
Compare the CH bond strength in ${{C}_{2}}{{H}_{6}},{{C}_{2}}{{H}_{4}}$ and ${{C}_{2}}{{H}_{2}}$.
Answer
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Hint: As we know that the order of increasing bond length is triple bond< double bond < single bond. And bond strength is inversely proportional to bond length therefore the order of increasing bond strength will be single bond < double bond < triple bond. Triple bond carrying compound will have the highest bond strength among the following.
Complete step by step solution:
From your chemistry lessons you have learned about the bond length and the bond strength of the molecule. Bond length is defined as the distance which is present between the centers of two covalently bonded atoms and the length of the bond is shown by the number of electrons that are bonded with it. If the bond order will be higher than the pull between the atoms will be stronger and therefore the bond length will also be stronger. Bond length increases as the % s character decreases because if the s character in the hybrid orbitals will be more than the size of the hybridized molecule will be less and as result the overlapping of the hybridized orbital will be more and the bond length will be less. Therefore the C-H bond length follows the order of:
\[s{{p}^{3}}>s{{p}^{2}}>sp\]
Thus the order of increasing bond length is triple bond< double bond < single bond. Whereas bond strength is defined as the strength which describes how strongly each atom is joined with the other atom and hence how much energy is required to break the bonds that are present between the two atoms. Bond strength is inversely proportional to the bond length. Therefore the C-H bond strength follows the order of:
\[sp>s{{p}^{2}}>s{{p}^{3}}\]
Thus the order of increasing bond strength will be single bond < double bond < triple bond. In the question we are asked to compare the CH bond strength in ${{C}_{2}}{{H}_{6}},{{C}_{2}}{{H}_{4}}$ and ${{C}_{2}}{{H}_{2}}$. As we know that in ${{C}_{2}}{{H}_{2}}$ two carbon atoms are joined through triple bond and in ${{C}_{2}}{{H}_{4}}$ two carbon atoms are joined through double bond whereas in ${{C}_{2}}{{H}_{6}}$two carbon atoms are joined through single bond.
Thus the correct order for the given compounds will be:
\[{{C}_{2}}{{H}_{2}}>{{C}_{2}}{{H}_{4}}>{{C}_{2}}{{H}_{6}}\]
Note: Bond energy is defined as the amount of energy required to break the covalent bond present between the two atoms; it is also called bond dissociation energy. Bond energy is also inversely proportional to the bond length, higher will be the bond energy or bond strength more will be the stability of the compound.
Complete step by step solution:
From your chemistry lessons you have learned about the bond length and the bond strength of the molecule. Bond length is defined as the distance which is present between the centers of two covalently bonded atoms and the length of the bond is shown by the number of electrons that are bonded with it. If the bond order will be higher than the pull between the atoms will be stronger and therefore the bond length will also be stronger. Bond length increases as the % s character decreases because if the s character in the hybrid orbitals will be more than the size of the hybridized molecule will be less and as result the overlapping of the hybridized orbital will be more and the bond length will be less. Therefore the C-H bond length follows the order of:
\[s{{p}^{3}}>s{{p}^{2}}>sp\]
Thus the order of increasing bond length is triple bond< double bond < single bond. Whereas bond strength is defined as the strength which describes how strongly each atom is joined with the other atom and hence how much energy is required to break the bonds that are present between the two atoms. Bond strength is inversely proportional to the bond length. Therefore the C-H bond strength follows the order of:
\[sp>s{{p}^{2}}>s{{p}^{3}}\]
Thus the order of increasing bond strength will be single bond < double bond < triple bond. In the question we are asked to compare the CH bond strength in ${{C}_{2}}{{H}_{6}},{{C}_{2}}{{H}_{4}}$ and ${{C}_{2}}{{H}_{2}}$. As we know that in ${{C}_{2}}{{H}_{2}}$ two carbon atoms are joined through triple bond and in ${{C}_{2}}{{H}_{4}}$ two carbon atoms are joined through double bond whereas in ${{C}_{2}}{{H}_{6}}$two carbon atoms are joined through single bond.
Thus the correct order for the given compounds will be:
\[{{C}_{2}}{{H}_{2}}>{{C}_{2}}{{H}_{4}}>{{C}_{2}}{{H}_{6}}\]
Note: Bond energy is defined as the amount of energy required to break the covalent bond present between the two atoms; it is also called bond dissociation energy. Bond energy is also inversely proportional to the bond length, higher will be the bond energy or bond strength more will be the stability of the compound.
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