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In $CH_3 − CH_2 − CH_2 − Br$, $C-Br$ bond is formed by the overlapping of

$\begin {array} {1 1} (A)\;sp^3-2pz & \quad (B)\;2sp^3-3pz \\ (C)\;2sp^3-2pz & \quad (D)\;2sp^3-4pz \end {array}$

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$Br$ lies in $ 4^th$ period , so Bromine has $4pz$ orbital thus $4pz$ orbital overlap with $sp^3$ hybrid orbital of $C$ adjescent to it.
Ans : (D)
answered Apr 9, 2014 by thanvigandhi_1
edited Apr 12, 2014 by archana.ghode.12_1

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