Extra Credit 45

Q24.6.7

For each complex, predict its structure, whether it is high spin or low spin, and the number of unpaired electrons present.

1. \([TiCl_{6}]^{3-}\)
2. \([CoCl_{4}]^{2-}\)

S24.6.7

1.

step 1: We take a look at the ligand and note that the central metal, Titanium, has 6 chlorine's bonded to it. Because there are 6 chlorine's attached to one titanium we can predict this to be an octahedral complex.

step 2: To predict whether this complex is high spin or low spin we need to first look at the spectrochemical series. This will tell us whether chlorine has a weak or strong field. A weak field corresponds to high spin whereas a strong field corresponds to low spin.

(weak) I− < Br− < S2− < SCN− < Cl− < NO3− < N3− < F− < OH− < C2O42− < H2O < NCS− < CH3CN < py < NH3

< en < bipy < phen < NO2− < PPh3 < CN− < CO (strong)

We see on the spectrochemical series that chlorine (\(Cl^{-}\)) is grouped with the weaker ligands so it is weak field ligand with relatively small splitting energy Δ . Because chlorine has a weak field, this octahedral complex will have high spin.

step 3: To find the number of unpaired electrons present we must first find the charge on Ti. From the given equation we know the overall charge of this complex to be a 3- charge. We see that there are 6 chlorines and each contain a 1- charge. Because there are 6 of them this would give us an overall charge of 6-. We know the overall charge of the complex is 3- and the overall charge of all the chlorines is 6-. In order to keep the overall complex charge at a 3- we have to assign Titanium a charge of 3+.

step 4: Now that we know Titanium has a charge of 3+ we can draw our crystal field splitting diagram. We look at the periodic table and see that Ti3+ holds only one d-electron.

step 5: We draw our energy diagram keeping in mind this is a high spin configuration. We fill in our one electron on the bottom orbital starting on the far left side and because this is our only electron, in this case, we are done. We see that there is only one unpaired electron because we only drew in one electron.

2.

step 1: We take a look at the ligand and note that the central metal, Cobalt, has 4 chlorine's bonded to it. Because there are 4 chlorine's attached to one cobalt metal we can predict this a tetrahedral complex.

step 2: Next we must look to the spectrochemical series to decide whether or not this complex is high spin or low spin. We see that the strength of the chlorine ligand field according to the series is a weak field. This means the complex will have a high spin electron configuration.

(weak) I− < Br− < S2− < SCN− < Cl− < NO3− < N3− < F− < OH− < C2O42− < H2O < NCS− < CH3CN < py < NH3

< en < bipy < phen < NO2− < PPh3 < CN− < CO (strong)

step 3: We must now find the charge on the Cobalt ion. We look at the complex and notice it has an overall 2- charge. We know that each chlorine ion will contain a 1- charge and because there are 4 of them they will contain an overall charge of 4-. Since the overall charge of the complex is 2- we have to assign a 2+ charge to cobalt.

step 4: Now that we have established that Cobalt has a 2+ charge we can look to the periodic table and find that it holds 7 electrons in its d-orbital.

step 5: We draw the energy diagram corresponding to a tetrahedral complex and begin filling in the lower line first with one electron in each orbital. We move to the top row and fill one electron in each orbital there. We have now drawn 5 electrons on the diagram, one in each orbital. Because we have 7 d-electrons we still have 2 more to place. We move back to the bottom row and pair the 2 unpaired electrons with one electron each of the opposite spin. We are now left with three unpaired electrons in the top row.

Q21.4.12

Write a nuclear reaction for each step in the formation of \(^{208}_{82}Pb\) from \(^{228}_{90}Th\) , which proceeds by a series of decay reactions involving the step-wise emission of α, α, α, α, β, β, α particles, in that order.

S21.4.12

step 1: We begin by writing the balanced equation for thorium undergoing alpha decay. When an element undergoes alpha decay 4 is subtracted from its mass number and 2 is subtracted from its atomic number. In this case, on the other side of the equation we would write \(^{4}_{2}\alpha+^{224}_{88}Ra\) to represent what is happening to the Thorium as it undergoes this sort of decay. The new mass number is 224 and the new atomic number is 88. We look to the periodic table to see what element has an atomic number of 88 and it's Radium. It is also important to keep in mind that both sides must equal each other. The atomic number and mass number on the left side of the equation should add up to the atomic and mass number on the right side of the equation.

