IB Chemistry SL R3.3 Electron sharing reactions questions

Exam-style multiple choice and written questions with full markschemes. Check multiple choice answers straight away, or type, photograph or draw a written answer and get it marked.

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Question 1

Multiple choiceMedium1 mark

Four oxygen-containing species are compared: OH, OH⁻, H₂O and H₃O⁺. Which analysis correctly identifies a radical and predicts its reactivity from its electron arrangement?

Show answer and explanation

Answer: C. Neutral OH has 6 + 1 = 7 valence electrons. Its Lewis representation has one unpaired electron, so it is a radical and is expected to be highly reactive.

A: omits the extra electron from the negative charge; OH⁻ has 8 valence electrons, all paired. B: counts H₂O correctly but mistakes paired lone-pair electrons for an unpaired electron. D: fails to subtract one electron for the positive charge; H₃O⁺ has 8 valence electrons, all paired.

Remember: add an electron for a negative charge and subtract one for a positive charge.

Question 2

Paper 2Medium6 marks

CFC-12, CF₂Cl₂, can reach the stratosphere. UV radiation can break one of its C—Cl bonds. In a simplified ozone-depletion sequence, a chlorine radical reacts with ozone to form a chlorine monoxide radical and oxygen. The chlorine monoxide radical then reacts with an oxygen radical: ClO• + O• → Cl• + O₂

(a)

Write the equation for the UV-induced breaking of one C—Cl bond in CF₂Cl₂, showing UV above the arrow. State whether this is an initiation, propagation or termination step.

[2]
(b)

Represent the bond being broken as C—Cl. Draw two single-barbed (fish-hook) arrows to show the movement of electrons during this bond breaking.

[1]
(c)

Write the balanced equation for the reaction of a chlorine radical with ozone described in the question.

[1]
(d)

Combine the reactions in the question and in part (c) to give the overall reaction. State why the chlorine radical can continue the sequence.

[2]

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