2.2 Intramolecular Force and Potential Energy

1. APoint V
   A student who thinks bonded atoms sit as close together as possible picks this. At V the nuclei are so close that their repulsion gives the pair a higher potential energy than the separated atoms have; the pair is not at its most stable separation.
2. BPoint W Correct
   The equilibrium bond length is the separation at which the potential energy of the pair is lowest. W is the minimum of the curve, at about 74 pm: at shorter distances repulsion between the nuclei raises the energy, and at longer distances the attraction is weaker.
3. CPoint X
   A student who thinks the bond length is where the potential energy is zero picks this. At X the pair has the same potential energy as the separated atoms; the minimum, at W, lies at a larger separation.
4. DPoint Z
   A student who thinks atoms are bonded as soon as they begin to attract picks this. At Z the attraction is weak and the energy is only slightly below zero; the atoms are held most stably at the minimum, W.

CED 2.2.A.1 · Read this in Fix

1. AC–C is the shortest, as one shared pair needs the least room between the nuclei
   A student who thinks extra shared electrons take up more room picks this, reasoning that the single bond, with the fewest shared electrons, needs the least space. The extra shared pairs attract both nuclei and draw them closer, so C–C (about 154 pm) is the longest of the three and C≡C (about 120 pm) the shortest.
2. BC–C is the shortest, as single bonds are the least reactive and so the strongest
   A student who takes reactivity as a sign of a weak bond picks this. The C–C bond has the smallest bond energy (about 348 kJ/mol, compared with about 614 and 839 kJ/mol) and is the longest of the three.
3. CAll have the same length, as each bond joins two carbon atoms of the same size
   A student who thinks bond length is fixed by the sizes of the atoms alone picks this. Bond length also depends on bond order: the three carbon–carbon bonds range from about 120 pm to about 154 pm.
4. DC≡C is the shortest, as three shared pairs pull the nuclei closest together Correct
   Between the same two atoms, a higher bond order gives a shorter bond: the shared electrons between the nuclei attract both nuclei, and more shared pairs pull them closer. C≡C is about 120 pm, C=C about 134 pm and C–C about 154 pm.

CED 2.2.A.2 · Read this in Fix

1. ACaO, because its formula unit has a greater mass than that of NaF
   A student who thinks heavier ions attract more strongly picks this. CaO does have the larger formula mass, but mass does not appear in Coulomb's law; the stronger attraction comes from the 2+ and 2− charges.
2. BNaF, because Na and F differ more in electronegativity than do Ca and O
   A student who ranks ionic attractions by electronegativity difference picks this. The difference is larger for Na and F, but the attraction between ions depends on their charges and separation, and the Ca²⁺ and O²⁻ charges are twice as large.
3. CCaO, because its ions carry larger charges than the ions of NaF do Correct
   With the ions about the same distance apart, the attraction depends on the charges. The force is proportional to the charge on each ion, so Ca²⁺ and O²⁻ (product of charges 4) attract about four times as strongly as Na⁺ and F⁻ (product 1).
4. DNeither, because positive and negative charges cancel in each compound
   A student who thinks the cancelling of charges in a neutral compound removes the effect of charge size picks this. Each compound is neutral overall, but the individual ions attract according to their own charges, which are larger in CaO.

CED 2.2.A.3.i · Read this in Fix

1. AEnergy is released, because the bond holds stored energy that is set free as it breaks
   A student who thinks energy is stored in bonds and released when they break picks this. The graph shows the potential energy rising as the atoms separate, so energy must be supplied, not released.
2. BEnergy is released, because −242 kJ/mol is more energy than the 0 kJ/mol at the end
   A student who reads a more negative number as a larger amount of energy picks this. −242 kJ/mol is lower than 0 kJ/mol, so going from the minimum to the separated atoms is an increase in potential energy.
3. CEnergy must be absorbed, because the potential energy rises as the atoms separate Correct
   Separating the atoms means moving from the minimum (−242 kJ/mol) toward large separation, where the potential energy approaches 0. The potential energy of the pair increases by about 242 kJ/mol, so that much energy must be supplied per mole of Cl₂: bond breaking absorbs energy.
4. DNo energy is needed, because the separated atoms have a potential energy of zero
   A student who takes the value at the end point for the change picks this. The energy needed is the change from −242 kJ/mol to 0 kJ/mol, an increase of 242 kJ/mol; zero is only the reference level for atoms that do not interact.

