CHEM 110 Exam 2 - Practice Test 1 - Solutionspenn.liontutors.com/uploads/Solutions_-_Practice_Tests_1,_2,__3... · CHEM 110 Exam 2 - Practice Test 1 - Solutions 1D ... The angle will

CHEM 110 Exam 2 - Practice Test 1 - Solutions

1D 1 has a triple bond. 2 has a double bond. 3 and 4 have single bonds. The stronger the bond, the shorter the length. 2A A 1:1 ratio means there must be the same amount of Carbons as there are Oxygens. A 2:1 ratio means there must be twice as many hydrogen’s as there are carbons. The only choice that satisfies these requirements is choice 1, since in choice 2 there are too many oxygens, and in choice 3 the formal charges on oxygen are -1. 3A The correct Lewis structures are imperative to answer this question with confidence. IF3 – the central iodine atom will have 3 single bonds and two lone pairs of electrons. The correct VSEPR is AX3E2, which corresponds to T-shaped. PCl5 - the central phosphorus atom will have 5 single bonds and no lone pairs of electrons. The correct VSEPR is AX5, which corresponds to trigonal bipyramidal. NH3 - the central nitrogen atom will have 3 single bonds and one lone pair of electrons. The correct VSEPR is AX3E1, which corresponds to trigonal pyramidal. 4(i)B Remember, the carbon we’re interested in also has a hydrogen bonded to it that’s not shown in the diagram. This gives it a total of three electron groups, all atoms, and so the shape is trigonal planar and the correct angle is about 120°. 4(ii)D

- The aldehyde is present since the carbon double bonded to the oxygen is also bonded to hydrogen.

- The amine simply requires a carbon bonded to nitrogen. - The alkene simply requires at least one double bond.

5D The nitrogen has three single bonds shown in the diagram. The correct Lewis structure for that area of the molecule would also show that N has a lone pair of electrons on it as well (remember that nitrogen likes to use 6 electrons for bonding and to keep 2 electrons for itself). Thus N has a total of 4 electron groups and must be sp3. 6D Again, the correct Lewis structures are essential to determining polarity. SF4 – This molecule has one lone electron pair on the central atom, which always means it’s polar. CH3Br – Here Br acts as the odd-man-out creating a polar bond that isn’t countered anywhere else in the molecule, thus it is polar. H2S – Very similar to water, the sulfur has two lone pairs of electrons, but since the molecule’s geometry is bent shaped, the lone pairs cannot cancel each other out, and thus it’s polar. XeF2 – Here the central atom has two single bonds and three lone pairs of electrons. The corresponding shape is linear, which is always nonpolar if the surrounding atoms (in this case the fluorines) are the same. Thus we have our correct answer. C4H9OH – As a rule, any molecule with an OH attached to it, is almost certainly going to be polar (and will also undergo hydrogen bonding). 7E Let’s analyze the 5 statements:

(a) This is almost true were it not for the lone pair of electrons that must reside on the N (remember N like to keep one lone pair whenever possible).

(b) There are actually 13 sigma bonds (one for every link between two atoms), and there are 5 pi bonds – 3 for the three double bonds, and 2 for the triple bond.

(c) The angle will depend on the geometry of the C atom that’s triple bonded to the N atom and single bonded to the other C atom that’s part of the benzene ring. This specific carbon only has two electron groups (both atoms) and is thus linear with bond angle 180°.

(d) This can only happen when there is a carbon-carbon double bond that isn’t part of a ringed structure. It cannot occur with triple bonds.

(e) The N atom has two electron groups (one is the carbon it’s bonded to, the other is the lone pair of electrons mentioned in answer choice A) and so its hybridization is sp.

8D This is one of the octet rule exceptions dealing with odd-numbered electrons. Whenever you have an odd number of electrons the central atom will usually be forced to have 7 electrons surrounding it rather than the usual 8. In the case of Nitrogen Dioxide, nitrogen has a single bond to one of the oxygens, a double bond to the other oxygen, and ONLY ONE lone electron that it keeps for itself. Thus the formal charge is: 5 – (1 + 3) = +1. 9B

The least amount of formal charge around each individual atom is accomplished through this structure best. 10(i)C The sigma bonds are always formed using the hybrid orbitals (except for H atoms which only use s-orbitals) and so we must find the hybridization of C, which has two electron groups, and therefore must be sp. 10(ii)B Pi bonds will always be formed using overlapping p-orbitals (not hybrid orbitals). 11D We first want to look at the Lewis structure (which shows three single bonds and a lone electron pair on the central atom), thus the shape is trigonal pyramidal and the ideal angle is 109.5°. However, in almost every case, the angle is less than that since the lone electron pair pushes together the other atoms, and so we should look for a value a little lower than 109.5, like 108° in this case.

