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GENERAL CHEM I BONDING AND MOLECULAR GEOMETRY PART II SOLUTIONS

OUR UNDERGRAD SERVICES SEMESTER PLAN COVERAGE CHEM 1211 & 1212, BIOL 1107

ON-CAMPUS PRIVATE TUTORING GENCHEM OCHEM

BIOL 1107

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Barium sulfate

Zinc(II) nitrate

Copper(II) sulfate

Aluminum chloride

Sodium hydroxide

Lead(II) chloride

Potassium nitrate

Magnesium hydroxide

Lithium chlorate

Nickel(II) sulfide

Calcium carbonate

Iron(II) oxide

Potassium sulfate

Silver(I) chloride

Calcium phosphate

Barium hydroxide

Sodium sulfide

Iron(II) chloride

Oxygen difluoride

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2. Practice: Draw all the structures of the following molecules, then draw the resonance hybrid.

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3.

+ H3O O2

In the upper left, hydroxide, H3O+, has just been rotated. No electrons have moved to different locations.

In the bottom left, this is isomerism, not resonance. For resonance to occur, the connectivity – the arrangement by which are connected in a molecule – must not change, but instead lone pairs and non-single bonds move around. Here, the chlorine and hydrogen atoms are connected differently.

In the bottom right, O2 is simply a molecule and shows only one version of itself. Honestly, I think that ended up there due to a formatting error haha.

4. CORRECTION: in an early version of this packet, the question said to assess only - bonds. It should have said “average of bonds that involve carbon.”

In carbon dioxide, CO2, there are two double bonds involving carbon. 4 푡표푡푎푙 푏표푛푑푠 = 2 2 푏표푛푑 푙표푐푎푡푖표푛푠

In carbon monoxide, CO, there is a single involving carbon. 3 푡표푡푎푙 푏표푛푑푠 = 3 1 푏표푛푑 푙표푐푎푡푖표푛

In methane, CH4, there are four single bonds involving carbon. 4 푡표푡푎푙 푏표푛푑푠 = 1 4 푏표푛푑 푙표푐푎푡푖표푛푠

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5.

with the resonance hybrid, we can actually see that the bond order will be 1.33, as the is shared among three different bond locations. To calculate, we pick any of the resonance structures and apply the same process as above.

4 푡표푡푎푙 푏표푛푑푠 = 1.33 3 푏표푛푑 푙표푐푎푡푖표푛푠 6.

Once we draw the structures, we realize that we have resonance for both molecules and that we’ll have fractional bond order. We pick a resonance structure for each molecule and calculate bond order.

- 4 푡표푡푎푙 푏표푛푑푠 Nitrate, NO3 : = 1.33 3 푏표푛푑 푙표푐푎푡푖표푛푠

- 3 푡표푡푎푙 푏표푛푑푠 Nitrite, NO2 : = 1.5 2 푏표푛푑 푙표푐푎푡푖표푛푠

Nitrite has higher bond order and therefore stronger bonds (and shorter bonds, as well)

7. H = (total energy of bonds in reactants) - (total energy of bonds in products)

REACTANTS 8 C-H bonds – (8 * 413) 2 C-C bonds – (2 * 348) 5 O=O bonds – (5 * 495) PRODUCTS 6 C=O bonds – (6 * 799) 8 O-H bonds – (8 * 463)

H = [ (8*413) + (2*348) + (5*495) ] – [ (6*799) + (8*463) ] all in kJ/mol -2023 kJ/mol Problems with the key? Questions in general? REACH OUT! (706) 372-4966 | www.nerddawgs.org | [email protected] 8. a) BeF2

2 regions of electron density, 0 lone pairs > linear electrical, linear molecular

All bond angles 180

No resonance structures

b) BCl3

3 regions of electron density, 0 lone pairs > trigonal planar electrical, trigonal planar molecular

All bond angles 120

No resonance structures

c) CCl4

4 regions of electron density, 0 lone pairs > tetrahedral electrical, tetrahedral molecular all bond angles 109.5

No resonance structures

d) PBr5

5 regions of electron density, 0 lone pairs > trigonal bypyramidal electrical, trigonal bypyramidal molecular

This molecular geometry has several different bond angles. The atoms in the equatorial positing have angles of 120 degrees from one another, and 90 degrees from the atoms in the axial positions. There is also an angle of 180 between the two axial atoms.

No resonance structures

e) SI6

6 regions of electron density, 0 lone pairs > octahedral electrical, octahedral molecular

All angles are 90 degrees, except the two axial atoms having a 180 angle in a straight line.

No resonance structures

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4 regions of electron density, 1 , tetrahedral electrical, trigonal pyramidal molecular

The bond angles will be less than the ideal bond angle of 109.5, because they are pushed closer together by repulsive forces from the lone pair. The angle will be about 107.5. You should be able to report that the angle is less that 109.5, regardless of how that option may be worded on an exam, etc.

No resonance structures

+ g) ClF4

5 regions of electron density, 1 lone pair, trigonal bipyramidal electrical, seasaw molecular

The ideal bond angles are 90 degrees between equatorial and axial atoms, and 120 between two equatorial atoms. Because of repulsive forces from the lone pair, these angles should actually be slightly smaller. If given the option, you should report that the angles are less than 90 and less than 120. Note that the effect of lone pairs on bond angles is typically assessed via tetrahedral electronic geometry, so a question may not take into account this effect on trigonal bypyramidal electronic geometry. Just know that the angles should be impacted, and choose that option if it is given.

No resonance structures

– h) SF5

6 regions of electron density, 1 lone pair, octahedral electrical, square pyramidal molecular

All bond angles will have an ideal bond angle of 90 degrees, though some should actually be slightly smaller due to repulsive forces from the lone pair in the axial position. There is no 180 bond angle, because there is only one in the axial position.

No resonance structures

9. H-F C-N C-H H-Cl

C-H < C-N < H-Cl < H-F Least polar Most polar

A carbon-hydrogen bond is actually nonpolar; this bond is used as a rule of thumb, such that bonds with a larger electronegativity difference than C-H are considered polar, and those with an equal or smaller difference are considered nonpolar.

Hydrogen-fluorine should be clearly the most polar, because hydrogen is low in electronegativity and fluorine is the most electronegative element.

The middle two bonds are less obvious. It is necessary to either have memorized some common electronegativity values, or have a chart to make this determination. Problems with the key? Questions in general? REACH OUT! (706) 372-4966 | www.nerddawgs.org | [email protected]

10.

CO2

• 2 regions of electron density, 0 lone pairs, linear molecular geometry • Linear CAN be nonpolar, IF all its bonds are the same (to the same element, in the same way (single/double/triple)) • Both bonds are made to the same element (oxygen), and both are double bonds • This is nonpolar

- IF2

• 5 regions of electron density, 3 lone pairs, linear molecular geometry • Linear CAN be nonpolar, IF all its bonds are the same (to the same element, in the same way (single/double/triple)) • Both bonds are made to the same element, and both are single bonds • This is nonpolar

CH2Br2

• 4 regions of electron density, 0 one pairs, tetrahedral geometry • Tetrahedral CAN be nonpolar, IF all its bonds are the same (to the same element, in the same way (single/double/triple)) • The bonds are being made do different elements, hydrogen and bromine. • This is polar.

Note that an extra of CH2Br2 was accidentally included on the first version of this packet.

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