Question:
Match List-I with List – II : List-I (Oxoacids of Sulphur) List-II (Bonds) A Peroxodisulphuric acid I Two S–OH, Four S=O, One S–O–S B Sulphuric acid II Two S–OH, One S=O C Pyrosulphuric acid III Two S–OH, Four S=O, One S–O–O–S D Sulphurous acid IV Two S–OH, Two S=O
Choose the correct answer from the options given below.
Match List-I with List – II :
| List-I (Oxoacids of Sulphur) | List-II (Bonds) | ||
| A | Peroxodisulphuric acid | I | Two S–OH, Four S=O, One S–O–S |
| B | Sulphuric acid | II | Two S–OH, One S=O |
| C | Pyrosulphuric acid | III | Two S–OH, Four S=O, One S–O–O–S |
| D | Sulphurous acid | IV | Two S–OH, Two S=O |
Choose the correct answer from the options given below.
Updated On: Oct 22, 2024
- A–III, B–IV, C–I, D–II
- A–I, B–III, C–IV, D–II
- A–I, B–III, C–IV, D–II
- A–I, B–III, C–II, D–IV
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The Correct Option is A
Solution and Explanation
A. Peroxodisulfuric acid - Two S–OH, Four S=O, One S–O–S
B. Sulphuric acid - Two S–OH, One S=O
C. Pyrosulphuric acid - Two S–OH, Four S=O, One S–O–O–S
D. Sulphurous acid - Two S–OH, Two S=O
B. Sulphuric acid - Two S–OH, One S=O
C. Pyrosulphuric acid - Two S–OH, Four S=O, One S–O–O–S
D. Sulphurous acid - Two S–OH, Two S=O
Therefore, The correct option is (A): A–III, B–IV, C–I, D–II
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Concepts Used:
P-Block Elements
- P block elements are those in which the last electron enters any of the three p-orbitals of their respective shells. Since a p-subshell has three degenerate p-orbitals each of which can accommodate two electrons, therefore in all there are six groups of p-block elements.
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- Group 14 Elements: Carbon family
- Group 15 Elements: Nitrogen family
- Group 16 Elements: Oxygen family
- Group 17 Elements: Fluorine family
- Group 18 Elements: Neon family