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Chapter 7: Problem 97

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 \(\mathrm{Li}, \mathrm{N}, \mathrm{Ni}, \mathrm{Te}, \mathrm{Ba}\), and \(\mathrm{Hg} .\) Which of these atoms would be expected to be paramagnetic, and how many unpaired electrons are present in each paramagnetic atom?

Short Answer

Expert verified
The paramagnetic elements and their number of unpaired electrons are as follows: - Li: 1 unpaired electron - N: 3 unpaired electrons - Ni: 2 unpaired electrons - Te: 2 unpaired electrons

Step by step solution

01

Determine the ground-state electron configurations of the given elements.

To do this, we need to refer to the periodic table. We will write the electron configurations as follows: - Li (Lithium): 1s虏2s鹿 - N (Nitrogen): 1s虏2s虏2p鲁 - Ni (Nickel): 1s虏2s虏2p鈦�3s虏3p鈦�4s虏3d鈦� - Te (Tellurium): 1s虏2s虏2p鈦�3s虏3p鈦�4s虏3d^104p鈦�5s虏4d^105p鈦� - Ba (Barium): 1s虏2s虏2p鈦�3s虏3p鈦�4s虏3d^104p鈦�5s虏4d^105p鈦�6s虏 - Hg (Mercury): 1s虏2s虏2p鈦�3s虏3p鈦�4s虏3d^104p鈦�5s虏4d^105p鈦�6s虏4f^145d^10
02

Identify if each element has unpaired electrons and count them if paramagnetic.

Now that we know the electron configurations, we can determine if each has unpaired electrons. This will indicate if they are paramagnetic or not. - Li (Lithium): 1s虏2s鹿 has one unpaired electron - paramagnetic. - N (Nitrogen): 1s虏2s虏2p鲁 has three unpaired electrons - paramagnetic. - Ni (Nickel): 1s虏2s虏2p鈦�3s虏3p鈦�4s虏3d鈦� has two unpaired electrons - paramagnetic. - Te (Tellurium): 1s虏2s虏2p鈦�3s虏3p鈦�4s虏3d^104p鈦�5s虏4d^105p鈦� has two unpaired electrons - paramagnetic. - Ba (Barium): 1s虏2s虏2p鈦�3s虏3p鈦�4s虏3d^104p鈦�5s虏4d^105p鈦�6s虏 has no unpaired electron - not paramagnetic. - Hg (Mercury): 1s虏2s虏2p鈦�3s虏3p鈦�4s虏3d^104p鈦�5s虏4d^105p鈦�6s虏4f^145d^10 has no unpaired electron - not paramagnetic.
03

List the paramagnetic elements and their number of unpaired electrons.

Based on our work in step 2, we can write the paramagnetic elements and their number of unpaired electrons as follows: - Li: 1 unpaired electron - N: 3 unpaired electrons - Ni: 2 unpaired electrons - Te: 2 unpaired electrons

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Most popular questions from this chapter

Answer the following questions assuming that \(m_{s}\) could have three values rather than two and that the rules for \(n, \ell\), and \(m_{\ell}\) are the normal ones. a. How many electrons would an orbital be able to hold? b. How many elements would the first and second periods in the periodic table contain? c. How many elements would be contained in the first transition metal series? d. How many electrons would the set of 4 forbitals be able to hold?

Answer the following questions based on the given electron configurations and identify the elements. a. Arrange these atoms in order of increasing size: \([\mathrm{Kr}] 5 s^{2} 4 d^{10} 5 p^{6} ;[\mathrm{Kr}] 5 s^{2} 4 d^{10} 5 p^{1} ;[\mathrm{Kr}] 5 s^{2} 4 d^{10} 5 p^{3}\) b. Arrange these atoms in order of decreasing first ionization energy: \([\mathrm{Ne}] 3 s^{2} 3 p^{5} ;[\mathrm{Ar}] 4 s^{2} 3 d^{10} 4 p^{3} ;[\mathrm{Ar}] 4 s^{2} 3 d^{10} 4 p^{5}\)

The electron affinities of the elements from aluminum to chlorine are \(-44,-120,-74,-200.4\), and \(-384.7 \mathrm{~kJ} / \mathrm{mol}\), respectively. Rationalize the trend in these values.

What is the physical significance of the value of \(\psi^{2}\) at a particular point in an atomic orbital?

Give the maximum number of electrons in an atom that can have these quantum numbers: a. \(n=4\) b. \(n=5, m_{\ell}=+1\) c. \(n=5, m_{s}=+\frac{1}{2}\) d. \(n=3, \ell=2\) e. \(n=2, \ell=1\)
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