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Chapter 7: Problem 101

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
The possible sets of quantum numbers for all electrons in a boron atom (1s²2s²2p¹) in the ground state are: Electron 1: \( (n = 1, l = 0, m_l = 0, m_s = +1/2) \) Electron 2: \( (n = 1, l = 0, m_l = 0, m_s = -1/2) \) Electron 3: \( (n = 2, l = 0, m_l = 0, m_s = +1/2) \) Electron 4: \( (n = 2, l = 0, m_l = 0, m_s = -1/2) \) Electron 5: \( (n = 2, l = 1, m_l = -1, m_s = +1/2) \) For a nitrogen atom (1s²2s²2p³) in the ground state, quantum numbers are the same as boron for electrons 1-4. The rest are: Electron 5: \( (n = 2, l = 1, m_l = -1, m_s = +1/2) \) Electron 6: \( (n = 2, l = 1, m_l = 0, m_s = +1/2) \) Electron 7: \( (n = 2, l = 1, m_l = +1, m_s = +1/2) \)

Step by step solution

01

Electron configuration of boron

To find the quantum numbers, we first need to determine the electron configuration of boron. The ground state electron configuration of boron is 1s²2s²2p¹.
02

Quantum numbers for boron

Now, we will determine the four quantum numbers for each electron in a boron atom: Electron 1 (1s²): - n = 1 - l = 0 - m_l = 0 - m_s = +1/2 Electron 2 (1s²): - n = 1 - l = 0 - m_l = 0 - m_s = -1/2 Electron 3 (2s²): - n = 2 - l = 0 - m_l = 0 - m_s = +1/2 Electron 4 (2s²): - n = 2 - l = 0 - m_l = 0 - m_s = -1/2 Electron 5 (2p¹): - n = 2 - l = 1 - m_l = -1, 0, or +1 (we choose -1 since this is the first electron in the 2p orbital) - m_s = +1/2 Now, let's move on to the nitrogen atom:
03

Electron configuration of nitrogen

To find the quantum numbers, we first need to determine nitrogen's electron configuration. The ground state electron configuration of nitrogen is 1s²2s²2p³.
04

Quantum numbers for nitrogen

Now, we will determine the four quantum numbers for each electron in a nitrogen atom: Electrons 1-4 are in the same orbitals as the boron, so their quantum numbers will be the same. Electron 5 (2p³): - n = 2 - l = 1 - m_l = -1 - m_s = +1/2 Electron 6 (2p³): - n = 2 - l = 1 - m_l = 0 (since m_l can't be the same for two electrons in the same orbital) - m_s = +1/2 Electron 7 (2p³): - n = 2 - l = 1 - m_l = +1 (since m_l can't be the same for two electrons in the same orbital) - m_s = +1/2

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