General Chemistry 1
Check the solutions on the questions and answer that you have decided correctly
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.
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°
C1 is tetrahedral ⇒ bond angle H-C1-C2 = 109.5°
4) C1-C2 is a single bond while C2-C3 is a double bond ⇒ bond length C1-C2 > bond length C2-C3
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?
1) MCH2Cl2 = MC + 2 x MCl2 + 2 x MH2 = 12.01 + 2 x 35.45 + 2 x 1.01 = 84.93 g.mol-1
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).
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.
2) C : 1s22s22p2
C2 : (σ1s)2(σ*1s)2(σ2s)2(σ*2s)2(π2p)4
3) Bond order: (6-2) / 2 = 2
4) no unpaired electrons ⇒ C2 is diamagnetic
5) N2 : (σ1s)2(σ*1s)2(σ2s)2(σ*2s)2(π2p)4(σ2p)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
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.
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 (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?
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: 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
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.
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
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)
· 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.
I1 (Cs) = 0.623 aJ
EA1(S) = - 0.332 aJ
EA2(S) = + 0.980 aJ
dMgS = 351 pm
Coulomb constant = 231 aJ.pm
Chemical equation: 2 Cs (g) + S (g) → Cs2S (s)
First step: Cs (g) → Cs+ (g) + e-
S(g) + e- → S- (g)
S- (g) + e- → S2- (g)
Ereaction = 2 x I1 (Cs) + EA1 (S) + EA2 (S) + Elattice = -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 108 m.s-1
1) E = = - 2.4221 x 10-1 aJ = -2.4221 x 10-19 J
2) Ephoton = 2.1799 ( – ) = 2.1799 (1 – ) = 1.9377 aJ = 1.9377 x 10-18 J
3) Ephoton-656 = hν = = = 3.028 x 10-19 J
4) Etotal ≥ 4.0 x 10-17 J
and Etotal = n x Ephoton-656 ⇒ n = Etotal / Ephoton-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 = K2CO3
n (K2CO3) =
M(K2CO3) = 2 M(K) + M(C) + 3M(O) = 138.21 g.mol-1
⇒ n (K2CO3) = = 1.577 mol
Number of K = N(K) = 2 x N(K2CO3) = n (K2CO3) x NA = 1.577 x 6.022 x 1023 = 9.467 x 1023 atoms.
The anion carbonate is stable because it has 3 different resonance forms ⇒ stabilization by resonance