BONDING BONDING CHEMISTRY CHEMISTRY BONDING BONDING CHEMISTRY CHEMISTRY

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Gyles Wade
5 years ago
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1 a diagram to show how a molecule NH 3 is attracted to a molecule H 2 O. Include all partial charges all lone pairs electrons in your diagram. Make sure you draw your diagram nice big in exam. Ensure you include correct number lone pairs all partial charges on both molecules. 5 ion. Trigonal bipyramidal 5 ion. 5 ion. There are no lone pairs around central atom which is most common mistake made with this structure.

2 an scl 3 molecule a Cl 3 + ion. Include any lone pairs electrons that influence. Name each. Pyramidal Bent The most common error is to include 2 lone pairs in scl 3. For Cl 3 + most common error is to draw a linear or cyclic structure. Make sure you draw a 3D not just a dot cross diagram. PH 3 reacts with H + to form PH 4+. PH 3 molecule state type bond formed between PH 3 H +. Explain how it is formed. The bond formed is a dative covalent bond. The P atom donates lone pair electrons to H + ion. 5 ion. 5 ion. For all dative covalent/coordinate bond questions you need to be specific state which atom/ion donates a pair electrons to which atom/ion instead giving a generic explanation.

3 a diagram to show hydrogen bonding between 2 HF molecules. Make sure you include all partial charges lone pairs in your diagram. : : It is easy to lose marks here if you forget any partial charges or leave out any lone pairs. : : : : uggest an ICl 4 - ion. 5 ion. 5 ion. Remember that lone pairs repel each or as far apart as possible, hence two lone pairs are 180 o apart in this structure.

PERIODICITY dd melting points P,, Cl r to graph: PERIODICITY The structure changes from giant covalent for i to simple molecular for P, Cl monatomic for r.

4 PERIODICITY dd melting points P,, Cl r to graph: PERIODICITY The structure changes from giant covalent for i to simple molecular for P, Cl monatomic for r. forms 8 rings, P forms P 4 structures Cl forms Cl 2 molecules which explains differences in boiling points. uggest a H 3 O + ion which is formed when H 2 O reacts with a H + ion. Name type bond form explain how it is formed. The bond formed is a dative covalent bond. The O atom donates a lone pair 5 ion. electrons 5 ion. to H + ion. For all dative covalent/coordinate bond questions you need to be specific state which atom/ion donates a pair electrons to which atom/ion instead giving a generic explanation.

5 how how particles are arranged in potassium. Then explain why potassium is malleable Potassium is malleable as layers can slide past each or. When asked to show particle arrangement in metals, always draw a minimum 6 particles in 2 rows in a regular pattern. how +ve charge metal ion; you do not need to include any electrons. BCl 3, a covalent compound, can react with a Cl - ion to form BCl 4-. BCl 4 - explain how bond is formed between Cl - BCl 3. Tetrahedral. The bond formed is a dative covalent bond. The Cl - ion donates a lone pair electrons to 5 ion. B atom. 5 ion. For all dative covalent/coordinate bond questions you need to be specific state which atom/ion donates a pair electrons to which atom/ion instead giving a generic explanation.

Lone pair-lone pair repulsion > lone-pair-bonding pair repulsion > bonding pair-bonding pair repulsion which results in a F- Cl-F angle <90 o.

6 Using valence electron pair repulsion ory, predict, bond angles in a molecule chlorine trifluoride (ClF 3 ). You are not required to unfamiliar s. For information: this is called a T. 3 bonding pairs 2 lone pairs repel as far apart as possible. Lone pair-lone pair repulsion > lone-pair-bonding pair repulsion > bonding pair-bonding pair repulsion which results in a F- Cl-F angle <90 o. structure sodium chloride. 5 ion. 5 ion. You are required to draw a 3D diagram for ionic structures clearly identify +ve ve ions (you should use actual element symbols for this purpose).

