Question

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

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

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 1s², 2s², 2p¹. Now, let's find out the set of quantum numbers for all the electrons. For the 1s² 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 2s² 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 2p¹ electron: - Electron 5: \(n = 2\), \(l = 1\), \(m_l = -1\), \(m_s = +\frac{1}{2}\)

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 1s², 2s², 2p³. Let's find the set of quantum numbers for all the electrons. For the 1s² 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 2s² 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 2p³ 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.