Chapter 3: Simple Bonding Theory

Chapter 3: Simple Bonding TheoryParas, Pidlaoan, Primero, Quitoriano, Rodriguez, Sanchez, Santos, Sarsagat, Silong, Tolete, Tulod3-3 Draw the resonance structures for the isoelectronic ions NSO- and SNO-, and assign formal charges. Which ion is likely to be more stable?NSO-Large formal charges like the -2 on N and +1 on S on structure a are very unlikely. Thus, structure b is more likely to be stable since it has smaller formal charges.

SNO-Negative formal charges must be on the most electronegative atoms. In SNO-, the most electronegative atom is oxygen. Structure b is better than structure a because the -1 formal charge is found on oxygen.

Since the negative formal charge of SNO- is found only oxygen, it has a better structure compared to NSO- which has formal charges on all its atoms.

3-3 Draw the resonance structures for the isoelectronic ions NSO- and SNO-, and assign formal charges. Which ion is likely to be more stable?3-6 Select from each set the molecule or ion having the smallest bond angle, and briefly explain your choice:NH3, PH3, or AsH3AsH3 has the smallest bond angle because arsenic is the largest central atom w/c minimizes the bond pair-bond pair repulsions. It also has the least electronegativity w/c allows electrons to be drawn out farther thus, lowering the repulsions.

b. O3+, O3, or O3-O3- has the smallest bond angle because the additional electron makes two positions for electron repulsion as it spend time on the central O.

c. (X S X) angleF is more electronegative than Cl which helps it to pull the electrons farther from S. This makes the F-S-F angle smaller.

d. NO2- or O3NO2- has a smaller bond angle. The N-O electronegativity difference pull the electrons away from nitrogen, making the angle smaller and reducing the bp-bp repulsion. 3-6 Select from each set the molecule or ion having the smallest bond angle, and briefly explain your choice:e. ClO3- or BrO3-BrO3- because Br is larger in size compared to Cl. Moreover, Cl is more electronegative which makes it to hold electron closer and increase bp-bp repulsion. f.

Since the LCP radius of F-S is smaller than Cl-S, the fluorine atoms can approach each other more closely. Thus, SOF2 has the smallest bond angle.

3-6 Select from each set the molecule or ion having the smallest bond angle, and briefly explain your choice:3-9 Give the Lewis dot structure and sketch the shapes of the following:a. ICl2- shape: linear

b. H3PO3 shape: distorted tetrahedral

c. BH4-shape: tetrahedral

d. POCl3shape: distorted tetrahedral

e.IO4-shape: tetrahedral

f.IO(OH)5shape:octahedral

3-9 Give the Lewis dot structure and sketch the shapes of the following:g. SOCl2shape:trigonal pyramidal

h. ClOF4-shape: square pyramidal

i. XeO2F2shape: trigonal planar

3-9 Give the Lewis dot structure and sketch the shapes of the following:j. ClOF2+shape: trigonal pyramidal

3-9 Give the Lewis dot structure and sketch the shapes of the following:3-12. Give the Lewis dot structure and shape fora. VOCl3shape: distorted tetrahedral Cl-V-Cl bond angle: 111Cl-V-O bond angle: 108b. PCl3shape: trigonal pyramidalCl-P-Cl bond angle: 100.4

3-12. Give the Lewis dot structure and shape forc. SOF4shape: distorted trigonal bipyramidalequatorial F-S-F bond angle: 100axial F-S-F bond angle: ~180d. ClO2-shape: trigonalO-S-O bond angle: 120

3-12. Give the Lewis dot structure and shape fore. ClO3-shape: Trigonal PyramidalCl-O- bond angle: 107f. P4O6shape: AdamantaneP-O-P bond angle: 127O-P-O bond angle: 99.5

Bond Order (ClO3-) Bond Order (ClO4-) = (1+2+2)/3=1.67= (1+2+2+2)/4= 1.75ClO4- has a higher bond order than ClO3-. Hence, a stronger interaction, as well as shorter double bonds, between the electrons of ClO4- compared to ClO3- will be observed.

3-15. Compare the bond orders as expected of ClO4- and ClO3-3-18. Carbon monoxide has a larger bond dissociation energy (1072 kJ/mol) than molecular nitrogen (945 kJ/mol). Suggest an explanation.The difference in electronegativity between C and O allows carbon monoxide to have about 76 kcal/mol bond energy contribution. This attraction between the slightly positive and negative ends present in CO strengthens the bond.

Molecular nitrogen, on the other hand, do not have this ionic contribution. A repulsion in the sigma bonding due to the short bond distance may have caused N2 to have a smaller bond dissociation energy compared to CO.

a. PCl5 is a stable molecule, but NCl5 is not.The central atom, N in NCl5 is too small compared to P that also uses 3d orbitals together with 3s and 3p due to its 10 electrons around the atom. Also, the use of 3s, 3p or 3d by N will need a very higher energy to be used effectively.

b. SF4 and SF6 are known, but OF4 and OF6 are not. The central atom, O in OF4 and OF6 is too small to have accessible orbitals beyond 2s and 2p. On the other hand, S is large enough to accommodate the 8 and 12 electrons around it.

3-21. Explain the following: -3-24. Predict the structure of I(CF3)Cl2. Do you expect the CF3 group to be in an axial or equatorial position? Why? The structure is a slightly distorted T-shape due to the lone pairs spreading out to take up the less hindered equatorial positions of the trigonal bipyramidal structure.

The CF3 group would sit in an equatorial position, along with the two lone pairs, as this will minimize the electrostatic repulsions of the three larger groups.

3-27. Although the C F distances and the F - C F bond angles differ considerably in F2C=CF2, F2CO, CF4, and F3CO- (CF distances: 131.9 to 139.2 pm; FCF bond angles: 101.3o to 109.5o), the F . . . F distance in all four structures is very nearly the same (215 to 218 pm). Explain, using the LCP model of Gillespie. 3.36

Bond angles and Distances:Steric NumberC-F (pm)FCF angle ()F-F (pm)F2C=CF23133.6109.2217.80F2CO3131.9107.6212.88CF44131.9109.5215.43F3CO-4139.2101.3215.28Solution:

sin(109.2/2)=(X/133.6) X=108.90pm

Since F-F distance, twice the value of X, F-F length: 217.80pm or approx. 218 pm

*Same equation is used for the other compounds, only the bond length and the bond angles differ.

Explanation:

The LCP model uses outer atoms as a guide to identifying the molecular shape of the molecule. It says that for a series of molecules having the same central atom, the nonbonded distances between the outer atoms are consistent, but the bond angles and bond lengths change. In the problem, the F-F distance values remains nearly constant or close to each other while the central atom, C moves to minimize repulsions.