CHEM (EXAM 4) flashcards

All 50 cards

1. Card 1

   Balance the following redox reaction in acidic solution using the half-reaction method: MnO₄⁻(aq) + C₂O₄²⁻(aq) → Mn²⁺(aq) + CO₂(g).

   First, separate into half-reactions: Oxidation: C₂O₄²⁻ → CO₂; Reduction: MnO₄⁻ → Mn²⁺. Balance atoms other than O and H, then O with H₂O, then H with H⁺, and finally charge with electrons. The balanced equation is 2MnO₄⁻(aq) + 5C₂O₄²⁻(aq) + 16H⁺(aq) → 2Mn²⁺(aq) + 10CO₂(g) + 8H₂O(l).

2. Card 2

   Define the terms 'electrochemical cell', 'voltaic (galvanic) cell', and 'electrolytic cell', and distinguish between them based on spontaneity and energy conversion.

   An electrochemical cell is a device that either generates electrical energy from a chemical reaction or uses electrical energy to drive a non-spontaneous chemical reaction. A voltaic (galvanic) cell is a type of electrochemical cell that generates electricity from a spontaneous chemical reaction. An electrolytic cell uses electrical energy to drive a non-spontaneous chemical reaction.

3. Card 3

   A galvanic cell is constructed with a zinc electrode in a 1.0 M Zn(NO₃)₂ solution and a copper electrode in a 1.0 M Cu(NO₃)₂ solution. Sketch this cell, identifying the anode, cathode, salt bridge, and the direction of electron flow.

   The sketch would show two beakers connected by a salt bridge. The zinc electrode (anode) is in the Zn(NO₃)₂ solution, and the copper electrode (cathode) is in the Cu(NO₃)₂ solution. Electrons flow from the zinc electrode (anode) through an external wire to the copper electrode (cathode). The salt bridge allows ion migration to maintain charge neutrality.

4. Card 4

   A current of 0.50 A flows through a wire for 30 minutes. Calculate the total charge that passes through the wire in Coulombs.

   Charge (Q) is calculated as current (I) multiplied by time (t). First, convert time to seconds: 30 minutes * 60 seconds/minute = 1800 seconds. Then, Q = I * t = 0.50 A * 1800 s = 900 C. Therefore, 900 Coulombs of charge pass through the wire.

5. Card 5

   Explain the significance of the standard hydrogen electrode (SHE) in determining standard electrode potentials for other half-reactions.

   The standard hydrogen electrode (SHE) serves as the reference electrode with an assigned standard electrochemical potential of 0.00 V. By connecting any other half-cell to the SHE, the measured cell potential directly corresponds to the standard electrode potential of that specific half-reaction, allowing for a standardized comparison of reduction potentials.

6. Card 6

   Given the standard reduction potentials: Ag⁺(aq) + e⁻ → Ag(s), E° = +0.80 V, and Fe³⁺(aq) + e⁻ → Fe²⁺(aq), E° = +0.77 V. Calculate the standard cell potential for a galvanic cell involving these two half-reactions and predict the direction of spontaneity.

   To form a galvanic cell, one half-reaction must be reversed to act as oxidation. Since Ag⁺ has a higher reduction potential, it will be reduced, and Fe²⁺ will be oxidized. The oxidation half-reaction is Fe²⁺(aq) → Fe³⁺(aq) + e⁻, with E° = -0.77 V. The standard cell potential E°cell = E°cathode - E°anode = +0.80 V - (+0.77 V) = +0.03 V. Since E°cell is positive, the reaction is spontaneous in this direction.

7. Card 7

   For a redox reaction, if the standard change in free energy (ΔG°rxn) is -150 kJ/mol, calculate the standard potential for the electrochemical cell (E°) at 25°C. Assume n = 2 for the reaction. (Faraday's constant F = 96485 C/mol).

