unit 7- periodicity and groups 2/7

Question 1

blocks of the periodic table

Answer 1

• H, He and earth metals are ‘s-block’
• groups 3 to 7 are ‘p-block’
• transition metals are ‘d-block’
• lanthanides and actinides are ‘f-block’

Question 2

atomic radius

Answer 2

• decreases along period as more protons are added but no extra shells, so positive charge is stronger
• increases down groups as more shells are added that increase effect of shielding against the positive charge

Question 3

Ionization energy

Answer 3

• energy required to remove one electrons from a mole of gaseous atoms
• on average increases along periods as decreasing atomic radius makes electrons harder to take
• increases every time atom is ionised as the greater overall positive charge makes the rest of the electrons harder to remove
• when an electron is removed from a higher sub shell the energy decreases as it is easier to remove electrons from higher sub shells and energy levels
• also takes less energy to remove electrons from paired orbitals, so the orbitals all lose one electron first for less energy, then the unpaired ones

Question 4

melting points

Answer 4

• melting points generally decrease along periods as bonds go from metallic to covalent to simple molecular (usually an increase with covalent bonds but overall there is a decrease)
• metallic bonds get stronger along periods as ions lose more electrons and have a greater electrostatic attraction
• Van der Waals forces decrease with atomic radius along period

Question 5

group 2 rules

Answer 5

• atomic radius increases down group as more electron shells
• 1st ionisation energy decreases down group as every atom has a higher energy level
• melting point decreases down group due to increased atomic radius as de-localised electrons in metallic bonds kept further away
• reactivity increases as 1st ionisation energy decreases
• reactions with water become more violent as speed of reaction increased with reactivity
• -1 ionic compounds become more soluble down group (e.g. hydroxides), -2 ionic compounds become less soluble (e.g. sulphates)

Question 6

group 2 uses

Answer 6

• Mg(OH)2 and Ca(OH)2 neutralise acidic soil in agriculture
• barium used to test for sulfate ions as it forms an insoluble precipitate when dilute HCl and BaCl2 is added (as Barium displaced whatever else is bonded to the sulfate ion)
• barium meals used to make stomach, intestines or oseophagus opaque to x-rays as barium sulfate is insoluble and doesnt let x-rays through
• Mg(OH)2 is sparingly soluble
• magnesium us used to extract titanium, as titanium chloride is reduced by Mg at 1000 degrees
• CaO or CaCO3 are used to remove SO2 from flue gases

Question 7

group 7 rules

Answer 7

• F2 is yellow, Cl 2 is green, Br2 is red-brown, I2 is grey
• boiling points increase down group as van der waals forces increase with radius
• electronegativity decreases down group due to shielding
• 1st ionsiation energy also decreases down group
• halogens become less oxidising down the group, so a halide will displace another halide only if it is below it in the group

Question 8

group 7 uses

Answer 8

• bleach is formed when cold, dilute NaOH reacts with Cl2 to form NaCl(I) and NaClO solution. As Cl2 is both oxidised and reduced it is a disproportionation reaction
• Chlorine reacts with water in a reversible reaction to form Cl- and ClO-, which is also disproportionation.
• ClO- kills bacteria, so Cl is used to treat water, although it can cause damage if in large amounts, and can react with organic compounds to make carcinogenic molecules, but the risk is lesser than untreated water
• halogens always displace other halogens lower down than them in the group

Question 9

halide ions

Answer 9

-all halides react with H2SO4 to give hydrogen halides
(NaCl + H2SO4 -> NaHSO4 + HCl)
-reducing strength increases down the group
-some halides are strong enough reducing agents to reduce H2SO4 to 2H2O and SO4
(NaBr + H2SO4 -> NaHSO4 + HBr, HBr + H2SO4 -> Br2 + SO2 + 2H2O)
-some are even stronger and reduce SO2 to S or H2S
(NaI + H2SO4 -> NaHSO4 + HI, HI + H2SO4 -> I2 + SO2 + 2H2O, 6HI + SO2 -> H2S + 3I2 + 2H2O)

Question 10

silver nitrate test

Answer 10

• add dilute nitric acid to remove unwanted ions
• add AgNO3 to form a silver halide precipitate
• silver chloride is white, silver bromide is cream, silver iodide is yellow
• adding ammonia further tests this as solubility decreases down the group (AgCl would be very soluble, AgBr would be insoluble)