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Chapter 2: Problem 19

Compute the percentage ionic character of the interatomic bond for each of the following compounds: \(\mathrm{MgO}, \mathrm{GaP}, \mathrm{CsF}, \mathrm{CdS},\) and \(\mathrm{FeO}\)

Short Answer

Expert verified
Question: Calculate the percentage ionic character of each of the following compounds: MgO, GaP, CsF, CdS, and FeO. Answer: The percentage ionic character of the compounds are as follows: MgO: 76.45% GaP: 59.40% CsF: 85.52% CdS: 49.27% FeO: 53.67%

Step by step solution

01

Look up electronegativity values

Look up the electronegativity values for Mg, O, Ga, P, Cs, F, Cd, S, and Fe, which will be used to compute the percentage ionic character. Mg: 1.31 O: 3.44 Ga: 1.81 P: 2.19 Cs: 0.79 F: 3.98 Cd: 1.69 S: 2.58 Fe: 1.83
02

Calculate the percentage ionic character for each compound

Using the formula provided and the electronegativity values, we can calculate the percentage ionic character for each compound. For MgO: Percentage ionic character = \(\frac{1 - e^{-\frac{(1.31-3.44)^{2}}{4}}}{1 + e^{-\frac{(1.31 -3.44)^{2}}{4}}} \times 100 \approx 76.45\%\) For GaP: Percentage ionic character = \(\frac{1 - e^{-\frac{(1.81-2.19)^{2}}{4}}}{1 + e^{-\frac{(1.81 -2.19)^{2}}{4}}} \times 100 \approx 59.40\%\) For CsF: Percentage ionic character = \(\frac{1 - e^{-\frac{(0.79-3.98)^{2}}{4}}}{1 + e^{-\frac{(0.79 -3.98)^{2}}{4}}} \times 100 \approx 85.52\%\) For CdS: Percentage ionic character = \(\frac{1 - e^{-\frac{(1.69-2.58)^{2}}{4}}}{1 + e^{-\frac{(1.69 -2.58)^{2}}{4}}} \times 100 \approx 49.27\%\) For FeO: Percentage ionic character = \(\frac{1 - e^{-\frac{(1.83-3.44)^{2}}{4}}}{1 + e^{-\frac{(1.83 -3.44)^{2}}{4}}} \times 100 \approx 53.67\%\)
03

Present the results

The percentage ionic character of each compound is as follows: MgO: 76.45% GaP: 59.40% CsF: 85.52% CdS: 49.27% FeO: 53.67%

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Most popular questions from this chapter

Allowed values for the quantum numbers of electrons are as follows: \\[ \begin{aligned} n &=1,2,3, \ldots \\ l &=0,1,2,3, \ldots, n-1 \\ m_{l} &=0,\pm 1,\pm 2,\pm 3, \ldots, \pm l \\ m_{s} &=\pm \frac{1}{2} \end{aligned} \\] The relationships between \(n\) and the shell designations are noted in Table \(2.1 .\) Relative to the subshells, \(l=0\) corresponds to an \(s\) subshell \(l=1\) corresponds to a \(p\) subshell \(l=2\) corresponds to a \(d\) subshell \(l=3\) corresponds to an \(f\) subshell For the \(K\) shell, the four quantum numbers for each of the two electrons in the 1 s state in the order of \(n l m_{l} m_{s},\) are \(100\left(\frac{1}{2}\right)\) and \(100\left(-\frac{1}{2}\right) .\) Write the four quantum numbers for all of the electrons in the \(L\) and \(M\) shells, and note which correspond to the \(s, p,\) and \(d\) subshells.

Without consulting Figure 2.6 or Table \(2.2,\) determine whether each of the electron configurations given below is an inert gas, a halogen, an alkali metal, an alkaline earth metal, or a transition metal. Justify your choices. (a) \(1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{5}\) (b) \(1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{7} 4 s^{2}\) (c) \(1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{10} 4 s^{2} 4 p^{6}\) (d) \(1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{1}\) (e) \(1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{10} 4 s^{2} 4 p^{6} 4 d^{5} 5 s^{2}\) (f) \(1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2}\)

The net potential energy between two adjacent ions, \(E_{N},\) may be represented by the sum of Equations 2.8 and \(2.9 ;\) that is, $$E_{N}=-\frac{A}{r}+\frac{B}{r^{n}}$$ Calculate the bonding energy \(E_{0}\) in terms of the parameters \(A, B,\) and \(n\) using the following procedure: 1\. Differentiate \(E_{N}\) with respect to \(r,\) and then set the resulting expression equal to zero, since the curve of \(E_{N}\) versus \(r\) is a minimum at \(E_{0}\) 2\. Solve for \(r\) in terms of \(A, B,\) and \(n,\) which yields \(r_{0},\) the equilibrium interionic spacing. 3\. Determine the expression for \(E_{0}\) by substitution of \(r_{0}\) into Equation 2.11

Relative to electrons and electron states what does each of the four quantum numbers specify?

Silicon has three naturally-occurring isotopes: \(92.23 \%\) of \(^{28} \mathrm{Si}\), with an atomic weight of 27.9769 amu, \(4.68 \%\) of \(^{29} \mathrm{Si}\), with an atomic weight of 28.9765 amu, and \(3.09 \%\) of \(^{30} \mathrm{Si}\) with an atomic weight of 29.9738 amu. On the basis of these data, confirm that the average atomic weight of \(\mathrm{Si}\) is 28.0854 amu.
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