InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
(a) Account for the following observations : (i) `SF_(1)` is easily hydrolysed whereas `SF_(6)` is not easily hydrolysed. (ii) Chlorine water is a powerful bleaching agent. (iii) Bi (V) is a stronger oxidising agent than Sb(V) (b) What happens when : (i) White phosphorus is heated with concentrated NaOH solution in an inert atmosphere of `CO_(2).` (ii) `XeF_(6)` undergoes partial hydrolysis. (Give the chemical equations involved). |
|
Answer» (a) (i) S atom in `SF_(4)` is not sterically protected as it is surrounded by only four F atoms, so attack of water molecules can take place easily. In contrast, S atom in `SF_(6)` is protected by six F atoms. Thus attack by water molecules cannot take place easily. (ii) Chlorine water produces nascent oxygen (causes oxidation) which is responsible for bleaching action. `CH_(2) + H_(2)O to 2HCI + [O]` (iii) Due to inert pair affect Bi(V) can accept a pair of electrons to form more stable Bi (III). (+ 3 oxidation state of Bi is more stable than its + 5 oxidation state). (b) (i) Phosphorus undergoes disproportionation reaction to form phosphine gas. `P_(4) + 3NaOH + 3H_(2)O to PH_(3) + 3NaH_(2) PO_(2)` (ii) On partial hydrolysis `XeF_(6)` gives oxyfluoride `XeOF_(4)` and HF. `XeF_(6) + H_(2)O to XeOF_(4) + 2HF` |
|