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

Short Answer

Expert verified
For the given molecules, we have the following Lewis structures, oxidation numbers, and formal charges: a) OCS: \(\mathrm{O=C=S}\), oxidation numbers: O (-2), C (+2), S (-1), formal charges: O (0), C (0), S (0). b) SOCl2: \(\mathrm{S=O-Cl-Cl}\), oxidation numbers: O (-2), S (+6), Cl (-1), formal charges: O (0), S (+4), Cl (0). c) BrO3-: \(\mathrm{Br-O-O=O=O}\), oxidation numbers: Br (+5), O (-2), formal charges: Br (+1), O (0). d) HClO2: \(\mathrm{H-O-Cl=O=O}\), oxidation numbers: O (-2), H (+1), Cl (+3), formal charges: O (0), H (0), Cl (+2).

Step by step solution

01

a) OCS

Step 1: Total number of valence electrons For OCS, the total number of valence electrons is \(6 + 4 + 6 = 16\). Step 2: Choose the central atom and draw the skeleton structure C has the lowest electronegativity, so it is the central atom. Hence, the skeleton structure is O=C=S. Step 3: Assign electrons around the atoms by following the octet rule Now, we complete the octet for each atom: O=C=S becomes: \(\mathrm{O=C=S}\) (10 valence electrons). Step 4: Calculate oxidation numbers and formal charges The oxidation numbers for O, C, and S are -2, +2, and -1, respectively. Formal charges for O, C, and S are 0, 0, and 0, respectively.
02

b) SOCl2

Step 1: Total number of valence electrons For SOCl2, the total number of valence electrons is \(6 + 6 + 7 \cdot 2 = 26\). Step 2: Choose the central atom and draw the skeleton structure S has the lowest electronegativity, so it is the central atom. Hence, the skeleton structure is S with two Cl atoms and one O atom bonded to it. Step 3: Assign electrons around the atoms by following the octet rule Now, we complete the octet for each atom: S-O-Cl -> S=O-Cl-Cl (26 valence electrons). Step 4: Calculate oxidation numbers and formal charges The oxidation numbers for O, S, and Cl are -2, +6, and -1 (each Cl), respectively. Formal charges for O, S, and Cl are 0, +4, and 0 (each Cl), respectively.
03

c) BrO3-

Step 1: Total number of valence electrons For BrO3-, the total number of valence electrons is \(7 + 6 \cdot 3 +1=26\). Step 2: Choose the central atom and draw the skeleton structure Br has the lowest electronegativity, so it is the central atom. Hence, the skeleton structure is Br with three O atoms bonded to it. Step 3: Assign electrons around the atoms by following the octet rule Now, we complete the octet for each atom: Br-O-O-O -> Br-O-O=O=O (26 valence electrons). Step 4: Calculate oxidation numbers and formal charges The oxidation numbers for Br and O are +5 and -2 (each O), respectively. Formal charges for Br and O are +1 and 0 (each O), respectively.
04

d) HClO2

Step 1: Total number of valence electrons For HClO2, the total number of valence electrons is \(1 + 7 + 6 \cdot 2=20\). Step 2: Choose the central atom and draw the skeleton structure Cl has the lowest electronegativity, so it is the central atom. The H is bonded to O. Hence, the skeleton structure is Cl with two O atoms bonded to it and one O bonded to H. Step 3: Assign electrons around the atoms by following the octet rule Now, we complete the octet for each atom: H-O-Cl-O -> H-O-Cl=O=O (20 valence electrons). Step 4: Calculate oxidation numbers and formal charges The oxidation numbers for O, H, and Cl are -2, +1, and +3, respectively. Formal charges for O, H, and Cl are 0, 0, and +2, respectively.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding the Octet Rule
The octet rule is a fundamental principle that helps explain the stability of atoms when they form chemical compounds. According to this rule, atoms are most stable when they have eight electrons in their valence shell, which is the outermost electron shell. This emulates the electron configuration of noble gases, known for their lack of reactivity due to having complete valence shells.

When atoms form compounds, they tend to share, donate, or accept electrons until they reach a stable octet. However, there are exceptions to this rule. For instance, hydrogen and helium are stable with two valence electrons, and some elements can have more than eight electrons due to expanded valence shells.

To illustrate the octet rule, let's examine the compound OCS. Here, oxygen (O) and sulfur (S) each strive to complete their octet. Carbon (C), being the central atom with the lowest electronegativity, forms double bonds with oxygen and sulfur, allowing all atoms to satisfy the octet rule. By completing their octets, these atoms achieve a lower energy state and, consequently, greater stability.
Navigating Oxidation Numbers
Understanding oxidation numbers is crucial for analyzing redox reactions and determining the electron distribution in compounds. An oxidation number is a hypothetical charge an atom would have if the compound was composed entirely of ions. This number indicates the degree of oxidation (loss of electrons) or reduction (gain of electrons) an atom undergoes during a reaction.

