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The state of hybridization of Boron in $BCl_3$

$(a)sp^3\qquad(b)\;sp^2\qquad(c)\;sp\qquad(d)\;spd^3$

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Boron has 3 unpaired e's in the exited state.
Hence (b) is the correct answer.
answered Nov 27, 2013 by sreemathi.v
 
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