Is there any change in the hybridisation of B and N atoms as a result of the following reaction?
BF3 + NH3 → F3B.NH3
The atomic number of boron is 5 & its electronic configuration in ground state is 1s2 2s2 2p1, & in excited state is 1s2 2s1 2p2 , this means its one s & 2 p orbital will take part in hybridization to form sp2 hybrid orbital.
Similarly N atom has atomic number of 7, with electronic configuration of 1s2 2s2 2p3 in its ground state & in it excited state it has sp3 hybridization.
Now according to the question Boron & Nitrogen in reactant stage has sp2 & sp3 hybridization respectively, but on the product side an adduct is formed, wherein Boron has changed it s hybridization to sp3 while Nitrogen remains in same sp3 hybridization
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(ii) whose value is independent of path
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(i) 34.216
(ii) 10.4107
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(i) used to determine heat changes
(ii) whose value is independent of path
(iii) used to determine pressure volume work
(iv) whose value depends on temperature only.
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You explained but no one understands no offence
i dunno understand that ???
How can we find hybridization of F3B.NH3
There is no change in both the case for hybridisation. This statement is given in a guide.....Which one is true? Why?