Presentation on theme: "P. 1 / 15 The two hydrogen atoms of each water molecule also form hydrogen bonds with oxygen atoms of nearby water molecules. hydrogen bond hydrogen bond."— Presentation transcript:
P. 1 / 15 The two hydrogen atoms of each water molecule also form hydrogen bonds with oxygen atoms of nearby water molecules. hydrogen bond hydrogen bond hydrogen atom oxygen atom
Reason: In ice, each water molecule is tetrahedrally bonded to 4 other water molecule by hydrogen bonds to form an open, cage-liked structure with very inefficient packing thus explains thelow density of ice. When ice melt, some of the hydrogen bonds originally present in ice are broken down. The open cage structure is partially collapsed so that the water molecules can pack more closely with one another. Molecules in liquid water, each water can on average form 2 hydrogen bonds per molecule. Thus water molecules are much closer packed together and hence higher density than ice. Ice floats on water 1. How many hydrogen bonds does each H 2 O molecule form with its neighbouring molecules in ice crystals? A.2 B.3 C.4 D.5 P.28
2. Electronegativity is P.12 A.the electrical conductivity of the negative electrode in a chemical cell. B.the ability of an atom to attract electrons in a covalent bond to itself. C.the ability of an anion to attract electrons from a cation to itself. D.the ability of a cation to attract electrons from an anion to itself.
3. Which of the following substances can be deflected by a charged rod? O = S H = F A. SF 6 B. CF 4 C. XeF 2 D. SF 2
4. What happens to the mass of each electrode as the cell operates? 1.Electrons flow from Zn to Cu 2.At Zinc plate: Zn Zn 2+ + 2e- (anode) 3.At copper plate: Cu 2+ + 2e Cu (cathode) Zn may have direct reaction with CuSO4, CU formed on it, but this is drawback only. Not a main reaction. Mass of anodeMass of cathode A.Increasesincreases B.Decreasesdecreases C.Decreasesincreases D.Increasesdecreases
e- flow from P to Q P > Q e- flow from P to S P > S P > Q > S e- flow from R to P R >P Ans: R > P > Q > S
MnO 4 - Mn 2+ X: gains electrons (reduction) (cathode) Fe 2+ Fe 3+ Y: loses electrons (negative) (2) Cations and anions in the electrolytes migrate into the salt bridge to complete the circuit and Cations and anions in the electrolytes ( ions in the salt bridge) balance the extra charges generated in each half cell. (1) The half equation for the change occurring at electrode Y is Fe 2+ + 2e - → Fe. (2) Cations and anions in the electrolytes migrate into the salt bridge to complete the circuit and balance the extra charges generated in each half cell. (3) Electrode Y is the positive electrode. (4) Electrode X is the cathode. A. (4) only B. (1) and (3) only C. (2) and (4) only D. (2) and (3) only
X X n+ + n e- X: loses electrons (oxidation) (anode) 2H + + 2e- H 2 Cu: gains electrons (reduction) (cathode) Which of the following statements concerning the above set-up is correct? A. Gas bubbles was formed at metal X. (from direct reaction of the draw back) B. Copper plate gradually dissolves. C. The position of copper in the electrochemical series is higher than that of X. D. The sulphuric acid turns less acidic
1. a) Explain the term “hydrogen bonding”. P. 24 0.5 m X X hydrogen bonding is a bonding/ covalent bond……
b) Draw the electron diagram of a XeOF 4 molecule. 1 m Draw the electron diagram of a IOH 3 molecule. -- will you put O as the centre atom??
c) Account for the following: (i) The boiling point of NH 3 is –36°C, but that of PH 3 is –87°C. The boiling point of NH 3 is higher than PH 3 There are strong hydrogen bondings and weak van der Waal’s forces between NH 3 molecules. 1M But there are weak van der Waal’s force between PH 3 molecules only. 1M More energy is required to break the forces between molecules of NH 3 than PH 3. (comparative question comparative answer)
(ii) CH 4 is a gas and SiH 4 is a solid at room temperature. They are both non-polar molecules. The molecular size of SiH 4 is larger than CH 4 1M Number of electrons in SiH 4 is more than in CH 4 Higher chance of uneven distribution of electrons in SiH 4 than in CH 4 Van der Waal’s Waal’s forces between moelcules of SiH 4 is stronger than that of CH 4 1M Molecules of SiH 4 are packed closer to each other than Molecules f CH 4 (comparative question comparative answer) SiO 2 and Si is in giant covalent structure. Don’t think that all compounds with Si e.g. SiH 4 SiCl 4, SiF 4 are also giant covalent structure. They are simple molecular sturctures! They are just molecules!!!
a)POCl 3 b)SCl 6 c)NH 3 BF 3 2. Draw the three-dimensional structure for each molecule below. State the shape of each molecule.