   The relationship is ΔG°rxn = -nFE°. Rearranging for E°: E° = -ΔG°rxn / (nF). E° = -(-150,000 J/mol) / (2 mol e⁻ * 96485 C/mol e⁻) = 150,000 J / 192,970 C = 0.777 V. The standard cell potential is approximately +0.78 V.

8. Card 8

   A concentration cell is set up with two Ag/Ag⁺ half-cells. One half-cell has [Ag⁺] = 0.10 M and the other has [Ag⁺] = 0.0010 M. Calculate the cell potential (E) at 25°C using the Nernst equation. (R = 8.314 J/(mol·K), F = 96485 C/mol).

   For a concentration cell, E°cell = 0 V. The Nernst equation is E = E° - (RT/nF)lnQ. Here, n=1. Q = [Ag⁺]anode / [Ag⁺]cathode. The more dilute solution is the anode. Q = 0.0010 M / 0.10 M = 0.01. E = 0 - (8.314 * 298 / (1 * 96485))ln(0.01) = -0.0257 * (-4.605) = 0.118 V. The cell potential is approximately 0.12 V.

9. Card 9

   Define a 'nuclide' and distinguish it from an 'isotope'. Provide an example for each.

   A nuclide is a general term for an atom characterized by a specific number of protons and neutrons in its nucleus. For example, carbon-12 (¹²C) is a nuclide. Isotopes are nuclides of the same element (same number of protons) but with different numbers of neutrons. Carbon-12 and carbon-14 (¹⁴C) are isotopes of carbon.

10. Card 10

    Balance the following redox reaction in basic solution: Cr(OH)₃(s) + ClO⁻(aq) → CrO₄²⁻(aq) + Cl⁻(aq).

    First, balance in acidic solution then convert to basic. The balanced acidic equation is 2Cr(OH)₃(s) + 3ClO⁻(aq) + 4H⁺(aq) → 2CrO₄²⁻(aq) + 3Cl⁻(aq) + 5H₂O(l). To convert to basic, add 4OH⁻ to both sides to neutralize H⁺. The final balanced basic equation is 2Cr(OH)₃(s) + 3ClO⁻(aq) + 4OH⁻(aq) → 2CrO₄²⁻(aq) + 3Cl⁻(aq) + 5H₂O(l).

11. Card 11

    Which of the following statements correctly describes the relationship between the anode and oxidation in an electrochemical cell? A) The anode is where reduction occurs. B) Oxidation occurs at the anode in both voltaic and electrolytic cells. C) The anode is always positively charged. D) Oxidation occurs at the cathode in electrolytic cells.

    B) Oxidation occurs at the anode in both voltaic and electrolytic cells. The anode is defined as the electrode where oxidation takes place, regardless of whether the cell is galvanic (spontaneous) or electrolytic (non-spontaneous).

12. Card 12

    A student wants to deposit 5.0 g of silver from a solution containing Ag⁺ ions using electrolysis. If a current of 2.0 A is supplied, how long (in hours) will it take to deposit the silver? (Molar mass of Ag = 107.87 g/mol, Faraday's constant F = 96485 C/mol).

    First, calculate moles of Ag: 5.0 g / 107.87 g/mol = 0.04635 mol Ag. Since Ag⁺ + e⁻ → Ag, 1 mole of electrons is needed per mole of Ag. Total charge Q = moles of e⁻ * F = 0.04635 mol * 96485 C/mol = 4472 C. Time t = Q / I = 4472 C / 2.0 A = 2236 seconds. Convert to hours: 2236 s / 3600 s/hr = 0.62 hours. It will take approximately 0.62 hours.

13. Card 13

    Describe the key components of a galvanic cell and explain the role of each component in generating electricity.

    A galvanic cell consists of two half-cells, each containing an electrode immersed in an electrolyte. The anode is where oxidation occurs, releasing electrons. The cathode is where reduction occurs, consuming electrons. An external wire connects the electrodes, allowing electron flow, while a salt bridge connects the electrolytes, maintaining charge neutrality by allowing ion migration.

