Question 1 Which of the following properties of the halogens increase from F to I? I. Atomic radius II. Melting point III. Electronegativity Answer: B (atomic radius & melting point) Reason: atomic radius increases down a group since more electron shells are being added each time; the melting point increases since the atoms get larger & hence there are more electrons to induce more temporary dipoles which leads to stronger Van der Waals’ forces.

Question 2 Which factors lead to an element having a low value of first ionisation energy? I. Large atomic radius II. High number of occupied energy levels III. High nuclear charge Answer: A (large atomic radius & high number of occupied energy levels) Reason: The larger the atom, the further away the outer (valence) electrons are from the nucleus; a large number of occupied energy levels gives rise to increased shielding; hence the outermost electron is more easily removed in the process: X(g)  X + (g) + e -

Question 3 What is the correct number of each particle in a fluoride ion, 19 F – ? protonsneutronselectrons A B C. 910 D (Total 1 mark) Answer: C Reason: the fluoride anion means that there is 1 more electron than proton to give the ion a single negative charge,; there are 10 neutrons as 10+9= 19 which is the mass number

Question 4 Which properties are typical of most non-metals in period 3 (Na to Ar)? I.They form ions by gaining one or more electrons. II.They are poor conductors of heat and electricity. III.They have high melting points. Answer: A (II & I only) Reason: Non-metals usually form anions by gaining electrons from metals; they are poor conductors due to only Van der Waals forces being present which cannot carry permanent charge

Question 6 Rubidium is an element in the same group of the periodic table as lithium and sodium. It is likely to be a metal which has a A.high melting point and reacts slowly with water. B.high melting point and reacts vigorously with water. C.low melting point and reacts vigorously with water. D.low melting point and reacts slowly with water. Answer: C Reason: It is a group 1 alkali metal

Question 7 When the following species are arranged in order of increasing radius, what is the correct order? A.Cl –, Ar, K + B.K +, Ar, Cl – C.Cl –, K +, Ar D.Ar, Cl –, K + Answer: D (Ar K + ) Reason: Atomic radius increases down a group & from right to left across a period. An increasing nuclear charge across a period causes the atom to contract by “pulling” the outer (valence) electrons inwards. On descending a group more shells are being added each time so the valence electrons become further away from the nucleus while at the same time they are more shielded by the extra shells.

Question 8 Which property decreases down group 7 in the periodic table? A.atomic radius B.electronegativity C.ionic radius D.melting point Answer: B (electronegativity) Reason: On descending the group, more shells are being added so the valence electrons are further away from the nucleus & are better shielded by the extra inner shells. This outweighs the effect of the increased nuclear charge & hence the attraction for the outer electron pair in a covalent bond is decreased.

Question 9 What is the correct sequence for the processes occurring in a mass spectrometer? A.vaporization, ionization, acceleration, deflection B.vaporization, acceleration, ionization, deflection C.ionization, vaporization, acceleration, deflection D.ionization, vaporization, deflection, acceleration Answer: A (vaporization, ionization, acceleration, deflection)

Question 10 Which of the reactions below occur as written? I.Br 2 + 2I – → 2Br – + I 2 II.Br 2 + 2Cl– → 2Br – + Cl 2 A.I only B.II only C.Both I and II D.Neither I nor II Answer: A Reason: Bromine is more reactive (stronger oxidising agent) than iodine but less reactive than chlorine & hence will only displace iodide ions from their salts & oxidise them to iodine. Reactivity decreases down group 7 due to the electron affinity decreasing since more shells are being added, causing the valence electrons to be further away from the nucleus & better shielded.

Question 11 Which general trends are correct for the oxides of the period 3 elements (Na 2 O to Cl 2 O)? I.Acid character decreases. II.Electrical conductivity (in the molten state) decreases. III. Bonding changes from ionic to covalent. Answer: C Reason: Electricity conductivity decreases across period 3 for the oxides since the bonding changes from ionic to covalent. Only ionic compounds can conduct electricity in the molten state; covalent compounds are uncharged species & hence are non-conductors. Acid character increases across the period.

Question 12 What increases in equal steps of one from left to right in the periodic table for the elements lithium to neon? A.the number of occupied electron energy levels B.the number of neutrons in the most common isotope C.the number of electrons in the atom D.the atomic mass Answer: C Reason: The atomic number (no. of protons) increases in increments of 1 across period 2; the number of electrons increases in the same manner since all atoms are neutral (the number of electrons must equal the number of protons).

Question 13 Which of the physical properties below decrease with increasing atomic number for both the alkali metals and the halogens? I.Atomic radius II.Ionization energy III.Melting point A.I only B.II only C.III only D.I and III only Answer: B (II only) Reason: The valence electrons are more easily removed on descending a group since they are further away from the nucleus & better shielded.

Question 14 Which is the correct description of polarity in F 2 and HF molecules? A.Both molecules contain a polar bond. B.Neither molecule contains a polar bond. C.Both molecules are polar. D.Only one of the molecules is polar. Answer: D (Only one of the molecules is polar) Reason: Only HF is polar since it is composed of 2 different atoms with very different electronegativity values.

