Inorganic Chemistry Assignment

Inorganic Chemistry Assignment Words: 659

Describe and explain the trend in the boiling points of the elements down group VII from Fluorine to Iodine. All of the halogens exist as diatomic molecules (F2, Cl2 and I2) the intermolecular attractions between the molecules hold them together. The larger the molecule (as it moves down the halogen group) the bigger the intermolecular forces are between electrons because with more rings the distance between each electron is larger. The larger elements such as Bromine and Iodine are larger molecules with a larger amount of intermolecular force between each electron.

This increase in intermolecular attraction means that the required heat energy to reach the boiling point increases. Fluorine has a significantly lower boiling point than the rest of the halides because it has a set of electrons that produces repulsion between them. This weakens the bond. Chlorine has a boiling point of -34. 6°C because its covalent bond is stronger than fluorine’s and so on until Iodine which has a boiling point of 184. 0°C because it has many more energy levels which requires more heat energy.

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As the intermolecular and the amount of energy shells increase so does the boiling point (increases down the group) (b) Describe what you would observe when aqueous Silver nitrate is added to separate aqueous solutions of Sodium chloride and sodium Bromide When the solution of sodium chloride is added to the aqueous solution of silver nitrate it would be displaced by the sodium chloride to form a silver chloride and sodium nitrate. This would happen because the chloride is a more reactive element and displaces the weaker element, which is the silver.

The silver chloride is insoluble and this would form a white precipitate. AgNO3 (aq) + NaCl (aq) NaNO3 (aq) + AgCl(s) Ag+ (aq) + Cl- (aq) AgCl (s) When you add the aqueous solution of silver bromide to the silver nitrate it will be displaced by the sodium bromide, as the bromide is the more reactive element in this reaction and will displace the less reactive, which is the silver. The reaction will produce a solution of sodium nitrate and silver bromide. The silver bromide is insoluble and this will form a white precipitate.

The precipitate formed by the silver bromide would be slightly darker than that of the silver chloride because it is further down in group 7. AgNO3 (aq) + NaBr (aq) NaNO3 (aq) + AgBr (s) Ag+ (aq) + Br- (aq) AgBr (s) (3) State the trend in the oxidising abilities of the elements down group 7 from chlorine to iodine. Explain how this trend can be shown by displacement reactions between halogens and halide ions in aqueous solutions. Illustrate your answer with appropriate observations and equations. The oxidizing abilities get lower as you go down group 7. Fluorine is a stronger agent than chlorine, bromine and iodine.

Once fluorine has taken electrons from the other halides to form fluoride ions, the other halides cant take back the electrons they donated because they don’t have strong enough oxidising abilities. Similarly chlorine is stronger than bromine and iodine. Chlorine can take their electrons to form chloride ions but bromine and iodine can’t take them back. They form a hydrated ion which is their aqueous state (they bond with water molecules) the formation of hydrated ions happens more easily at the top of group 7 and harder at the bottom. This trend exists with the oxidising abilities as well.

Chlorine displaces bromine and Iodine in aqueous solutions because it has a higher oxidising ability than them. These equations show this With (Cl2) and (Br) ions and also (Cl2) and (I) ions. Cl2 (aq) + 2KBr (aq) —> 2KCl (aq) + Br2 (aq) Cl2 (aq) + 2KI (aq) —> 2KCl (aq) + I2 (aq) Bibliography http://www. chemguide. co. uk/inorganic/group7/properties. html#top http://www. chemguide. co. uk/atoms/bonding/vdw. html#top http://old. iupac. org/didac/Didac%20Eng/Didac03/Content/R06. htm http://au. encarta. msn. com/media_121632732/electrode_potential_of_redox_reactions. html http://www. chemicalelements. com/