# Final

General Chemistry 1

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The Lewis structure of but-2-ene (C_{4}H_{8}) is shown below:

1) Draw 3 structural isomers of but-2-ene.

2) What is the hybridization of C_{1} and C_{2} and the molecular shape around these atoms?

3) What is the bond angle C_{1}-C_{2}-C_{3}?

What is the bond angle H-C_{1}-C_{2}?

4) Compare the bond length C_{1}-C_{2} and C_{2}-C_{3}.

1)

2) C_{1} ⇒ 4 electron groups ⇒ sp^{3} ⇒ tetrahedral

C_{2} ⇒ 3 electron groups ⇒ sp^{2} ⇒ trigonal planar

3) C_{2} is trigonal planar ⇒ bond angle C_{1}-C_{2}-C_{3 }= 120°

C_{1 }is tetrahedral ⇒ bond angle H-C_{1}-C_{2} = 109.5°

4) C_{1}-C_{2 }is a single bond while C_{2}-C_{3} is a double bond ⇒ bond length C_{1}-C_{2} > bond length C_{2}-C_{3}

Dichloromethane has a molecular formula of CH_{2}Cl_{2}.

1) What is the molar mass of dichloromethane?

2) Draw the structure of dichloromethane using wedge and dash notation.

3) Is dichloromethane chiral? Explain.

4) Dichloromethane is synthesized in two steps: methane (CH_{4}) reacts with chlorine to form chloromethane, which reacts again with chlorine to give dichloromethane.

Write the chemical equation of these two steps. What is the byproduct of this reaction?

1) M_{CH2Cl2} = M_{C} + 2 x M_{Cl2} + 2 x M_{H2} = 12.01 + 2 x 35.45 + 2 x 1.01 = 84.93 g.mol^{-1}

2)

3) No, dichloromethane is superimposable with its mirror image (no four different substituents on the central carbon + plan of symmetry).

4) Chlorine is a diatomic molecule Cl_{2}

CH_{4} + Cl_{2} → CH_{3}Cl + HCl

CH_{3}Cl + Cl_{2} → CH_{2}Cl_{2} + HCl

The byproduct is HCl (hydrochloric acid).

1) Draw the molecular orbital diagram of C_{2}.

2) What is the electron configuration of C and C_{2}?

3) Draw a Lewis structure of C_{2} consistent with the molecular orbital diagram.

4) What are the magnetic properties of C_{2}?

5) Compare the bond length and bond energy of C_{2} and N_{2}.

1)

2) C : 1s^{2}2s^{2}2p^{2}

C_{2} : (σ_{1s})^{2}(σ*_{1s})^{2}(σ_{2s})^{2}(σ*_{2s})^{2}(π_{2p})^{4}

3) Bond order: (6-2) / 2 = 2

4) no unpaired electrons ⇒ C_{2} is diamagnetic

5) N_{2} : (σ_{1s})^{2}(σ*_{1s})^{2}(σ_{2s})^{2}(σ*_{2s})^{2}(π_{2p})^{4}(σ_{2p})^{2}

Bond order: (8-2) / 2 = 3

N_{2} is a triple bond while C_{2} is a double bond

⇒ bond length of N2 < bond length of C2

⇒ bond energy of N_{2} > bond energy of C_{2}

Consider the carbonyl sulfide OCS where the carbon atom is the central atom.

1) Draw the three resonance structures for this molecule.

2) What is the most stable resonance form? Explain.

3) A C=O double bond has an average bond length of 124 pm. What can you say about the CO bond length of the carbonyl sulfide?

4) Is OCS polar? If yes, draw the dipole moment of the most stable resonance form.

1)

2)

No formal charges

3) In the most stable resonance form of OCS, CO is a double bond so its length is close to 124 pm.

Electronegativity of O > electronegativity of S. The second more stable resonance form is :

CO is a single bond in this form

⇒ The CO bond length of the carbonyl sulfide is < 124 pm.

4) Since the electronegativity of O, C and S are different, OCS is polar ⇒ the bond dipole of CO > bond dipole of CS.

Propene (C_{3}H_{6}) gas reacts with oxygen to produce carbon dioxide and water vapor.

1) Write the balanced equation for this reaction.

2) Calculate the change in enthalpy ΔH^{0} for this reaction?

3) Is this an endothermic or exothermic reaction?

Data:

Bond energy (BE):

H-H = 436 kJ.mol^{-1}

C-H = 413 kJ.mol^{-1}

C-C = 347 kJ.mol^{-1}

C=C = 607 kJ.mol^{-1}

C-O = 358 kJ.mol^{-1}

C=O = 805 kJ.mol^{-1}

O-O = 204 kJ.mol^{-1}

O=O = 498 kJ.mol^{-1}

O-H = 464 kJ.mol^{-1}

1) Unbalanced: C_{3}H_{6} (g) + O_{2} (g) → CO_{2} (g) + H_{2}O (g)

Balanced: 2 C_{3}H_{6} (g) + 9 O_{2} (g) → 6 CO_{2} (g) + 6 H_{2}O (g)

2) ΔH^{0} = 12 BE (C-H) + 2 BE (C=C) + 2 BE (C-C) + 9 BE (O=O) -12 BE (C=O) - 12 BE (H-O)

= -3,882 kJ.mol^{-1}

3) ΔH^{0} < 0 ⇒ exothermic

An element has the following electron configuration: 1s^{2}2s^{2}2p^{6}3s^{2}3p^{5}

1) What is this element?

