Which of the following statements are correct?
A. Chlorine can show a maximum oxidation state of \(+3\)
B. Deacon's process is oxidation of \(HCl\)in presence of \(CuCl_2\)
C. Chlorine on reacting with cold \(NaOH\) gives \(NaOCl\) not \(NaOCl_3\)
D. \(ClF_5\) has T shape
E. \(HCl\) is a stronger acid than \(HF\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is option 3. B, C, E only.
Let us examine each statement in detail to clarify which are correct and which are not.
Statement A: Chlorine can show a maximum oxidation state of \(+3\)
Chlorine is in Group 17 of the periodic table (halogens) and has seven valence electrons. The possible oxidation states for chlorine range from \(-1\) to \(+7\).
Maximum oxidation state: Chlorine can reach an oxidation state of \(+7\) in compounds like perchloric acid (\(HClO_4\)) or perchlorates (\(ClO_4^-\)).
Conclusion: The statement is incorrect because chlorine's maximum oxidation state is \(+7\), not \(+3\).
Statement B: Deacon's process is oxidation of \(HCl\) in presence of \(CuCl_2\)
Deacon's process is an industrial method for producing chlorine gas (\(Cl_2\)) by oxidizing hydrogen chloride (\(HCl\)) with oxygen (\(O_2\)) in the presence of a catalyst such as cupric chloride (\(CuCl_2\)). The reaction is:
\(4HCl\, \ + \, \ O_2 \overset{CuCl_2}{\longrightarrow}\, \ 2Cl_2 \, \ + \, \ 2H_2O\)
Conclusion: The statement is correct.
Statement C: Chlorine on reacting with cold \(NaOH\) gives \(NaOCl\) not \(NaOCl_3\)
When chlorine reacts with cold and dilute sodium hydroxide (\(NaOH\)), it undergoes a disproportionation reaction:
\(\text{Cl}_2 + 2 \, \text{NaOH} \rightarrow \text{NaOCl} + \text{NaCl} + \text{H}_2\text{O}\)
The product is sodium hypochlorite (\(NaOCl\)), not sodium perchlorate (\(NaOCl_3\)).
Conclusion: The statement is correct.
Statement D: \(ClF_5\) has T shape
\(ClF_5\) is a molecule where chlorine (central atom) is bonded to five fluorine atoms. The molecular geometry is based on the
VSEPR theory (Valence Shell Electron Pair Repulsion theory). Chlorine in \(ClF_5\) has 7 valence electrons, 5 of which are used for bonding with fluorine atoms, leaving 1 lone pair. The presence of one lone pair results in a square pyramidal structure, not a T-shape.
Conclusion: The statement is incorrect.
Statement E: \(HCl\) is a stronger acid than \(HF\)
The strength of an acid is generally determined by its ability to dissociate in water to produce hydrogen ions (\(H^+\)). HCl is a stronger acid than HF because:
The bond between \(H\) and \(Cl\) is weaker and longer than the bond between \(H\) and \(F\), making it easier for \(HCl\) to dissociate. The chloride ion (\(Cl^-\)) is larger and better stabilizes the negative charge than the fluoride ion (\(F^-\)), making dissociation of \(HCl\) more favorable.
Conclusion: The statement is correct.
Thus, the correct statements are B, C, E.