MCAT Chemistry Review

+3

C. Neutrons are slightly heavier than protons, which are much heavier than electrons.

It is the shell. A shell is equivalent to a row in the periodic table. It is designated n and can take the value of any positive number.

The subshell and can take any integral value from 0 to n-1. It has the designation of l. Each subshell has a letter designation: 0=s, 1=p, 2=d, 3=f

It is the orbital, or the region surrounding an atom's nucleus where an electron is most likely to be. It can be any integral value from -l to l including zero. There are 2l+1 orbitals in the l subshell.

It is the electron spin state. It is experimentally derive and it can either be +1/2 or -1/2 (they either spin one direction or the other direction).

• Using the Aufbau principle (follow the arrows)
•        ↙
• 2   1s ↙  ↙
• 2   2s 2p ↙  ↙
• 8   3s 3p 3d ↙  ↙
• 8   4s 4p 4d 4f ↙
• 18 5s 5p 5d 5f
• 18 6s 6p 6d

Add the numbers in the left column for number of electrons (at most) at each stage.

C.

It increases as you move to the right and upward of the periodic table. So does ionization energy and electron affinity.

Alkali metals

Alkaline earth metals

Halogens

Noble gases

Columns 1 and 2 (including Helium)

Groups 13-18 (IIIA-VIIIA)

Groups 3-12 (IB-VIIIB) - the transition metals

B. Choices A and B represent true statements, however, these statements do not answer the question. Choice C is true, since the second ionization energy of any atom is higher than the first ionization energy. However, it fails to address why a +2 cation is not formed, since it would imply that a +1 cation is the most common ion for scandium. Choice D is the correct answer, since the most common cation formed from scandium is the +3 ion. Therefore, the +2 ion is not formed because if it was formed ,it would be immediately transformed into the +3 ion, which corresponds to a noble gas configuration for scandium.

D. Metallic character has to do with the tendency of an atom to give up electrons. A low first ionization energy indicates a valence electron that can be easily lost. Electron affinity is the willingness of an atom to gain an electron. While it may seem straightforward to assume that an atom that is less willing to give up an electron must be more willing to take an electron, it is not quite so clear. By knowing the first ionization energy of an atom, we can say nothing about the electron affinity of that atom.

B. This requires integration of organic chemistry knowledge with inorganic chemistry knowledge. The MCAT will frequently require this. First, to eliminate choices B and D, we need only to consider the fact that in removing electrons, cations are made, not anions. From organic chemistry, remember that a tertiary cation is more stable than a secondary cation, a secondary cation is more stable than a primary cation, and a primary cation is more stable than a methyl cation.

C. Iron has an atomic number of 26. Therefore, Fe²⁺ has 24 electrons and Fe³⁺ has 23 electrons. An atom, or ion, with an odd number of electrons cannot be diamagnetic (all electrons paired). For this reason, choices B and C are eliminated. Since iron is filling a d subshell, it has five orbitals to fill before it will electron pair. Fe²⁺ has six electrons past the nearest noble gas configuration. It is irrelevant whether two of these electrons are paired in the 4s orbital, or one elctron is in the 4s orbital, or no electrons are in the 4s orbital. In any case, the remaining 4, 5, 6 electrons, respectively, will not all pair in the 3d orbitals. Therefore, Fe²⁺ is paramagnetic, and the answer must be D.

A. The electron configuration given is the ground state electron configuration for a calcium atom. This electron configuration also represents possible electron configurations for the dipositive titanium ion and the pentapositive manganese ion. In its ground state, the potassium atom (K) has only one electron in the 4s subshell; therefore, its electron configuration is 1s²2s²2p⁶3s²3p⁶4s¹.