\[^{228}_{90}Th\to ^{4}_{2}\alpha+^{224}_{88}Ra\]

step 2: We are told the element undergoes alpha decay again. We begin with the radium on the left side of the equation. Since it undergoes alpha decay it loses 4 in the mass number which turns to 220 and 2 in its atomic number which turns to 86. We look to the periodic table to find which element has an atomic number of 86 and find that it is radon. If we were to add both mass numbers and atomic numbers on the right side of the equation they would add up to the mass number and atomic number on the left side of the equation.

\[^{224}_{88}Ra\to ^{4}_{2}\alpha+^{220}_{86}Rn\]

step 3: We proceed the same way for this next step. Again we rewrite radon on the left side of the equation and what it becomes after undergoing alpha decay. Because its mass number is 220 after undergoing alpha decay it will be 216. Its atomic number will decrease from 86 to 84. We now look to the periodic table to see what elements atomic number is 84 and it is Polonium. Again, both sides of the equation must be balanced and equal one another.

\[^{220}_{86}Rn\to ^{4}_{2}\alpha+^{216}_{84}Po\]

step 4: We now begin with Polonium on the left side. After it undergoes alpha decay the new element is Lead. We follow the same pattern in calculating the new mass and atomic number as well as in making sure the equation is balanced.

\[^{216}_{84}Po\to ^{4}_{2}\alpha+^{212}_{82}Pb\]

step 5: Following the pattern asked in the question the next decay is Beta. Because beta decay is different than alpha decay the process will slightly differ. Beta decay has no mass number and has -1 as its atomic number. When an element undergoes beta decay its mass number does not change but its atomic number increases by one. In this case, lead would undergo beta decay. On the left side of the equation we rewrite lead with its atomic and mass number and on the right side of the equation we write what happens when it undergoes beta decay. On the right side we add beta to the new element which has a mass number of 212 and a new atomic number of 83 which, after looking at the periodic table we can see is Bismuth. Again, both side of the equation equal one another.

\[^{212}_{82}Pb\to ^{0}_{-1}\beta+^{212}_{83}Bi\]

step 6: Following the same pattern, because we are told this undergoes beta decay again we put \(^{212}_{83}Bi\) on the left side of the equation and write our new element on the right side. Since the mass number doesn't change it remains 212 and we add 1 to the atomic number to obtain a new atomic number of 84. The element that has an atomic number of 84 is Polonium.

\[^{212}_{83}Bi\to ^{0}_{-1}\beta+^{212}_{84}Po\]

step 7: In the last step we are told the element undergoes alpha decay again. We place \(^{212}_{84}Po\) on the left side of the equation. On the right side we add alpha and the new element. The new element's mass number will be 212 minus 4 which is 208 and the the new atomic number will be 84 minus 2 which is 82. We look to the periodic table and see that Lead has an atomic number of 82. Thorium was supposed to be transformed into \(^{208}_{82}Pb\) which is exactly what we have.

\[^{212}_{84}Po\to ^{4}_{2}\alpha+^{208}_{82}Pb\]

Q12.3.8

The rate constant for the radioactive decay of \(^{14}C\) is 1.21 × \(10^{-4}\) \(year^{-1}\). The products of the decay are nitrogen atoms and electrons (beta particles):

\[^{6}_{14}C⟶^{6}_{14}N+e^{-}\]

\[rate=k[^{6}_{14}C]\]

What is the instantaneous rate of production of N atoms in a sample with a carbon-14 content of \(6.5 × 10^{-9}\) M?

S12.3.8

step 1: We look at the given equation, and can say that the rate of consumption of carbon is equal to the rate of production of the nitrogen.

step 2: We are given our k value is 1.21 × \(10^{-4}\) \(year^{-1}\). We use this value in our rate equation. Because we know our molarity of Carbon is \(6.5 × 10^{-9}\) M we can multiply the two to obtain the rate.

step 3: Our new equation is \(rate=1.21 × 10^{-4}year^{-1}\) x \(6.5 × 10^{-9}\)M which we find that rate is equal to 7.9 x \(10^{-13}\)\(year^{-1}\).

Q17.6.4

Suppose you have three different metals, A, B, and C. When metals A and B come into contact, B corrodes and A does not corrode. When metals A and C come into contact, A corrodes and C does not corrode. Based on this information, which metal corrodes and which metal does not corrode when B and C come into contact?