3. Explain why buckminsterfullerene cannot conduct electricity and it is a soft solid in terms of its structure. Buckminsterfullerene is in simple molecular structure, so between each C 60, there are weak van der Waal’s forces between molecules. 1M It does not conduct electricity, because of the absence of mobile electrons in the structure. 1M P.32
4. Draw how does one carbon atom of buckminsterfullerene bond with other carbon atoms. Each carbon atom is bonded with other C atoms with 2 single bonds and 1 double bond.
salt bridge graphite electrode X AgNO 3 (aq) NaI(aq) graphite electrode Y 5. Graphite is inert!! It cant react at all!!! it is wrong to say electrode X or Y dissolves! C CO 2 XNO 3 - NO xAg Ag + + e - xNa Na + + e - xI 2 2I - + 2e- a)(i)Write ionic half equations for reactions taking place at electrodes X and Y respectively. (ii)State any observable changes at electrodes X and Y respectively. X: Ag + (aq) + e - Ag (s) 1M Y: I - (aq) I 2 (aq) + 2e- 1M X: silvery solid is formed. (NOT silver solid /Ag is formed) 1M Y:solution turns from colourless to brown. 1M
b)If the silver nitrate solution is replaced by acidified potassium permanganate solution, the voltage measured becomes larger. Explain why. salt bridge graphite electrode X AgNO 3 (aq) NaI(aq) graphite electrode Y KMnO 4 - /H + Do you rememeber MnO 4 - is at the 2nd last one in the e.c.s.? That means it is a very strong O.A. The position of permanganate ion is lower than silver ion in the e.c.s, so permanganate ion is a stronger oxidizing agent than silver ion.
The position of permanganate ion is lower than silver ion in the e.c.s, so permanganate ion is a stronger oxidizing agent than silver ion. MnO 4 - and I - is further apart than Ag + and I -. (1M)
c) Explain why concentrated sodium chloride solution could not be used as the electrolyte in the salt bridge. salt bridge graphite electrode X AgNO 3 (aq) NaI(aq) graphite electrode Y Con NaCl – Na +, Cl - All ions in the cell: Na +, Cl -, Ag +, NO 3 -, Na +, I - Na +, Cl - do not undergo redox reaction with Ag +, NO 3 -, Na +, I - Any insoluble salt formed with the ions in salt bridge mixed with the 2 solutions in the 2 half cells? All ions in the cell: Na +, Cl-, Ag +, NO 3 -, Na +, I - NaNO 3 NaI AgCl NaCl The Cl- in concentrated sodium chloride solution reacts with Ag + in AgNO 3, and formed insoluble AgCl, (1M) which block the salt bridge which ions cannot pass from one half cell to another.
a) What is the direction of current flow? b) Write the half chemical equations for the reactions occur at magnesium and lead electrodes respectively. a)Electrons flow from more reactive metal to least reactive metal electrode in the external circuit. Pb Mg in the external circuit. (1M) (Not Mg Pb) H+H+ SO 4 2- b) Mg : Mg (s) Mg 2+ (aq) + 2e- (1M) Pb : Pb 2+ + 2e- Pb (wrong) 2e- + 4H + + SO 4 2- SO 2 +2H 2 O (wrong) 2e- + 2H + (aq) H 2 (g) (1M)
c) A student set up the following chemical cell and found that the bulb light up for a short while and then got dimmer. Explain why the brightness of the bulb got dimmer. H+H+ H+H+ H+H+ H+H+ H+H+ H+H+ H 2 gas bubbles Insoluble PbSO 4 Pb directly reacts with H 2 SO 4 and formed insoluble PbSO 4, (1M) And the hydrogen gas bubbles formed on Pb, (1M) Increase the resistance of H+ to gain electrons at Pb electrode. Which lowers the current flow in the external circuit. 2e- + 2H + (aq) H 2(g) (wrong)Mg directly reacts with H 2 SO 4,(1M) And the hydrogen gas bubbles formed on Mg, (1M) Increase the resistance of cell.
7.For the following reaction, write half-equations and an overall ionic equation for the redox reaction. State all possible observation(s). Reaction between acidified potassium dichromate solution and potassium sulphite solution. 6e - + 14H + (aq) + Cr 2 O 7 2- (aq) 2Cr 3+ (aq) + 7H 2 O (l) H 2 O (l) + SO 3 2- (aq) SO 4 2- (aq) + 2H + (aq) + 2e- 8H + (aq) + 3SO 3 2- (aq) + Cr 2 O 7 2- (aq) 3SO 4 2- (aq) + 2Cr 3+ (aq) + 4H 2 O (l) Solution turns from orange to green. (1M)
8.Copper can be extracted from an ore by adding concentrated sulphuric acid and filtering off the insoluble material. a)Write a chemical full equation for the reaction between copper in the ore and concentrated sulphuric acid. b)Identify the reducing agent. Explain your answers in terms of changes in oxidation number. a)No need to write half equations first! Directly write full equations for metal reacts with con sulphuric acid Metal + con sulphuric acid salt + SO 2 + H 2 O Cu (s) + 2H 2 SO 4 (l) CuSO 4 (s) + SO 2 (g) + 2H 2 O (l) 1M b) Cu. It is oxidized. 1M The oxidation number of Cu increased from 0 to +2. 1M
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