Extra 2

S8B.1

\(\ce{A}\) is disappear at the rate of \(-8.2 \times 10^{-4} \; M/s\).

a. r = -1Δ\(\ce{A}\)/3Δt

r = -(\(-8.2 \times 10^{-4} \; M/s\))/3

r = (\(2.7 \times 10^{-4} \; M/s\))

b. r = -1Δ\(\ce{B}\)/2Δt

(\(2.7 \times 10^{-4} \; M/s\)) = -1Δ\(\ce{B}\)/2Δt

Δ\(\ce{B}\)/Δt = (\(-5.5 \times 10^{-4} \; M/s\))

c. r = 1Δ\(\ce{C}\)/1Δt

Δ\(\ce{C}\)/Δt = (\(2.7 \times 10^{-4} \; M/s\))

Q8B.11

The initial rate of the reaction

\[\ce{2A + B \rightarrow 3C + D}\]

can be determined by the following table. Using the given information to:

1. Find the order of reaction with respect to \(\ce{A}\) and to \(\ce{B}\).
2. Solve the overall reaction order.
3. Solve for the rate constant, \(\ce{k}\)

S8B.11

a. [A] = (0.15/0.3)^x = (3.72 x 10^-3 / 7.04 x 10^-3) [B] = (0.123/0.246)^y = (3.72 x 10^-3 / 1.408 x 10^-2)

(0.5)^x = (0.5) (0.5)^y = 0.26

x = 1 y = 2

[A] is first order [B] is second order

b. 1 + 2 = 3

the overall reaction order is 3^rd order

c. r = k[A][B]²

3.72 x 10^-3 = k (0.15) (0.123)²

k = 1.639/M²s

Q8B.17

The first order reaction \(\ce{A \rightarrow products}\) has \(\mathrm{t_{1/2} = 180\, s}\):

1. What percent of a sample of \(\ce{A}\) remains unreacted 720 s after a reaction has been started?
2. What is the rate of reaction when \(\mathrm{[A] = 0.25\, M}\)?

S8B.17

a. k = 0.693/t_1/2

k = 0.693/180 = 0.00385

(A/A_o) = e^-kt

(A/A_o) = e^{-0.00385 x 720}

(A/A_o) = 0.0625 = 6.25% remains

b. r = k[A]

r = 0.00385(0.25)

r = 9.625 x 10^-4 M/s

Q4.23

Construct a cell diagram from the given reaction couple: Cr³⁺+ e^-→ Cr²⁺ E^o_cell = -0.41 V ; 2H⁺ + 2e^-→ H₂ E^o_cell = 0V

S4.23

Cr³⁺+ e^-→ Cr²⁺ = anode 2H⁺ + 2e^-→ H₂ = cathode

Pt (s) l Cr³⁺(aq) , Cr²+ (aq) ll H⁺ (aq) l H₂ (g) l Pt (s)

Q5B.2

The complexes \([NiCl_4]^{2–}\) and \([Ni(CN)_4]^{2–}\) are paramagnetic and diamagnetic, respectively. What does this tell you about their structures?

S5B.2

Their magnetism lets us know about their structure and spin. Since \([NiCl_4]^{2–}\) is paramagnetic and has a weak field ligand, this tells us that the complex has a high spin.

And since \([Ni(CN)_4]^{2–}\) is diamagnetic and has a strong field ligand, this tells us that the complex has a low spin.

Q5B.3

Consider a pair of isomeric cationic complexes having the molecular formula \([Co(en)_2Br_2]ClO_4\). One is optically active but the other is not.

1. Write structural formulas showing each isomeric complex but only one of the enantiomers. en = \(H_2NCH_2CH_2NH_2\).
2. What is the oxidation state of cobalt in these complexes? How many \(d\) electrons are found in each complex?

S5B.3

a.

b. Cobalt's oxidation state is -3 in both complexes

Co(III) has 6 d electrons.

There are 6 d electrons in each complex

Q5B.4

When the d orbitals of the central metal ion are split in energy in an octahedral ligand field, which orbitals are raised least in energy?

1. \(d_{xy}\) and \(d_{x^2-y^2}\)
2. \(d_{xy}\), \(d_{xz}\) and \(d_{yz}\)
3. \(d_{xz}\) and \(d_{yz}\)
4. \(d_{xz}\), \(d_{yz}\) and \(d_{z^2}\)
5. \(d_{x^2-y^2}\) and \(d_{z^2}\)

S5B.4

When the d orbitals come into contact with an octahedral ligand field, the \(d_{xy}\), \(d_{xz}\) and \(d_{yz}\) orbitals are raised the least because they are not disrupted as much by the ligands and don't have much overlap.

