Molecular Structure Intermolecular Forces Saturday Study Session 2

Molecular Structure & Intermolecular Forces Saturday Study Session #2 3 rd Class

Lewis Structures Lewis structures are representations of molecules showing all electrons, bonding and nonbonding.

Draw the Lewis structure for CH 2 Cl 2 OR

Draw the Lewis Structure for NO+

Draw the Lewis Structure for Xe. F 2 • *Xe can have more than an octet of electrons!

From the Lewis Structure we can determine: Electron geometry Molecular geometry Hybrid Orbital Polarity Intermolecular bond

Molecular Shapes • The shape of a molecule plays an important role in its reactivity. • By noting the number of bonding and nonbonding electron pairs, we can easily predict the shape of the molecule.

Electron Domains • The central atom in this molecule, A, has four electron domains. • We can refer to the electron pairs as electron domains. • In a double or triple bond, all electrons shared between those two atoms are on the same side of the central atom; therefore, they count as one electron domain.

Valence-Shell Electron-Pair Repulsion Theory (VSEPR) “The best arrangement of a given number of electron domains is the one that minimizes the repulsions among them. ”

If all electron domains are bonds the molecular shapes are like these below:

If some of the electron domains are unshared pairs of electrons then the molecular shapes are as indicated in this chart and the on the following slide.

Let’s practice molecular geometry

Molecular Geometry answers 1. CH 2 Cl 2 2. NO+ 3. Xe. F 2 1. Tetrahedral 2. Linear 3. Linear, but its electron geometry is octahedral due to 3 unshared pairs of e-

Bond Angles 120 o 45 o

Bond Angles for molecules without lone pairs of electrons.

Nonbonding Pairs and Bond Angle • Nonbonding pairs are physically larger than bonding pairs. • Therefore, their repulsions are greater; this tends to decrease bond angles in a molecule.

Multiple Bonds and Bond Angles • Double and triple bonds place greater electron density on one side of the central atom than do single bonds. • Therefore, they also affect bond angles.

Bond and Molecular POLARITY

Polar Covalent Bonds Intramolecular • Though atoms often form compounds by sharing electrons, the electrons are not always shared equally. • Fluorine pulls harder on the electrons it shares with hydrogen than hydrogen does. • Therefore, the fluorine end of the molecule has more electron density than the hydrogen end.

Polar Covalent Bonds • The greater the difference in electronegativity, the more polar is the bond.

Polarity • Just because a molecule possesses polar bonds does not mean the molecule as a whole will be polar.

Polarity of Molecules By adding the individual bond dipoles, one can determine the overall dipole moment for the molecule.

Polarity

Molecular Polarity Polar Molecules Non-Polar Molecules • Must have some polar • Polar or nonpolar bonds (∆EN > 1. 7) • Dipoles cancel • Overall net dipole • Insoluble in Water • Soluble in Water • Look for – Symmetrical – Asymmetry molecule – -OH, -NH 2 groups

Practice Molecular Polarity • Which molecule is more polar? 1. CS 2 or SF 2 NONPOLAR 2. BH 3 or NH 3 NONPOLAR

Which is more polar? 3. Benzene POLAR NONPOLAR OR Glucose

Molecules Stick Together • All molecules have some attractive forces for each other. • Polar molecules have more types of attractive forces than do nonpolar molecules. • These attractive forces are called INTERMOLECULAR FORCES (IMF)

Inter vs. Intra molecular forces Inter (between molecules) • London dispersion forces • Dipole-dipole forces • Hydrogen bonds Intra (inside molecules) • Ionic bonds • Covalent bonds • Metallic bonds

Dipole-Dipole forces

Hydrogen bonding

Which IMFs are present? 1. CF 4 1. London dispersion forces 2. BF 3 2. London dispersion forces, Dipole – dipole forces 3. NH 3 3. London dispersion forces, Hydrogen bonding 4. London dispersion forces, Dipole – dipole forces (in water, weak H bonding) 4. H 2 CS

Effects of IMFs • States of matter • Phase changes –Melting points –Boiling points –Vapor pressure

States of Matter (Increasing) Molecular Interactions ARE Intermolecular Forces

Particles getting farther apart means they are overcoming intermolecular forces by adding energy.