CED 2.2.A.1 · Read this in Fix

1. ACl₂, because its potential-energy well is the deeper of the two wells Correct
   The bond energy is the depth of the minimum below the energy of the separated atoms: about 242 kJ/mol for Cl₂ and about 155 kJ/mol for F₂. The Cl–Cl bond is longer, because Cl has the larger core, yet it is the stronger bond.
2. BF₂, because its bond is shorter, and shorter bonds are stronger
   A student who thinks the shorter of any two bonds is the stronger one picks this. Shorter-and-stronger goes with a higher bond order between the same atoms; here the graph shows the shorter F–F bond with the shallower well.
3. CF₂, because its lowest potential energy is the larger of the two numbers
   A student who takes the bond energy as the signed value at the minimum picks this. −155 kJ/mol is a larger number than −242 kJ/mol, but the bond energy is the depth of the well: about 155 kJ/mol for F₂ and about 242 kJ/mol for Cl₂.
4. DNeither, because both are single bonds with equal bond energy
   A student who thinks every single bond has the same bond energy picks this. Both are single bonds, but the graph shows wells of different depths; bond energy depends on the atoms as well as the bond order.

Working Read each minimum: F₂ about −155 kJ/mol at about 142 pm; Cl₂ about −242 kJ/mol at about 199 pm. Bond energy = depth of the minimum below 0 (the energy of the separated atoms): F₂ about 155 kJ/mol, Cl₂ about 242 kJ/mol. Cl₂ has the larger bond energy although its bond is the longer one.

CED 2.2.A.1 · Read this in Fix

1. AThe increase in potential energy from the curve's lowest point to complete separation Correct
   The bond energy is the energy required to separate the atoms: the increase in potential energy from the bottom of the curve (the minimum) to the energy of the atoms far apart. It is a positive quantity, equal to the depth of the minimum below the energy of the separated atoms.
2. BThe value of the potential energy at the minimum, including its negative sign
   A student who copies the coordinate of the minimum as the bond energy picks this. The bond energy is the energy that must be absorbed, a positive amount equal to the depth of the minimum below the energy of the separated atoms.
3. CThe energy released as the atoms move from the minimum to a large separation
   A student who thinks breaking a bond releases stored energy picks this. Moving from the minimum to large separation raises the potential energy, so energy is absorbed, not released.
4. DThe value of the potential energy when the two atoms are very far apart
   A student who takes the value at one point for the change between two points picks this. The potential energy of separated atoms is the reference level (usually 0); the bond energy is the difference between it and the minimum.

Working Bond energy = energy required to separate the bonded atoms = (potential energy of the separated atoms) − (potential energy at the minimum). With the separated atoms at 0, this is the depth of the minimum, a positive quantity.

CED 2.2.A.1 · Read this in Fix

1. AOne-half as great in Y as in X
   A student who treats the force as inversely proportional to the distance (not its square) picks this. Coulomb's law has r² in the denominator, so doubling r divides the force by 4.
2. BExactly as great in Y as in X
   A student who thinks only the ion charges matter picks this. The charges are equal, but the force also depends on the distance between the ion centers, which is twice as great in Y.
3. CTwice as great in Y as in X
   A student who thinks larger ions attract more strongly picks this, scaling the force up with the size of the ions. The larger ions keep the charges farther apart, so the force is smaller, one-fourth as great.
4. DOne-fourth as great in Y as in X Correct
   The force varies as 1/r². With the same charges and twice the distance, the force is (1/2)² = 1/4 as great: larger ions attract more weakly.

