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White phosphorus is a tetra atomic solid...

White phosphorus is a tetra atomic solid `[P_(4)(s)]` at room temperature and on strong heating in absence of oxyge, it polymerizes into red phosphorus as:
`/_\H=-104kJ//"mol"` of `P_(4)`
The enthalpy of sublimation `[P_(4)(s)toP_(4)(g)]` white is `59KJ//"mol"` and enthalpy of atomization is `1265 KJ//"mol"` of `P(g)`
The average `P-P` bond enthalpy in `P_(4)` molecule is:

A

`102KJ`

B

`201KJ`

C

`104KJ`

D

`120KJ`

Text Solution

Verified by Experts

The correct Answer is:
B

b
`P_(4)(s)toP_(4)(g) /_\H=59KJ`
`P_(4)(s)to4P(g) /_\H=1265KJ`
`P_(4)(g)to 4P(g) /_\H=1206KJ`
Average `P-P` bond enthalpy` =1206/6=20KJ`
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White phosphorus is a tetra atomic solid [P_(4)(s)] at room temperature and on strong heating in absence of oxyge, it polymerizes into red phosphorus as: /_\H=-104kJ//"mol" of P_(4) The enthalpy of sublimation [P_(4)(s)toP_(4)(g)] white is 59KJ//"mol" and enthalpy of atomization is 1265 KJ//"mol" of P(g) The P-P bond enthalpy in red phosphorus joining the two tetrahedral is:

White phosphorus (P_4) has

White phosphorus (P_(4)) has

White phosphorus (P_4) does not contain

White phosphorus is a tetra-atomic solid P_(4) (s)at room temperature. find bond enthalpy (P-P) in kJ//mol. Given : DeltaH_("sublimation") of P_(4) (s)= 61 kL//mol DeltaH_("atomisation") of P_(4) (s)= 1321 kJ//mol]

White phosphorus is a tetra atomic solid P_(4) (s) at room temperature. Find average (P-P) bond enthalpy (in KJ/mol). {:("Given",DeltaH_("sublimation") "of" P_(4)(s)=59 KJ//mol,),(,DeltaH_("atomizatiom") "of" P_(4)(s) = 1265 KJ//mol,):}

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