Question

Explain the following with proper reason : 

(i) Fluorine cannot be prepared from fluorides by chemical oxidation. 

(ii) Anhydrous HCl is a bad conductor of electricity while aqueous HCl is a good conductor. 

(iii) Halogens are strong oxidising agents. 

(iv) The bleaching action of chlorine is permanent while that of sulphur dioxide is temporary. 

(v) Iodine dissolves more in KI solution than in water..

Answer 1

(i) The standard reduction potential of fluorine is maximum. 

 

Thus, it cannot be oxidised by any other reagent. F^– ion is very stable due to small size and high electronegativity of fluorine atom. 

(ii) In anhydrous state, HCl is a covalent molecule. In aqueous solution, HCl combines with water molecule to form H₃O⁺ and Cl ^– ions. 

HCl + H₂O → H₃O⁺ + Cl ^– 

(iii) Halogens act as strong oxidising agents because they have high tendency to accept electron, i.e., they have high electron affinity values. Their reduction potentials are high (positive) and decrease from F to I. Thus, oxidising nature decreases from F to I. 

(iv) Chlorine bleaching action is due to oxidation while that of sulphur dioxide is due to reduction. Hence, the substance bleached by SO₂ is reoxidised by the oxygen of the air to its original state. 

(v) I₂ is a covalent molecule. thus, its solubility is less in polar solvent, i.e., water. Potassium iodide combines with iodine and forms a polyhalide which is an ionic compound. Being ionic, KI₃ is more soluble. 

KI + I₂ → KI₃ (K⁺ I₃^– )