NZPQA. New Zealand Pretend Qualifications Authority. Level 2 Chemistry KEY NOTES LEWIS DIAGRAMS

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1 NZPQA New Zealand Pretend Qualifications Authority Level 2 Chemistry (2.4): Demonstrate understanding bonding, structure, properties and energy changes. + WALK-TRUGS + KEY NTES LEWIS DIAGRAMS a full luorine ams are one electron short shell (E.C. = 2,7) so y form one covalent bond with carbon. The shared pair electrons forms a very strong bond. INCLUDING GST CNDITINS APPLY (c) no lone pair electrons. 2S3 Write equilibriu Δr m expression. 2 xygen molecules contain a double bond. Note how each oxygen am has 8 electrons from 2 lone pairs and a double bond. S3 If concentration temperature, calculate S2 was mol L-1 and was mol L-1 at concentration 2 that S. (d) (e) Example - A water molecule (2) (f) A Calculate tal valence electrons available: Total = 8 2 x = 2 and 1 x = 6 B Total if all ams have full outer shells: D Calculate number covalent bonds 4 / 2 = 2 = C /2 Justify your answer Describe effect Justify your answer temperature on (h) Add lone pairs electrons complete oxygen s octet. Add number bonds you calculated for D (i) that "look" linear but are in fact bent! Exam questions ten show molecules pressure on position Describe effect Justify your answer removing S from system on 3 position equilibriu m. Removing S from system will3 lower its concentration. Equilibrium will shift replace it by favouring forward reaction. Trigonal Planar TIME Solution Y and starch, with or without a small 2+ amount Cu The following experiments A cross drawn on paper were carried out, and times taken for cross Temperature / C Experiment 1 No Cu2+ present 2 No Cu2+ present 3 Cu2+ present disappear recorded. Time for Cross Disappear reaction in 3 occur faster than in Experiment 2 and Experiment Elaborate on why reactions Experiment 1. following words or terms. In your answer, include catalyst effective temperature activation energy collisions B 180 An increase in pressure will be partially undone if equilibrium shifts right by favouring forward reaction, reby decreasing number gas particles. Equilibrium shifts right. (h) above. Be Linear timing Equilibrium will shift right. 120 with powder. reaction is Y with starch present. When mixing solution X and solution A clock reaction involves blue-black in colour. flask. ver time, complete solution turns X and solution Y in a conical reaction between solution when viewed from above. A student carried out this under flask disappeared cross on piece paper Add solution X and start equilibrium. (f) above. EXAM WARNING Beryllium and boron form molecules with less than an octet (e.g. Be2, B3). acid would react faster Equilibrium will shift left. Le Chatelier's Principle states that position equilibrium will shift partially undo an imposed change. As reaction is exormic increasing temp. favours reverse reaction and so equilibrium will shift left. increasing m. There are three moles gas particles on left and two on right. Explain why hydrochloric position equilibriu (d) above. Number covalent bonds = 2 (ii) A large number so products are favoured. Rearrange equilibrium expression find S3 and substitute values. increasing Number shared electrons = 4 = B - A: 12-8 = 4 Draw central oxygen am surrounded by two hydrogen ams. (g) Total = 12 electrons : 2 valence electrons x 2 = 4. : 8 valence Describe effect bond. electrons. Using involves working out number valence The first step in drawing Lewis diagrams five, group 16 has six. 1 has one valence electron, group 15 has group number is easiest. or example, group y rarely react. eight valence electrons which explains why Group 18 elements ( Noble gases) have C Shared electrons Products on reactants as denominar. Mole ratios become powers raise concentration values. 3 Nitrogen molecules contain a triple bond. Note how each nitrogen am has 8 electrons from one lone pair and a triple N Reaction rates (lumps) and powder QUESTIN NE (2013) in form marble chips with calcium carbonate a chemical reaction. ydrochloric acid was reacted facrs affecting rate (a) an experiment investigate (crushed marble chips) in investigated. Identify facr being (i) p, Walk-through kj mol-1 If Kc = 1300 at a given temperature, circle concentration: species that would have highest S2 ydrogen ams require only one electron form a full valence shell two electrons. They form single covalent bonds and have NCEA Exam Questions - ium when six electrons are double bonded. A triple bond occurs Molecules that share 4 electrons (two pairs) (three pairs) are shared. Examples include: PREVIUS EXAMS Walk-through - Equilibr reaction: 2S2(g) + 2(g) (a) Carbon has formed four covalent bonds, With each bond having two electrons in it. a full valence shell formed by bonds, carbon does not have lone pair electrons. N C lone luorine ams are drawn with three pairs electrons show that each am lone has a full valence shell. Each has three pairs and a pair forming a bond. $12 or following location any lone where covalent bonds are located, and The Lewis diagram a molecule shows pairs electrons. NLY 1. PTCPYING PRIBITED Key notes and worked examples PTCPYING PTCPYING PRIBITED PRIBITED NCEA-style questions with walk-throughs, hints & tips. 71 our years previous NCEA exam questions. VIEW/RDER/PURCASE NLINE AT or Assessor s use only Achievement Criteria Achievement Achievement with Merit Achievement with Excellence Demonstrate understanding bonding, Demonstrate in-depth understanding Demonstrate comprehensive structure, properties and energy bonding, structure, properties and understanding bonding, structure, changes. energy changes. properties and energy changes. verall Level Performance

2 QUESTIN NE: BNDING, STRUCTURE AND PRPERTIES. 1. Sodium bromide (NaBr) has a melting point 770 C. In terms structure and bonding within compound, explain why it has such a high melting point. 2. Sodium bromide (NaBr) is soluble in water (90 g / 100 ml). Discuss sodium bromide dissolving in water in terms structure and bonding solute and solvent.

