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Resonance is defined as the movement of electrons through π (pi) bonds. π bonds are any bonds beyond a single bond. For instance a double bond has 1 π bond and a triple bond has 2 π bonds. If there are only single bonds, there are no π bonds, therefore there is no resonance.
a) 𝑁O2
b) 𝑆O3
2−
c) O3
d) CH4
Methane is the only compound with only single bonds. Moving electrons in this structure results in broken bonds – this is another useful way to think of structures that can and can’t have resonance.
e) O2
You may have thought oxygen only has one structure but, because it has a double bond, there is a minor structure that can occur.
Problem 9. Which of the following statements is/are correct for the formate ion HCO2 ?
Answer: C
Solution: the oxidation number of the C atom is +2, there are only 2 plausible contributing structures, and HCO2
= 18 e- = AX3 = trigonal planar
Problem 10. a) Answer:
b) Answer:
Sigma bonds = 4
Pi bonds = 1
c) Answer: A single bond contains one σ bond, whereas a double bond consists of one σ and one п bond. σ bonds are covalent bonds between electron pairs in the area between two atoms. п bonds are a covalent bond between parallel p orbitals and the electron pairs shared are below & above the line joining the 2 atoms.
C N
d) Answer: There are 3 electron pairs around the central C atom, and thus it is sp2 hybridized. Thus
Center Cl has 2 bonding pairs and 2 lone pairs of electrons sp3 hybridized & tetrahedral formation. Since there are 2 lone pairs of electrons, the structure is bent-shaped.
Problem 12. Draw the Lewis structure for the peroxymonosulfate ion, H-O-O-SO3-, and estimate the H-O-O bond angle.
Answer: < 109
Solution:
The center oxygen in H-O-O has 4 pairs of electrons (2 lone pairs of electron and 2 bonding pairs of electrons); as such, this part of the molecule has a tetrahedral structure. The bond angle is less than
109.5 due to the 2 lone pairs of electrons that bend the structure more than bonding pairs of electrons. The tetrahedral formation means the electrons are distributed in sp3 orbitals.
Solution: The most ionic bond will have the most electronegative species and the least electronegative species. Electronegativity increases as you move across and up the periodic table. F is the most electronegative and Li, here, is the least E.N.
Problem 3. Answers:
NH3– Br: dipole – induced dipole, London dispersion
ClBr– ClBr: dipole – dipole, London dispersion
CO2– CO: induced dipole – dipole, London dispersion
b) All 3 molecules are non-polar, so only London dispersion forces
London dispersion forces increase in magnitude as MW increase
As IM forces increase, boiling point increases
Br2 has highest MW, so has strongest IM forces and highest boiling point
bp(N2) < bp(O2) < bp(Br2)
c) Compound Type of intermolecular forces
CH4 Non-polar; London dispersion forces
CH3OH Polar; London dispersion forces & H-bonding
CCl4 Non-polar; London dispersion forces that are greater
than those in CH4 or CH3OH
The greater the intermolecular forces present, the lower the vapor pressure.
There is more interaction between the atoms of carbon tetrachloride due to stronger London dispersion forces, which increase the boiling point with respect to methane. Chlorine is more polarizable than hydrogen because it has MANY more electrons that are also further from the nucleus, resulting in stronger London forces.
Problem 5. Answer: B
Solution: HF is more likely to form hydrogen bonds with water because of the large dipole present in the HF molecule. This allows the hydrogen bonds to form more readily.
Problem 6. Solution: B
Methanol experiences hydrogen bonding due to the polar nature of the molecule.
Problem 7. Solution: D
The lowest boiling point will occur in non-polar compounds. The only non-polar compound above is N2.
Problem 8. Solution: C
Increasing the temperature increases the kinetic energy of the molecules in the liquid causing the molecules to move more quickly and increase the vapor pressure.
Problem 9. Solution: A) H2O
The boiling point depends on the type of intermolecular forces present.
H-bonding > dipole-dipole > London dispersion forces.
CH4 has the lowest boiling point as there are only London dispersion forces present.
OHH H
SH F
H NHH
HH H
H
H
H-bonding H-bonding H-bonding H-bonding London forces
A chiral carbon is one that has 4 different groups attached to it via σ bonds. A carbon containing double bonds does not count as asymmetric carbons since there are only 3 σ bonds.