Energy In Energy Out Must overcome IMFs cause particles to congregate

Effects of IMFs on properties • Greater IMFs = higher melting and boiling points and lower vapor pressure. • Greater Molar Mass = more electrons = greater IMFs • Volatile substances have high VP due to low IMFs • Greater IMFs = high ΔHvap

MC Question 1 In which of the following processes are covalent bonds broken? A) I 2(s) → I 2(g) B) CO 2(s) → CO 2(g) C) Na. Cl(s) → Na. Cl(l) D) C(diamond) → C(g) E) Fe(s) → Fe(l)

Question 1 Answer • Correct answer is D • Diamond is a covalently bonded network crystal. In order to form a gas the covalent bonds must be broken. • A and B the molecules remain intact, only IMF are “broken” • C is held together with ionic bonds • E is held together with metallic bonds

MC Question 2 The structural isomers C 2 H 5 OH and CH 3 OCH 3 would be expected to have the same values for which of the following? (Assume ideal behavior. ) A) Gaseous densities at the same temperature and pressure B) Vapor pressures at the same temperature C) Boiling points D) Melting points E) Heats of vaporization

Question 2 Answer • Correct answer is A • Density is a function of mass and volume. Isomers have the same molecular mass, same volume can be assumed. • All other answers the values change according to differences in IMFs. • C 2 H 5 OH has hydrogen bonding but CH 3 OCH 3 does not.

MC Question 3 X: CH 3–CH 2–CH 3 Y: CH 3–CH 2–CH 2–OH Z: HO–CH 2–CH 2–OH Based on concepts of polarity and hydrogen bonding, which of the following sequences correctly lists the compounds above in the order of their increasing solubility in water? A) Z < Y < X B) Y < Z < X C) Y < X < Z D) X < Z < Y E) X < Y < Z

Question 3 Answer • Correct answer is E • The pure hydrocarbon butane X, is the least polar, thus has the lowest solubility in water. • The presence of an –OH group on butanol Y, makes it more soluble than butane, but less soluble than the 1, 3 -propanediol • Z, that contains two –OH groups. Aren’t you glad you don’t have to name all of them?

MC Question 4 Hydrogen Halide Normal Boiling Points, °C HF +19 HBr – 67 HCl HI – 85 – 35 The relatively high boiling point of HF can be correctly explained by which of the following? A) HF gas is more ideal. B) HF is the strongest acid. C) HF molecules have a smaller dipole moment. D) HF is much less soluble in water. E) HF molecules tend to form hydrogen bonds.

Question 4 Answer • Correct answer is E • Hydrogen bonding occurs when a H is bound to a “highly electronegative atom” (F, N or O) • The bonded H is attracted to an unshared electron pair or another highly electronegative atom on a neighboring molecule.

MC Question 5 Which of the following gases deviates most from ideal behavior? (Ideal gases assume no interparticle attractions) A) SO 2 B) Ne C) CH 4 D) N 2 E) H 2

Question 5 Answer • Correct answer is A • Deviations occur due to molecular volume (larger molecules have more mass as well) and attractive forces. • The more electrons present, the more polarizable a molecule, thus the greater the London dispersion (induced dipole-induced dipole) attractive forces become. • SO 2 has a higher molecular mass, more electrons and is more polarizable than the other answer choices.

MC Question 6 Molecular iodine would be most soluble in: A) water B) carbon tetrachloride C) vinegar (acetic acid and water) D) vodka (ethanol and water) E) equally soluble in all four

Question 6 Answer • Correct answer is B • Molecular iodine is a nonpolar molecule. • Carbon tetrachloride is the only nonpolar solvent listed.

FR Question 1 Explain each of the following in terms of the electronic structure and/or bonding of the compounds involved. (a)At ordinary conditions, HF (normal boiling point = 20°C) is a liquid, whereas HCl (normal b. p. = -114°C) is a gas.

FRQ 1 Answer • HF exhibits hydrogen bonding but HCl does not. Both molecules have dispersion forces (HCl slightly greater than HF) but the hydrogen bonds are stronger in HF (F very highly electronegative) and require more energy to overcome to allow HF molecules to leave the liquid state and enter the gaseous state.