A. The nitrogen molecule will not have an overall dipole moment, nor will it have a bond dipole moment. The two nitrogens are equivalent, and they share a pure covalent bond. Methane (CH₄) will not have an overall dipole moment due to the symmetry of the molecule; however, it will have an individual bond dipole in each C-H bonds. These dipoles result from the difference in electronegativity of carbon and hydrogen. It is important to recognize that carbon is slightly more electronegative than hydrogen, so each dipole points in the direction of the carbon atom. Sodium has an electronegativity of 0.92 on the Pauling scale (relative to fluorine, which has an electronegativity of 4.0). The electronegativity of hydrogen is 2.20. The difference in electronegativity between these two atoms is therefore 1.27. Recalling that an electronegativity difference of greater than 1.7 is needed for an ionic bond, it can be stated that this is a polar covalent bond.  The electronegativity difference between hydrogen and oxygen is 1.24. This is close to the difference between sodium and hydrogen, but not quite as large. (You do not need to know the values of electronegativity for the MCAT, but must recognize that electronegativity is a periodic trend that increases from bottom to top and left to right. Also, imagine hydrogen as falling between boron and carbon in the periodic table. Using this scenario, without resorting to electronegativity values, it can be seen that Na-H is the most polar bond in the above question.

D. Because silver is more electronegative than sodium, we expect sodium to maintain itself as a 'naked' cation (not accepting electron density from the water). This leads to a large ion-dipole interaction between the sodium and the water. It is true that the sodium cation experiences ion-dipole interactions with the water; however, it does not provide the complete explanation of the problem at hand, since the silver cation also experiences ion-dipole interactions with the water. What is a polar molecule; as such, it will dissolve polar entities (the more polar, the better). The extremely low electronegativity of sodium makes the sodium cation a better "positive charge" than the silver cation.

• 1) ion-ion
• 2) ion-dipole
• 3) dipole-dipole
• 4) ion-induced dipole
• 5) dipole-induced dipole
• 6) induced dipole-induced dipole
• Note: even molecules that do not possess a permanent dipole have interatomic or intermolecular forces that are created by induced dipole-induced dipole interactions. These forces, also known as London dispersion forces, are the weakest intermolecular forces.

• A hydrogen bond consists of two components: a hydrogen atom attached to an electronegative atom X and a second electronegative atom Y that contains a lone electron pair. Since the hydrogen that is attached to X is participating in a net dipole with X, the hydrogen posses a net partial positive charge. The electrons that form the bond between the hydrogen and the electronegative atom spend a majority of time associated with the electronegative atom
• Y: →H-X ⇒ Y---H−X
• Note: On the MCAT, the term hydrogen bond is generally restricted to situations in which hydrogen interacts, as shown, with nitrogen (N), fluorine (F) or oxygen (O). Hydrogen bonding, and the degree of hydrogen bonding that occurs within a given substance, tends to prevent its molecules from undergoing physical separation, and this tends to increase its boiling point.

D. is correct. Both molecules can form hydrogen bonds with themselves, since they both contain a hydrogen attached to an electronegative oxygen atom, and they both contain an electronegative oxygen with lone pairs to donate to the hydrogen. The hydrogen bonds I water are much stronger than the hydrogen bonds in ethanol, which makes the boiling point of water higher than the boiling point of ethanol. Ethanol does have a higher molecular weight than water and is expected to have a higher boiling point but does not solve the problem that has been presented.

• B. is correct. If we consider the Lewis do structures of each of these molecules, we see:
•            ..   ..    ..   ..     .. ..            
•    H-H  N=N    O=O    :F-F:
•                       ..   ..     .. ..
• Both hydrogen gas and fluorine gas have single bonds. Oxygen gas has a double bond. Nitrogen gas has a triple bond.

• ammonium - NH₄⁺
• nitrate - NO₃^-
• carbonate - CO₃^2-
• sulfate - SO₄^2-
• phosphate - PO₄^3-

C. is correct. Envision a structure with a central phosphorus atom and four attached oxygen atoms. Since the phosphate ion has a net charge of -3, only one of these oxygens will have a double bond to phosphorus in any resonance structure, and each of the other three oxygens will form a single bond with phosphorus. Since there are four different oxygen atoms that could form the double bond, there are four resonance structures.