S17.6.4

step 1: Corrosion is a galvanic process by which metals works as a oxidizing agent with reducing agents, such \(O_2\), in the natural world. During this process, metals which are better oxidizing agent will corrode first.

step 2: We are told when A and B come into contact B corrodes which means B is a better oxidizing agent than A.

step 3: We are told when A and C come into contact A corrodes which means A is a better oxidizing agent than C.

step 4: Because B is a better oxidizing agent than A and A a better oxidizing agent than C, then B will a better oxidizing agent than C. Thus B and C come into contact we can assume that B will corrode first.

Q20.2.16

Classify each reaction as an acid–base reaction, a precipitation reaction, or a redox reaction, or state if there is no reaction; then complete and balance the chemical equation:

1. \(Zn(s) + HCl(aq)\) →
2. \(HNO_{3}(aq) + AlCl_{3}(aq)\) →
3. \(K_{2}CrO_{4}(aq) + Ba(NO_{3})_{2}(aq)\) →
4. \(Zn(s) + Ni^{2+}(aq) → Zn^{2+}(aq) + Ni(s)\)

S20.2.16

1.

step 1: Balance the equation. Whatever we have on the left side must equal what we put on the right side. Our balanced equation should look like:

\(Zn(s) + 2HCl(aq)\) → \(ZnCl_{2}(aq)+H_{2}(g)\)

We have one Zinc , 2 Hydrogens and 2 Chlorines on each side of the equation. Because Zinc's oxidation state is 2+ we have to pair it with two chlorines to even out the charge since each chlorine has a charge of 1-. The overall charge on the compound now is 0 and because we had to put a coefficient in front of the HCl to have two chlorines we automatically have 2 hydrogens as well.

step 2: We can identify this reaction as a redox reaction. We do this by separating the aqueous substances to find a net reaction. We observe a transfer of charge between Zinc and Hydrogen in this separated reaction ultimately changing the charge on both. If necessary, you can separate both zinc and hydrogen into oxidation and reduction half reactions to see this transfer more clearly. The charge change on both is why this would be classified as a redox reaction.

separation into ions:

\[Zn(s) + 2H^{+}(aq) + 2Cl^{-}(aq) → H_{2}(g) + Zn^{2+}(aq) + 2Cl^{-}(aq)\]

2.

step 1: Balance the equation. Whatever we have on the left side must equal what we put on the right side. Our balanced equation should look like:

\[3HNO_{3}(aq) + AlCl_{3}(aq) →3HCl(aq)+Al(NO_{3})_{3}(aq)\]

We have an equal number of each element on each side of the equation. As there is no electron transferred, this is not a redox reaction.

step 2: Both \(HCl(aq)\) and \(Al(NO_{3})_{3}(aq)\) are soluble in water according to Solubility Rules. This is not a precipitation reaction

step 3: Even though \(HNO_{3}(aq)\) is a acid, \(AlCl_{3}(aq)\) is not a base but a acidic salt. This is not an acid–base reaction.

step 4: We can state that there is no reaction.

3.

step 1: Balance the equation. Whatever we have on the left side must equal what we put on the right side. Our balanced equation should look like:

\[K_{2}CrO_{4}(aq) + Ba(NO_{3})_{2}(aq) →BaCrO_{4}(s)+2K(NO_{3})(aq)\]

We now add up the elements on each side and find that there should be an equal number of each on both sides. As there is no electron transferred, this is not a redox reaction.

step 2: We can classify this reaction as a precipitate reaction because if we refer to Solubility Rules, we see that \(BaCrO_{4}\) is insoluble. Most chromates are considered insoluble and following that rule a solid will form on the right side of this equation meaning this reaction is a precipitate reaction.

4.

step 1: \(Zn(s) + Ni^{2+}(aq) → Zn^{2+}(aq) + Ni(s)\) is our given equation. We take note that the reaction is already balanced with one Zinc and one Nickel element on each side.

step 2: We see that each elements charge changes classifying it as a redox reaction.

Q20.5.11

The chemical equation for the combustion of butane is as follows:

\[C_4H_{10}(g)+\frac{13}{2}O_2(g)→4CO_2(g)+5H_2O(g)\]

This reaction has ΔH° = −2877 kJ/mol. Calculate E°cell and then determine ΔG°. Is this a spontaneous process? What is the change in entropy that accompanies this process at 298 K?

