Compute the percents ionic character of the interatomic bonds for the following compounds: TiO2, ZnTe, CsCl, InSb, and MgCl2.
The Correct Answer and Explanation is :
To calculate the percent ionic character of a bond, we use the equation: % Ionic Character=(1−e(−(χA−χB)24))×100\% \text{ Ionic Character} = \left(1 – e^{\left(\frac{- ( \chi_A – \chi_B )^2}{4}\right)}\right) \times 100
where:
- χA\chi_A and χB\chi_B are the electronegativities of the atoms in the bond.
- The exponent accounts for the difference in electronegativity squared, divided by 4.
Electronegativity Values (Pauling Scale):
- Ti (Titanium): 1.54
- O (Oxygen): 3.44
- Zn (Zinc): 1.65
- Te (Tellurium): 2.10
- Cs (Cesium): 0.79
- Cl (Chlorine): 3.16
- In (Indium): 1.78
- Sb (Antimony): 2.05
- Mg (Magnesium): 1.31
Calculations:
- TiO₂:
χTi=1.54,χO=3.44\chi_{\text{Ti}} = 1.54, \chi_{\text{O}} = 3.44
Δχ=3.44−1.54=1.90\Delta\chi = 3.44 – 1.54 = 1.90
% Ionic Character=(1−e(−(1.90)24))×100=55.0%\% \text{ Ionic Character} = \left(1 – e^{\left(\frac{- (1.90)^2}{4}\right)}\right) \times 100 = 55.0\% - ZnTe:
χZn=1.65,χTe=2.10\chi_{\text{Zn}} = 1.65, \chi_{\text{Te}} = 2.10
Δχ=2.10−1.65=0.45\Delta\chi = 2.10 – 1.65 = 0.45
% Ionic Character=(1−e(−(0.45)24))×100=7.6%\% \text{ Ionic Character} = \left(1 – e^{\left(\frac{- (0.45)^2}{4}\right)}\right) \times 100 = 7.6\% - CsCl:
χCs=0.79,χCl=3.16\chi_{\text{Cs}} = 0.79, \chi_{\text{Cl}} = 3.16
Δχ=3.16−0.79=2.37\Delta\chi = 3.16 – 0.79 = 2.37
% Ionic Character=(1−e(−(2.37)24))×100=67.5%\% \text{ Ionic Character} = \left(1 – e^{\left(\frac{- (2.37)^2}{4}\right)}\right) \times 100 = 67.5\% - InSb:
χIn=1.78,χSb=2.05\chi_{\text{In}} = 1.78, \chi_{\text{Sb}} = 2.05
Δχ=2.05−1.78=0.27\Delta\chi = 2.05 – 1.78 = 0.27
% Ionic Character=(1−e(−(0.27)24))×100=3.7%\% \text{ Ionic Character} = \left(1 – e^{\left(\frac{- (0.27)^2}{4}\right)}\right) \times 100 = 3.7\% - MgCl₂:
χMg=1.31,χCl=3.16\chi_{\text{Mg}} = 1.31, \chi_{\text{Cl}} = 3.16
Δχ=3.16−1.31=1.85\Delta\chi = 3.16 – 1.31 = 1.85
% Ionic Character=(1−e(−(1.85)24))×100=53.2%\% \text{ Ionic Character} = \left(1 – e^{\left(\frac{- (1.85)^2}{4}\right)}\right) \times 100 = 53.2\%
Explanation:
The percent ionic character of a bond depends on the electronegativity difference between the two atoms. A higher difference results in a more ionic bond, while a smaller difference results in a more covalent bond.
- CsCl has the highest ionic character (67.5%) because it consists of a highly electropositive metal (Cs) and a highly electronegative non-metal (Cl), leading to a strong electrostatic attraction.
- TiO₂ and MgCl₂ have moderately high ionic characters (~55%) since both involve a metal and a highly electronegative element (oxygen or chlorine).
- ZnTe and InSb have the lowest ionic characters (~3.7%–7.6%), meaning their bonds are mostly covalent.
This calculation is important in materials science to determine the physical and electrical properties of compounds. Highly ionic compounds (e.g., CsCl, MgCl₂) tend to have high melting points and solubility in water, while covalent compounds (e.g., InSb) are often semiconductors due to their lower ionic character.