Question 15 For which element are the group number and the period number the same? A.Li B.Be C.B D.Mg Answer: B (Be) Reason: Beryllium is in both group 2 & period 2

Question 22i Crocetin consists of the elements carbon, hydrogen and oxygen. Determine the empirical formula of crocetin, if 1.00 g of crocetin forms 2.68 g of carbon dioxide and g of water when it undergoes complete combustion. Moles C = Moles CO 2 = 2.68 g/44.01 g mol -1 = mol Moles H = 2 x moles H 2 O = 2 x (0.657 g/ g mol -1 ) = mol Mass of C = mol x g mol -1 = g Mass of H = mol x 1.01 g mol -1 = g The remainder must be the mass of oxygen in crocetin Mass of O = = g Moles O = 0.195/16 = CarbonHydrogenOxygen Moles Mole ratio0.0609/ / / Emperical formula Empirical formula: C 5 H 6 O

Question 22ii Determine the molecular formula of crocetin given that mole of crocetin has a mass of 98.5 g Molar mass of crocetin = 98.5 g / mol = 328 g mol -1 Total molecular mass of EF, (C 5 H 6 O) = Molar mass of MF/ Molar mass of EF =328/82.11 = 4 Molecular formula: C 20 H 24 O 4

Question 23i Nitrogen is found in period 2 and group 5 of the periodic table. (i)Distinguish between the terms period and group. Period is a horizontal row in the periodic table & a group is a vertical column

Question 23ii State the electron arrangement of nitrogen and explain why it is found in period 2 and group 5 of the periodic table. Electron arrangement: 2,5 Found in period 2 as electrons are in 2 energy levels/shells Found in group 5 as 5 outer/valence electrons

Question 24ai A sample of germanium is analysed in a mass spectrometer. The first and last processes in mass spectrometry are vaporization and detection. State the names of the other three processes in the order in which they occur in a mass spectrometer. Ionization, acceleration, deflection

Question 24aii For each of the processes named in (a) (i), outline how the process occurs Ionization: sample bombarded with high energy/high- speed electrons Acceleration: use electric field/oppositely charged plates Deflection: magnetic field

Question 24bi The sample of germanium is found to have the following composition: Isotope 70 Ge 72 Ge 74 Ge 76 Ge Relative abundance /  Define the term relative atomic mass. Average/weighted mean of masses of all isotopes of an element; Relative to 1 atom of 12 C

Question 24bii Calculate the relative atomic mass of this sample of germanium, giving your answer to two decimal places. A r = (70 x 0.226) + (72 x ) + (74 x ) + (76 x ) = 72.89

Question 25a State the meaning of the term electronegativity. the ability of an element/atom/nucleus to attract a bonding pair of electrons in a covalent bond

Question 25b State and explain the trend in electronegativity across period 3 from Na to Cl. Electronegativity increases across the period; The number of protons/nuclear charge increases

Question 28ai Explain the following statements. (a) The first ionization energy of sodium is less than that of magnesium. Na has a lower nuclear charge/number of protons; electrons being removed from the same energy level/shell

Question 28aii Explain the following statements. (ii) The first ionization energy of sodium greater than that of potassium The electrons in sodium are closer to the nucleus/in a lower energy level/ Na has less shielding effect hence harder to remove the valence electrons

Question 28b Explain the following statements. The electronegativity of chlorine is higher than that of sulfur. Chlorine has a higher nuclear charge; Attracts the electron pair/electrons in the bond more strongly

Question 29i Explain why the ionic radius of chlorine is less than that of sulfur. Chlorine has an extra proton/greater nuclear charge; Same amount of shielding (2 inner shells); Hence outer electrons attracted more strongly

Question 29ii Explain what is meant by the term electronegativity and explain why the electronegativity of chlorine is greater than that of bromine. Electronegativiy is the ability of an atom to attract a bonding pair of electrons in a covalent bond chlorine has a smaller radius/electrons are closer to the nucleus/; hence repelled by fewer inner electrons/decreased shielding effect

Question 30 A toxic gas, A, consists of 53.8% nitrogen and 46.2% carbon by mass. At 273 K and 1.01×105 Pa, g of A occupies 462 cm3. Determine the empirical formula of A. Calculate the molar mass of the compound and determine its molecular structure. Empirical formula of A: CN CarbonNitrogen Moles46.2/12.01 = /14.01 = 3.85 Mole ratio11

Question 30 (cont’d) Moles A = volume of A/ molar volume =0.462dm 3 / 22.4dm mol -3 = Molar mass A = Mass/moles = 1.048g / mol = 50.8 g mol -1 M r of empirical formula (CN) = = Molar mass of A/M r of empirical formula = 50.8/26.02 = 2.00 Hence Molecular formula of A: C 2 N 2 Possible structure: N ≡ C−C ≡ N (Cyanogen)