Chemistry Solutions 9701/11/O/N/15

Ref: 9701/11/O/N/15 #13

Rat poison needs to be insoluble in rain water but soluble at the low pH of stomach contents. What is a suitable barium compound to use for rat poison?

A. barium carbonate
B. barium chloride
C. barium hydroxide
D. barium sulfate

Solution:
A. barium carbonate

Barium chloride and barium hydroxide are both soluble in rain water and also acids, so B and C are incorrect. Barium sulfate is insoluble in both water and acids, so D is incorrect as well. Barium carbonate is insoluble in water, but dissolves in acids to give off carbon dioxide.
BaCO3 + 2HCl -> BaCl2 + CO2 + H2O


Ref: 9701/11/O/N/15 #15

The melting points of the Period 3 elements sodium to aluminium are shown in the table.

Which factor explains the increase in melting points from sodium to aluminium?

A. the changes in first ionisation energy from sodium to aluminium
B. the increase in electronegativity from sodium to aluminium
C. the increase in the Ar of the elements from sodium to aluminium
D. the increase in the number of outer electrons in each atom from sodium to aluminium

Solution:
D. the increase in the number of outer electrons in each atom from sodium to aluminium

Strength of the metallic bonds depends on the number of delocalised electrons in the metal. The greater the number of delocalised electrons, the greater the strength of the metallic bonds, the higher the melting and boiling points. Take note that melting and boiling points only depend on metallic bonds for metals, intermolecular forces for non-metallic compounds (permanent dipoles, Van der Waals' forces/temporary induced dipoles, hydrogen bonds), ionic bonds for ionic compounds.
Melting and boiling points DO NOT depend on the strength of covalent bonds.


Ref: 9701/11/O/N/15 #19

Element X reforms a pollutant oxide Y. Y can be futher oxidised to Z. Two students made the following statements.

Student P  "The molecule of Y contains lone pairs of electrons."
Student Q  "The oxidation number of X increases by 1 from Y to Z."

X could be carbon or nitrogen or sulfur

Which student(s) made a correct statement?

A. P only
B. Q only
C. Both P and Q
D. neither P nor Q

Solution:
A. P only
-Nitrogen monoxide has 3 lone pairs , one from nitrogen , two from
oxygen. Hence student P is correct. If nitrogen was element X, it would form nitrogen monoxide/nitric oxide which would be oxidised to nitrogen dioxide. In this process, the oxidation number of nitrogen increases by 2, hence student Q is incorrect.

 -If carbon was element X, it would form carbon monoxide which has lone pairs. Carbon monoxide is not oxidised in the atmosphere. Hence student P is correct and student Q is incorrect.

- If sulfur was element X, it would form sulfur dioxide which has 6 lone pairs. Hence P is correct. Sulfur dioxide would be oxidised to sulfur trioxide in the atmosphere.In this process, the oxidation number of sulfur increases by 2, hence student Q is incorrect.







Ref: 9701/11/O/N/15 #20

How many isomeric esters have the molecular formula C4H8O2?

A. 2
B. 3
C. 4
D. 5

Solution:
C. 4

Usually for isomeric esters, the first step is to list out the alcohol and carboxylic acid components, then consider the isomers for each component. In this case, only propanol has isomers, which are propan-1-ol, and propan-2-ol.

Methyl propanoate CH3CH2CO2CH3
Ethyl ethanoate CH3CO2CH2CH3
Propyl methanoate HCO2CH2CH2CH3
2-methyl-ethyl methanoate HCO2CH(CH3)CH3


Ref: 9701/11/O/N/15 #21

A new jet fuel has been produced that is a mixture of different structural isomers of compound Q.

 

Which skeletal formula represents a structural isomer of Q?

Solution:
C

Once you draw out the compounds and look carefully, you will see that A, D and Q are actually the same compound (notice that the parts circled orange are the same). A and D are laterally inverted. Hence A and D are incorrect. B has a total of 13 carbon atoms in its compound whereas Q has 14 carbon atoms. Hence B is incorrect. C is a structural isomer of Q, so C is the answer.

*note that the compounds are drawn for illustration purposes, it would be safer to draw the compound with CH3 branches tetrahedrally rather than V-shaped in structured questions.


Ref: 9701/11/O/N/15 #23

Compound Q contains 3 double bonds per molecule.

 

 Which bond, X or Y, will be ruptured by hot, concentrated acidified KMnO4 and how many lone pairs of electrons are present in one molecule of Q ?

Solution:
B

Each oxygen atom has 2 lone pairs. Hence molecule Q has 6 lone pairs. Bond X will be broken when the alkene molecule undergoes vigorous oxidation with hot concentrated acidified KMnO4. Y is a ketone, therefore it cannot be oxidised.



Ref: 9701/11/O/N/15 #31

Which statements about orbitals in a krypton atom are correct ?
1. The 1s and 2s orbitals have the same energy as each other but different sizes.
2. The 3rd energy level ( n=3) has 3 subshells and 9 orbitals
3. the 3d subshell has 5 orbitals that have the same energy as each other in an isolated atom

Solution:
C  2 and 3

1 is incorrect because the 2s orbital is at a higher energy level than 1s orbital . The 2s orbital is larger than the 1s orbital.
2 is correct because the 3rd energy level consist of the 3s, 3p and 3d subshells. The 3s subshell has 1 orbital. The 3p subshell has 3 orbitals and the 3d subshell has 5 orbitals.
3 is correct because the 5 orbitals are degenerates and have the same energy level.


Ref: 9701/11/O/N/15 #34


Which statements correctly describe an effect of a rise in temperature on a gas phase reaction.
1. More particles now have energies greater than the activation energy  .
2. The energy distribution profile changes with more particles having the most probable energy.
3. The activation energy of the reaction is decreased.


Solution:
D. 1 only
Statement 1 is correct because the peak of the Boltzmann Distribution curve is lowered and the whole graph shifts to the right. The activation energy remains the same, so more particles have energies more than or equal to the activation energy.
Statement 2 is wrong because most probable energy = energy possessed by the greatest number of molecules. (Refer to peaks in the attached diagram.) At higher temperature, the peak is lowered. Less particles have the most probable energy at a higher temperature.
Statement 3 is wrong because the activation energy of the reaction remains the same.

Ref: 9701/11/O/N/15 #38
Each of the compounds below is treated separately with excess NaBH4. The product of the reaction is then heated with excess concentrated H2SO4.
In each case, one ore more products are formed with molecular formula C7H10.
Which compounds give only one final product with the molecular formula C7H10?

Solution:
C. 2 and 3 only

1 produces two different compounds with the molecular formula C7H10, hence it is incorrect.
2 produces one compound only. The diagram shows the products being laterally inverted --- when flipped, they are actually the same compound.
3 produces one compound only, because a double bond cannot be formed between C1 and C2. This is because the H atom from C1 is eliminated along with the branched -OH group to form a double bond already.