Question 23M.2.HL.TZ1.9
| Date | May 2023 | Marks available | [Maximum mark: 7] | Reference code | 23M.2.HL.TZ1.9 |
| Level | HL | Paper | 2 | Time zone | TZ1 |
| Command term | Calculate, Explain, Formulate, Suggest | Question number | 9 | Adapted from | N/A |
Explain why a colorimeter set at a wavelength of 500 nm is not suitable to investigate reactions of Zn2+ compounds. Use section 3 of the data booklet.
[2]
Zn2+ does not form coloured compounds/ has a complete d subshell/orbital ✓
500 nm/«the setting on the colorimeter» in visible region AND no absorbance will be seen ✓
Nitrogen (II) oxide radicals (NO•) catalyse the decomposition of ozone (O3).
Formulate equations showing how NO• acts as a catalyst in this reaction.
[2]
«O3 (g) → O2 (g) + O• (g)»
NO• (g) + O3 (g) → NO2• (g) + O2 (g) ✓
NO2• (g) + O3 (g) → NO• (g) + 2O2 (g)
OR
NO2• (g) + O• (g) → NO• (g) + O2 (g)✓
Accept radicals without • if consistent throughout.
Chlorine also forms free radicals; the bond enthalpy for Cl2 is 4.02 × 10−19 J.
Calculate the minimum frequency of light needed to break this bond.
Use sections 1 and 2 of the data booklet.
[1]
«v = E/h = 4.02 × 10−19/6.63 × 10−34 =» 6.06 × 1014 «Hz» ✓
Calculate the formal charge on each atom in the two Lewis structures of the NO2• (g) radical.
[1]
✓
Lewis structure A is more stable. Suggest, giving one reason, whether the formal charge model supports this.
[1]
No
AND
Structure B has all atoms of formal charge 0 ✓