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Chapter 7: Problem 31

Consider the isoelectronic ions \(\mathrm{F}^{-}\) and \(\mathrm{Na}^{+}\). (a) Which ion is smaller? (b) Using Equation 7.1 and assuming that core electrons contribute 1.00 and valence electrons contribute 0.00 to the screening constant, \(S,\) calculate \(Z_{\text {eff }}\) for the \(2 p\) electrons in both ions. (c) Repeat this calculation using Slater's rules to estimate the screening constant, \(S .(\mathbf{d})\) For isoelectronic ions, how are effective nuclear charge and ionic radius related?

Short Answer

Expert verified
(a) The \(\mathrm{Na}^{+}\) ion is smaller than the \(\mathrm{F}^{-}\) ion. (b) Using the given values for the screening constant, we find that \(Z_{\text {eff }}=7\) for a \(2p\) electron in \(\mathrm{F}^{-}\) and \(Z_{\text {eff }}=9\) for a \(2p\) electron in \(\mathrm{Na}^{+}\). (c) Using Slater's rules, we obtain \(Z_{\text {eff }}=5.2\) for a \(2p\) electron in \(\mathrm{F}^{-}\) and \(Z_{\text {eff }}=6.5\) for a \(2p\) electron in \(\mathrm{Na}^{+}\). (d) For isoelectronic ions, the ionic radius and effective nuclear charge are inversely related.

Step by step solution

01

(a) Comparing Ionic Radii of Isoelectronic Ions

For isoelectronic ions, the ion with more protons will have a larger nuclear charge, resulting in a stronger attraction between the nucleus and the electrons. This causes the electron cloud to be pulled in more tightly, making the ion smaller. The \(\mathrm{F}^{-}\) ion has 9 protons while the \(\mathrm{Na}^{+}\) ion has 11 protons. Therefore, the \(\mathrm{Na}^{+}\) ion is smaller.
02

(b) Calculate \(Z_{\text {eff }}\) using given values for \(S\)

We will use Equation 7.1: \(Z_{\text {eff }}=Z-S\), where \(Z\) is the atomic number (number of protons) and \(S\) is the screening constant. For \(\mathrm{F}^{-}\): \(Z = 9\) For the \(2p\) electron, there are 2 core electrons contributing 1.00 to \(S\) and 6 valence electrons contributing 0.00 to \(S\). So, \(S = 2(1.00) + 6(0.00) = 2\). \(Z_{\text {eff }}=9-2=7\) for a \(2p\) electron in \(\mathrm{F}^{-}\). For \(\mathrm{Na}^{+}\): \(Z = 11\) For the \(2p\) electron, there are 2 core electrons contributing 1.00 to \(S\) and 8 valence electrons contributing 0.00 to \(S\). So, \(S=2(1.00) + 8(0.00) = 2\). \(Z_{\text {eff }}=11-2=9\) for a \(2p\) electron in \(\mathrm{Na}^{+}\).
03

(c) Calculate \(Z_{\text {eff }}\) using Slater's Rules

According to Slater's rules for \(2p\) electrons, the screening constant can be determined as follows: \(S=0.85\times\)(number of electrons in the same shell with principal quantum number \(n-1\)) + \(0.35\times\)(number of electrons in the same shell) For \(\mathrm{F}^{-}\): \(S=0.85\times(2) + 0.35\times(6) = 1.7 + 2.1 = 3.8\) \(Z_{\text {eff }}=9-3.8=5.2\) for a \(2p\) electron in \(\mathrm{F}^{-}\) using Slater's rules. For \(\mathrm{Na}^{+}\): \(S=0.85\times(2) + 0.35\times(8) = 1.7 + 2.8 = 4.5\) \(Z_{\text {eff }}=11-4.5=6.5\) for a \(2p\) electron in \(\mathrm{Na}^{+}\) using Slater's rules.
04

(d) Relationship between Effective Nuclear Charge and Ionic Radius

For isoelectronic ions, as the effective nuclear charge \(Z_{\text {eff }}\) increases, the attraction between the nucleus and the electrons becomes stronger. This results in the electron cloud being pulled in more tightly, reducing the ionic radius. Therefore, the ionic radius and effective nuclear charge are inversely related for isoelectronic ions.

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