A. is correct. Since aluminum is a group 13 element, it has a valence state of 3⁺. Sulfate ion should be memorized as SO₄^2- so it has a charge of 2^-. From this it can be seen that two aluminum ions (3⁺ x 2 = 6⁺) are needed for every three sulfate ions (2^- x 3 = 6^-).

• Linear
• _       _
• O=C=O

• Trigonal Planar
•                  _
•                 |F|
•                  |
•                  B
•                /  
•             |F|   |F|

• Bent (Trigonal Bent)
•         ..
•         N
•      //  
•    |O|  |O|

• Tetrahedral
•         H
•         |
•         C
•       / | 
•     H  H   H
• The middle H on the bottom should be coming out of the C toward the reader.

• Trigonal Pyramidal
•          ..
•          N
•       / |  
•     H  H   H
• The middle H on the bottom should be coming out of the N toward the reader.

• Bent (Tetrahedral, Bent)
•      ..
•      O:
•    / /
•   H H
• The second H should be coming out of the O toward the reader.

• 1) Trigonal Bipyramidal
•              Cl
•               |
•         Cl - P - Cl
•             /  |
•          Cl   Cl
• 2) Seesaw
•               F
•               |
•             : S - F
•             /  |
•           F    F
• 3) T-shaped
•               F
•               |
•         F - Cl - F
•           ..    ..
• 4) Linear
•              ..
•         F - Xe - F
•            ..    ..

• 1) Octahedral
•            F    F
•            |   /
•       F - S - F
•         /  |
•       F    F
• 2) Square Pyramidal
•            F    F
•            |   /
•       F - Br - F
•         /  ..
•       F
• 3) Square Planar
•                   F
•             ..  /
•       F - Xe - F
•          /  ..
•         F

D. is correct. Mn - 7 valance electrons; since three are allocated to bonding with the three bromine atoms, there are four nonbonding electrons, or two lone pairs.

• Synthesis: A + B ->AB
• Decomposition (analysis): AB -> A + B
• Single replacement: AB -> AX + B
• Double replacement: AX + BY -> AY + BX
• Oxidation-reduction: may take the form of any of the four other classes, and is when the oxidation number of an atom changes during a reaction; a redox reaction.

C. is correct. This reaction is an example of single replacement reaction.

C. is correct. This reaction is an example of a synthesis reaction.

A. is correct. This reaction is an example of a double replacement reaction.

• A. is correct. Balance the equation
•      FeBr₃ + NaOH → Fe(OH)₃ + NaBr
•      FeBr₃ + NaOH → Fe(OH)₃ + 3NaBr
•      FeBr₃ + 3NaOH → Fe(OH)₃ + 3NaBr
• Now that the equation is balanced, it can be seen that the ratio of Fe(OH)₃ to NaBr is 1:3.

It is the reagent that is present in the smallest quantity (based on equivalents). It limits the reaction.

• First, convert gram masses into moles:
•   28gN₂/28g/mol = 1 mole N₂
•   4gH₂/2g/mol = 2 mole H₂
• Next, divide each molar quantity by the appropriate stoichiometric coefficient to arrive at a number of "equivalents" (the amount of each substance that is needed to balance the reaction):
•   N₂ + 3H₂ → 2NH₃
•       1mol N₂/1mol/equiv = 1 equivalent N₂
•       2mol H₂/3mol/equiv = 0.667 equivalent H₂
• Since there are fewer equivalents of hydrogen gas than of nitrogen, hydrogen is the limiting reagent.
•    To find ammonia produced, multiply the number of equivalents of the limiting reagent by the stoichiometric coefficient of ammonia.
•   0.667equiv x 2mol NH₃/equiv = 1.333mol NH₃
•   1.333mol x 17g/mol = 22.667g NH₃

