Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
NEET2017Chemistryd and f Block ElementsActual

Name the gas that can readily decolourise acidified KMnO 4 solution.

Options

  1. ACO 2
  2. BSO 2
  3. CNO 2
  4. DP 2 O 5

Correct answer

B. SO 2

Step-by-step solution

SO 2 readily decolourises acidified KMnO 4 . K Mn + 7 O 4 + S + 4 O 2 → MnSO 4 Colourless + H 2 SO 4 + K 2 SO 4 Potassium permagnate gives the colouless solution with sulphurdioxide.

Practice d and f Block Elements on Quantrex Academy →

More from d and f Block Elements

Although +3 oxidation state is most common in lanthanoids, cerium still shows +4 oxidation state because : 2026The lanthanide ion having four unpaired electrons is (Given : Atomic numbers of Ce =58 , Nd =60 , Tb =65 and Ho =67 ) 2026The green paramagnetic species formed by heating KMn O₄ at 513 K is 2026The calculated 'spin-only' magnetic moment of Ti ²⁺ (3d^2) is : 2026Match List I with List II : List I (Transition metal/compound/complex) List II (Catalytic Role) A. V ₂ O ₅ I. Preparation of ammonia from N ₂/ H ₂ mixture B. Fe II. Polymerisation 2026Given below are two statements : One is labelled as Assertion A and the other is labelled as Reason R . Assertion A : Generally, 3d transition metals have high melting points. Reas 2026Given below are two statements : Statement I : Ferromagnetism is considered as an extreme form of paramagnetism. Statement II : The number of unpaired electrons in a Cr ²⁺ ion ( Z 2025Which of the following set of ions act as oxidising agents? 2024 Full d and f Block Elements list All NEET PYQs