14. Card 14

    Consider a metal M. If M does not dissolve in acid (e.g., HCl), what can be inferred about its standard reduction potential relative to H⁺/H₂?

    If metal M does not dissolve in acid, it means that the reaction M(s) + nH⁺(aq) → Mⁿ⁺(aq) + n/2 H₂(g) is non-spontaneous. This implies that the standard reduction potential of Mⁿ⁺/M is greater than (more positive than) the standard reduction potential of H⁺/H₂, which is 0.00 V. Therefore, M is less easily oxidized than hydrogen.

15. Card 15

    The equilibrium constant (K) for a redox reaction at 25°C is 1.0 x 10¹⁰. If n = 2 for this reaction, calculate the standard cell potential (E°) for the electrochemical cell. (R = 8.314 J/(mol·K), F = 96485 C/mol).

    The relationship is E° = (RT/nF)lnK. E° = (8.314 J/(mol·K) * 298 K / (2 mol e⁻ * 96485 C/mol e⁻)) * ln(1.0 x 10¹⁰). E° = (0.02569 V / 2) * (23.026) = 0.012845 V * 23.026 = 0.2958 V. The standard cell potential is approximately 0.296 V.

16. Card 16

    A galvanic cell operates under nonstandard conditions. The reaction involves the transfer of 3 electrons, and the measured cell potential (E) is 0.85 V. If the standard cell potential (E°) is 0.92 V at 25°C, calculate the reaction quotient (Q).

    Using the Nernst equation: E = E° - (RT/nF)lnQ. Rearranging for lnQ: lnQ = (nF/RT)(E° - E). lnQ = (3 * 96485 C/mol * (0.92 V - 0.85 V)) / (8.314 J/(mol·K) * 298 K). lnQ = (289455 * 0.07) / 2477.572 = 20261.85 / 2477.572 = 8.178. Q = e⁸·¹⁷⁸ ≈ 3560. The reaction quotient Q is approximately 3.6 x 10³.

17. Card 17

    Define 'electromotive force' (EMF) and 'cell potential'. How are these terms related to the spontaneity of a redox reaction?

    Electromotive force (EMF) is the maximum potential difference between the two electrodes of a galvanic cell when no current is flowing, representing the driving force for electron flow. Cell potential (Ecell) is the potential difference between the electrodes under specific conditions (standard or nonstandard). A positive cell potential (or EMF) indicates a spontaneous redox reaction, while a negative value indicates a non-spontaneous reaction.

18. Card 18

    Write the nuclear symbol for an alpha particle and explain its composition.

    The nuclear symbol for an alpha particle is ⁴₂He or ⁴₂α. An alpha particle consists of two protons and two neutrons, which is identical to the nucleus of a helium-4 atom. It carries a +2 charge.

19. Card 19

    An electrolytic cell is used to plate nickel onto a metal object. If 1.5 A of current is passed through a NiSO₄ solution for 45 minutes, what mass of nickel (in grams) is deposited? (Molar mass of Ni = 58.69 g/mol, Faraday's constant F = 96485 C/mol).

    First, calculate total charge Q = I * t. Time in seconds: 45 min * 60 s/min = 2700 s. Q = 1.5 A * 2700 s = 4050 C. For Ni²⁺ + 2e⁻ → Ni, n=2 moles of electrons per mole of Ni. Moles of e⁻ = Q / F = 4050 C / 96485 C/mol = 0.04197 mol e⁻. Moles of Ni = 0.04197 mol e⁻ / 2 = 0.020985 mol Ni. Mass of Ni = 0.020985 mol * 58.69 g/mol = 1.23 g. Approximately 1.23 grams of nickel are deposited.

20. Card 20

    What is the primary difference between a voltaic (galvanic) cell and an electrolytic cell in terms of energy conversion and spontaneity?

