Question 20N.2.hl.TZ0.6

$«E=\frac{745 000 \mathrm{J} {\mathrm{mol}}^{-1}}{6.02\times {10}^{23} {\mathrm{mol}}^{-1}}=»1.24\times {10}^{-18} \mathrm{J}$ ✔

$«E=h\nu »$
$«1.24\times {10}^{-18} \mathrm{J}=6.63\times {10}^{-34} \mathrm{J} \mathrm{s}\times \mathrm{\nu }»$
$\mathrm{\nu }=1.87\times {10}^{15} «{\mathrm{s}}^{-1}/\mathrm{Hz}»$ ✔

Award [2] for correct final answer.
Award [1] for 1.12 × 10³⁹ «Hz».

orange light is absorbed «and the complementary colour is observed» ✔

Any TWO from:
partially filled d-orbitals ✔
«ligands/water cause» d-orbitals «to» split ✔
light is absorbed as electrons move to a higher energy orbital «in d–d transitions»
OR
light is absorbed as electrons are promoted ✔
energy gap corresponds to «orange» light in the visible region of the spectrum ✔

cathode/negative «electrode» AND ${\mathrm{Cu}}^{2+}$ reduced «at that electrode» ✔

Accept cathode/negative «electrode» AND copper forms «at that electrode».