12E Electronegativity increases going up and to the right of the periodic table; fluorine is the most electronegative element. The bond with the greatest difference in electronegativity between both atoms will be the most polar. B and F are separated by the most space compared to the other bonds, so that will be the most polar bond. 13C The central C is double bonded to O and single bonded to each Cl. As usual, there are no lone electrons on the C. Thus, C has three electron groups, and is sp2. 14D For this problem, we must add up all the number of each type of element present in the diagram, remembering to include all the necessary hydrogens. This gives us the molecular formula of C10H12. All we do now is find the ratio of atoms to one another and we’ve got the empirical formula: C5H6. 15B

Due to the -3 charge, this is the best way for the formal charges to be spread out among the atoms. So, we only have 1 electron pair on As. 16E Remember, the surest way to be certain if they are structural isomers is if they have the exact same number and type of atoms. Analysis of the options shows that i, ii and iv each has 6 carbons and 12 hydrogens.

17D sp3d2 corresponds to a central atom with 6 electron groups, and that can only mean octahedral. 18C A single bond contains one sigma bond. A double bond contains one pi bond and one sigma bond. A triple bond contains two pi bonds and one sigma bond. There are four C-H bonds not written in the structure because it is assumed the will be recognized; they are at the four empty corners of the 6-membered ring. A total of 15 single bonds, 5 double bonds, and two triple bonds lead to 22 sigma bonds and 9 pi bonds. 19D A structure with delocalized bonding is one that has resonance. This is a common occurrence in compounds with double bonds. The one double bond in SO3 can be between sulfur and any three of the oxygens, so the hybrid structure shown below is the true representation of the structure.

20E Sigma bonds result from the head-on overlap of orbitals along the internuclear axis. Sigma bonds have more overlap of the orbitals than pi bonds, so they are stronger interactions that pi bonds. A triple bond contains one sigma and two pi bonds. Atoms cannot rotate along pi bonds because it would decrease the overlap between the parallel orbitals. The sideways overlap of two parallel, unhybridized p orbitals is an example of a pi bond, so E is true.

21D Halogens can take part in addition reactions at room temperature by attacking the double or triple bonds of organic compounds. A pi bond becomes uncoupled, and two new sigma bonds form.

22A (Not on Test) Noble gases are entirely uncharged, and do not have polar covalent bonds, ionic bonds, or dipole moments. The only intermolecular forces that exist in noble gases are London dispersion forces. Although they are very weak forces, they will cause noble gases to liquefy at very low temperatures. 23D A structure with free rotation around all its bonds would have all sigma bonds (single bonds) and zero pi bonds (double and triple bonds).

I. II.

III. 24E The overal dipole moment of a molecule is the vector sum of the individual dipole moments of each bond, taking into account the molecule’s geometry. HF is a linear molecule that has two atoms with the largest difference in electronegativities, so it is the most polar molecule and therefore the molecule with the greatest dipole moment. 25A The total number of electron domains attached to an atom determines its hybridization. Electron domains can be either a bond or a pair of nonbonding electrons. NO2

- has three electron domains, which are two bonds between N and an O and one nonbonding electron pair, so it has three sp2 hybrid orbitals. CO2 and O3 have two and three electron domains, respectively, so CO2 has two sp hybrid orbitals and O3 has three sp2 hybrid orbitals.

26E Octahedral is the electron domain geometry of sp3d2 hybridization. 27A COOH is a carboxylic acid functional group and an ester is represented below.

28E Remember that valence electrons correspond directly to the group number of the element. Since Br and F are both in group 7, we add 7 electrons for Br and 28 electrons for the four Fs. We also add one more electron for the negative charge. The total is 36 electrons.

