AS Chemistry - Group 7

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State the trend down group 7
Group 17 elements are all simple molecules  there are weak induced dipole-dipole attractions between halogen molecules  going down Group 17 boiling point increases
1 of 10
Explain the trend down group 7
the number of electrons in each halogen molecule increases  the strength of induced dipole-dipole attractions increases  more energy is required to break stronger intermolecular forces
2 of 10
State and explain the trend in electronegativity going down Group 7
decreases  there are more full inner shells with more electron shielding  bonding electrons are further from nucleus  nuclear charge outweighed  nuclear attraction for the bonding electrons decreases
3 of 10
State and explain the trend in reactivity of Group 7 elements down the group
 reactivity decreases  electron being added is further from nucleus  there are more full inner shells with greater electron shielding  nuclear attraction for the electron being added decreases
4 of 10
State the colours of Cl, Br and I in aq solution
Cl=yellow Br=orange I=brown
5 of 10
State the colours of Cl, Br and I in organic solvents
Cl=yellow Br=orange I=purple
6 of 10
If chlorine water is tested with blue litmus paper...
it goes red then white as the chloric ion is a bleach
7 of 10
Cl2(g) + H2O(l) =
HCl(aq) + HClO(aq)
8 of 10
Cl2(g) + 2NaOH(aq) = (BLEACH)
H2O(l) + NaCl(aq) + NaClO(aq)
9 of 10
3Cl2(g) + 6NaOH(aq) = (WEEDKILLER)
3H2O(l) + 5NaCl(aq) + NaClO3(aq)
10 of 10

Other cards in this set

Card 2

Front

Explain the trend down group 7

Back

the number of electrons in each halogen molecule increases  the strength of induced dipole-dipole attractions increases  more energy is required to break stronger intermolecular forces

Card 3

Front

State and explain the trend in electronegativity going down Group 7

Back

Preview of the front of card 3

Card 4

Front

State and explain the trend in reactivity of Group 7 elements down the group

Back

Preview of the front of card 4

Card 5

Front

State the colours of Cl, Br and I in aq solution

Back

Preview of the front of card 5
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