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Chapter 8: Problem 109

When molten sulfur reacts with chlorine gas, a vile-smelling orange liquid forms that has an empirical formula of SCl. The structure of this compound has a formal charge of zero on all elements in the compound. Draw the Lewis structure for the vile-smelling orange liquid.

Short Answer

Expert verified
The final Lewis structure for the vile-smelling orange liquid with the empirical formula SCl and a formal charge of zero on all elements is: S = Cl : :

Step by step solution

01

Determine the Number of Valence Electrons

Sulfur (S) is in group 16 and has 6 valence electrons, while Chlorine (Cl) is in group 17 and has 7 valence electrons. As the empirical formula is SCl, we have one atom of Sulfur and one atom of Chlorine in our molecule. The total number of valence electrons for the compound is: Total Valence Electrons = (S_valence_electrons) + (Cl_valence_electrons) = 6 + 7 = 13
02

Create the Initial Structure

We start by placing the least electronegative atom, sulfur (S), in the center and put the other atom (chlorine) next to it. We place a single bond between sulfur and chlorine, representing two shared electrons. S - Cl
03

Distribute Remaining Valence Electrons

We have already used two valence electrons for the single bond between S and Cl. Hence, we need to distribute the remaining 11 electrons (13 - 2 = 11). Start with the outer atom (Cl) by adding electron pairs until it achieves an octet. S - Cl : : : Sulfur (S) also needs to achieving its octet. However, adding an electron pair to sulfur will lead to more than eight electrons in its outer shell. To get around this, we can create a double bond with chlorine.
04

Create the Double Bond

To achieve an octet for both atoms without violating the octet rule for sulfur, we can move one electron pair from chlorine to form a double bond with sulfur. This will give eight electrons to both sulfur and chlorine. S = Cl : :
05

Check the Formal Charges and Complete the Lewis Structure

Now, we check the formal charges of each atom. The formal charge for sulfur (S) and chlorine (Cl) is: Formal_charge_S = [Valence_electrons_S] - [Non_bonding_electrons_S] - [0.5*Bonding_electrons_S] = 6 - 0 - (0.5*4) = 0 Formal_charge_Cl = [Valence_electrons_Cl] - [Non_bonding_electrons_Cl] - [0.5*Bonding_electrons_Cl] = 7 - 4 - (0.5*4) = 0 As both formal charges equal to zero, we have achieved the desired Lewis structure with a formal charge of zero on all elements. Final Lewis Structure: S = Cl : :

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Most popular questions from this chapter

Order the following species with respect to carbon–oxygen bond length (longest to shortest). $$\mathrm{CO}, \quad \mathrm{CO}_{2}, \quad \mathrm{CO}_{3}^{2-}, \quad \mathrm{CH}_{3} \mathrm{OH}$$ What is the order from the weakest to the strongest carbon–oxygen bond? $\left(\mathrm{CH}_{3} \mathrm{OH} \text { exists as } \mathrm{H}_{3} \mathrm{C}-\mathrm{OH} .\right)$

For each of the following groups, place the atoms and/or ions in order of decreasing size. a. \(\mathrm{V}, \mathrm{V}^{2+}, \mathrm{V}^{3+}, \mathrm{V}^{5+}\) b. \(\mathrm{Na}^{+}, \mathrm{K}^{+}, \mathrm{Rb}^{+}, \mathrm{Cs}^{+}\) c. \(\mathrm{Te}^{2-}, \mathrm{I}^{-}, \mathrm{Cs}^{+}, \mathrm{Ba}^{2+}\) d. \(\mathrm{P}, \mathrm{P}^{-}, \mathrm{P}^{2-}, \mathrm{P}^{3-}\) e. \(\mathrm{O}^{2-}, \mathrm{S}^{2-}, \mathrm{Se}^{2-}, \mathrm{Te}^{2-}\)

Think of forming an ionic compound as three steps (this is a simplification, as with all models): (1) removing an electron from the metal; (2) adding an electron to the nonmetal; and (3) allowing the metal cation and nonmetal anion to come together. a. What is the sign of the energy change for each of these three processes? b. In general, what is the sign of the sum of the first two processes? Use examples to support your answer. c. What must be the sign of the sum of the three processes? d. Given your answer to part c, why do ionic bonds occur? e. Given your above explanations, why is NaCl stable but not $\mathrm{Na}_{2} \mathrm{Cl} ? \mathrm{NaCl}_{2} ?$ What about MgO compared to \(\mathrm{MgO}_{2} ? \mathrm{Mg}_{2} \mathrm{O} ?\)

Use bond energies to estimate \(\Delta H\) for the combustion of one mole of acetylene: $$\mathrm{C}_{2} \mathrm{H}_{2}(g)+\frac{5}{2} \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)$$

Which of the following incorrectly shows the bond polarity? Show the correct bond polarity for those that are incorrect. a. \(^{\delta+} \mathrm{H}-\mathrm{F}^{\delta-} \quad\) d. $\delta^{+} \mathrm{Br}-\mathrm{Br}^{\delta-}$ b. \(^{\delta+} \mathrm{Cl}-\mathrm{I}^{\delta-} \qquad\) e. \(\quad\) e. $\quad ^{\delta+}\mathrm{O}-\mathrm{P}^{\delta-}$ c. \(\quad \delta+\mathrm{Si}-\mathrm{S}^{\delta-}\)
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