    A voltaic (galvanic) cell converts chemical energy into electrical energy through a spontaneous redox reaction, meaning ΔG is negative and Ecell is positive. Conversely, an electrolytic cell uses electrical energy to drive a non-spontaneous redox reaction, meaning ΔG is positive and Ecell is negative, effectively converting electrical energy into chemical energy.

21. Card 21

    Represent the following electrochemical cell using cell notation: A galvanic cell with a lead electrode in a 1.0 M Pb(NO₃)₂ solution and a silver electrode in a 1.0 M AgNO₃ solution. Lead is oxidized.

    The cell notation should represent the anode (oxidation) on the left and the cathode (reduction) on the right, separated by a salt bridge (||). Since lead is oxidized, it's the anode. The notation is: Pb(s) | Pb²⁺(aq, 1.0 M) || Ag⁺(aq, 1.0 M) | Ag(s).

22. Card 22

    A concentration cell is designed using two half-cells with copper electrodes. One half-cell contains 0.010 M Cu²⁺ and the other contains 1.0 M Cu²⁺. Explain how this cell generates a potential difference and identify which half-cell acts as the anode.

    A concentration cell generates a potential difference because of the difference in ion concentrations, driving the system towards equilibrium. The half-cell with the lower concentration of Cu²⁺ (0.010 M) will act as the anode, where oxidation (Cu → Cu²⁺ + 2e⁻) occurs to increase the Cu²⁺ concentration. The half-cell with the higher concentration (1.0 M) acts as the cathode, where reduction (Cu²⁺ + 2e⁻ → Cu) occurs to decrease its Cu²⁺ concentration.

23. Card 23

    How does increasing the concentration of reactants in the cathode half-cell affect the cell potential (E) of a galvanic cell, according to the Nernst equation?

    According to the Nernst equation, E = E° - (RT/nF)lnQ. Increasing the concentration of reactants in the cathode half-cell would decrease the value of the reaction quotient (Q), assuming products concentrations remain constant or decrease. A smaller Q value leads to a less positive lnQ, which in turn makes the term (RT/nF)lnQ smaller (or more negative), thus increasing the overall cell potential (E). This shift drives the reaction more towards products, making the cell more efficient.

24. Card 24

    Define 'oxidation' and 'reduction' in terms of electron transfer and changes in oxidation states.

    Oxidation is the loss of electrons by a substance, resulting in an increase in its oxidation state. Reduction is the gain of electrons by a substance, resulting in a decrease in its oxidation state. These two processes always occur simultaneously in a redox reaction.

25. Card 25

    Write the symbols for a proton, neutron, and electron, including their mass number and atomic number.

    The symbol for a proton is ¹₁p or ¹₁H. The symbol for a neutron is ¹₀n. The symbol for an electron is ⁰₋₁e or ⁰₋₁β. These symbols indicate the mass number (superscript) and the charge or atomic number (subscript).

26. Card 26

    An electrolytic cell is used to deposit 5.00 g of copper from a CuSO4 solution. If a current of 2.00 A is applied, how long in minutes will it take to deposit the copper? (Molar mass of Cu = 63.55 g/mol, Faraday's constant = 96485 C/mol e-)

    First, calculate moles of copper: 5.00 g / 63.55 g/mol = 0.07868 mol Cu. The half-reaction for copper deposition is Cu2+ + 2e- → Cu, so 2 moles of electrons are needed per mole of copper. Moles of electrons = 0.07868 mol Cu * 2 mol e-/mol Cu = 0.15736 mol e-. Total charge Q = moles of e- * Faraday's constant = 0.15736 mol e- * 96485 C/mol e- = 15187 C. Time t = Q / I = 15187 C / 2.00 A = 7593.5 seconds. Converting to minutes: 7593.5 s / 60 s/min = 126.6 minutes.

27. Card 27

    Define a chelate and explain the significance of chelating agents in coordination chemistry.