CHEM 110 Exam 2 - Practice Test 2 – Solutions 1B The electron configuration of carbon is 1s22s22p2, so to form the four single, symmetric bonds in CH4 it must promote one of the electrons in the 2s orbital to the 2p orbital. Then, the three half filled p orbitals mix with the half filled s orbital to form the expected four sp3 hybrid orbitals. Here is a list of possible molecules the other answer choices could represent: A. BF3 C. PF5 D. H20 E. NH3 2D Structural isomers are compounds with the same molecular formula, but different atomic connectivity in their structures. The molecular formula of the molecule shown and the molecule in I and II is C5H11NO2, but the molecular formula of the molecule in III is C5H9NO2. 3C Carbon 1 has three electron domains, so it has sp2 hybridization. Carbon 2 has four electron domains, so it has sp3 hybridization. 4C Remember “penta-“ means 5, “tetra-“ means 4, “tri-: means 3, “di-” means 2, and “mono-“ means 1. Of the choices only IF5 has six electron domains, so it has an octahedral electron domain geometry.

5C Polarity is a measure of differing electronegativity values for the two elements that form the bond. Another way to look at this is just to measure how far apart the elements are from each other on the periodic table. Using this method, we get Mg – F > N – F > Br – Br. Note: Bonds involving the same element (like Br – Br) have zero polarity and are called non-polar.

6D The empirical formula of a molecule is the most reduced whole number of atoms of each element present in a compound. The molecular formula of the compound given is C18H24O2, so the most reduced whole number form is found by dividing each subscript by two. 7D The only elements that can violate the octet rule (meaning that the central atom does not have exactly eight electrons surrounding it) are those in periods 3 and higher as well as B and Be. With that in mind you don’t even need to draw Lewis structures to know the answer must be BH3. 8E Atom 1 is sp3 hybridized, atom 2 is sp2 hybridized, and atom 3 is sp2 hybridized. The ideal bond angle of sp3 hybridization is 109.5° and the ideal angle of sp2 hybridization is 120°. However, atom 1 has two nonbonding electron pairs as electron domains, which have a greater force of repulsion than the electrons in bonds, so the actual angles will be less than 109.5°. Both atom 2 and 3 have a double bond as one of its electron domains, which also has a greater force of repulsion than the single bonds, so the actual angles will be less than 120°. 9C Trigonal bipyramidal molecular geometry corresponds to a central atom with five electron domains attached to it, none of which are lone electron pairs. PCl5 is the only compound that fits this structural setup.

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CHEM 110 Exam 2 - Practice Test 3 – Solutions 1E Even though the overall charge of the phosphate ion is -3, the central atom, phosphorus, has a formal charge of +1 in this Lewis structure. Formal charge = # valence e’s - (# unshared valence e’s + 1/2 # shared e’s) Phosphorous has 5 valence electrons (Group 5A) and is bonded to four oxygens with single bonds, and has no lone pairs. So, Formal charge = 5 - (0 + 1/2*4) = +1 The formal charges of each individual atom add up to the total charge of the ion. This is a good way to check your answer. 2D

While it appears as if there is an alcohol group, the -OH is part of the carboxylic acid functional group, which gets priority over labeling it as an alcohol. Similarly, there are no ketone groups because both C=O’s have higher priority classifications (carboxylic acid and ester). 3A There are three cases of violations to the octet rule (8 e’s surrounding atom): expanded valence shells (>8 e’s), electron deficient molecules (<8 e’s), and odd-electron molecules. NO2 NO3

- SiBr4 CN-

H2CO

All central atoms have 8 valence e’s, except for NO2. This is the case in which the molecule has an odd number of electrons.

Carboxylic acid group Ester group Ketone group Alcohol group

4A Determine the hybridization of the carbons by counting the number of electron groups the carbon has. The ideal bond angles are sp (2 EGs): 180° sp2 (3 EGs): 120° sp3 (4 EGs): 109.5° The carbon attached to the alcohol group has 4 electron groups (2 H’s not drawn on the figure), so it is sp3 hybridized and has an ideal bond angle of 109.5°. The aldehyde carbon has three electron groups, so the ideal bond angle is 120°. 5D First, draw the Lewis structure.

