l = -1\), \(m_s = +\frac{1}{2}\) For a nitrogen atom in the ground state, the set of quantum numbers for all the electrons are: - Electron 1: \(n = 1\), \(l = 0\), \(m_l = 0\), \(m_s = +\frac{1}{2}\) - Electron 2: \(n = 1\), \(l = 0\), \(m_l = 0\), \(m_s = -\frac{1}{2}\) - Electron 3: \(n = 2\), \(l = 0\), \(m_l = 0\), \(m_s = +\frac{1}{2}\) - Electron 4: \(n = 2\), \(l = 0\), \(m_l = 0\), \(m_s = -\frac{1}{2}\) - Electron 5: \(n = 2\), \(l = 1\), \(m_l = -1\), \(m_s = +\frac{1}{2}\) - Electron 6: \(n = 2\), \(l = 1\), \(m_l = 0\), \(m_s = +\frac{1}{2}\) - Electron 7: \(n = 2\), \(l = 1\), \(m_l = 1\), \(m_s = +\frac{1}{2}\)

Step by step solution

01

Boron's Electron Configuration and Quantum Numbers

First, we need to determine the electron configuration of boron in the ground state. Boron has 5 electrons, so its electron configuration is 1s虏, 2s虏, 2p鹿. Now, let's find out the set of quantum numbers for all the electrons. For the 1s虏 electrons: - Electron 1: \(n = 1\), \(l = 0\), \(m_l = 0\), \(m_s = +\frac{1}{2}\) - Electron 2: \(n = 1\), \(l = 0\), \(m_l = 0\), \(m_s = -\frac{1}{2}\) For the 2s虏 electrons: - Electron 3: \(n = 2\), \(l = 0\), \(m_l = 0\), \(m_s = +\frac{1}{2}\) - Electron 4: \(n = 2\), \(l = 0\), \(m_l = 0\), \(m_s = -\frac{1}{2}\) For the 2p鹿 electron: - Electron 5: \(n = 2\), \(l = 1\), \(m_l = -1\), \(m_s = +\frac{1}{2}\)
02

Nitrogen's Electron Configuration and Quantum Numbers

Next, let's determine the electron configuration of nitrogen in the ground state. Nitrogen has 7 electrons, so its electron configuration is 1s虏, 2s虏, 2p鲁. Let's find the set of quantum numbers for all the electrons. For the 1s虏 electrons: - Electron 1: \(n = 1\), \(l = 0\), \(m_l = 0\), \(m_s = +\frac{1}{2}\) - Electron 2: \(n = 1\), \(l = 0\), \(m_l = 0\), \(m_s = -\frac{1}{2}\) For the 2s虏 electrons: - Electron 3: \(n = 2\), \(l = 0\), \(m_l = 0\), \(m_s = +\frac{1}{2}\) - Electron 4: \(n = 2\), \(l = 0\), \(m_l = 0\), \(m_s = -\frac{1}{2}\) For the 2p鲁 electrons: - Electron 5: \(n = 2\), \(l = 1\), \(m_l = -1\), \(m_s = +\frac{1}{2}\) - Electron 6: \(n = 2\), \(l = 1\), \(m_l = 0\), \(m_s = +\frac{1}{2}\) - Electron 7: \(n = 2\), \(l = 1\), \(m_l = 1\), \(m_s = +\frac{1}{2}\) In conclusion, we have identified the possible sets of values of the four quantum numbers for all the electrons in a boron atom and a nitrogen atom if each is in the ground state.

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