    A chelate is a complex ion formed when a polydentate ligand binds to a central metal ion at two or more donor atoms, forming a ring-like structure. Chelating agents are significant because they form highly stable complexes due to the chelate effect, which is an entropy-driven phenomenon. This stability makes them useful in various applications, such as metal detoxification, water softening, and as catalysts.

28. Card 28

    A sample of an ancient artifact contains 12.5% of its original carbon-14. Given that the half-life of carbon-14 is 5730 years, calculate the age of the artifact.

    The first-order integrated rate law is ln(Nt/N0) = -kt, and the half-life equation is t1/2 = ln(2)/k. First, calculate the rate constant k: k = ln(2) / 5730 years = 0.693 / 5730 years = 1.209 x 10^-4 years^-1. Now, use the integrated rate law with Nt/N0 = 0.125: ln(0.125) = -(1.209 x 10^-4 years^-1) * t. Solving for t: -2.079 = -(1.209 x 10^-4 years^-1) * t, so t = 2.079 / (1.209 x 10^-4 years^-1) = 17196 years. Alternatively, 12.5% is 1/8, which means 3 half-lives have passed (100% -> 50% -> 25% -> 12.5%). So, age = 3 * 5730 years = 17190 years.

29. Card 29

    What is the primary difference between structural isomers and stereoisomers in coordination compounds?

    Structural isomers have the same molecular formula but different connectivity of atoms, meaning the atoms are bonded in different sequences. Stereoisomers, on the other hand, have the same molecular formula and the same connectivity of atoms, but they differ in the spatial arrangement of their atoms. Examples of structural isomers include coordination and linkage isomers, while stereoisomers include geometric and optical isomers.

30. Card 30

    A research team is attempting to synthesize a novel coordination compound. They combine a transition metal ion with a ligand and observe that the resulting solution is intensely colored. Which theory best explains the origin of this color and what factors influence it?

    The color of transition metal complexes is best explained by Crystal Field Theory (CFT). In CFT, the d-orbitals of the central metal ion are split into different energy levels by the electric field created by the surrounding ligands. When visible light shines on the complex, electrons absorb specific wavelengths of light to jump from lower to higher energy d-orbitals. The observed color is the complementary color of the light absorbed, and it is influenced by the nature of the ligand (weak-field vs. strong-field), the oxidation state of the metal, and the geometry of the complex, all of which affect the magnitude of the crystal field splitting.

31. Card 31

    An engineer needs to electroplate a steel part with chromium. If they want to deposit 10.0 g of chromium using a current of 5.00 A, how long should the process run in hours? (Molar mass of Cr = 52.00 g/mol, Faraday's constant = 96485 C/mol e-, Cr is deposited from Cr3+ solution).

    First, calculate moles of chromium: 10.0 g / 52.00 g/mol = 0.1923 mol Cr. The half-reaction for chromium deposition from Cr3+ is Cr3+ + 3e- → Cr, so 3 moles of electrons are needed per mole of chromium. Moles of electrons = 0.1923 mol Cr * 3 mol e-/mol Cr = 0.5769 mol e-. Total charge Q = moles of e- * Faraday's constant = 0.5769 mol e- * 96485 C/mol e- = 55655 C. Time t = Q / I = 55655 C / 5.00 A = 11131 seconds. Converting to hours: 11131 s / (3600 s/hr) = 3.09 hours.

32. Card 32

    Which of the following describes a nuclide? A) An atom with a specific number of protons and electrons. B) An atom with a specific number of neutrons and electrons. C) An atom characterized by a specific number of protons and neutrons. D) An atom that has lost or gained electrons.

    C) A nuclide is an atom characterized by a specific number of protons (atomic number) and neutrons (leading to a specific mass number). It specifies a particular nuclear species.

33. Card 33

    Explain the concept of high spin versus low spin complexes in the context of Crystal Field Theory for octahedral complexes.