FR Question 2 (a) Identify the type(s) of intermolecular attractive forces in: (i) pure glucose (ii) pure cyclohexane

(b) Glucose is soluble in water but cyclohexane is not soluble in water. Explain.

FRQ 2 answer • ALL molecules have London dispersion forces. • Glucose has hydrogen bonding because its hydrogens are bonded to oxygen but cyclohexane does not. Cyclohexane’s hydrogens are bonded only to carbon. • Glucose can also form hydrogen bonds with water increasing its solubility, while hexane can not hydrogen bond with water.

FR Question 3 Consider the two processes represented below. Process 1: H 2 O(l) → H 2 O(g) ∆H° = +44. 0 k. J mol Process 2: H 2 O(l) → H 2(g) + 1/2 O 2(g) ∆H° = +286 k. J mol (i) For each of the two processes, identify the type(s) of intermolecular or intramolecular attractive forces that must be overcome for the process to occur.

FRQ 3 Answer • Process 1 requires overcoming London dispersion forces and hydrogen bonding. • Process 2 requires overcoming much stronger covalent bonds.

(ii) Indicate whether you agree or disagree with the statement in the box below. Support your answer with a short explanation. When water boils, H 2 O molecules break apart to form hydrogen molecules and oxygen molecules Water Boiling: H 2 O(l) → H 2 O(g) molecules remain intact Water molecules breaking apart: H 2 O(l) → H 2(g) + 1/2 O 2(g) This process requires electrolysis

FR Question 4 Explain each of the following in terms of atomic and molecular structures and/or intermolecular forces. (a) Solid K conducts an electric current, whereas solid KNO 3 does not. (b) The normal boiling point of CCl 4 is 77°C, whereas that of CBr 4 is 190°C. (c) Iodine has a greater boiling point than bromine even though the bond energy in bromine is greater than the bond energy in iodine

FRQ 4 Answer (a) Solid K has metallic bonds with a mobile sea of electrons allowing current to flow. The electrons in KNO 3 are localized in ionic and covalent bonds and are not allowed to move throughout the material. (b) CCl 4 has a lower boiling point than CBr 4 because CBr 4 has more electrons and greater London dispersion forces.

(c) Bond energies measure the strength of the covalent bonds in the diatomic molecules I 2 and Br 2. The boiling points depend upon intermolecular forces. Both molecules have only London dispersion forces. Iodine has more electrons per molecule than bromine causing the higher boiling point for iodine.

FR Question 5 Use appropriate chemical principles to account for each of the following observations. In each part, your response must include specific information about both substances. (a) At 25°C and 1 atm, F 2 is a gas, whereas I 2 is a solid. (b) The melting point of Na. F is 993°C, whereas the melting point of Cs. Cl is 645°C. (c) Ammonia, NH 3 , is very soluble in water, whereas phosphine, PH 3 , is only moderately soluble in water.

FRQ 5 Answer (a) F 2 molecules are smaller and have fewer electrons than I 2 molecules. The only intermolecular forces present in both molecules are London dispersion forces. Fluorine has fewer IMF allowing the molecules to overcome the attractive forces at 25 o. C and move apart into the gas phase, while iodine molecules with more electrons have greater London forces keeping the molecules very close together as a solid.

(b) Coulomb’s Law states that the magnitude of the attractive forces between two charged particles is equal to the product of the charges of the particles divided by the square of the distance between the particles. Na. F and Cs. Cl are both ionic compounds. In order to melt their ionic bonds must be broken. The bonds in Na. F are stronger than the bonds in Cs. Cl because Na and F are smaller atoms than Cs and Cl respectively. This means that their nuclei are closer the other atoms’ electron cloud, increasing the strength of the attractive forces between atoms. Also the charges of Na and Cs are the same as are the charges of F and Cl. Using Coulomb’s Law in both cases would result in the same number in the numerator of his equation but with a smaller value in the denominator for Na. F than for Cs. Cl resulting in stronger forces for Na. F and thus a higher melting point.

(c) Ammonia can form hydrogen bonds with water but phosphine can not. The ability to form hydrogen bonds along with the polarity that both molecules exhibit increases the solubility of ammonia in water as compared to that of phosphine.