• D. is correct.
• First, write out a balanced equation for the reaction:
• CaCO₃ + 2NaOH → Ca(OH)₂ + Na₂CO₃
• Next, calculated equivalents for each reactant:
• 100g CaCO₃ x 1mol/100g = 1 mol CaCO₃
• 1mol CaCO₃ x 1 eq/1mol = 1 equiv CaCO₃
• 100g NaOH x 1mol/40g = 2.5 mol NaOH
• 2.5mol NaOH x 1 eq/2mol = 1.25 equiv NaOH
• Since CaCO₃ is available in the least number of equivalents, it is the limiting reagent.

• +1
• +2
• +3
• -1
• -2, -1
• +1, -1

• 1) F^-
• 2) Cl^-
• 3) Br^-
• 4) I^-
• 5) O^2-

• 1) S^2-
• 2) N^3-
• 3) OH^-
• 4) NO₃^-
• 5) ClO4^-

• 1) CO₃^2-
• 2) SO₄^2-
• 3) PO₄^3-
• 4) CH₃CO₂^-

Highly charged species (ex: PO₄^3- →  ALPO₄) have a greater force of attraction thus are much less soluble in water than those with little charge (NaCl)

• 1) Na⁺
• 2) Li⁺
• 3) K⁺
• 4) NH₄⁺
• 5) H₃O⁺

• 1) H⁺
• 2) Ca²⁺
• 3) Mg²⁺
• 4) Fe²⁺
• 5) Fe³⁺

• Oxygen has an oxidation number of -2.
• Potassium (being a group I) has an oxidation number of +1
•   K  2 x +1 = +2
•   O  7 x -2 = -14
•              = -12 total
• since the sum must be zero, Cr must balance
•   Cr 2 x __ = +12; so Cr must be +6

• A. is correct
• H = +1 x 1 = +1
• O = -2 x 4 = -8
•              = -7
• So, I = +7 to balance

• First, determine which atoms have been oxidized and which have been reduced.
• K₂  Cr₂  O₇ + H  I  →  K  I + Cr  I₃ + I₂ + H₂  O
• +1  +6  -2  +1  -1    +1 -1  +3  -1   0    +1  -2
• Note that chromium is reduced and iodine oxidized 
•     Cr⁶⁺ + 3e^- → Cr³⁺      and
•     2I^1- → I₂ + 2e^-
• These two reactions are not balanced in terms of electrons:
•   2 x (Cr⁶⁺ + 3e^- → Cr³⁺)
•   3 x (2I^1- → I₂ +2e^-)
• giving
•   2Cr⁶⁺ + 6I^1- → 2Cr³⁺ + 3I₂
• Therefore
•   K₂Cr₂O₇ + 6HI → KI +2CrI₃ + 3I₂ + H₂O
• Now balance for potassium, oxygen, and hydrogen:
•   K₂Cr₂O₇ + 14HI → 2KI + 2CrI₃ + 3I₂ + 7H₂O

• A. is correct.
• Cu + HNO₃ → Cu(NO₃)₂ + H₂O + NO
• (0) (+1)(+5)(-2) →(+2)(+5)(-2) (+1)(-2) (+2)(-2)
• Look at the redox:
• 3(Cu → Cu²⁺ + 2e^-) - oxidation
• ?(N⁵⁺ + 3e^- → N²⁺) - reduction
• However, not all of the N⁵⁺ was reduced (which is why the ? is used in the above equation).
• 3Cu + 8HNO₃ → 3Cu(NO₃)₂ + 4H₂O + 2NO

For each elementary process, or step, there is an energy barrier that must be overcome in order for that step to proceed. This energy barrier is termed the activation energy for that step. The higher the activation energy of a step, the less likely that step is to occur and, therefore, the slower that step will be. For any mechanism, the step with the highest activation energy is the slowest step. The slowest step in a series of steps will limit the overall progression from reactants to products, so this step is termed the rate-determining step.