In most cases, the oxidation number for a pure element is zero. For ions, it is equal to the charge of the ion. For covalent compounds, there are specific rules, for example, hydrogen usually has an oxidation number of +1, and oxygen typically has -2, with specific exceptions. In compounds like SOCl2, we can determine that sulfur (S) has a +6 oxidation number due to oxygen's known oxidation state and chlorine's typical -1 state. This calculation helps establish the electron count for each atom and predicts how atoms will react during chemical processes.
The Significance of Formal Charges
Formal charge is a concept used to estimate the distribution of electrons among atoms in a molecule. It is the hypothetical charge you would assign to an atom in a molecule, assuming that electrons in all chemical bonds are shared equally between atoms, regardless of relative electronegativity. When determining the most probable Lewis structure for a molecule, formal charges are instrumental because the structure with the lowest possible formal charges on each atom is often the most stable.

To calculate an atom's formal charge, use the formula: formal charge = (valence electrons) - (non-bonding electrons) - (half the bonding electrons). For example, in OCS, the formal charge of carbon and sulfur is zero when considering the number of valence electrons and their shared bonding in the structure. A molecule like HClO2 has formal charges that guide the placement of electrons and ensure the most stable structure with respect to the octet rule. Understanding how to compute and apply formal charges facilitates a better comprehension of molecular geometry and reactivity patterns.

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Most popular questions from this chapter

One scale for electronegativity is based on the concept that the electronegativity of any atom is proportional to the ionization energy of the atom minus its electron affinity: electronegativity \(=k(I-E A)\), where \(k\) is a proportionality constant. (a) How does this definition explain why the electronegativity of \(\mathrm{F}\) is greater than that of \(\mathrm{Cl}\) even though \(\mathrm{Cl}\) has the greater electron affinity? (b) Why are both ionization energy and electron affinity relevant to the notion of electronegativity? (c) By using data in Chapter 7 , determine the value of \(k\) that would lead to an electronegativity of \(4.0\) for \(F\) under this definition. (d) Use your result from part (c) to determine the electronegativities of \(\mathrm{Cl}\) and \(\mathrm{O}\) using this scale. (e) Another scale for electronegativity defines electronegativity as the average of an atom's first ionization energy and its electron affinity. Using this scale, calculate the electronegativities for the halogens, and scale them so fluorine has an electronegativity of 4.0. On this scale, what is Br's electronegativity?

(a) Triazine, \(\mathrm{C}_{3} \mathrm{H}_{3} \mathrm{~N}_{3}\), is like benzene except that in triazine every other \(\mathrm{C}-\mathrm{H}\) group is replaced by a nitrogen atom. Draw the Lewis structure(s) for the triazine molecule. (b) Estimate the carbon-nitrogen bond distances in the ring.

Draw the dominant Lewis structures for these chlorine-oxygen molecules/ions: \(\mathrm{ClO}, \mathrm{ClO}^{-}, \mathrm{ClO}_{2}^{-}, \mathrm{ClO}_{3}^{-}, \mathrm{ClO}_{4}^{-} .\)Which of these do not obey the octet rule?

Write electron configurations for the following ions, and determine which have noble-gas configurations: (a) \(\mathrm{Cd}^{2+}\), (b) \(\mathrm{P}^{3-}\), (c) \(\mathrm{Zr}^{4+}\), (d) \(\mathrm{Ru}^{3+}\), (e) \(\mathrm{As}^{3-}\), (f) \(\mathrm{Ag}^{+}\).

(a) Consider the lattice energies for the following compounds: \(\mathrm{BeH}_{2}, 3205 \mathrm{~kJ} / \mathrm{mol} ; \mathrm{MgH}_{2}, 2791 \mathrm{~kJ} / \mathrm{mol} ; \mathrm{CaH}_{2}, 2410 \mathrm{~kJ} / \mathrm{mol} ;\) \(\mathrm{SrH}_{2}, 2250 \mathrm{~kJ} / \mathrm{mol} ; \mathrm{BaH}_{2}, 2121 \mathrm{~kJ} / \mathrm{mol}\). Plot lattice energy versus cation radius for these compounds. If you draw a line through your points, is the slope negative or positive? Explain. (b) The lattice energy of \(\mathrm{ZnH}_{2}\) is \(2870 \mathrm{~kJ} / \mathrm{mol}\). Based on the data given in part (a), the radius of the \(\mathrm{Zn}^{2+}\) ion is expected to be closest to that of which group \(2 \mathrm{~A}\) element?

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