IB Chemistry HL Paper 2 Question Bank

1.) The most common hydride of phosphorus is called Phosphine. It has a formula PH3
a.) Draw a Lewis structure of Phosphine

Answer:

Since Phosphine is PH3, one phosphorus atom is in the middle connected by 3 hydrogen atoms around it. The line above the phosphorus atom shows the lone pair of electrons that has not been bonded with any other atom.

b.) State the hybridization of the phosphorus atom in Phosphine and its molecular structure
Answer:
The hybridisation of the phosphorus atom in phosphine is sp3, which means it has three bonds with hydrogen atoms. PH3 also has a trigonal pyramidal structure. The trigonal pyramid geometry is formed when the central atom is attached to three atoms and contains one lone pair.

c.) Determine if whether Phosphine would act as a Lewis base or a Lewis acid or neither
Answer:
Phosphine would act as a Lewis base because it is capable of donating a pair of electrons. A Lewis base is an electron-pair donor. PH3 has a pair of nonbonding electrons and can act as a donor.

d.) Define a Bronsted Lowry base
Answer:
A Bronsted-Lowry base is a substance which accepts a proton or H+ ion from the other compound and forms conjugated acid.

2.) The properties of the elements can be deduced from where they are placed on the periodic table.
a.) Explain why the atomic radius reduces as you go across Period 3.
Answer:
The nuclear charge or the number of protons increase as you go across Period 3. Therefore, when the nucleon number increases, it causes a stronger pull on the outer electrons. The electrostatic force (force acting between the electrons and the nucleus) increases therefore causing the size (atomic radius) of the atom to reduce.

b.) The periodic table also shows the relationship between the properties of elements and their electronic configuration
i.) Define electronegativity
Answer:
The tendency of an atom in a molecule to attract the shared pair of electrons towards itself is known as electronegativity.

ii.) State two reasons why metal ions have a smaller radius than metal atoms
Answer:
The ions tend to have a greater net positive charge. They would be pulling fewer electrons with the same number of protons therefore causing the ionic radius to decrease further.

iii.) The ionic radius of P3- is larger than that of the ionic radius of Si4+ Outline why.
Answer:
P3- has one more energy level than Si4+ so its valence electrons will be further from the nucleus and it will have a larger ionic radius.

3.) Hydrazine and ethene are hydrides of adjacent elements in the periodic table.
a.) The boiling point of hydrazine happens to be much higher than that of ethene. Outline why this is the case with reference to their intermolecular forces.
Answer:
N2H4 is a polar molecule with London dispersion forces, dipole-dipole forces, and hydrogen bonding between molecules, whereas C2H6 is nonpolar and only has London dispersion forces between molecules. It takes more energy to overcome the stronger IMFs in hydrazine, resulting in a higher boiling point.

b.) A reaction occurs between hydrazine and oxygen. It is represented by the following equation:
N2H4 (l) + O2 (g)) → N2 (g) + 2H2O (l)

i.) Define entropy
Answer:
Entropy is the measure of the distribution of available energy among the particles.

ii.) Calculate the change in entropy. Comment on the magnitude
Answer:
Change in entropy = Total sum of entropy products – total sum of entropy reactants
Change in entropy = (191 + 2(69.9)) – (121 + 205) = +4.8 J/Kmol
A positive (+) entropy change means there is an increase in disorder.

4.) Consider the following reaction that is in equilibrium
2SO2 (g) + O2 (g) 2SO3 (g) Enthalpy = -198 kJ/mol
a.) Explain how the conditions can be altered to increase the yield of the products
Answer:
Since it is an exothermic reaction, the temperature of the reaction can be reduced. Additionally, the pressure on the yield of the sulfur trioxide can be increased (when pressure increases on the product side of the reaction with fewer moles, the yield also increases).

b.) State how the catalyst would affect the forward and reverse reaction, position of the equilibrium and the value of the equilibrium constant
Answer:
A catalyst would increase the rate of reaction of both the forward and backward reaction equally. Since it increases equally, and because catalysts don’t affect the yield produced in the reaction, there is no effect on the position of the equilibrium. Consequently, it does not affect the equilibrium constant either.

c.) Electroplating is one of the important applications of electrolysis.
a.) State the composition of the electrode and the electrolytes used in silver electroplating process
Answer:
Cathodes i.e. the negative electrodes, are the objects that are plated in the silver electroplating process. Objects that are made of steel such as a spoon. As for the anodes i.e. the positive electrodes, it is silver. This silver will be coated on the steel spoon. The electrolyte is Sodium argentocyanide solution or silver nitrate.

ii.) Outline two differences between a voltaic cell and an electrolytic cell
Answer:
A voltaic cell converts chemical energy to electrical energy whereas an electrolytic cell converts electrical energy to chemical energy. Furthermore, in an electrolytic cell, the cathode is the negative electrode and the anode is the positive electrode. However, it is the opposite for a voltaic cell – the anode is the negative and the cathode is positive.