An important feature of elementary processes; while an overall chemical reaction may not be considered to be reversible, given the order and combination of mechanical steps, ALL elementary processes are reversible. It is this reversibility of elementary processes that is termed the principle of microscopic reversibility. It says that for every forward process, the reverse process also occurs in exactly the reverse manner.

A transition state is a high-energy species found at the peak of the reaction curve. Intermediates are found in the troughs of reaction curves.

C. is correct. Kinetics is not concerned with the stoichiometry of the reaction under investigation. Kinetics is affected by catalysts and other reaction conditions, such as temperature and pressure. Kinetics is concerned with the mechanism of reactions. Mechanisms of reactions are comprised of steps, or elementary processes, which are all reversible.

D. is correct. The rate of reaction is given by either the rate of disappearance of reactants in a given time or the rate of formation of products in a given time. Choices Δ[NO]/2Δt and Δ[O₂]/Δt would be rate of reaction expressions for the reverse reaction. The last choice is nonsense.

• Rate = k[A]^x[B]^y
• where [A] and [B] are the concentrations of A and B, respectively The value of x is the order of the reaction with respect to A and the value of y is the order of the reaction with respect to B. The values of x and y must be experimentally determined and bear no relationship to the stoichiometric numbers a and b.
• The overall order of this reaction is given by the sum of x and y.

• The specific rate constant is denoted by the letter k in the following equation
•  rate = k[A]^x[B]^y and
• The value of k is unique to each reaction at a given temperature.
• The value of k will change if the temperature is changed.
• The value of k does not change with time.
• The value of k is not dependent on the concentrations of either reactants or products.
• The value of k must be determined experimentally.
• The units of k depend on the overall order of the reaction.
• ** The magnitude of the change in the rate constant for different temperatures is DIRECTLY PROPORTIONAL to the activation energy of the reaction -- the larger the activation energy, the greater the change in rate constant for the same temperature change.

D. is correct. Be careful of the double negative questions on the MCAT. Try rephrasing the question to:"Which of the following affects k?"

• B. is correct. Given the equation
• 
• Because temperature is measured in Kelvin, the temperature term must always be positive when temperature increases (T₂>T₁). By definition, the activation energy, E_a, is always positive, as is the gas constant, R. Therefore, for increasing temperature, the right side of the Arrhenius equation is always positive. The equation can now be reduced to
•                log K=P
• where K=k₂/k₁ and P is meant to designate the positive value of the right side of the equation.
• Solving for K: K=10^P and
• since 10^P is always greater than I, K must always be greater than I, implying that the numerator (k₂) must always be greater than the denominator (k₁).

A. is correct. Catalysts operate by altering the activation energy of a reaction. The alteration in activation energy is achieved by providing an alternative mechanism -- via altered elementary processes. Catalysts do not affect equilibrium.

B. is correct. This question tests your understanding of a catalyst: they speed up the rate of a reaction (kinetics), they decrease the activation energy, they do not affect Keq, they are not used up in a reaction, and finally, they do not affect thermodynamics, ΔG°

• D. is correct. This question will be fully explored below using LDA as an example.
• This question presents two important points: (i) it describes a compound called LDA as hindered (= bulky) and (ii) its conjugate acid has a pK_a of about 40. Of course, pK_a equals -log K_a and K_a is the acid dissociation constant. A K_a which is extremely low (approx. 10-40), indicating that the conjugate acid is verrrrry weak, would give a pK_a of about 40. All to say, weak conjugate acid = strong base!
• Translation of this question: what does a strong bulky base like to do? Before we answer, what does a strong small base like to do in organic chemistry? They tend to engage in nucleophilic substitution reactions (i.e. OH-, CN-). A big bulky base usually has difficulty accessing carbon for a nucleophilic reaction. Instead, it takes the easy way out: it plucks protons off a molecule like oranges off a tree! Of course some H⁺'s are easier to remove than others. This depends on the stability of the suggested product. For example, removing a H⁺ from the methoxy substituent would create a primary carbanion (verrrry unstable) with no real stable options to get rid of the negative charge. On the other hand, examine the carbon in the ring which is attached to the methyl group. If you remove the only hydrogen attached to that 'ringed' carbon (= a-hydrogen which is happy to be plucked, you get a tertiary carbanion. Furthermore, you get a logical ensuing reaction: the negative charge is quickly attracted to the carbonyl carbon which is delta positive thus forming a double bond. Simultaneously, the pi electrons from the carbonyl group are kicked up to the oxygen creating an enolate anion (almost identical to keto-enol tautomerism,).