    In octahedral complexes, d-orbitals split into t2g (lower energy) and eg (higher energy) sets. The difference between high spin and low spin depends on the balance between the crystal field splitting energy (Δo) and the pairing energy (P). High spin complexes occur when Δo is small (weak-field ligands), and electrons prefer to occupy separate orbitals in both t2g and eg sets before pairing up, maximizing the number of unpaired electrons. Low spin complexes occur when Δo is large (strong-field ligands), and electrons preferentially pair up in the lower energy t2g orbitals before occupying the eg orbitals, minimizing the number of unpaired electrons.

34. Card 34

    A geologist finds a rock sample containing uranium-238 and lead-206. If the ratio of 206Pb to 238U (daughter to parent) is 0.50, and the half-life of 238U is 4.47 x 10^9 years, what is the age of the rock? Assume all 206Pb came from 238U decay.

    The ratio of daughter to parent (D/P) can be related to the number of half-lives. If D/P = 0.50, this means that for every 1 part of 238U remaining, there is 0.5 part of 206Pb formed. So, the original amount of 238U was P + D = 1 + 0.5 = 1.5 parts. The fraction of 238U remaining is Nt/N0 = P / (P+D) = 1 / 1.5 = 0.6667. Using the first-order integrated rate law: ln(Nt/N0) = -kt. First, calculate k: k = ln(2) / t1/2 = 0.693 / (4.47 x 10^9 years) = 1.55 x 10^-10 years^-1. Then, ln(0.6667) = -(1.55 x 10^-10 years^-1) * t. -0.405 = -(1.55 x 10^-10 years^-1) * t. So, t = 0.405 / (1.55 x 10^-10 years^-1) = 2.61 x 10^9 years.

35. Card 35

    Draw and identify the geometric isomers for the square planar complex [Pt(NH3)2Cl2].

    For a square planar complex with the formula MA2B2, two geometric isomers are possible: cis and trans. In the cis isomer, the two identical ligands (e.g., NH3) are adjacent to each other (at 90 degrees), while in the trans isomer, they are opposite each other (at 180 degrees). [Drawing: One isomer showing two NH3 groups next to each other and two Cl groups next to each other. Another isomer showing two NH3 groups opposite each other and two Cl groups opposite each other.]

36. Card 36

    A chemist wants to determine the oxidation state of the metal in the coordination compound [Co(en)2Cl2]Cl, where 'en' represents ethylenediamine, a bidentate ligand. How would they approach this, and what would be the oxidation state?

    To determine the oxidation state, one must consider the charges of the ligands and the overall charge of the complex. Ethylenediamine (en) is a neutral ligand (charge = 0). Chloride (Cl) has a charge of -1. The counterion outside the brackets is Cl-, indicating the complex ion [Co(en)2Cl2] has a +1 charge. Let x be the oxidation state of Co. Then, x + 2(0) + 2(-1) = +1. So, x - 2 = +1, which means x = +3. The oxidation state of cobalt is +3.

37. Card 37

    During an experiment involving an electrolytic cell, a student observes that the mass of metal deposited is less than expected. What are two possible reasons for this discrepancy based on the principles of electrolysis?

    Two possible reasons for less mass deposited than expected include: (1) Inefficient current usage due to side reactions: Other redox reactions might be occurring at the electrodes, consuming some of the current without contributing to the desired metal deposition. (2) Impurities in the electrolyte or competing ions: If other metal ions are present in the solution that are easier to reduce, they might deposit preferentially or simultaneously, leading to a lower yield of the target metal. Also, if the current is not maintained constant or if there's a short circuit, the actual charge passed might be less than calculated.

38. Card 38

    Define the term 'ligand' and provide an example of a monodentate and a bidentate ligand.

    A ligand is a molecule or ion that forms a coordinate covalent bond with a central metal atom or ion by donating one or more pairs of electrons. A monodentate ligand donates one electron pair and binds at one site, such as Cl- (chloride ion) or H2O (water). A bidentate ligand donates two electron pairs and binds at two sites, such as ethylenediamine (en) or oxalate (C2O4^2-).