Problem 17. Answer: Alkene, alkyne, ketone, alcohol
Notes:
Alkene (double bond), alkyne (triple bond),
ketone (contains a carbonyl), alcohol (contains a hydroxyl)
NOTE: Hydroxyl (-OH) and carbonyl (C=O) ARE NOT functional groups. They are pieces of functional groups. For instance, an alcohol (R-OH) contains a hydroxyl. A ketone (R2C = O) contains a carbonyl. A carboxylic acid (R-COOH) contains BOTH a hydroxyl and carbonyl.
Problem 18. Answer: B
An ester group is one that contains RCOOR ( ) which is not present
Problem 19. Answer:
CH3CH2CHCH2CH2CH3
OH
H3C Cl
C
CC
C
CC
Cl
OH
H
H
H
H
C C OH3C
H
NH3
O
**
*The middle two structures are achiral
Problem 20. Consider the following molecules A, B, C, D and E.
Answers:
i. A and D or B and D
ii. A and B
iii. NONE
Problem 21. Answers:
(i) A and F
(ii) C and E
(iii) B, C, D and E are not optically active, thus, no rotation of plane polarized light. A and F are optically active but form racemic mixture. Thus, no rotation of plane polarized light.
On reason nylon is so strong is that it is a high density polymer meaning it has long non-branching chains of high molecular weight. In the absence of the special amid functional group these chains could pack together tightly in a uniform structure and be fairly strong. However, nylon’s amide functional groups enable it to form hydrogen bonds between the nitrogen hydrogen of one strand and the carbonyl oxygen of an adjacent strand. These hydrogen bonds allow for tight packing and greatly increase its strength:
Solution: H will have the highest 1s orbital energy because it has the least positive charge on the nucleus.
Problem 17. Answer: a) V5+ < Ti4+ < Sr2+ < Br-
The electron configurations are...
# of protons
Sr2+: [Kr] 38
Br-: [Kr] 35
V5+: [Ar] 23
Ti4+: [Ar] 22
Sr2+ and Br- has the same number of electrons; however, Zeff is greater for Sr2+ due to a greater # of protons, resulting in a greater + charge that pulls the electrons closer to the nucleus. Thus, Sr2+ is smaller than Br-.
V5+ and Ti4+ has the same number of electrons; however, Zeff is greater for V5+ due to a greater # of protons. Thus V5+ is smaller than Ti4+.
Sr2+ and Br- are larger than V5+ and Ti4+ because there are more electrons held in a larger subshell.
(smallest) V5+ < Ti4+ < Sr2+ < Br- (largest)
b) Cl > Br > I
Problem 18. Solutions
a) Electron affinity becomes more negative from left to right because Cl has higher Zeff than S.
b) For k, valence electron is in 4s orbital while valence electron in Li is in 2s orbital. 4s orbital is larger than 2s.
c) Ionization of Xe removes 5p electron while ionization of Kr removes 4p electron. Ionization energy
Problem 19. a) Mg, Ionization energy increase as you move up a group and it also increases as you move right across a period.
b) Mg, Radius increase as you move left along a period and as you move down a group.
c) Ca<Be<P<Cl<O
Electronegativity increases as you up a group and as you move right across a period. F is the most electronegative element and Cs is the least electronegative element.
Problem 20. Answer:
a) [Ar] or [Ne]3s23p6 b) S2- c) S2- d) Ca2+ e) S2- < Ar < Ca2+
Solution:
a) [Ar] or [Ne]3s23p6 b) Ar, Ca2+ and S2- all have the same number of electrons; however, Ar has 18 protons, Ca2+ has
20, and S2- has 16. Thus S2- has the least Zeff since it has the smallest charge pulling on the electrons.
c) The species with the least favorable electron affinity is S2- because it has the smallest Zeff (a smaller positive charge to pull electrons towards the nucleus).
d) Ca2+ has the greatest Zeff because it has the greatest number of protons pulling on the 18 electrons, resulting in the electrons being closer to the nucleus.