A. is correct. The sum of the mole fractions of each species present must be unity (= 1). Therefore, the partial pressure of O₂ is 1 - (1/6 + 1/2) = 1/3. The partial pressure of a species is given by the product of its mole fraction and the total pressure of the system. Therefore, the answer is given by 1/3 x 1 atm = 0.33 atm, approximately.

• D. is correct. Reaction I is exothermic (ΔH is negative ∴ heat is released). The reaction could be written as:2SO₂(g) + O₂(g) ↔ 2SO₃(g) + Heat
• Adding heat (increasing the temperature) adds to the right hand side of the equilibrium forcing a shift to the left in order to compensate. The reverse occurs by decreasing the temperature thus creating a shift to the right which would produce more SO₃(g) and more heat is released.

• D. is correct. Since we are told to consider the first ionization of H₂SO₃ as if from a strong acid (Reaction II), we assume that the first proton completely dissociates. However, the second proton ionizes as if from a very weak acid (Reaction III). After all, a Ka ≈ 10-6 means that the product of the reactants is about 1 000 000 (one million!) times greater than the product of the products (which includes the second proton; for Ka see CHM 6.1). The preceding fact combined with the imprecision of the available multiple choice answers means that our answer can be estimated by assuming that H2SO3 acts as a strong monoprotic acid like HCl.
• Therefore, one proton completely dissociates while the second proton's concentration is relatively negligible; thus [H+] = 0.01 mol dm-3 (1 dm-3= 1 L-1, see App. B.2) The pH is equal to the negative logarithm of [H+] = -log (0.01) = -log (10-2) = 2.

• A. is correct. The passage states that the refractive index is equal to the ratio of the velocities of light in a vacuum and in the medium.
• n_medium = c / v_medium
• Since the value for n_medium is 1.00 when the medium is air, the velocity of light in air (v_medium) must be approximately that of light in a vacuum (that is, "c").

• B. is correct.  Since MgCO₃ ↔ Mg₂+ + CO₃^2-,
• K_sp = [Mg2+][CO₃^2-] = s x s = s² where s is the solubility.
• From the table provided, s = 1.30 x 10-3 mol L-1. Thus K_sp = [Mg2+][CO32-] = s2 = (1.30 x 10-3)2 = 1.69 x 10-6 ≈ 1.7 x 10-6. {Mathfax: (13)2= 169 ∴ (1.3)2 = 1.69; also, to increase speed you should at least be able to recognize the squares of numbers between 1-15, i.e. (12)² = 144; rules of exponents.

• A. is correct. More moles.
• Consider these two equations:
• (i) CaSO₄ ↔ Ca₂⁺ + SO₄^2-
• (ii) Ca(OH)₂ ↔ Ca₂+ + 2OH^-
• On the Surface: for the same K_sp, the more ions produced, means the more soluble
• (Ca(OH)₂ produces 3 ions as opposed to just 2). Though temperature is indeed related to solubility, it would affect both compounds in a similar way so having 3 ions will still mean that it is more soluble than just 2 ions.
• Going Deeper: If [Ca₂⁺] = s = solubility of salt, then:
• For (i): s² = K_sp, therefore s = square root of K_sp
• For (ii): (2s)² x s = K_sp, therefore s = cube root of (1/4 x K_sp)Since K_sp is always less than 1 for sparingly soluble salts (i.e. see the table provided in the problem), the value for s obtained in (ii) necessarily is greater than that obtained in (i). {For fun(!), work out values for 's' in (i) and (ii) using any value for K_sp less than 1}

B. is correct. SrCO₃ is actually less soluble than the salt below it (see the table in the question) despite the fact that Ba is situated lower down Group II in the periodic table. Thus the explanation given in the question text has to be considered to explain the stability or extremely low solubility of SrCO₃.