So, the AXmEn formula is AX4E1 because there are 4 atoms attached to the central atom, and there is one lone pair on the central atom. Sulfur is able to break the octet rule since it is in the third row of the periodic table. The molecular geometry of AX4E1 is seesaw. Seesaw molecular geometries are polar, as seen in the 3D structure. Lone pairs are more electronegative (contribute more to polarity) than any atomic electron domain. The lone pair is not cancelled out by a lone pair opposite to it, so SF4 is polar. 6C First, find the ideal bond angles by identifying the hybridization of the central atoms to the bonds. The oxygen in bond angle 1 is sp3 because there are 4 electron groups (two single bonds and two lone pairs that are not shown). The nitrogen in bond angle 2 is also sp3 hybridized because there are 4 electron groups (three single bonds and one lone pair not shown). The carbon in bond angle 3 is sp2 hybridized because there are three electron groups (a double bond and two single bonds). The ideal bond angle for sp3 hybridized central atoms is 109.5° and the ideal bond angle for sp2

hybridized central atoms is 120°. The question isn’t asking about ideal bond angles, however. Since the oxygen has two lone pairs, these highly electronegative groups push the atomic groups away, making the angle between atomic groups less than the ideal 109.5°. Nitrogen will also have bond angles of less than 109.5°, but it will be greater than 1 because there is only one lone pair on the nitrogen.

2D structure 3D structure

7C Draw the lewis structures for each and write the AXmEn formula and assign the associated molecular geometry.

So, i and ii are correctly assigned. Always remember exceptions to the octet rule! 8E Dipole induced-dipole interaction is when a polar molecule is attracted to a nearby non polar molecule. Acetaminophen is a polar molecule because it has electronegative groups and has no symmetry. So, since acetaminophen is polar, when interacting with another polar acetaminophen molecule, there will be no dipole induced-dipole interaction. Hydrogen bonding is a specific type of dipole-dipole interaction. Since acetaminophen is polar it is able to have dipole-dipole interaction with another acetaminophen molecule. Hydrogen bonding occurs when N, O, or F atoms are bonded to a hydrogen. So, this will be present because in acetaminophen there are N-H and O-H. London dispersion forces are always present between any two molecules. 9B Amine group: At each vertex there is a carbon, and each carbon must have 4 “lines” coming out of it. Nitrogen must have 3 “lines” coming out of it if it has a formal charge of 0. So, the formula is C6H7N.

AX4E2 : Square planar AX3E1 : Trigonal pyramidal AX4E1 : Seesaw

10B Draw out possible Lewis structures for NCO- and assign formal charges to each of the atoms. The Lewis structures below obey the octet rule; there are other Lewis structures that do not. The most stable Lewis structure will only involve 0, -1, and +1 charges. This eliminates Lewis structure 3. So, the question is if the -1 charge will go on the oxygen or nitrogen. If there are two Lewis structures that have the same charges but on different atoms, put the charge on the more electronegative atom. Oxygen is more electronegative than nitrogen, so Lewis structure 1 is the best. 11B It is necessary to draw out possible resonance structures for each ion. NO3

-

NO2-

NO+

So, in NO3- and NO2

- the π electrons are delocalized (spread throughout the ion). The average bond order between nitrogen and oxygen can be calculated by adding the total number of bonds (e.g. double bond=2 bonds) and dividing by the number of oxygen atoms nitrogen is bonded to. NO3

- : 4 bonds/3 oxygen atoms = 4/3 bond order NO2

- : 3 bonds/2 oxygen atoms = 3/2 bond order NO+ only has one structure that is all-octet, so its bond order is 3. With increasing bond order, the bonds get shorter because the atoms are bonded closer together. So, B is the correct answer.

0 0 -1

-1

0 0

+1 0 -2

12C This problem seems very complicated at first, but the molecules can be simplified by recognizing that an atom directly across from the same atom will cancel out the polarity effects due to symmetry. So, the following simplifications can be made.

Fluorine is more electronegative than chlorine, so it contributes more to polarity. Also, the closer the atoms are to each other, the less they cancel out with each other, and the more polar the molecule will be. For example, iv is polar, but the chlorines are almost opposite to each other. The polarity in the right and left directions is cancelled out, but there is still a net polarity in the up direction. To contrast, in ii the chlorines are more in the same direction, so there is a greater net dipole in the right direction. i is not polar at all because there are no electronegative groups. Both position and atomic groups must be considered. So, the answer is C. 13B The best Lewis structure will give each atom an octet and have the fewest charges overall. Since it’s an uncharged molecule it is likely that the formal charge of each atom will be 0. For these criteria, nitrogens will have 3 bonds (e.g. double bond=2 bonds) and 1 lone pair, carbons will have 4 bonds, and oxygens will have 2 bonds and 2 lone pairs. The only structure that does this is B. 14D Isomers have the same molecular formula but different connectivity. i and iii both have the same molecular formula as propionic acid. 15E First note that the formal charges are not written on the molecule. The chlorines both have formal charges of 0, and the sulfur has the formal charge of -1. But, the molecule is not charged overall because if it were, there would be a bracket around it with the charge. So, this means that “X” will have the formal charge of +1 to give the molecule a net charge of 0. The only element that will give it a formal charge of +1 is argon, Ar. 16B Find the overall molecular weight of caffeine by first finding its molecular formula, C8H10N4O2. 12.011*8 + 1.0079*10 + 14.007*4 + 15.999*2 = 194.2 g/mol Then divide the mass of nitrogen in caffeine ([4 nitrogen atoms*14.007 g]/194.2) = 0.289 —> 28.9% nitrogen.