S20.5.11

step 1: We begin by calculating \(\Delta{G}\) using our given enthalpy change value. We use the equation \(\Delta{G}=\Delta{H}-T\Delta{S}\). We have our \(\Delta{H}\) value but in order to find our missing \(\Delta{G}\) value we need to find what \(\Delta{S}\) is equal to. We refer to a table of standard entropy values:

step 2: We calculate the standard entropy of the reaction based on the values above. On our products side we have 4 Carbon dioxide molecules and 5 water molecules. We look the table and see carbon dioxide's value is -393.5 and because we have 4 of them we multiply the standard entropy value by 4. We follow the same pattern for water and see that the water molecule's value is -242 and because there are 5 we multiply -242 by 5. To obtain the total entropy on the products side we add the two values together which would look like:

\[\Delta{S}_{product} = 4*S_{CO_2} + 5*S_{H_2O} = (-393.5*4)+(-242*5)=-2659 kJ/mol\]

We follow the same steps in obtaining the standard entropy on the reactants side and we get the following equation for the reactants:

\[\Delta{S}_{reactant} = (-125x1)+(0x\frac{13}{2})=-125kJ/mol\]

step 3: Since we now have both standard entropies on both sides of the equation we take the entropy values and use the formula products-reactants. We get:

\[\Delta{S} = \Delta{S}_{product} -\Delta{S}_{reactant} = -2659-(-125)=-2534kJ/mol\]

step 4: Now that we've found our \(\Delta{S}\) value we can plug it into the equation from above.

\[\Delta{G}=\Delta{H}-T\Delta{S}\]

\[\Delta{G}=-2877kJ/mol-(298Kx-2534kJ/mol)\]

At this point we stop and check that our enthalpy and entropy values are in the same units and we use the standard temperature value in Kelvin.

\[\Delta{G}\)=752,255kJ/mol

step 5: Now that we have obtained our \(\Delta{G}\) value we proceed to find the E°cell value. To do this we can use the equation:

\(\Delta{G}\)=-nFE°cell

The only thing we're missing in order to find this value is the number of electrons transferred between the two half-reactions. We begin by separating this equation into oxidation and reduction half reactions. We first separate the reduction reaction and balance:

\(\frac{13}{2}O_2(g)→5H_2O(g)\)

We multiply each coefficient by 2 to get rid of the fraction and obtain a new half reaction of:

\(52e^{-}\)+\(52H^{+}\)+\(13O_2(g)→10H_2O(g)+16H_2O(g)\) since all coefficients are divisible by 2 we divide them by 2 and get:

\(26e^{-}\)+\(26H^{+}\)+\(\frac{13}{2}O_2(g)\)→\(5H_2O(g)+13H_2O(g)\)

We find \(26e^{-}\) transferred in this half reaction.

step 6: We balance the oxidation half reaction:

\(5H_2O(g)+C_4H_{10}(g)→4CO_2(g)+14e^{-}\)+\(14H^{+}\)

step 7: Our overall balanced reaction seems to be:

\[\frac{13}{7}C_4H_{10}(g)+\frac{13}{2}O_2(g)→\frac{13}{7}CO_2(g)+\frac{78}{7}H_2O(g)\]

step 8: We plug our value of \(26e^{-}\) into the \(\Delta{G}\)=-nFE°cell equation converting our \(\Delta{G}\) value to Joules and it turns out to look like:

752,255kJ/mol=-26x96485C/molxE°cell we find E°cell=-299.87V

step 9: Because we obtained an E°cell value is negative we can conclude it is non-spontaneous.

Q12.5.17

Use the PhET Reactions & Rates interactive simulation to simulate a system. On the “Single collision” tab of the simulation applet, enable the “Energy view” by clicking the “+” icon. Select the first A+BC⟶AB+CA+BC⟶AB+C reaction (A is yellow, B is purple, and C is navy blue). Using the “angled shot” option, try launching the A atom with varying angles, but with more Total energy than the transition state. What happens when the A atom hits the BC molecule from different directions? Why?

S12.5.17

step 1: We use the simulation to launch the A atom in different directions. We observe that when the A atom hits the BC molecule it does not react and bounces off in a different direction.

In this experiment, first two rules of 14.6: Collision Theory are satified. The A atom and the BC atom collide with more total energy than the transition state. However, not all A atoms will react with BC atoms. This shows us the third rule, which is that orientation of a molecule is critical to reaction kinetics and indicates whether or not a reaction will occur. It also shows us that A does not possess enough energy to react even though it collides in the proper orientation to.