B. is correct. A non-volatile solute causes a lowering of the vapor pressure of the solvent and hence an elevation of the boiling point. The preceding is a colligative property.

B is correct. This answer is a typical phase diagram for a substance. The exceptions to this rule include substances which exhibit larger intermolecular forces than usual, for example hydrogen bonding. These include water and ammonia, which would yield a phase diagram such as that in answer phase diagram A.

B. is correct. The text says that the layers in graphite can move relative to one another. Hence, it can be used as a lubricant. No evidence is provided for any of the other properties as they relate to the layered structure of graphite.

• B. is correct. On the Surface: Don’t convert mL to litres because it will cancel out.
• x = molarity of Mg₃(PO₄)₂ --> Solve for x
• (.005M Mg²⁺)(5mL)+(3x)(15mL) =
• (20mL)(.05M Mg²⁺) 
• Note the 3 in front of x because there are 3 potential Mg²⁺ generated from each Mg₃(PO₄)₂ in aqueous solution.
• .025 + 45x = 1 
• 45x = 0.975 thus x = 0.022
• Going Deeper: a detailed calculation . . . Let's begin with the total number of moles of Mg²⁺ present in the final solution: 0.05 moles/L x 0.02 L = 0.001 moles of Mg²⁺. Next, let's look at the number of moles of Mg²⁺ obtained from MgCl₂: 0.005 moles/L x 0.005 L = 0.000025 moles of Mg²⁺. Now we know the number of moles of Mg²⁺ we need supplied from Mg₃(PO₄)₂: (0.001 - 0.000025) moles = 0.000975 moles. Thus from the 15 mL of Mg₃(PO₄)₂ we need 0.000975 moles of Mg²⁺. But each mole of Mg₃(PO₄)₂ contains 3 moles of Mg²⁺ Therefore, the concentration of Mg₃(PO₄)₂ = [(0.000975 moles)/(0.015 L)] x 1/3 = 0.022 mol/L= 2.2 x 10^-2 M.{Note: the math is much faster if you notice that the answer to the second step is negligible thus you can skip the third step and the final calculation is easier}

• D. is correct. The reduced species of the electrochemical equilibrium with the most negative E^o value is the strongest reducing agent. Memory aside (!), it is of value to note that a reducing agent reduces the other substance, thus a reducing agent is oxidized. Note that only answer choices Cr²⁺ and Mn²⁺ are oxidized (= lose electrons). When you write the two relevant equations as oxidations, instead of reductions like the table provided, you will note that only answer Cr²⁺ has a positive E^o value indicating the spontaneous nature of the reaction. The table provided demonstrates half-reactions written as reduction potentials. In order to write the oxidation, simply reverse the reaction and change the sign of E^o:
• Oxidation: Cr²⁺ ↔ Cr³⁺ + e^- E^o = 0.410

A. is correct. As one moves across the periodic table, the atomic radius decreases as a result of the increasing effective nuclear charge (without an increase in the number of atomic orbitals). In other words, the nucleus becomes more and more positive from left to right on the periodic table resulting in the drawing of negatively charged orbital electrons nearer and nearer to the nucleus. As a result, atoms will accept electrons more readily as we go across the periodic table and the electron affinity (EA) becomes more negative (less positive). For example, halogens have very negative EA values because of their strong tendencies to form anions. Alkaline earths have positive EA values.