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

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

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

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General Chemistry 1

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Question 1

The Lewis structure of but-2-ene (C₄H₈) is shown below:

1) Draw 3 structural isomers of but-2-ene. 2) What is the hybridization of C₁ and C₂ and the molecular shape around these atoms? 3) What is the bond angle C₁-C₂-C₃? What is the bond angle H-C₁-C₂? 4) Compare the bond length C₁-C₂ and C₂-C₃.

Solution

2) C₁ ⇒ 4 electron groups ⇒ sp³ ⇒ tetrahedral C₂ ⇒ 3 electron groups ⇒ sp² ⇒ trigonal planar 3) C₂ is trigonal planar ⇒ bond angle C₁-C₂-C₃= 120° C₁is tetrahedral ⇒ bond angle H-C₁-C₂ = 109.5° 4) C₁-C₂is a single bond while C₂-C₃ is a double bond ⇒ bond length C₁-C₂ > bond length C₂-C₃

Question 2

Dichloromethane has a molecular formula of CH₂Cl₂. 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 (CH₄) 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?

Solution

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 Cl₂ CH₄ + Cl₂ → CH₃Cl + HCl CH₃Cl + Cl₂ → CH₂Cl₂ + HCl The byproduct is HCl (hydrochloric acid).

Question 3

1) Draw the molecular orbital diagram of C₂. 2) What is the electron configuration of C and C₂? 3) Draw a Lewis structure of C₂ consistent with the molecular orbital diagram. 4) What are the magnetic properties of C₂? 5) Compare the bond length and bond energy of C₂ and N₂.

Solution

2) C : 1s²2s²2p² C₂ : (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 ⇒ C₂ is diamagnetic 5) N₂ : (σ_1s)²(σ*_1s)²(σ_2s)²(σ*_2s)²(π_2p)⁴(σ_2p)² Bond order: (8-2) / 2 = 3 N₂ is a triple bond while C₂ is a double bond ⇒ bond length of N2 < bond length of C2 ⇒ bond energy of N₂ > bond energy of C₂

Question 4

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.

Solution

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.

Question 5

Propene (C₃H₆) 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 ΔH⁰ 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

Solution

1) Unbalanced: C₃H₆ (g) + O₂ (g) → CO₂ (g) + H₂O (g) Balanced: 2 C₃H₆ (g) + 9 O₂ (g) → 6 CO₂ (g) + 6 H₂O (g) 2) ΔH⁰ = 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) ΔH⁰ < 0 ⇒ exothermic

Question 6

An element has the following electron configuration: 1s²2s²2p⁶3s²3p⁵ 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.

Solution

1) Chlorine Cl. 2) 7 valence electrons (3s²3p⁵). 3) -1. It needs one electron to fill the p block and have the same electron configuration as argon. 4) Mg ⇒ Mg²⁺ and Cl ⇒ Cl^- The formula of the ionic compound is MgCl₂ (magnesium chloride). 5) All the halogen atoms have the same number and type of valence electrons: F, Br, I

Question 7

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)

Solution

· 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

Question 8

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:

I₁(Cs) = 0.623 aJ EA1(S) = - 0.332 aJ EA₂(S) = + 0.980 aJ

d_MgS= 351 pm Coulomb constant = 231 aJ.pm

Solution

Chemical equation: 2 Cs (g) + S (g) → Cs₂S (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 I₁(Cs) + EA₁(S) + EA₂(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.

Question 9

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 10⁸ m.s^-1

Solution

1) E =

\frac{- 2.1799 (aJ)}{n^{2}}

= - 2.4221 x 10^-1aJ = -2.4221 x 10^-19 J

2) E_photon = 2.1799 ( $\frac{1}{{n_{f}}^{2}}$ – $\frac{1}{{n_{i}}^{2}}$ ) = 2.1799 (1 – $\frac{1}{9}$ ) = 1.9377 aJ = 1.9377 x 10^-18 J

3) E_photon-656 = hν = $\frac{hc}{λ}$ = $\frac{(6.626 \times 10^{- 34}) \times (2.998 \times 10^{8})}{656 \times 10^{- 9}}$ = 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.

Question 10

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?

Solution

Potassium carbonate = K₂CO₃

n (K₂CO₃) = $\frac{m (K_{2} {CO}_{3})}{M (K_{2} {CO}_{3})}$

M(K₂CO₃) = 2 M(K) + M(C) + 3M(O) = 138.21 g.mol^-1

⇒ n (K₂CO₃) = $\frac{218.0}{138.21}$ = 1.577 mol

Number of K = N(K) = 2 x N(K₂CO₃) = n (K₂CO₃) x N_A = 1.577 x 6.022 x 10²³ = 9.467 x 10²³atoms.

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

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