39. Card 39

    A sample contains 8.0 x 10^16 atoms of a radioactive isotope with a half-life of 15 days. How many atoms of the isotope will remain after 45 days?

    First, determine the number of half-lives that have passed: Number of half-lives = Total time / Half-life = 45 days / 15 days = 3 half-lives. After each half-life, the amount of the isotope is halved. So, after 1 half-life: 8.0 x 10^16 / 2 = 4.0 x 10^16 atoms. After 2 half-lives: 4.0 x 10^16 / 2 = 2.0 x 10^16 atoms. After 3 half-lives: 2.0 x 10^16 / 2 = 1.0 x 10^16 atoms. Alternatively, Nt = N0 * (1/2)^(t/t1/2) = 8.0 x 10^16 * (1/2)^3 = 8.0 x 10^16 * 0.125 = 1.0 x 10^16 atoms.

40. Card 40

    Explain how an electrolytic cell works to drive a nonspontaneous chemical reaction.

    An electrolytic cell drives a nonspontaneous chemical reaction by using an external source of electrical energy (like a battery or power supply) to force electrons to flow in the non-preferred direction. In this cell, the anode is where oxidation occurs and is connected to the positive terminal of the power supply, while the cathode is where reduction occurs and is connected to the negative terminal. The applied voltage overcomes the natural potential difference of the nonspontaneous reaction, making the unfavorable redox process proceed.

41. Card 41

    Write the electron configuration for the transition metal ion Mn2+ and determine if it is diamagnetic or paramagnetic in a high-spin octahedral complex.

    Manganese (Mn) has an atomic number of 25, so its ground state electron configuration is [Ar] 3d5 4s2. For Mn2+, two electrons are removed from the 4s orbital, resulting in the electron configuration [Ar] 3d5. In a high-spin octahedral complex, the five d electrons will occupy the t2g and eg orbitals with maximum unpaired electrons. This means all five d electrons will be unpaired (t2g^3 eg^2). Since there are unpaired electrons, Mn2+ in a high-spin octahedral complex is paramagnetic.

42. Card 42

    Distinguish between coordination isomers and linkage isomers, providing an example for each.

    Coordination isomers are structural isomers where the ligands are exchanged between the complex cation and complex anion. For example, [Co(NH3)6][Cr(CN)6] and [Cr(NH3)6][Co(CN)6] are coordination isomers. Linkage isomers are structural isomers where a ligand capable of binding through more than one donor atom (an ambidentate ligand) attaches to the metal ion through a different atom. For example, [Co(NH3)5(NO2)]Cl2 (nitro-isomer, N-bound) and [Co(NH3)5(ONO)]Cl2 (nitrito-isomer, O-bound) are linkage isomers.

43. Card 43

    A nuclear power plant accidentally releases a radioactive isotope with a half-life of 2.0 years. If the initial concentration of the isotope in a nearby water source was 1000 Bq/L, how long will it take for the concentration to drop to 125 Bq/L?

    The concentration decreases by half for every half-life. From 1000 Bq/L to 125 Bq/L: 1000 -> 500 (1 half-life), 500 -> 250 (2 half-lives), 250 -> 125 (3 half-lives). Therefore, it will take 3 half-lives. Total time = 3 * 2.0 years = 6.0 years. Alternatively, using the formula Nt = N0 * (1/2)^(t/t1/2): 125 = 1000 * (1/2)^(t/2.0). 0.125 = (1/2)^(t/2.0). Since 0.125 = (1/2)^3, we have 3 = t/2.0, so t = 6.0 years.

44. Card 44

    Which of the following is an example of an alpha particle? A) 0/-1 e B) 0/+1 e C) 4/2 He D) 1/0 n

    C) An alpha particle is equivalent to a helium nucleus, consisting of two protons and two neutrons, hence its symbol 4/2 He.

