
The ratio of the number if $ sp,\text{ }s{{p}^{2}} $ and $ s{{p}^{3}} $ carbons in the compound given below is $ H_{2}^{{}}C=C=CH-CH_{3}^{{}} $
(A) $ 1:2:1 $
(B) $ 2:1:1 $
(C) $ 1:1:2 $
(D) $ 1:2:3 $
Answer
415.2k+ views
Hint: We know that hybridization is a concept used in organic chemistry to explain chemical bonding in cases where the valence bond theory does not provide satisfactory clarification. This theory is especially useful to explain the covalent bonds in organic molecules.
Complete answer:
Each carbon atom has two unpaired electrons (in the $ 2px $ and $ 2py $ orbitals). Based on the valence theory, only two hydrogen molecules could be paired to the two unpaired electrons of the carbon atom and there will be a formation of only $ 2\text{ }C-H $ bonds in the molecule. This will lead to an incomplete octet in the $ 2nd $ orbital of the carbon molecule ( $ 2pz $ orbital is unfilled) and so the molecule should be unstable. However, we see that actually the methane molecule is extremely stable in nature and has $ 4\text{ }C-H $ bonds and not two. Thus, the valence theory doesn’t explain the covalent bond of the methane molecule. Hybridization concept explains the formation of identical $ 4\text{ }C-H $ bonds and the tetrahedral shape of the molecule. The s orbitals of the hydrogen atom overlap with one sp hybrid orbital of each of the carbon atoms forming the $ 2\text{ }C-H $ bonds. The $ C-C $ covalent bond is formed by overlapping between the $ sp-sp $ orbitals of the two hybrid carbon atoms.
Here we know that;
$ \underset{s{{p}^{2}}\text{ }}{\mathop{{{H}_{2}}C}}\,=\underset{sp}{\mathop{C}}\,=\underset{s{{p}^{2}}\text{ }}{\mathop{CH}}\,-\underset{\text{ }s{{p}^{3}}}{\mathop{C{{H}_{3}}}}\, $
$ \therefore sp:s{{p}^{2}}:s{{p}^{3}}=1:2:1 $
Therefore, the correct answer is option A.
Note:
Remember that the atoms of different elements try to attain electronic configuration like noble gas atoms or to complete their octet by chemical bonding. In other words, atoms of all main group elements tend to bond in such a way that each atom has eight electrons in its valence shell so that the atoms will attain electronic configuration like noble gases.
Complete answer:
Each carbon atom has two unpaired electrons (in the $ 2px $ and $ 2py $ orbitals). Based on the valence theory, only two hydrogen molecules could be paired to the two unpaired electrons of the carbon atom and there will be a formation of only $ 2\text{ }C-H $ bonds in the molecule. This will lead to an incomplete octet in the $ 2nd $ orbital of the carbon molecule ( $ 2pz $ orbital is unfilled) and so the molecule should be unstable. However, we see that actually the methane molecule is extremely stable in nature and has $ 4\text{ }C-H $ bonds and not two. Thus, the valence theory doesn’t explain the covalent bond of the methane molecule. Hybridization concept explains the formation of identical $ 4\text{ }C-H $ bonds and the tetrahedral shape of the molecule. The s orbitals of the hydrogen atom overlap with one sp hybrid orbital of each of the carbon atoms forming the $ 2\text{ }C-H $ bonds. The $ C-C $ covalent bond is formed by overlapping between the $ sp-sp $ orbitals of the two hybrid carbon atoms.
Here we know that;
Number of sigma $ \left( \sigma \right) $ bonds | Hybridization |
$ 4 $ | $ s{{p}^{3}} $ |
$ 3 $ | $ s{{p}^{2}} $ |
$ 2 $ | $ sp $ |
$ \underset{s{{p}^{2}}\text{ }}{\mathop{{{H}_{2}}C}}\,=\underset{sp}{\mathop{C}}\,=\underset{s{{p}^{2}}\text{ }}{\mathop{CH}}\,-\underset{\text{ }s{{p}^{3}}}{\mathop{C{{H}_{3}}}}\, $
$ \therefore sp:s{{p}^{2}}:s{{p}^{3}}=1:2:1 $
Therefore, the correct answer is option A.
Note:
Remember that the atoms of different elements try to attain electronic configuration like noble gas atoms or to complete their octet by chemical bonding. In other words, atoms of all main group elements tend to bond in such a way that each atom has eight electrons in its valence shell so that the atoms will attain electronic configuration like noble gases.
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