7 Using valence electron pair repulsion ory, predict, bond angles in a molecule OF 2. 2 bonding pairs 2 lone pairs repel as far apart as possible. Lone pair-lone pair repulsion > lone-pair-bonding pair repulsion > bonding pair-bonding pair repulsion which results in a F- O-F angle o (bent ) You are required to recognise s that are based on textbook examples. Using valence electron pair repulsion ory, predict, bond angles in, a molecule IF 5. 5 bonding pairs 1 lone pair repel as far apart as possible. lone-pair-bonding pair repulsion > bonding pair-bonding pair repulsion 5 ion. 5 ion. which results in a reduced angle slightly <90 o. You are not required to unfamiliar s.

8 Using valence electron pair repulsion ory, predict, bond angles in a molecule GeF 2. 2 bonding pairs 1 lone pair repel as far apart as possible. Lone-pair-bonding pair repulsion > bonding pairbonding pair repulsion which results in a F-Ge-F angle less than 120 o ( o, bent ) You are required to recognise s that are based on textbook examples. Using valence electron pair repulsion ory, predict, bond angles in BH bonding pairs electrons repel each or as far apart as possible to form a tetrahedral with a bond angle o 5 ion. 5. ion. Note that fourth bond is a coordinate bond as H - ion has donated an electron pair to B atom.

9 Using valence electron pair repulsion ory, predict, bond angles in an ion PH. 2 bonding pairs 2 lone pair repel as far apart as possible. Lone-pair- lone pair repulsion > lone-pair to bonding pair repulsion > bonding pairbonding pair repulsion which results in a H-P-H angle o (bent ) You are required to recognise s that are based on textbook examples. Using valence electron pair repulsion ory, predict, bond angles in COCl 2. 3 bond regions ( no lone pairs electrons) repel each or as far apart as possible to form a trigonal planar with a 5 ion. 5 bond ion. angle 120 o. For se questions simple count number lone pairs number bonding regions around central atom to determine bond angle.

TOMIC TRUCTURE Label mass spectrometer. TOMIC TRUCTURE Keep it simple stick to main labels. ny additional information you provide might be contradictory result in deduction marks.

10 TOMIC TRUCTURE Label mass spectrometer. TOMIC TRUCTURE Keep it simple stick to main labels. ny additional information you provide might be contradictory result in deduction marks. TOMIC TRUCTURE TOMIC TRUCTURE Plot approximate second ionisation energies period 3 elements Mg to. 2 nd IE/kJmol -1 Mg l i P 2 nd IE/kJmol -1 5 ion. Mg 5 ion. l i P This is a tricky question as you will need to work out electron configurations first carefully assess which orbital second electron is removed from.

11 TOMIC TRUCTURE TOMIC TRUCTURE rubidium sample contains 2.5 times more 85 Rb than 87 Rb. Calculate relative atomic mass Rb to 2 decimal places. (2.5x85)+ (1x87)/3.5 = 85.6 Instead % abundance, a ratio is provided. The method is same as for abundance provided in %. TOMIC TRUCTURE TOMIC TRUCTURE Identify element that has a 4+ ion with an electron configuration 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 Germanium (Ge). If you add 4 extra electrons, electron configuration Ge atom will be 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 4s 2 4p 2 5 ion. 5 ion. Remember that orbitals are emptied in order from highest energy level to lowest (refore 4s orbital is emptied before 3d orbital).

12 TOMIC TRUCTURE Give electron configuration Cu +. TOMIC TRUCTURE 1s 2 2s 2 2p 6 3s 2 3p 6 4s 0 3d 10 Remember that electron configuration Cu is 3d 10 4s 1. The one electron removed to form Cu + is 4s 1 electron. TOMIC TRUCTURE TOMIC TRUCTURE Explain presence peaks at m/z= in a mass spectrum Cu-63 Cu-65. Explain why se peaks are very small. The peak at m/z=31.5 is due to a 63 Cu 2+ ion. The peak at m/z=32.5 is due to a 65 Cu 2+ ion. s more energy is required to remove a second electron, few 2+ ions are formed. 5 ion. 5 ion. Remember that 63:2 = 31.5:1 which is where peak for 2+ ion occurs on mass spectrum.

13 TOMIC TRUCTURE TOMIC TRUCTURE Calculate average isotopic mass magnesium: Mg-24: 78.99%; Mg-25: 10.00%; Mg-26: 11.01% (78.99 x 24)+ (10.00 x 25) + (11.01 x 26)/100 = lways check what abundances add up to; don t assume it is always 100. TOMIC TRUCTURE TOMIC TRUCTURE Write an equation, including state symbols, for ionisation Germanium that requires minimum energy. Ge (g) Ge + (g) + e - 5 ion. 5 ion. The minimum energy implies that enough energy is available to remove first electron only.

