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Chapter 9: Problem 40

Classify the following bonds as ionic, polar covalent, or covalent, and give your reasons: (a) the \(\mathrm{SiSi}\) bond in \(\mathrm{Cl}_{3} \mathrm{SiSiCl}_{3},\) (b) the \(\mathrm{SiCl}\) bond in \(\mathrm{Cl}_{3} \mathrm{SiSiCl}_{3}\), (c) the CaF bond in \(\mathrm{CaF}_{2}\), (d) the \(\mathrm{NH}\) bond in \(\mathrm{NH}_{3}\).

Short Answer

Expert verified
a) Si-Si bond is covalent. b) SiCl bond is polar covalent. c) CaF bond is ionic. d) NH bond is polar covalent.

Step by step solution

01

Determine the Si-Si bond

The Si-Si bond is between two identical atoms, therefore, the difference in electronegativity is zero, which indicates a covalent bond. Bonds between identical atoms are always covalent as the electron pair is shared equally.
02

Determine the Si-Cl bond

Silicon (Si) has an electronegativity value of 1.90 and Chlorine (Cl) has an electronegativity value of 3.16. The difference in electronegativity is 1.26. Since the difference is more than 0.5 but less than 1.7, this indicates a polar covalent bond. Polar covalent bonds are formed when the electron sharing between atoms is unequal.
03

Determine the Ca-F bond

Calcium (Ca) has an electronegativity value of 1.00 and Fluorine (F) has an electronegativity value of 3.98. The difference in electronegativity is 2.98, which is greater than 1.7. This suggests an ionic bond. Ionic bonds are formed when there is a complete transfer of electron(s) from a metal to a non-metal.
04

Determine the N-H bond

Nitrogen (N) has an electronegativity value of 3.04 and Hydrogen (H) has an electronegativity value of 2.20. The difference of 0.84 suggests a polar covalent bond.

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Most popular questions from this chapter

Write Lewis structures for the following molecules and ions: (a) \(\mathrm{OF}_{2},\) (b) \(\mathrm{N}_{2} \mathrm{~F}_{2},\) (c) \(\mathrm{Si}_{2} \mathrm{H}_{6},\) (d) \(\mathrm{OH}^{-}\), (e) \(\mathrm{CH}_{2} \mathrm{ClCOO}^{-},\) (f) \(\mathrm{CH}_{3} \mathrm{NH}_{3}^{+}\)

Write Lewis structures for the following molecules: (a) ICl, (b) \(\mathrm{PH}_{3}\), (c) \(\mathrm{P}_{4}\) (each \(\mathrm{P}\) is bonded to three other \(\mathrm{P}\) atoms (d) \(\mathrm{H}_{2} \mathrm{~S}\), (e) \(\mathrm{N}_{2} \mathrm{H}_{4}\), (f) \(\mathrm{HClO}_{3}\), (g) \(\mathrm{COBr}_{2}\) (C is bonded to \(\mathrm{O}\) and \(\mathrm{Br}\) atoms).

Four atoms are arbitrarily labeled \(D, \mathrm{E}, \mathrm{F},\) and \(\mathrm{G} .\) Their electronegativities are as follows: \(\mathrm{D}=3.8\), \(\mathrm{E}=3.3, \mathrm{~F}=2.8,\) and \(\mathrm{G}=1.3 .\) If the atoms of these elements form the molecules \(\mathrm{DE}, \mathrm{DG}, \mathrm{EG},\) and \(\mathrm{DF}\) how would you arrange these molecules in order of increasing covalent bond character?

A rule for drawing plausible Lewis structures is that the central atom is invariably less electronegative than the surrounding atoms. Explain why this is so. Why does this rule not apply to compounds like \(\mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{NH}_{3} ?\)

In 1998 scientists using a special type of electron microscope were able to measure the force needed to break a single chemical bond. If \(2.0 \times 10^{29} \mathrm{~N}\) was needed to break a \(\mathrm{C}-\mathrm{Si}\) bond, estimate the bond enthalpy in \(\mathrm{kJ} / \mathrm{mol} .\) Assume that the bond had to be stretched by a distance of 2 A \(\left(2 \times 10^{-10} \mathrm{~m}\right)\) before it is broken.
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