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Chapter 8: Problem 31

Using Lewis symbols and Lewis structures, diagram the formation of \(\mathrm{SiCl}_{4}\) from \(\mathrm{Si}\) and \(\mathrm{Cl}\) atoms.

Short Answer

Expert verified
The Lewis symbols for Si and Cl are: Si has 4 valence electrons and Cl has 7 valence electrons, represented as dots around the chemical symbols. The Lewis structure for SiCl4 is formed by silicon sharing its valence electrons with each of the four chlorine atoms, forming single covalent bonds and satisfying the octet rule for both atoms. The Lewis structure for SiCl4 is: \(\hspace{11mm} \mathrm{Cl} \) \(\hspace{7.5mm} // \backslash\) \(\mathrm{Cl} - \mathrm{Si} - \mathrm{Cl}\) \(\hspace{7.5mm} // \backslash\) \(\hspace{11mm} \mathrm{Cl} \)

Step by step solution

01

Determine the Lewis symbols for Si and Cl

Begin by identifying the valence electrons of silicon (Si) and chlorine (Cl). Silicon has 4 valence electrons and is located in group 14, while chlorine has 7 valence electrons and is located in group 17. Represent these valence electrons as dots around the chemical symbols. The Lewis symbols for Si and Cl are: Si: \(\cdot \hspace{1mm} \mathrm{Si} \hspace{1mm} \cdot\) \: \: \: \: \: \: \: \: \: \: \: \|/ Cl: \(\cdot \hspace{0.5mm} \mathrm{Cl} :\)
02

Combine the atoms to form SiCl4

Now, silicon will form covalent bonds by sharing its valence electrons with each of the four chlorine atoms to satisfy the octet rule for both atoms. To illustrate this bonding, show the electron sharing between the atoms and represent the bonded electrons as lines. The Lewis structure for SiCl4 is: \(\hspace{11mm} \mathrm{Cl} \) \(\hspace{7.5mm} // \backslash\) \(\mathrm{Cl} - \mathrm{Si} - \mathrm{Cl}\) \(\hspace{7.5mm} // \backslash\) \(\hspace{11mm} \mathrm{Cl} \) In this Lewis structure, the silicon atom shares a pair of electrons with each of the four chlorine atoms, forming single covalent bonds. Each chlorine atom has now satisfied the octet rule, having four bonded electrons and three lone pair electrons. The silicon atom also satisfies the octet rule by sharing its four valence electrons with the chlorine atoms.

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Most popular questions from this chapter

(a) What is the octet rule? (b) How many electrons must a sulfur atom gain to achieve an octet in its valence shell? (c) If an atom has the electron configuration \(1 s^{2} 2 s^{2} 2 p^{3}\), how many electrons must it gain to achieve an octet?

Arrange the bonds in each of the following sets in order of increasing polarity: (a) \(\mathrm{C}-\mathrm{F}, \mathrm{O}-\mathrm{F}, \mathrm{Be}-\mathrm{F}\) (b) \(\mathrm{O}-\mathrm{Cl}, \mathrm{S}-\mathrm{Br}, \mathrm{C}-\mathrm{P}\) (c) \(\mathrm{C}-\mathrm{S}, \mathrm{B}-\mathrm{F}, \mathrm{N}-\mathrm{O}\)

(a) Use bond enthalpies to estimate the enthalpy change for the reaction of hydrogen with ethene: $$\mathrm{H}_{2}(g)+\mathrm{C}_{2} \mathrm{H}_{4}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)$$ (b) Calculate the standard enthalpy change for this reaction, using heats of formation. Why does this value differ from that calculated in (a)?

To address energy and environmental issues, there is great interest in powering vehicles with hydrogen rather than gasoline. One of the most attractive aspects of the "hydrogen economy" is the fact that in principle the only emission would be water. However, two daunting obstacles must be overcome before this vision can become a reality. First, an economical method of producing hydrogen must be found. Second, a safe, lightweight, and compact way of storing hydrogen must be found. The hydrides of light metals are attractive for hydrogen storage because they can store a high weight percentage of hydrogen in a small volume. One of the most attractive hydrides is \(\mathrm{NaAlH}_{4}\), which can release \(5.6 \%\) of its mass as \(\mathrm{H}_{2}\) upon decomposing to \(\mathrm{NaH}(\mathrm{s})\), Al(s), and \(\mathrm{H}_{2}(\mathrm{~g})\). NaAlH \(_{4}\) possesses both covalent bonds, which hold polyatomic anions together, and ionic bonds. (a) Write a balanced equation for the decomposition of \(\mathrm{NaAlH}_{4} .\) (b) Which element in \(\mathrm{NaAlH}_{4}\) is the most electronegative? Which one is the least electronegative? (c) Based on electronegativity differences, what do you think is the identity of the polyatomic anion? Draw a Lewis structure for this ion.

Which ionic compound is expected to form from combining the following pairs of elements: (a) barium and fluorine, (b) cesium and chlorine, (c) lithium and nitrogen, (d) aluminum and oxygen?
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