IJC H1 CHEM P1 Worked Solutions

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2  IJC 2015 Prelim 1/8872/01 [Turn over Section A For each question there are four possible answers, A, B, C, and D. Choose the one you consider to be correct. 1 An exciting chemical demonstration is the ‘barking dog’. An equation which describes the reaction is shown. 6NO + 2CS2 3N 2 + 2CO + 2SO2 + 1 4 S8 Carbon disulfide, CS2, is a liquid with a density of 1.26 g cm3. Which mass of sulfur would be formed from 5.00 cm 3 of CS2 if the reaction proceeded with 80% yield? [Mr of CS2 = 76.2; S8 = 256.8] A 1.67 g B 2.12 g C 2.65 g D 3.32 g Mass of CS2 = 1.26 x 5 = 6.30g Amount of CS2 = 6.30 / (12.0 + 32.1 + 32.1) = 0.08268 mol Amount of S = 0.08268 / 2 x ¼ = 0.01033 Mass of S formed = 0.01033 x 80/100 x 32.1 x 8 = 2.12g 2 The following half-reactions occur when potassium iodate( V), K IO3, in hydrochloric acid solution oxidises iodine to ICl2 . IO3  + 2Cl + 6H+ + 4e ICl2  + 3H2O I2 + 4Cl 2ICl2  + 2e What is the ratio of IO3  to I2 in the balanced chemical equation for the overall reaction? A 1: 1 B 1: 2 C 1:4 D 2 :1 IO3  + 2Cl + 6H+ + 4e ICl2  + 3H2O I2 + 4Cl 2ICl2  + 2e -------------------x2 IO3  + 6H+ + 2I2 + 10Cl 4ICl2  + ICl2  + 3H2O Answer = 1:2 3 Use of the Data Booklet is relevant to this question. In which pairs do both species have different number of unpaired electrons? A O and C l+

3  IJC 2015 Prelim 2/8872/01 [Turn over B F+ and Ga C P and Ne+ D Ga+ and Ne O: 1s2 2s2 2p4 -- 2 unpaired electrons Cl+ : 1s2 2s2 2p6 3s2 2p4 -- 2 unpaired electrons F + : 1s2 2s2 2p4 -- 2 unpaired electrons Ga- : 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p2 -- 2 unpaired electrons P: 1s2 2s2 2p6 3s2 3p4 -- 2 unpaired electrons Ne+ : 1s2 2s2 2p5 -- 1 unpaired electrons Ga + : 1s2 2s2 2p6 3s2 3p6 3d10 4s2 -- 0 unpaired electrons Ne: 1s2 2s2 2p6 -- -- 0 unpaired electrons

4  IJC 2015 Prelim 2/8872/01 [Turn over 4 The first seven ionisation energies of an element M are shown in the sketch graph. What could be the identity of element M? A aluminium B magnesium C phosphorus D silicon Biggest jump is from the 4th IE to the 5th IE. This shows that the 5th electron is removed from an inner quantum shell. Hence, there are 4 valence electrons and the element will be in group IV. 5 Ions of the two most common isotopes of zinc are shown below: 64 30 2+Zn 66 30 2+Zn Which of the following statements is correct? A Both Zn 2+ ions have the same number of electrons but different number of protons. B Both Zn2+ ions have the same electron configuration 1s22s22p63s23p63d84s2. C The 64 30 2Zn  ion has fewer neutrons in its nucleus than the 66 30 2Zn  ion. D The 66 30 2+Zn ion will be deflected more than the 64 30 2Zn  ion in an electric field of the same strength. A) Isotopes: Same number of protons but different number of neutrons. B) Since they have same number of elec trons, they will have same electronic configuration. However, the electronic configuration should be 1s22s22p63s23p63d10. C) 64 30 2+Zn Number of neutrons = 64 – 30 = 34

5  IJC 2015 Prelim 2/8872/01 [Turn over 66 30 2+Zn Number of neutrons = 66 – 30 = 36 D) Charge over mass ratio of 64 30 2+Zn is smaller than charge over mass ratio of 66 30 2+Zn . 66 30 2+Zn should deflect less.

6  IJC 2015 Prelim 2/8872/01 [Turn over 6 Refer to the molecule below. Which one of the following shows the correct information with regard to the various labelled bond angles? smallest bond angle  largest bond angle A c d e B d a c C e b a D b e d a = 105° , b = 107°, c = 180°, d = 120°, e = 109.5° 7 Three substances, X, Y and Z, have the physical properties shown in the table. substance melting point / °C boiling point / °C density / g cm-3 conductivity (solid) conductivity (liquid) X Y Z 114 993 650 184 1695 1755 4.93 2.56 1.74 poor poor good poor good good What could be the identities of X, Y and Z? X Y Z A I2 SiO 2 Ca B I2 NaF Mg C Si Mg NaF D Ca KCl SiO2 a b c e d

7  IJC 2015 Prelim 2/8872/01 [Turn over Substance X has low mp and bp and poor conductivity, this shows that X will be a simple molecular structure. Substance Y has high mp and bp with good conductivity when it is in the liquid (molten) state, this shows that Y will have a giant ionic structure. Substance Z has high mp and bp with good conductivity in both solid and liquid state, this shows that it has a giant metallic structure. 8 Consider the following four compounds. 1 (CH 3)3CH 2 CH 3CH2CH2CH3 3 CH 3CH2CH2OH 4 CH3CH2Cl What is the order of decreasing boiling point of the compounds? A 1 2 4 3 B 3 4 1 2 C 3 4 2 1 D 4 3 2 1 3 will have the highest boiling point due to hydrogen bonding followed by 4 with pd – pd. 1 and 2 will only have Td – Id as the forces of attraction which is the weakest IMF. 1 will have the lowest Td – Id as it is spherical in shape while 2 is linear in shape so the surface area of contact between molecules 1 will be smaller. 9 What does the following equation represents? H2(g) + O2(g) H 2O2(g) A standard enthalpy change of combustion of H2 B standard enthalpy change of formation of H2O2 C sum of standard enthalpy change of formation of H 2O2 and the enthalpy change of vaporisation of H2O2. D sum of standard enthalpy change of combustion of H 2 and the enthalpy change of vaporisation of H2O2.

8  IJC 2015 Prelim 2/8872/01 [Turn over A) Std enthalpy change of combustion of H 2  H2 + ½ O2  H2O B) standard enthalpy change of formation of H 2O2  H2 + O2  H2O2 C) D) Cycle will be incomplete to use Hess Law 10 I