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Chapter 8: Problem 133

Which member of the following pairs would you expect to be more energetically stable? Justify each choice. a. \(\mathrm{NaBr}\) or \(\mathrm{NaBr}_{2}\) b. \(\mathrm{ClO}_{4}\) or \(\mathrm{ClO}_{4}\) c. \(\mathrm{SO}_{4}\) or \(\mathrm{XeO}_{4}\) d. \(\mathrm{OF}_{4}\) or \(\mathrm{SeF}_{4}\)

Short Answer

Expert verified
The more energetically stable compounds are: a. NaBr, as NaBr2 doesn't form a balanced chemical formula. b. ClO4 is the same compound in both cases, so there is no comparison. c. SO4 is more stable due to resonance, whereas XeO4 is less stable because noble gas compounds are generally less stable. d. OF4 is more energetically stable than SeF4 due to less electron-electron repulsion.

Step by step solution

01

Determine the formulas possible

Sodium (Na) is an alkali metal that forms a +1 cation and bromine (Br) is a halogen that forms a -1 anion. A balanced chemical formula comprises ions that result in equal positive and negative charges.
02

Compare and choose the stable compound

Since Na has a +1 charge and Br has a -1 charge, they combine in a 1:1 ratio to form NaBr. NaBr2 is not a possible compound due to an excess of Br.
03

Justification

Hence, NaBr is more energetically stable than NaBr2 as NaBr2 doesn't form a balanced chemical formula. #Step 2: Comparing ClO4##
04

Identifying the compounds

There is an error in the exercise. Both given compounds are the same, ClO4.
05

Justification

Since both are the same compounds, there is no comparison to be made for stability. #Step 3: Comparing SO4 and XeO4#
06

Determine the charge on the compounds

SO4 is the sulfate ion with a 2- charge. XeO4 is xenon tetroxide, which is a neutral compound.
07

Compare the bonding and stability

SO4 has resonance structures, which stabilize the negative charge and hence add to its stability. XeO4 involves covalent bonds between Xenon, a noble gas, and oxygen.
08

Justification

Sulfate ion (SO4) is more energetically stable due to resonance, whereas XeO4 is less stable due to the instability of noble gas compounds. #Step 4: Comparing OF4 and SeF4#
09

Identify the total number of valence electrons in both compounds

In OF4, oxygen has 6 valence electrons and each fluorine atom has 7 valence electrons. There are a total of 6 + 4 × 7 = 34 valence electrons in this compound. In SeF4, selenium has 6 valence electrons and each fluorine atom also has 7 valence electrons. There are a total of 6 + 4 × 7 = 34 valence electrons in this compound as well.
10

Determine the bonding structure

In OF4, oxygen forms 4 single bonds with fluorine atoms, while in SeF4, selenium forms 4 single bonds with fluorine atoms and has one lone pair on itself.
11

Compare the electron pair repulsion

In OF4, there are 4 bond pairs and no lone pairs, whereas in SeF4, there are 4 bond pairs and 1 lone pair. Lone pairs repel more than bond pairs, and hence greater repulsion is present in SeF4 than in OF4.
12

Justification

As there is greater electron-electron repulsion in SeF4, OF4 is more energetically stable than SeF4.

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Most popular questions from this chapter

What noble gas has the same election configuration as each of the ions in the following compounds? a. cesium sulfide b. strontium fluoride c. calcium nitride d. aluminum bromide

For each of the following, write an equation that corresponds to the energy given. a. lattice energy of \(\mathrm{NaCl}\) b. lattice energy of \(\mathrm{NH}_{4} \mathrm{Br}\) c. lattice energy of \(\mathrm{MgS}\) d. \(\mathrm{O}=\mathrm{O}\) double bond energy beginning with \(\mathrm{O}_{2}(g)\) as a reactant

Use the following data to estimate \(\Delta H\) for the reaction \(\mathrm{S}^{-}(g)+\) \(\mathrm{e}^{-} \rightarrow \mathrm{S}^{2-}(g)\). Include an estimate of uncertainty. $$ \begin{array}{|lcccc|} \hline & & \text { Lattice } & & \Delta H_{\text {sub }} \\ & \Delta \boldsymbol{H}_{\mathrm{t}}^{\circ} & \text { Energy } & \text { I.E. of } \mathbf{M} & \text { of M } \\ \hline \mathrm{Na}_{2} \mathrm{~S} & -365 & -2203 & 495 & 109 \\ \mathrm{~K}_{2} \mathrm{~S} & -381 & -2052 & 419 & 90 \\ \mathrm{Rb}_{2} \mathrm{~S} & -361 & -1949 & 409 & 82 \\ \mathrm{Cs}_{2} \mathrm{~S} & -360 & -1850 & 382 & 78 \\ \hline \end{array} $$ $$ \begin{aligned} \mathrm{S}(s) & \longrightarrow \mathrm{S}(g) & \Delta H &=227 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{S}(g)+\mathrm{e}^{-} & \longrightarrow \mathrm{S}^{-}(g) & \Delta H &=-200 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$ Assume that all values are known to \(\pm 1 \mathrm{~kJ} / \mathrm{mol}\).

An alternative definition of electronegativity is Electronegativity \(=\) constant (I.E. \(-\) E.A.) where I.E. is the ionization energy and E.A. is the electron affinity using the sign conventions of this book. Use data in Chapter 7 to calculate the (I.E. \(-\) E.A.) term for \(\mathrm{F}, \mathrm{Cl}, \mathrm{Br}\), and \(\mathrm{I}\). Do these values show the same trend as the electronegativity values given in this chapter? The first ionization energies of the halogens are \(1678,1255,1138\), and \(1007 \mathrm{~kJ} / \mathrm{mol}\), respectively. (Hint: Choose a constant so that the electronegativity of fluorine equals 4.0. Using this constant, calculate relative electronegativities for the other halogens and compare to values given in the text.)

For each of the following groups, place the atoms and/or ions in order of decreasing size. a. \(\mathrm{Cu}, \mathrm{Cu}^{+}, \mathrm{Cu}^{2+}\) b. \(\mathrm{Ni}^{2+}, \mathrm{Pd}^{2+}, \mathrm{Pt}^{2+}\) c. \(\mathrm{O}, \mathrm{O}^{-}, \mathrm{O}^{2-}\) d. \(\mathrm{La}^{3+}, \mathrm{Eu}^{3+}, \mathrm{Gd}^{3+}, \mathrm{Yb}^{3+}\) e. \(\mathrm{Te}^{2-}, \mathrm{I}^{-}, \mathrm{Cs}^{+}, \mathrm{Ba}^{2+}, \mathrm{La}^{3+}\)
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