Working F ∝ q₁q₂/r². The charges are unchanged and r_Y = 2r_X, so F_Y/F_X = (r_X/r_Y)² = (1/2)² = 1/4. The force in Y is one-fourth as great.

CED 2.2.A.3.ii · Read this in Fix

1. A4.0
   A student who thinks only the charges matter picks this: 4 × 1 = 4.0. The ions in pair 2 are also closer together, which makes the attraction 2.25 times stronger again.
2. B4.5
   A student who uses the total charge of the two ions instead of the product picks this: (2 + 2)/(1 + 1) = 2, and 2 × 2.25 = 4.5. The force depends on q₁q₂, which rises from 1 to 4.
3. C6.0
   A student who treats the force as inversely proportional to the distance, not its square, picks this: 4 × 1.5 = 6.0. The distance enters as r², giving a factor of 2.25, not 1.5.
4. D9.0 Correct
   The force is proportional to the product of the charges and to 1/r². The charge product rises from 1 to 4, and the distance ratio 0.30/0.20 = 1.5 squared is 2.25, so the force is 4 × 2.25 = 9.0 times as great.

Working F ∝ q₁q₂/r². F₂/F₁ = [(2 × 2)/(1 × 1)] × (0.30 nm/0.20 nm)² = 4 × 2.25 = 9.0.

CED 2.2.A.3 · Read this in Fix

1. AKBr will melt at the higher temperature, because KBr has the greater formula mass
   A student who thinks heavier ions attract more strongly picks this. Mass does not appear in Coulomb's law; KBr has the weaker attraction between its ions and melts at the lower temperature (about 734 °C, compared with about 858 °C for KF).
2. BKF will melt at the higher temperature, because F⁻ is smaller than Br⁻ Correct
   Both compounds contain 1+ and 1− ions, so the difference comes from ion size. F⁻ is smaller than Br⁻, so the ion centers in KF are closer together and the ions attract more strongly; more energy is needed to separate them. Measured melting points are about 858 °C for KF and 734 °C for KBr.
3. CKBr will melt at the higher temperature, because the larger Br⁻ attracts K⁺ more
   A student who thinks larger ions attract more strongly picks this. Br⁻ has the same 1− charge as F⁻ but is larger, so the K⁺ and Br⁻ centers are farther apart and their attraction is weaker.
4. DBoth will melt at about the same temperature, because the ion charges are equal
   A student who thinks the attraction depends only on the ion charges picks this. The charges are equal, but the distance between the ion centers is smaller in KF, so KF has the stronger attraction and the higher melting point.

CED 2.2.A.3.ii · Read this in Fix

1. AH–Cl is the longest, because the Cl nucleus has the fewest protons to attract the shared pair
   A student who thinks more protons always pull the shared electrons closer picks this, reasoning that Cl, with the fewest protons of the three halogens, holds the shared pair least tightly. Cl also has the fewest occupied shells and so the smallest core: H–Cl (about 127 pm) is the shortest of the three bonds.
2. BAll are equal in length, because each is a single bond made of one shared pair of electrons
   A student who thinks every single bond has the same length picks this. The bond order is the same, but the halogen atoms differ in the size of their cores, so the lengths differ: about 127 pm, 141 pm and 161 pm.
3. CH–I is the longest, because the I atom has the largest core of the three halogens Correct
   The three bonds have the same bond order, so the size of the atoms' cores decides. I has more occupied electron shells than Br, and Br more than Cl, so its core is the largest and the shared pair lies farthest from its nucleus. Measured bond lengths are about 127 pm (H–Cl), 141 pm (H–Br) and 161 pm (H–I).
4. DH–Cl is the longest, because the Cl atom is the lightest and attracts the H atom the least
   A student who thinks heavier atoms attract more strongly picks this, reasoning that the lightest halogen holds the H atom least closely. Mass plays no part in the attractions within a bond; Cl has the smallest core of the three halogens, so H–Cl (about 127 pm) is the shortest bond.

CED 2.2.A.2 · Read this in Fix