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Give brief about structure of ethye (Acetylene). |
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Answer» Solution :Molecular formula of ethyne is `C_(2)H_(2)`. It is a first member of alkyne series. Ethyne has sp HYBRIDIZATION of exited carbon and both the carbon possess non-hybridized 2p orbitals. Two sp orbitals are linear in shape and sp orbitas are `bot 2p_(x) bot 2p_(y)`. Bond formation in ethyne (`sigma`-bond) : Carbon carbon sigma `(sigma)` bond is obtained by the head-on overlapping of the two sp hybridised orbitals of the two carbon atoms. The remaining sp hybridized orbital of each carbon atoms undergoes overlapping along the internuclear axis with the 1s orbital of each of the two hydrogen atoms forming two C-H sigma bonds. H-C-C bond angle is of `180^(@)`.   Each carbon has two unhybridised p-orbitals which are perpendicular to each other as well as to the plane of the C-C sigma bond. `sigma`-bond : H-C-C-H  Arrangement of `pi`-bond : Two carbon atoms are perpendicular to each other and H-C-C-H are 2p orbitals of sigma bond edges. These 2p orbitals are parallel to each other. `2p_(x)||2p_(x)` and `2p_(y)||2p_(y)`. These `2p_(x)-2p_(x)` and `2p_(y)-2p_(y)` overlap by side by side and form two `pi`-bonds. The following diagram justifies these :  So the simple structure of ethyne is as following.  CHARACTERISTICS of ethyne structure : "Ethyne is linear molecule" `angleH - C-C=180^(@)` due to sp carbon. Ethyne has one C-C `sigma` and two `C-H sigma-` bond. Also it has two C-C `pi`-bonds. `C-=C` bond enthalpy `= 823 KJ mol^(-1)` C=C bond enthalpy `= 681 kJ mol^(-1)` C-C bond enthalpy `= 348 kJ mol^(-1)` `therefore` So order of bond strength is : `C-=C gt C=Cgt C-C` Bond length`alpha 1` / bond order `therefore` So order of bond length is : `C-=C lt C=C lt C-C` `(("120 pm"),("ethyne")) lt (("133 pm"),("ethene")) lt (("154 pm"),("ethane"))` In ethylene, cloud of two electrons of `pi`-bond is arranged as a cyclinderical symmetry around the axis of INTER nuclear two carbon. |
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