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Explain bond formation in BCl_(3). Explain it is symmetrical trigonal molecute. |
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Answer» Solution :In BORON in ground state electron configuration is B (Z = 5 ) : [He] `2s^(2)2p^(1) ` in exited state its configuration will be B* [He] `2s^(1) 2p_(y)^(1) 2p_(z)^(1)`. One electron of 2s enter in 2p ORBITAL. In exited state. B has three half filled 2s, `2p_(y), 2p_(z)` orbital these orbitals overlap and FORM three `SP^(2)` hybrid orbitals. Such three sp hybrid orbitals arrange in planar trigonal shape at `120^(@)` angle These three `sp^(2)` hybrid orbitals overap with three half filled 3p orbital of chlorine and form three B - Cl o bond. So `BCl_(3)` has trigonal planer shape & Cl - B - Cl `120^(@)` angle as follow: 
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