14 MOUNT OF UBTNCE ZCl 4 (l) + 2H 2 O(l) ZO 2 (s) + 4HCl(aq) g ZCl 4 was added to water. ZO 2 was removed resulting solution was made up to 250 cm cm 3 portion this solution was titrated against mol dm 3 NaOH 21.7 cm 3 were required to reach end point. Calculate number moles HCl produced number moles ZCl 4 present in sample. Calculate relative molecular mass, M r, ZCl 4. Find relative atomic mass Z hence its identity. MOUNT OF UBTNCE To help you see through all information provided, I strongly recommend using highlighters in exam crossing f any values you have used. Mole calculations appear challenging, but essentially only ever use 3 principles: moles = concentration x volume/1000; moles = mass/mr; mole ratio from equation. moles NaOH used = vol / 1000 conc = 21.7/ = (range to ) moles HCl in 25 cm 3 = moles HCl in 250 cm 3 = x 10 = Ratio ZCl4 : HCl = 1: 4 moles ZCl4 = / 4 = Mr = mass / no. Moles = 1.304g/ = r = (4 x 35.5) = 72.7 Therefore element Z is Germanium

15 MOUNT OF UBTNCE M 2 CO 3 + 2HCl 2MCl + CO 2 + H 2 O 0.394g M 2 CO 3 were reacted with 21.7cm moldm -3 HCl. Calculate number moles HCl M 2 CO 3 used. Calculate relative molecular mass, M r, M 2 CO 3. Find relative atomic mass M hence its identity. MOUNT OF UBTNCE moles HCl used = vol / 1000 conc = 21.7/ = Ratio M2CO3 : HCl = 1: 2 moles M2CO3 = /2 = Mr = mass / no. Moles = 0.394g/ = 138 r (M2)= (CO3 = 60) = 78 r (M) = 78/2 = 39 Therefore element is Potassium To help you see through all information provided, I strongly recommend using highlighters in exam crossing f any values you have used. Mole calculations appear challenging, but essentially only ever use 3 principles: moles = concentration x volume/1000; moles = mass/mr; mole ratio from equation.

16 MOUNT OF UBTNCE MgCO 3 + 2HCl MgCl 2 + CO 2 + H 2 O 1.25g impure MgCO 3 were reacted with 70cm 3 0.4moldm -3 HCl. fter reaction some acid was left over. Titration with NaOH required 25.3cm 3 0.4moldm -3 NaOH to neutralise left over acid. Calculate % MgCO 3 in impure sample. MOUNT OF UBTNCE moles HCl reacted = vol / 1000 conc = 70/ = moles NaOH used = vol / 1000 x conc = 25.3/1000 x 0.4 = moles HCl left over = ; moles HCl reacted with MgCO3 = = Ratio MgCO3 : HCl = 1: 2; moles MgCO3 = /2 = Mass = Mr x no. Moles = 84.3 x = 0.75g Percentage = 0.75/1.25 = 60% To help you see through all information provided, I strongly recommend using highlighters in exam crossing f any values you have used. Mole calculations appear challenging, but essentially only ever use 3 principles: moles = concentration x volume/1000; moles = mass/mr; mole ratio from equation. Whenever a chemical is added in excess so some is left over you will need to find actual number moles that reacted subtract this value from total number moles that were added.

