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) CH3CH=CH2,
(iii) (CH3)2CO,
(iv) CH2=CHCN,
(v) C6H6
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(iii) CH4
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(a) NaH2PO4
(b) NaHSO4
(c) H4P2O7
(d) K2MnO4
(e) CaO2
(f) NaBH4
(g) H2S2O7
(h) KAl(SO4)2.12 H2O
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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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a) What is the initial effect of the change on vapour pressure?
b) How do rates of evaporation and condensation change initially?
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Calculate the energy required for the process
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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?