45. Card 45

    A pharmaceutical company is developing a new drug that requires the precise deposition of a thin layer of silver onto a sensor. If they need to deposit 250 mg of silver using a constant current of 0.15 A, how many hours will this process take? (Molar mass of Ag = 107.87 g/mol, Faraday's constant = 96485 C/mol e-).

    First, convert mass to grams: 250 mg = 0.250 g Ag. Calculate moles of silver: 0.250 g / 107.87 g/mol = 0.002318 mol Ag. The half-reaction for silver deposition is Ag+ + e- → Ag, so 1 mole of electrons is needed per mole of silver. Moles of electrons = 0.002318 mol e-. Total charge Q = moles of e- * Faraday's constant = 0.002318 mol e- * 96485 C/mol e- = 223.7 C. Time t = Q / I = 223.7 C / 0.15 A = 1491.3 seconds. Converting to hours: 1491.3 s / (3600 s/hr) = 0.414 hours.

46. Card 46

    Explain the relationship between crystal field splitting and the magnetic properties of transition metal complexes.

    Crystal field splitting (Δo or Δt) dictates the energy difference between d-orbitals. This splitting, in conjunction with pairing energy, determines whether a complex will be high spin or low spin, which in turn affects the number of unpaired electrons. Complexes with unpaired electrons are paramagnetic and are attracted to a magnetic field, while complexes with all paired electrons are diamagnetic and are repelled by a magnetic field. Strong-field ligands lead to large splitting, often resulting in low-spin complexes with fewer unpaired electrons (or none), while weak-field ligands lead to small splitting, favoring high-spin complexes with more unpaired electrons.

47. Card 47

    Define the term 'nuclide' and distinguish it from 'isotope'.

    A nuclide is a specific type of atom characterized by a specific number of protons and neutrons in its nucleus. It refers to a particular nuclear species. An isotope, on the other hand, refers to atoms of the same element (same number of protons) but with different numbers of neutrons. Therefore, all isotopes of an element are nuclides, but the term nuclide is broader, encompassing any specific nuclear composition, whether it's an isotope or not.

48. Card 48

    Write a balanced nuclear reaction for the beta decay of carbon-14.

    In beta decay, a neutron in the nucleus is converted into a proton, and an electron (beta particle) is emitted. The mass number remains the same, but the atomic number increases by one. So, for carbon-14 (14/6 C), the reaction is: 14/6 C → 14/7 N + 0/-1 e. Here, carbon-14 transforms into nitrogen-14 with the emission of a beta particle.

49. Card 49

    A chemist is analyzing a coordination compound with the formula [Cr(H2O)4Cl2]Cl. What is the coordination number of the central metal ion, and what is its oxidation state?

    The coordination number is the number of ligands directly bonded to the central metal ion. In [Cr(H2O)4Cl2]Cl, there are four H2O ligands and two Cl ligands directly bonded to Cr, making the coordination number 4 + 2 = 6. To find the oxidation state, consider H2O as neutral (0 charge) and Cl as -1. The counterion Cl outside the brackets means the complex ion [Cr(H2O)4Cl2] has a +1 charge. Let x be the oxidation state of Cr: x + 4(0) + 2(-1) = +1. So, x - 2 = +1, which means x = +3. The oxidation state of chromium is +3.

50. Card 50

    Consider the complex ion [Fe(CN)6]3-. Given that CN- is a strong-field ligand, predict the number of unpaired electrons and whether the complex is high spin or low spin.

    Iron (Fe) has an atomic number of 26, so its electron configuration is [Ar] 3d6 4s2. For Fe3+, three electrons are removed (two from 4s, one from 3d), resulting in [Ar] 3d5. In an octahedral complex, the 3d orbitals split into t2g and eg. Since CN- is a strong-field ligand, it causes a large crystal field splitting (Δo), favoring a low-spin configuration. The five electrons will fill the lower energy t2g orbitals first. Therefore, the electron configuration will be (t2g)^5 (eg)^0. This results in one unpaired electron in the t2g set. The complex is low spin.

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