Question

please no copy paste...
Do qu.no. 2 & 3​

Answer 1

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2) i) H2O is a liquid at room temperature because it is highly associated via intermolecular hydrogen bonding. H2S, due to large size of sulphur cannot form hydrogen bonding. Therefore, H2S is a discrete molecule and the intermolecular forces of attraction are weak Van der Waal's forces. Hence, H2S is a gas at room temperature.

ii) Fluorine is the most electronegative element so it cannot show positive oxidation state.

3) i) XeF2 + PF5 --> [XeF]+ [PF6]-

Reason:- PF5 is a fluoride ion acceptor so it yields cationic species with xenon fluorides.

ii) Xe (g) + 2F2 (g) ----> XeF4 (s) at 873 K.

iii) Cl2 + 5F2 ---> 2ClF5 at 300°C

iv) 6XeF4 + 12H2O ---> 2XeO3 + 24HF + 4Xe + 3O2

Hope It Will Help.❤️

Answer 2

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2) i) H2O is a liquid at room temperature because it is highly associated via intermolecular hydrogen bonding. H2S, due to large size of sulphur cannot form hydrogen bonding. Therefore, H2S is a discrete molecule and the intermolecular forces of attraction are weak Van der Waal's forces. Hence, H2S is a gas at room temperature.

ii) Fluorine is the most electronegative element so it cannot show positive oxidation state.

3) i) XeF2 + PF5 --> [XeF]+ [PF6]-

Reason:- PF5 is a fluoride ion acceptor so it yields cationic species with xenon fluorides.

ii) Xe (g) + 2F2 (g) ----> XeF4 (s) at 873 K.

iii) Cl2 + 5F2 ---> 2ClF5 at 300°C

iv) 6XeF4 + 12H2O ---> 2XeO3 + 24HF + 4Xe + 3O2