5.) The following reaction exists:
Cl2 (g) + H2O (l) HClO (aq) + HCl (aq)
a.) The products yielded are hypochlorous acid and hydrochloric acid. Hypochlorous acid is considered a weak acid. Define a weak acid.
Answer
A weak acid is an acid that only partially dissociates into its constituent ions when it is dissolved in water or an aqueous solution.

b.) Define a conjugate base and state the formula of the conjugate base of the hypochlorous acid
Answer
A conjugate base is a substance formed when an acid loses a hydrogen ion. The conjugate base of the hypochlorous acid is ClO-

c.) Calculate the concentration of the H+ ions in a HClO solution with a pH of 3.61
Answer
Formula: [H+] = 10-pH = 10-3.61 = 2.4547 x 10-4 mol/dm3

6.) Iron exists as several isotopes.
a.) State the type of spectroscopy that could be used to identify the relative abundances and briefly outline how it works.
Answer
Relative abundances can be determined experimentally using a technique called mass spectrometry. A mass spectrometer ionizes atoms and molecules with a high-energy electron beam and then deflects the ions through a magnetic field based on their mass-to-charge ratios. The mass spectrum of a sample shows the relative abundances of the ions on the y-axis and their ratios on the x-axis.
b.) Explain why zinc compounds are not coloured like typical transition metals
Answer
The zinc ion has completely filled d orbitals and no free electrons. Therefore, zinc is not a transition element. Transition elements require partially filled d orbitals to promote the electron into this level.

c.) Transition metals have the property of forming complex ions. Define the term ligand and discuss the bonding between the transition metal and their ligands in terms of the acid-base theory.
Answer
A ligand is an ion or molecule, which donates a pair of electrons to the central metal atom or ion to form a complex with coordinate bonds. The transition metal ion acts as a Lewis acid, and the ligand acts as a Lewis base.

7.) Cobalt is a transition metal.
a.) Among its many ions, Co3+ is a common one. Draw the orbital diagram for a Co3+ ion.
Answer

b.) Explain why [Co(NH2)6]Cl3 is a coloured complex
Answer
The complex has a partially filled d orbital. Therefore, the d orbitals are split into two sets of different energies. Frequencies of visible light are absorbed by the electrons when there is electron transition between these to split energy levels (movement from lower to higher d levels). So, the color due to the remaining frequencies are seen in the coloured complex.

8.) State the hybridization of the carbon I and II atoms in but-2-ene. The formula for but-2-ene is shown below

Answer
Hybridization of carbon I: sp3
Hybridization of carbon II: sp2

9.) The following reaction is in equilibrium
2NO (g) N2O2 (g) Enthalpy is < 0
a.) What is the effect of increasing temperature of the N2O2
Answer
Since the reaction is exothermic, increasing the temperature would reduce the yield of the product, N2O2 – the backward reaction is favored.
b.) This reaction follows a two-step mechanism:
First step: 2NO (g) N2O2 (g) – fast
Second step: N2O2 (g) + O2 (g) 2NO2 (g) – slow
Deduce the rate expression for the mechanism
Answer
The Rate constant for the first step is Kc = [N2O2] / [NO]2
[N2O2] = Kc [NO]2
From the second step, which is the slow step, rate = k1 [N2O2][O2]
Replacing [N2O2] with [NO]2 with reference to the fast reaction
Rate = k2 K[NO]2[O2] = k[NO]2[O2]

10.) This question is about carbon and chlorine compounds
a.) A reaction happens between Ethane (C2H6) and the chlorine in the sunlight. State and briefly explain what this reaction is and the name of the mechanism that occurs during the reaction
Answer
It is a substitution reaction and the mechanism is free-radical.
Ethane reacts with chlorine by free radical halogenation in the presence of sunlight when chlorine breaks down to form two chlorine radicals. The chlorine radical reacts with ethane to give ethane radical which reacts with other chlorine to generate a halogenated product C2Cl6.

b.) A possible product that results from the reaction between ethane and chlorine has the following composition in terms of percentage mass:
Carbon: 24.27% Hydrogen: 4.08% Chlorine: 71.65%
Determine the empirical formula
Answer
Carbon: Atomic mass = 12.01 C = 24.27 / 12.01 = 2.021 2 moles
Hydrogen: Atomic mass = 1.01 C = 4.08 / 1.01 = 4.04 4 moles
Chlorine: Atomic mass = 35.45 C = 71.65 / 35.45 = 2.021 2 moles
Hence, C2H4Cl2 is the molecular formula
Empirical formula is the simplified molecular formula = CH2Cl