Problem 64 Draw the Lewis structures for ea... [FREE SOLUTION] (2024)

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Chapter 8: Problem 64

Draw the Lewis structures for each of the following molecules or ions.Identify instances where the octet rule is not obeyed; state which atom ineach compound does not follow the octet rule; and state how many electronssurround 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.

Short Answer

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In summary:- NO does not obey the octet rule as nitrogen is surrounded by six electrons.- BF3 does not obey the octet rule as boron is surrounded by six electrons.- ICl2- does not obey the octet rule as iodine is surrounded by ten electrons.- OPBr3 does not obey the octet rule as phosphorus is surrounded by ten electrons.- XeF does not obey the octet rule as xenon is surrounded by ten electrons.

Step by step solution

(a) Drawing Lewis Structure for NO

To draw the Lewis structure for the NO molecule, find the total number of valence electrons. Nitrogen has five valence electrons, and oxygen has six. The total count is 11 valence electrons. Next, place nitrogen and oxygen atoms next to each other since nitrogen has fewer lone pairs. Connect them with a single bond and distribute remaining electrons as lone pairs.NO Lewis Structure: O=N:\( \,\,\,\)The nitrogen atom is surrounded by six electrons, violating the octet rule. Therefore, NO does not obey the octet rule.

(b) Drawing Lewis Structure for BF3

Boron has three valence electrons, and each fluorine atom has seven valence electrons. Overall, there are 24 valence electrons. Place boron at the center and surround it with three fluorine atoms. Connect boron and each fluorine atom with single bonds and distribute the remaining electrons.BF3 Lewis Structure: F |B - F | F The boron atom is surrounded by six electrons, violating the octet rule. Therefore, BF3 does not obey the octet rule.

(c) Drawing Lewis Structure for ICl2-

Iodine has seven valence electrons, chlorine has seven valence electrons, and there is one extra electron from the negative charge, so the total count is 22 valence electrons. Place iodine at the center and surround it with two chlorine atoms. Connect iodine and each chlorine atom with single bonds and distribute the remaining electrons as lone pairs.ICl2- Lewis Structure: Cl |I - Cl ｜ -The iodine atom is surrounded by ten electrons, violating the octet rule. Therefore, ICl2- does not obey the octet rule.

(d) Drawing Lewis Structure for OPBr3

Oxygen has six valence electrons, phosphorus has five valence electrons, and each bromine atom has seven valence electrons. Overall, there are 32 valence electrons. Place phosphorus at the center, with oxygen above and three bromine atoms surrounding it. Connect each atom with single bonds and distribute the remaining electrons as lone pairs.OPBr3 Lewis Structure: O ｜Br - P - Br ｜ Br The phosphorus atom is surrounded by ten electrons, violating the octet rule. Therefore, OPBr3 does not obey the octet rule.

(e) Drawing Lewis Structure for XeF

Xenon has eight valence electrons, and fluorine has seven valence electrons. The total count is 15 valence electrons. Place xenon at the center and connect it to the fluorine atom with single bonds and distribute the remaining electrons as lone pairs.XeF Lewis Structure: F ｜ XeThe xenon atom is surrounded by ten electrons, violating the octet rule. Therefore, XeF does not obey the octet rule.

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Most popular questions from this chapter

Ammonia reacts with boron trifluoride to form a stable compound, as we saw inSection 8.7 . (a) Draw the Lewis structure of the ammonia-boron trifluoridereaction product. (b) The B-N bond is obviously more polar than the\(\mathrm{C}-\mathrm{C}\) bond. Draw the charge distribution you expect on the \(\mathrm{B}-\mathrm{N}\) bond within the molecule (using the delta plus anddelta minus symbols mentioned in Section 8.4\()\) . ( ) Boron trichloride alsoreacts with ammonia in a similar way to the trifluoride. Predict whether the \(B-N\) bond in the trichloride reactionproduct would be more or less polar than the \(B-N\) bond in the trifluorideproduct, and justify your reasoning.Write Lewis structures for the following: (a) \(\mathrm{H}_{2} \mathrm{CO}\)(both \(\mathrm{H}\) atoms are bonded to \(\mathrm{C} ),(\mathbf{b})\mathrm{H}_{2} \mathrm{O}_{2},(\mathbf{c}) \mathrm{C}_{2} \mathrm{F}_{6}\)(contains \(\mathrm{a} \mathrm{C}-\mathrm{C}\) bond), \((\mathbf{d})\mathrm{AsO}_{3}^{3-},\) (e) \(\mathrm{H}_{2} \mathrm{SO}_{3}(\mathrm{H}\) isbonded to \(\mathrm{O})\) \((\mathbf{f}) \mathrm{NH}_{2} \mathrm{Cl}\)Which of the following statements about electronegativity is false? (a)Electronegativity is the ability of an atom in a molecule to attract electrondensity toward itself. (b) Electronegativity is the same thing as electronaffinity. (c) The numerical values for electronegativity have no units. (d)Fluorine is the most electronegative element. (e) Cesium is the least electronegative element.Under special conditions, sulfur reacts with anhydrous liquid ammonia to forma binary compound of sulfur and nitrogen. The compound is found to consist of69.6\(\% \mathrm{S}\) and 30.4\(\% \mathrm{N} .\) Measurements of its molecularmass yield a value of 184.3 \(\mathrm{g} / \mathrm{mol}\) . The compoundoccasionally detonates on being struck or when heated rapidly. The sulfur andnitrogen atoms of the molecule are joined in a ring. All the bonds in the ringare of the same length. (a) Calculate the empirical and molecular formulas forthe substance. (b) Write Lewis structures for the molecule, based on theinformation you are given. (Hint: You should find a relatively small number ofdominant Lewis structures.) (c) Predict the bond distances between the atomsin the ring. (Note: The \(S-S\) distance in the \(S_{8}\) ring is 2.05 A.) ( d.)The enthalpy of formation of the compound is estimated to be 480 \(\mathrm{kJ}/ \mathrm{mol}^{-1} . \Delta H_{f}^{9}\) of \(\mathrm{S}(g)\) is 222.8\(\mathrm{kJ} / \mathrm{mol} .\) Estimate the average bond enthalpy in thecompound.Write Lewis structures that obey the octet rule for each of the following, andassign 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)

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