(i) $[\ce{Ne}] 3s^2 3p^5$ ✔

(ii) «single» shared pair «of electrons» between the two $\ce{Cl}$ atoms ✔

(iii) both $\ce{Cl2}$ and $\ce{Br2}$ are non-polar molecules with «only» London/dispersion forces between molecules ✔

$\ce{Br2}$ has more electrons «greater molar mass» than $\ce{Cl2}$

(iv) $\ce{Cl2(aq) + 2Br-(aq) -> 2Cl-(aq) + Br2(aq)}$

colour change: «almost» colourless to orange/yellow ✔

(v) chlorine is a weaker oxidizing agent than fluorine / more difficult to reduce than fluorine ✔

(vi) Product at anode: chlorine / $\ce{Cl2(g)}$

Product at cathode: cobalt / $\ce{Co(s)}$ ✔

(i) [1]; do not accept the full, non-condensed electron configuration. (ii) [1]; do not accept just 'covalent'. (iii) M1: both molecules are non-polar with only London/dispersion forces; M2: bromine has more electrons, so stronger dispersion forces and a higher boiling point. (iv) M1: correct equation; M2: correct colour change; accept a more specific shade of orange/yellow. (v) [1]; accept a valid answer referring to oxidizing power decreasing down group 17, or to E⊖ values; do not accept just 'fluorine is more reactive' with no comparison of oxidizing/reducing power. (vi) [1] for each correct product; award [1 max] for both products correct but assigned to the wrong electrode; do not accept ions.