2) How many valence electrons does this element have?

3) What charge do you expect for an ion from this element?

4) This element can form an iomic compound with magnesium. What is the formula and the name of this compound?

5) Give one example of another element with the same number and type of valence electrons.

1) Chlorine Cl.

2) 7 valence electrons (3s^{2}3p^{5}).

3) -1. It needs one electron to fill the p block and have the same electron configuration as argon.

4) Mg ⇒ Mg^{2+} and Cl ⇒ Cl^{-}

The formula of the ionic compound is MgCl_{2} (magnesium chloride).

5) All the halogen atoms have the same number and type of valence electrons: F, Br, I

In each part identify the element or subatomic particle that (best) fits the description:

· Has atomic mass (approximately) 80.

· Has (approximately) 80 particles in the nucleus, on average.

· Has negative charge (subatomic particle).

· Has “negligible” mass

· Located in the outer part of the atom.

· The lightest element in period 4.

· The lightest nonmetal in period 3.

· The lightest nonmetal in group 6A (16)

· Br – Bromine

· Br – Bromine (The atomic weight is due to the nuclear particles – protons and neutrons)

· electron

· electron

· electron

· K – Potassium

· Si – Silicon or P – Phosphorus (Si is a semi-metal, but on the non-metal side.)

· O - Oxygen

Determine the overall change in energy for the formation of solid cesium sulfide from gaseous cesium and sulfur, knowing that the lattice of CsS = -3.470 aJ.

Data:

I_{1 }(Cs) = 0.623 aJ

EA1(S) = - 0.332 aJ

EA_{2}(S) = + 0.980 aJ

d_{MgS }= 351 pm

Coulomb constant = 231 aJ.pm

Chemical equation: 2 Cs (g) + S (g) → Cs_{2}S (s)

First step: Cs (g) → Cs^{+} (g) + e^{-}

Second step:

S(g) + e^{-} → S^{-} (g)

S^{-} (g) + e^{-} → S^{2-} (g)

E_{reaction} = 2 x I_{1 }(Cs) + EA_{1 }(S) + EA_{2 }(S) + E_{lattice} = -1.576 aJ

We do not use the coulomb energy since the final compound is a solid and not a gas of an ionic compound.

We study the atomic emission spectra of hydrogen.

1) What is the energy of an electron in the third orbital (n =3)?

2) Calculate the energy of the n=3 to n=1 atomic transition.

3) What is the energy of a photon emitted at λ = 656 nm?

4) The retina of the human eye can detect light with energy ≥ 4.0 x 10^{-17} J.

What is the minimum number of photons at λ = 656 nm that must be emitted for a person to see the red emission line?

Data: Rydberg constant = -2.1799 aJ, h = 6.626 x 10^{-34} J.s, c = 2.998 x 10^{8} m.s^{-1}

1) E = $\frac{-2.1799\left(\mathrm{aJ}\right)}{{\mathrm{n}}^{2}}$ = - 2.4221 x 10^{-1 }aJ = -2.4221 x 10^{-19} J

2) E_{photon} = 2.1799 ($\frac{1}{{{\mathrm{n}}_{\mathrm{f}}}^{2}}$ – $\frac{1}{{{\mathrm{n}}_{\mathrm{i}}}^{2}}$) = 2.1799 (1 – $\frac{1}{9}$) = 1.9377 aJ = 1.9377 x 10^{-18} J

3) E_{photon-656} = hν = $\frac{\mathrm{hc}}{\mathrm{\lambda}}$ = $\frac{\left(6.626\times {10}^{-34}\right)\times \left(2.998\times {10}^{8}\right)}{656\times {10}^{-9}}$ = 3.028 x 10^{-19} J

4) E_{total} ≥ 4.0 x 10^{-17} J

and E_{total} = n x E_{photon-656 }⇒_{ }n = E_{total }/ E_{photon-656 }

n ≥ 4.0 x 10^{-17} / 3.028 x 10^{-19}

n ≥ 132.1

The minimum number of photons is 133.

You have 218.0 g of potassium carbonate.

How many moles of potassium carbonate do you have?

How many potassium atoms are there?

Draw a Lewis structure of carbonate. Is the anion carbonate stable?

Potassium carbonate = K_{2}CO_{3}

n (K_{2}CO_{3}) = $\frac{\mathrm{m}\left({\mathrm{K}}_{2}{\mathrm{CO}}_{3}\right)}{\mathrm{M}\left({\mathrm{K}}_{2}{\mathrm{CO}}_{3}\right)}$

M(K_{2}CO_{3}) = 2 M(K) + M(C) + 3M(O) = 138.21 g.mol^{-1}

⇒ n (K_{2}CO_{3}) = $\frac{218.0}{138.21}$ = 1.577 mol

Number of K = N(K) = 2 x N(K_{2}CO_{3}) = n (K_{2}CO_{3}) x N_{A} = 1.577 x 6.022 x 10^{23} = 9.467 x 10^{23 }atoms.

The anion carbonate is stable because it has 3 different resonance forms ⇒ stabilization by resonance