17 MOUNT OF UBTNCE 1.5g hydrated sodium carbonate (Na 2 CO 3.xH 2 O) was dissolved in 50cm 3 water made up to 250cm 3. 25cm 3 this solution was titrated against 0.1moldm -3 NaOH which 10.6cm 3 were required. Calculate value for x in Na 2 CO 3.xH 2 O. Na 2 CO 3 + 2HCl 2NaCl + H 2 O + CO 2. MOUNT OF UBTNCE To help you see through all information provided, I strongly recommend using highlighters in exam crossing f any values you have used. Mole calculations appear challenging, but essentially only ever use 3 principles: moles = concentration x volume/1000; moles = mass/mr; mole ratio from equation. These water crystallization questions always follow this pattern. Work out mass carbonate first; from this work out mass water convert this to moles. Then work out carbonate : water ratio to get x. moles NaOH reacted = vol / 1000 conc = 10.6/ = moles HCl in 25cm 3 = moles HCl in 250cm 3 = x 10 = Ratio Na2CO3 : HCl = 1: 2; moles Na2CO3 = /2 = Mass = Mr x no. Moles = 106 x = g Mass xh2o in Na2CO3.xH2O = = g Moles H2O in g = /18 = Mole ration Na2CO3 : H2O = : = 1:10; x = 10

18 MOUNT OF UBTNCE Lead reacts with nitric acid according to following equation: 3Pb + 8HNO 3 3Pb(NO 3 ) 2 + 2NO + 4H 2 O. What volume 1.5moldm -3 acid is required to react with 9.00g lead? What volume NO gas is produced at 101kPa 298K? (Gas constant R = 8.31 JK -1 mol -1 ) MOUNT OF UBTNCE Moles Pb used = mass/r = 9.00/207 = Ratio Pb:HNO3 = 3:8 Moles HNO3 = x 8/3 = Volume HNO3 = moles x 1000 / conc = x 1000 / 1.5 = 76.4cm 3 Ratio HNO3 : NO = 8:2 = 4:1 Moles NO = /4 = Volume NO = nrt/p = x 8.31 x 298/ = m 3 To help you see through all information provided, I strongly recommend using highlighters in exam crossing f any values you have used. Mole calculations appear challenging, but essentially only ever use 3 principles: moles = concentration x volume/1000; moles = mass/mr; mole ratio from equation. When using ideal gas equation remember that pressure is measured in Pa, not kpa, volume is measured in m 3, not cm 3. (1cm 3 = 1x10-6 m 3 ; 1m 3 = cm 3 )

19 MOUNT OF UBTNCE Limestone reacts with nitric acid according to following equation: CaCO 3 + 2HNO 3 Ca(NO 3 ) 2 + CO 2 + H 2 O. What mass limestone is required to neutralise 35cm moldm -3 acid. Ca(NO 3 ) 2 decomposes when heated strongly according to equation Ca(NO 3 ) 2(s) 2CaO (s) + 4NO 2(s) + O 2(s). t 101kPa 30 o C a total volume 4.00x10-3 m 3 gases will be produced. How many moles gas are produced in total? How many moles oxygen will be produced? MOUNT OF UBTNCE Moles acid reacted = volume/1000 x conc = 35/1000 x = Ratio CaCO3 : HNO3 = 1: 2 Moles CaCO3 = /2 = Mass CaCO3 = Mr x moles = x = 0.42g Moles gas produced = PV/RT = Pa x 4.00x10-3 m 3 /8.31 x 303K = 0.16 moles ccording to equation 5 moles gas are produced in total. 1mole this is oxygen (=1/5) Moles oxygen = 0.16 / 5 = moles. To help you see through all information provided, I strongly recommend using highlighters in exam crossing f any values you have used. Mole calculations appear challenging, but essentially only ever use 3 principles: moles = concentration x volume/1000; moles = mass/mr; mole ratio from equation. When using ideal gas equation remember that pressure is measured in Pa, not kpa, volume is measured in m 3, not cm 3. (1cm 3 = 1x10-6 m 3 ; 1m 3 = cm 3 )

20 MOUNT OF UBTNCE Nitric acid (concentrated) reacts with magnesium to form a nitrogen oxide compound. If this nitrogen oxide compound is made 30.4% by mass nitrogen, what is its empirical formula? MOUNT OF UBTNCE Nitrogen Oxygen 30.4% 69.6% r=14 r= / / : 2 EF = NO 2 Empirical formula questions are same at GCE level. lways make sure you state actual formula at end question. If you leave it at ratio, you will lose 1 mark overall (provided it is correct ratio).

2 Answer all the questions.

2 Answer all the questions. 1 A sample of the element boron, B, was analysed using a mass spectrometer and was found to contain two isotopes, 10 B and 11 B. (a) (i) Explain the term isotopes. Complete