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 \(NaClO_3\)
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
(A) Chlorine can show a maximum oxidation state of +3 — Incorrect. Chlorine uses its vacant 3d orbitals to show +1, +3, +5 and +7; the maximum is +7 (as in \( \ce{HClO4} \), \( \ce{Cl2O7} \)).
(B) Deacon’s process is oxidation of HCl in presence of CuCl2 — Correct. 4HCl + \( \ce{O2} \) → \( \ce{2Cl2} \) + \( \ce{2H2O} \), with \( \ce{CuCl2} \) as catalyst at 723 K.
(C) Chlorine on reacting with cold NaOH gives NaOCl not NaClO3 — Correct. With cold, dilute alkali: \( \ce{Cl2} \) + 2NaOH → NaCl + NaOCl + \( \ce{H2O} \). Chlorate forms only with hot, concentrated alkali: \( \ce{3Cl2} \) + 6NaOH → 5NaCl + \( \ce{NaClO3} \) + \( \ce{3H2O} \).
(D) ClF5 has T shape — Incorrect. Cl in \( \ce{ClF5} \) has 5 bond pairs and 1 lone pair (sp3d2), so the shape is square pyramidal. The T shape belongs to \( \ce{ClF3} \) (3 bond pairs, 2 lone pairs).
(E) HCl is a stronger acid than HF — Correct. The H–Cl bond (431 kJ mol−1) is much weaker than H–F (574 kJ mol−1), so HCl ionises more readily.
Note: This question is from the pre-rationalised NCERT book (2022-23). The topic is not there in the latest NCERT book (2026-27), and it is also out of the CUET 2026 syllabus.
NCERT (Pre-rationalised edition, 2022-23):
Manufacture of Chlorine