i ii iii iv

17E The elements of interest in caffeine for lone pairs are nitrogen and oxygen; uncharged carbons will have no lone pairs. Oxygens with two bonds and a full octet will have two lone pairs, while nitrogens with three bonds and a full octet will have one lone pair. Therefore, there are 8 lone pairs in caffeine. 18B Isomers have the same molecular formula but different connectivity. Geometric isomers arise because there is no free rotation about a double bond. To see if there is a geometric isomer, switch two of the groups on the same carbon, and then see if it is the same molecule or not. i and iii will both have geometric isomers. ii will not have a geometric isomer. Because on the one carbon the groups are identical if the groups are switched on the other carbon, you can see that it is the same as it was before. 19D Resonance structures arise when there are multiple Lewis structures that can be drawn. The connectivity of the atoms needs to be the same. So, iii does not show resonance structures because the connectivity isn’t the same. These molecules are isomers of each other. The figure to the right shows the electron-pushing mechanism. 20A The “formula” for seesaw molecular geometry is AX4E. So, there is one lone pair. Assuming the molecule is not charged, an atom that has 4 single bonds and one lone pair would need to have 6 valence electrons to achieve a formal charge of 0. Therefore, sulfur could be the central atom because it is in group 6. 21E Determine the hybridization of the oxygens by determining the number of electron groups that the oxygens have. Reminder: single/double/triple bond/lone pair = 1 electron group. The oxygen on the very left and the oxygen on the very right have two single bonds and two lone pairs that are not shown, so they have 4 electron groups. Atoms with 4 electron groups are sp3 hybridized. The oxygen at the top of the structure has a double bond and two lone pairs not shown. So there are 3 electron groups. Atoms with 3 electron groups are sp2 hybridized.

22D Single bonds contribute 1 σ, double bonds contribute 1 σ and 1 π, and triple bonds contribute 1 σ and 2 π. So, counting up all the single and double bonds (including C-H’s not explicitly drawn out) gives 23 σ and 4 π. Don’t forget the sneaky hydrogen on the carbon adjacent to the amine group! 23E First, we must determine the hybridization of the carbons that are bonded by a σ-bond. The right carbon has 3 electron groups, so it is sp2 hybridized. The carbon on the left has 4 electron groups (don’t forget the sneaky hydrogen), so it is sp3 hybridized. So these hybridized orbitals overlap to form the σ-bond. 24B First, the molecule must have a geometric isomer. A and C do not. In A, switching the positions of the groups on the left carbon will not change the molecule. The carbons in C are only attached to each other and either F or Br, so there can’t be a geometric isomer that has the same chemical formula but different connectivity. D is not polar as is and also cannot be polar if rearranged because it is just a hydrocarbon (hydrocarbons are not polar). So we are left with B and D. Let’s switch the groups on one side of each. So, B will have a polar geometric isomer. The -OH’s in E’s geometric isomer are directly across from each other and cancel out the polarity effect. 25E Delocalized π electrons arise when there are multiple resonance structures that can be drawn which switch the position of double bond(s). Recall that resonance structures are not actually how the molecule looks at any given time. In actuality, the electrons are delocalized, which means they are not bound to a position but are spread over the entire molecule. This reduces the charges of the individual atoms and gives the molecule extra stability. Delocalized π electrons can be denoted by dashed electrons on the Lewis structure. Phosphate, sulfite, sulfate, and carbonate all have double bonds whose positions change between resonance structures. This is effectively a combination of all the Lewis structures. C2H2 does have π electrons because of the triple bond, but these electrons are confined to that triple bond; there are no possible resonance structures.

Phosphate Sulfite Sulfate Carbonate C2H2