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Explain the bond formation by sp orbitals. OR Explain the bond formation in BeCl_(2) explain why BeCl_(2) is linear. |
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Answer» SOLUTION :In `BeCl_(2), "Be [He]" 2s^(2)`.And form exited `Be^(*) ["He"] 2s^(1) 2p^(1)`. The divalency of Be in exited STATE the one electron ARRANGE in 2s and one in 2p orbit. One 2s and one 2p orbital of exited Be undergo hybridization and form two sp hybrid orbital. These two sp hybrid orbitals are oriented in opposite direction formatting an angle of `180^(@)`. Cl atom [NE]` 3s^(2) 3p_(x)^(2) 3p_(y)^(2) 3p_(z)^(1)`. Every sp orbital overlap with `3p_(z)` orbital of chlorine (Cl) with z axis and form two Be - Cl `sigma` bond.  Each of the sp hybridised orbital overlaps with the 2p orbital of chlorine axially and form two Be-Cl sigma bond and `BeCl_(2)` form with LINEAR shape. |
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