One bit of evidence that the quantum mechanical model is “correct” lies in the magnetic properties of matter. Atoms with unpaired electrons are attracted by magnetic fields and thus are said to exhibit paramagnetism. The degree to which this effect is observed is directly related to the number of unpaired electrons present in the atom. Consider the ground-state electron configurations for Li, N, Ni, Te, Ba, and Hg. Which of these atoms would be expected to be paramagnetic, and how many unpaired electrons are present in each paramagnetic atom?

Short Answer

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The paramagnetic atoms among the given elements are: - Lithium (Li) with 1 unpaired electron - Nitrogen (N) with 3 unpaired electrons - Nickel (Ni) with 2 unpaired electrons - Tellurium (Te) with 2 unpaired electrons Barium (Ba) and Mercury (Hg) are not expected to be paramagnetic due to the absence of unpaired electrons.

Step by step solution

01

Electron Configurations

To determine the ground-state electron configurations, we must follow the Aufbau principle and organize the electrons according to the increasing energy levels and arrows (↑, ↓) representing the electrons and their spins. - Lithium (Li) - Atomic number 3: Electron configuration is 1s² 2s¹ - Nitrogen (N) - Atomic number 7: Electron configuration is 1s² 2s² 2p³ - Nickel (Ni) - Atomic number 28: Electron configuration is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁸ - Tellurium (Te) - Atomic number 52: Electron configuration is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁴ - Barium (Ba) - Atomic number 56: Electron configuration is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² - Mercury (Hg) - Atomic number 80: Electron configuration is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² 4f¹⁴ 5d¹⁰
02

Identifying Unpaired Electrons

We can now examine the electron configurations to find the presence of unpaired electrons, responsible for the paramagnetism: - Lithium (Li): 1s² 2s¹ - 1 unpaired electron - Nitrogen (N): 1s² 2s² 2p³ - 3 unpaired electrons - Nickel (Ni): 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁸ - 2 unpaired electrons - Tellurium (Te): 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁴ - 2 unpaired electrons - Barium (Ba): 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² - 0 unpaired electrons - Mercury (Hg): 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² 4f¹⁴ 5d¹⁰ - 0 unpaired electrons
03

Final Answer

According to the electron configurations identified in Step 2, the following atoms are expected to be paramagnetic: - Lithium (Li) with 1 unpaired electron - Nitrogen (N) with 3 unpaired electrons - Nickel (Ni) with 2 unpaired electrons - Tellurium (Te) with 2 unpaired electrons Barium (Ba) and Mercury (Hg) have no unpaired electrons, therefore they are not expected to exhibit paramagnetism.

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Most popular questions from this chapter

Give the maximum number of electrons in an atom that can have these quantum numbers: a. \(n=0, \ell=0, m_{\ell}=0\) b. \(n=2, \ell=1, m_{\ell}=-1, m_{s}=-\frac{1}{2}\) c. \(n=3, m_{s}=+\frac{1}{2}\) d. \(n=2, \ell=2\) e. \(n=1, \ell=0, m_{\ell}=0\)

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