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Chapter 8: Problem 51

Write Lewis structures that obey the octet rule for each of the following, and assign oxidation numbers and formal charges to each atom: (a) OCS, (b) SOCl_ \(_{2}(S\) is the central atom), \((\mathbf{c}) \mathrm{BrO}_{3}^{-},(\mathbf{d}) \mathrm{HClO}_{2}(\mathrm{H}\) is bonded to O)

Short Answer

Expert verified
(a) OCS Lewis Structure: O=C=S Formal Charges: Oxygen: +1, Carbon: +2, Sulfur: +1 Oxidation Numbers: Oxygen: -2, Carbon: +4, Sulfur: -2 (b) SOCl$_2$ Lewis Structure: ``` Cl | O = S-Cl ``` Formal Charges: Oxygen: +1, Sulfur: +2, Chlorine: 0 Oxidation Numbers: Oxygen: -2, Sulfur: +6, Chlorine: -1

Step by step solution

01

1. Total Number of Valence Electrons

Oxygen has 6 valence electrons, carbon has 4, and sulfur has 6. So the total number of valence electrons is 6 + 4 + 6 = 16.
02

2. Skeleton of the Molecule

Arrange the atoms and form skeletal bonds. Carbon is the central atom. Connect the two atoms in order: O-C-S.
03

3. Distribute Valence Electrons

We will now distribute the remaining valence electrons around the non-central oxygen and sulfur atoms: Oxygen has 6 - 2 = 4 electrons (two lone pairs) remaining. Sulfur has 6 - 2 = 4 electrons (two lone pairs) remaining. The Lewis structure for OCS is now: ``` O=C=S ```
04

4. Assign Formal Charges and Oxidation Numbers

To calculate the formal charge for each atom, use the formula: \[Formal\_Charge = Valence\_Electrons - (Bonds + Lone\_Electrons)\] Oxygen: 6 - (1 + 4) = 1 Carbon: 4 - (2 + 0) = 2 Sulfur: 6 - (1 + 4) = 1 Oxidation number: Oxygen: -2 Carbon: +4 Sulfur: -2 Now, let's move on to the second compound. (b) SOCl\(_2\) (S is the central atom)
05

1. Total Number of Valence Electrons

Sulfur has 6 valence electrons, oxygen has 6, and chlorine has 7. As there are two chlorine atoms, the total number of valence electrons is 6 + 6 + 7(2) = 26.
06

2. Skeleton of the Molecule

Sulfur is the central atom. Connect the central S atom to O and the two Cl atoms: S-O and S-Cl.
07

3. Distribute Valence Electrons

Distribute the remaining valence electrons around the non-central oxygen and chlorine atoms: Oxygen has 6 - 2 = 4 electrons (two lone pairs) remaining. Each chlorine atom has 7 - 1 = 6 electrons (three lone pairs) remaining. The Lewis structure for SOCl\(_2\) is now: ``` Cl | O = S-Cl ```
08

4. Assign Formal Charges and Oxidation Numbers

To calculate the formal charge for each atom, use the formula: Oxygen: 6 - (1 + 4) = 1 Sulfur: 6 - (4 + 0) = 2 Both Chlorine atoms: 7 - (1 + 6) = 0 (since the formal charges are the same for both Cl atoms) Oxidation number: Oxygen: -2 Sulfur: +6 Chlorine: -1 Complete the remaining two Lewis structures, formal charges, and oxidation numbers for the other compounds following the same steps.

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Most popular questions from this chapter

Draw the Lewis structures for each of the following molecules or ions. Identify instances where the octet rule is not obeyed; state which atom in each compound does not follow the octet rule; and state how many electrons surround these atoms: (a) \(\mathrm{NO},(\mathbf{b}) \mathrm{BF}_{3},(\mathbf{c}) \mathrm{ICl}_{2}^{-},(\mathbf{d}) \mathrm{OPBr}_{3}(\) the \(\mathrm{P}\) is the central atom), (e) XeF.

Consider the element silicon, Si. (a) Write its electron configuration. (b) How many valence electrons does a silicon atom have? (c) Which subshells hold the valence electrons?

Construct a Born-Haber cycle for the formation of the hypothetical compound NaCl , where the sodium ion has a \(2+\) charge (the second ionization energy for sodium is given in Table 7.2 . (a) How large would the lattice energy need to be for the formation of \(\mathrm{NaCl}_{2}\) to be exothermic? (b) If we were to estimate the lattice energy of \(\mathrm{NaCl}_{2}\) to be roughly equal to that of \(\mathrm{MgCl}_{2}(2326 \mathrm{kJ} / \mathrm{mol}\) from Table 8.1\(),\) what value would you obtain for the standard enthalpy of formation, \(\Delta H_{f}^{\circ},\) of \(\mathrm{NaCl}_{2} ?\)

Consider the ionic compounds KF, NaCl, NaBr, and LiCl. (a) Use ionic radil (Figure 7.8) to estimate the cation-anion distance for each compound. (b) Based on your answer to part (a), arrange these four compounds in order of decreasing lattice energy. (c) Check your predictions in part (b) with the experimental values of lattice energy from Table \(8.1 .\) Are the predictions from ionic radii correct?

(a) What is the trend in electronegativity going from left to right in a row of the periodic table? (b) How do electronegativity values generally vary going down a column in the periodic table? (c) True or false: The most easily ionizable elements are the most electronegative.
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