Ionization energy is the energy required to remove an electron. It is most difficult to remove an electron from Ca2+, as this will involve the removal of a core electron instead of a valence electron, and Ca2+ has the greatest Zeff. It is easier to remove an electron from S2- versus Ar because S2- has a smaller Zeff. S2- < Ar < Ca2+
a) IE1 for K > IE1 for Ca because Zeff increases from left to right across a period, so Ca has a higher Zeff and it is therefore harder to remove an electron. IE2 for K > IE2 for Ca because the ionization process for Ca leads to the formation of the stable noble gas configuration (Ca2+) while the ionization reaction for K requires the destruction of a stable noble gas configuration. The former will always be lower in energy.
b)
K(g)+ → K(g)
2+ + e
Ca(g)+ → Ca(g)
2+ + e
c) The larger the negative charge, the greater the repulsion between electrons, the larger the radius. These are all atoms that have the configuration of Kr, so only charge influences radius. Therefore, Rb+
< Br < Se2
The ionization energy decreases as you go down Group 1 because as n increases it is easier to remove the outer electron. Radius increases as you go down Group A because as n increases, the radius increases.
a) When AB is put into aqueous solution, it will dissociate into A- and B+ ions rather than A+ and B-. A has a greater ionization energy, and thus is less able to form A+ than B forming B+. Further, electron affinity for A is larger, and thus the reaction A A- is more favorable than B B-.
b) A has a greater electronegativity; A has a larger electron affinity value and thus releases more energy when it gains an electron, making the reaction more favorable than B gaining an electron.
c) A will be more to the right on the periodic table than B, since it has a higher ionization energy and larger electron affinity. Because A is further to the right, it has a greater Zeff and thus will be smaller than B.
d) Since nonmetals are on the right hand side of the periodic table, and thus have more favorable electron affinity due to a greater Zeff, element A is the nonmetal.
This gives the probability of finding an electron in a region of space.
Problem 26. Answer: O2- has 10 electrons therefore Ne is isoelectronic with it.
The electron configuration is 1s22s22p6
Problem 27. Answer: D
Solution: The electron configuration for N is 1s22s22p3. The 1s and 2s levels will have be full, each consisting of 2 electrons with opposite spins (which rules out diagram A). The 2p level will have 3 electrons, one in each orbital (which rules out diagram E) and C). The best diagram will show all three electrons in the 2p orbitals with the same spin (i.e. either all facing up or down). Therefore, diagram D) is the best representation of an N atom in its ground state.
Problem 28. Answer: C
Solution: Sodium, Na, should be 1s2 2s2 2p6 3s1
Problem 29. Answer:
a) [Ar]3d1. The electronic configuration of Ti is [Ar]4s23d2. Electrons are removed from the s subshell first when transition metals, such as Ti, changes to an ionic state. Thus, the electron configuration of Ti3+ is [Ar]3d1.
The electron configuration for As is 1s22s22p63s23p64s23d104p3
The valence electrons for the element As all reside in the 4th level so n = 4. The valence electrons only exist in s or p sublevels so l = 0 or 1. In the s sublevel there is only one orbital while in the p sublevel, there are 3 orbitals, therefore the value of ml = 0 for the s sublevel and ml = -1, 0, +1 for the p sublevel. ms = +1/2 or –1/2 since the spin of the electron in any orbital can only be up or down. The only answer that falls in range of all the possible choices is D.
Problem 31. Answer: C
Solution: 𝑙 = 0, 1, … (𝑛 − 1), therefore if n = 3, then 𝑙 cannot equal 3.
a) n is the principal quantum number and can be any whole integer number. l is the secondary quantum number = 0 to n-1. ml is the third quantum number = -l to +l. Set II) is invalid because ml cannot equal -2 if l is 1. Set III) is invalid because ml cannot equal 3 if l is 2. Set IV) is invalid because l cannot equal 1 if n is 1.
Therefore, N will more strongly pulled into an in-homogeneous magnetic field.
Problem 40.
Atomic Number Configuration Species State Paramagnetic Or Diamagnetic
1 1s2 H Ground Diamagnetic
7 1s22s22p3 N Ground Paramagnetic
17 1s22s22p63s23p6 Cl Ground Diamagnetic
11 1s22s22p63p1 Na Excited Paramagnetic
22 1s22s22p63s23p64s2 Ti2+ Excited Diamagnetic
13 1s22s22p6 Al3+ Ground Diamagnetic
Note: Ti2+ is in an excited state because it would normally lose its 4s electrons first before its 3d electrons. The given configuration is correct for the neutral atom but NOT the cation.