Q5B.5

Which one of the following statements is FALSE?

1. In an octahedral crystal field, the d electrons on a metal ion occupy the \(e_g\) set of orbitals before they occupy the \(t_{2g}\) set of orbitals.
2. Diamagnetic metal ions cannot have an odd number of electrons.
3. Low spin complexes can be paramagnetic.
4. In high spin octahedral complexes, \(\Delta_{o}\) is less than the electron pairing energy, and is relatively very small.
5. Low spin complexes contain strong field ligands.

S5B.5

b. In an octahedral crystal field, the d electrons on a metal ion occupy the \(e_g\) set of orbitals before they occupy the \(t_{2g}\) set of orbitals.

The orbital splitting with octahedrals causes the \(d_{xy}\), \(d_{xz}\) and \(d_{yz}\) orbitals to split from the \(d_{x^2-y^2}\) and \(d_{z^2}\) orbitals. The latter are the \(e_g\) orbitals, and they are disrupted the most, so they move up and have higher energy, and the other group, \(t_{2g}\) orbitals including \(d_{xy}\), \(d_{xz}\) and \(d_{yz}\) orbitals are not affected much, so they lower in energy. So these orbitals get filled by electrons first.

Q5B.6

Consider the complex ion \([Mn(OH_2)_6]^{2+}\) with 5 unpaired electrons. Which response includes all the following statements that are true, and no false statements?

1. It is diamagnetic.
2. It is a low spin complex.
3. The metal ion is a d⁵ ion.
4. The ligands are weak field ligands.
5. It is octahedral.

Options:

1. I, II
2. III, IV, V
3. I, IV
4. II, V
5. III, IV

S5B.6

\([Mn(OH_2)_6]^{2+}\) - H₂O is a weak field ligand, so it has a small splitting, so it would be high spin.

b. III, IV, V

Since Mn lost 2 electrons, it would become a d⁵ ion.

H₂O is a weak field ligand.

Since the coordination number here is six, this is an octahedral complex.

Q10.20

How much energy, in megaelectronvolts, is released in the following nuclear reaction?

\(\mathrm{\ce{^{241}_{95}Am} \rightarrow \ce{^{237}_{93}Np} + {^4_2He}}\)

The nuclidic masses are \(\mathrm{\ce{^{241}_{95}Am}=241.056829\, u}\), \(\mathrm{\ce{^{237}_{93}Np}=237.0481734\, u}\), and \(\mathrm{^4_2He=4.00260\, u}\).

S10.20

\(\mathrm{\ce{^{241}_{95}Am} \rightarrow \ce{^{237}_{93}Np} + {^4_2He}}\)

241.056829 → 237.0481734 + 4.0026

products - reactants = (237.0481734 + 4.0026) - 241.056829

-0.006056 u

[(-0.006056 u) x (931.5 MeV/c²)/1u] (2.998 x 10⁸)²

-5.07028 x 10¹⁷ MeV

Q10.21

Calculate energy released in nuclear reaction:

\[\mathrm{\ce{^{11}_6C} + {^4_2He} \rightarrow \ce{^{14}_7N} + {^1_1H}}\]

given the following nuclear masses:

• \(\mathrm{\ce{^{11}_6C} = 10.9\,u}\)
• \(\mathrm{^4_2He = 3.21\,u}\)
• \(\mathrm{\ce{^{14}_7N} = 13.104\,u}\)
• \(\mathrm{^1_1H = 1.01\,u}\)

S10.21

\[\mathrm{\ce{^{11}_6C} + {^4_2He} \rightarrow \ce{^{14}_7N} + {^1_1H}}\]

10.9 + 3.21 → 13.104 + 1.01

products - reactants

(13.104 + 1.01) - (10.9 + 3.21) = 6.424u

E = mc²

E = [(6.424 u ) x (1.6605 x 10^-27 kg/u)] (2.998 x 10⁸)²

E = 9.588 x 10^-10 J

Q10.22

Which isotope is more likely to be the most abundant in nature? Explain.

1. \(\ce{^{32}_{18}Ar}\) vs. \(\ce{^{52}_{18}Ar}\)
2. \(\ce{^{89}_{38}Sr}\) vs. \(\ce{^{90}_{38}Sr}\)

S10.22

1. \(\ce{^{32}_{18}Ar}\) would be more stable because the proton to neutron ratio is much smaller than the other isotope. The atomic mass is also closer to Ar's average atomic mass.
2. \(\ce{^{90}_{38}Sr}\) would be more stable because there is an even number of protons and neutrons.