3 3. (a) Lewis structures for two molecules are given below. Molecule 2 S C 4 Lewis diagram S C or each molecule, name shape molecule and give a reason for your answer. 2 S Shape: Reason: C 4 Shape: Reason: State polarity each se molecules. (i) 2 S Polar Non-polar (ii) C 4 Polar Non-polar (c) or each molecule, give a justification for your choice.

4 4. The diagrams below show 3-D structural representations diamond and graphite. Diamond and graphite are both made up carbon ams, but se ams are arranged differently in each solid. Structure diamond Structure graphite (a) Describe electrical conductivity and hardness diamond and graphite. Diamond Electrical conductivity: ardness: Graphite Electrical conductivity: ardness: Discuss electrical conductivity and hardness both diamond and graphite, using your knowledge structure and bonding.

5 QUESTIN TW: ENERGY CANGES 1. Classify following reactions, by writing in box below word Exormic or Endormic. Exormic or Endormic C 2 4(g) + 2 (g) C 2 5 (g) Δ r = 48.0 kj mol 1 2 (g) 2 (l) C 4(g) + 2 (g) C (g) + 3 2(g) 206 kj energy is absorbed. When zinc powder reacts with copper sulfate solution, temperature rises. 2. When a 12.2 g sample ammonia is burned, 275 kj energy is released. Calculate energy released for reaction below, when four moles ammonia are burned. 4N 3(g) + 3 2(g) 2N 2(g) (l) M(N 3 ) = 17.0 g mol 1 3. Methane reacts with oxygen according following reaction: C 4(g) (g) C 2 (g) (g) Use following bond enthalpies calculate Δ r for this reaction: Bond Bond enthalpy (kj mol -1 ) =C 745 = 498 C- 412

6 QUESTIN NE Suggested Answers - Chemistry 2.4 Exam 1. Sodium bromide is an ionic substance. It is made up sodium cations (Na + ) and bromide anions (Br - ). As a solid se cations and anions are held ger by strong electrostatic forces attraction between oppositely charged ions. or solid melt se strong bonds must be overcome. As bonds are strong a lot heat energy is required overcome attractive forces between ions. 2. Sodium bromide is an ionic substance. It is made up sodium cations (Na + ) and bromide anions (Br - ). As a solid se cations and anions are held ger by strong electrostatic forces attraction between oppositely charged ions. Water molecules are polar, meaning y have a slight positive charge on one end molecule ( hydrogen end), and a slight negative charge on or ( oxygen am). When sodium bromide dissolves cations and anions are removed from crystal lattice and become surrounded by water molecules. The sodium cations have negative charge on oxygen facing m, and positively charged hydrogen faces bromide anions. 3. (a) 2 S is bent. There are four regions negative charge surrounding central sulfur am. These repel for maximum separation in a tetrahedral shape. There are two bonding regions and two lone pairs electrons so molecule is bent, with a bond angle approximately 109. (c) C 4 is tetrahedral. There are four regions negative charge surrounding central carbon am. These repel for maximum separation in a tetrahedral shape. All four regions are bonding regions, re are no lone pairs electrons, so molecule is tetrahedral with a bond angle approximately S is polar C 4 is non-polar 2 S molecules have 2 polar covalent bonds, with negative dipole on more electronegative sulfur am. The molecule is bent so it is asymmetrical in regard se bonds. The bond dipoles do not cancel so re is a molecular dipole so 2 S is polar. C 4 molecules have 4 polar covalent bonds, with negative dipole on more electronegative carbon am. The molecule is tetrahedral so it is symmetrical and bond dipoles cancel. There is no molecular dipole so C 4 is non-polar. 4. (i) Diamond does not conduct electricity and is hard. Graphite does conduct electricity and is st. QUESTIN TW (ii) Diamond consists C ams each covalently bonded four or C ams, forming a 3-D tetrahedral arrangement. Graphite consists C ams each covalently bonded three or C ams in a 2-D (or layered) arrangement with weak intermolecular forces attraction between layers or sheets. 1. Exormic Exormic Endormic Exormic In diamond, covalent bonds between carbon ams are very strong and hold ams in place, making it difficult break bonds. Therefore, diamond is a very hard substance. owever in graphite, although bonds between covalently bonded carbon ams in layers are strong, forces between layers are weak, resulting in layers sliding over each or. Therefore, graphite is a st substance. In diamond, all valence electrons in each carbon am are involved in bonding or carbons. There are no mobile electrons carry charge. Therefore, diamond is unable conduct. owever, in graphite each carbon am is bonded three ors in layers and has one valence electron which is free move. These delocalised electrons result in ability graphite conduct electricity. 2. n(n 3 ) in 12.2 g sample = 12.2 / 17.0 = mol (3 s.f.) 4 mol releases / = kj (3 s.f.) 3. Bonds broken: 4 x C- = kj 2 x = = 996 kj Total = kj Bonds formed: 2 x C= = kj 4 x - = kj Total = kj NLY $12 INCLUDING GST CNDITINS APPLY Difference = = kj r = -698 kj mol -1