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hydrogen bond

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. 1. How many hydrogen bonds does each H 2 O molecule form with its neighbouring molecules in ice crystals?.

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hydrogen bond

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  1. 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

  2. 1. How many hydrogen bonds does each H2O molecule form with its neighbouring molecules in ice crystals? P.28 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 A. 2 B. 3 C. 4 D. 5

  3. 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.

  4. 3. Which of the following substances can be deflected by a charged rod? O = S H = F A. SF6 B. CF4 C. XeF2 D. SF2

  5. 4. What happens to the mass of each electrode as the cell operates? Electrons flow from Zn to Cu At Zinc plate: Zn  Zn2+ + 2e- (anode) At copper plate: Cu2+ + 2e Cu (cathode) Zn may have direct reaction with CuSO4, CU formed on it, but this is drawback only. Not a main reaction.

  6. e- flow from P to S e- flow from P to Q P > S P > Q > S P > Q Ans: R > P > Q > S e- flow from Rto P R >P

  7. Fe2+Fe3+ Y: loses electrons (negative) MnO4- Mn2+ X: gains electrons (reduction) (cathode) (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 Fe2+ + 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

  8. 2H+ + 2e-  H2 Cu: gains electrons (reduction) (cathode) X Xn+ + n e- X: loses electrons (oxidation) (anode) 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

  9. 1. a) Explain the term “hydrogen bonding”. P. 24 X hydrogen bonding is a bonding/ covalent bond…… 0.5 m 0.5 m

  10. b) Draw the electron diagram of a XeOF4 molecule. 1 m Draw the electron diagram of a IOH3 molecule. -- will you put O as the centre atom??

  11. c) Account for the following: (i) The boiling point of NH3 is –36°C, but that of PH3 is –87°C. The boiling point of NH3 is higher than PH3 There are strong hydrogen bondings and weak van der Waal’s forces between NH3 molecules. 1M But there are weak van der Waal’s force between PH3molecules only. 1M More energy is required to break the forces between molecules of NH3 than PH3. (comparative question comparative answer)

  12. (ii) CH4 is a gas and SiH4 is a solid at room temperature. SiO2 and Si is in giant covalent structure. Don’t think that all compounds with Si e.g. SiH4 SiCl4, SiF4are also giant covalent structure. They are simple molecular sturctures! They are just molecules!!! They are both non-polar molecules. The molecular size of SiH4 is larger than CH41M Number of electrons in SiH4 is more than in CH4 Higher chance of uneven distribution of electrons in SiH4 than in CH4 Van der Waal’s Waal’s forces between moelcules of SiH4 is stronger than that of CH4 1M Molecules of SiH4 are packed closer to each other than Molecules f CH4 (comparative question comparative answer)

  13. 2. Draw the three-dimensional structure for each molecule below. State the shapeof each molecule.

  14. 3. Explain why buckminsterfullerene cannot conduct electricityand it is a soft solid in terms of its structure. Buckminsterfullerene is in simple molecular structure, so between each C60, 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

  15. 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.

  16. salt bridge graphite electrode Y graphite electrode X AgNO3(aq) NaI(aq) 5. XNO3- NO xNa Na+ + e- xI2 2I- + 2e- xAg Ag+ + e- Graphite is inert!! It cant react at all!!! • it is wrong to sayelectrode X or Y dissolves! • C CO2 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) I2(aq)+ 2e- 1M X: silvery solid is formed. (NOT silver solid /Ag is formed) 1M Y:solution turns from colourless to brown.1M

  17. salt bridge graphite electrode Y graphite electrode X AgNO3(aq) NaI(aq) • If the silver nitrate solution is replaced by acidified potassium permanganate • solution, the voltage measured becomes larger. Explain why. KMnO4-/H+ Do you rememeber MnO4- 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.

  18. 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. MnO4- and I-is further apart than Ag+ and I-. (1M)

  19. salt bridge graphite electrode Y graphite electrode X AgNO3(aq) NaI(aq) c) Explain why concentrated sodium chloride solution could not be used as the electrolyte in the salt bridge. Con NaCl – Na+, Cl- • All ions in the cell: Na+, Cl- , Ag+, NO3- , Na+, I- • Na+, Cl-do not undergo redox reaction withAg+, NO3- , 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+, NO3- , Na+, I- NaNO3 NaI AgCl NaCl The Cl- in concentrated sodium chloride solution reacts with Ag+ in AgNO3, and formed insoluble AgCl, (1M) which block the salt bridge which ions cannot pass from one half cell to another.

  20. a) What is the direction of current flow? b) Write the half chemical equations for the reactions occur at magnesium and lead electrodes respectively. • 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) b) Mg : Mg(s) Mg2+(aq) + 2e-(1M) Pb : Pb2+ + 2e-  Pb (wrong) 2e- + 4H+ + SO42-  SO2 +2H2O (wrong) 2e- + 2H+(aq)  H2(g) (1M) SO42- H+

  21. 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. (wrong)Mg directly reacts with H2SO4,(1M) And the hydrogen gas bubbles formed on Mg, (1M) Increase the resistance of cell. H2 gas bubbles Insoluble PbSO4 H+ 2e- + 2H+(aq)  H2(g) H+ H+ H+ H+ H+ Pbdirectly reacts with H2SO4 and formed insoluble PbSO4, (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.

  22. For the following reaction, write half-equations and anoverall ionic equation • for the redox reaction. State allpossible observation(s). • Reaction between acidified potassium dichromate solution and • potassium sulphite solution. • 6e- + 14H+ (aq) +Cr2O72- (aq)  2Cr 3+ (aq)+ 7H2O (l) (1M) • H2O(l)+ SO32- (aq)  SO42-(aq)+ 2H+(aq) + 2e- (1M) (1M) • 8H+(aq) +3SO32- (aq) +Cr2O72- (aq)  3SO42-(aq)+ 2Cr 3+(aq)+ 4H2O (l) (1M) • Solution turns from orange to green.

  23. 8.Copper can be extracted from an ore by adding concentrated sulphuric acid and • filtering off the insoluble material. • Write a chemical full equation for the reaction between copper in the ore and • concentrated sulphuric acid. • 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 + SO2 + H2O Cu (s) + 2H2SO4 (l)  CuSO4 (s) + SO2 (g)+ 2H2O (l) 1M b) Cu. It is oxidized. 1M The oxidation number of Cu increased from 0 to +2. 1M

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