What is the bond order of $ C $ in $ HCN $ ?
Answer
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Hint: Bond order is the number of chemical bonds which are formed between a pair of atoms and indicates the stability of a bond. It also indicates the type and strength of covalent bonds between the atoms. Bond order is inversely proportional to bond length that means when bond order is increased, the bond length decreases.
Complete Step By Step Answer:
Bond order is defined as the number of covalent bonds between two atoms in a molecule which is equal to half the difference between the number of electrons in bonding molecular orbital and antibonding molecular orbitals. Mathematically, the expression for determination of bond order can be calculated as follows:
$ B.O=\dfrac{{{N}_{b}}-{{N}_{a}}}{2} $
Where, $ {{N}_{b}} $ represents the number of electrons in bonding molecular orbitals while $ {{N}_{a}} $ represents the number of electrons in antibonding molecular orbitals.
We know that molecular orbital diagram of a triatomic linear molecule is very complicated and as the given molecule i.e., $ HCN $ has a tendency to exist in its ionic form which is $ {{H}^{+}} $ and $ C{{N}^{-}} $ ions. So, we can calculate the bond order of carbon atoms by drawing the molecular orbital diagram for cyanide ions.
The molecular orbital diagram for cyanide ion will be as follows:
As per the given molecular orbital diagram, the number of electrons in bonding molecular orbital $ =10 $ and number of electrons in antibonding molecular orbitals $ =4 $ . So, the bond order will be as follows:
$ B.O=\dfrac{10-4}{2}\Rightarrow 3 $
Thus, we can conclude that the bond order of $ C $ in $ HCN $ is 3.
Note:
It is important to note that the bond order can also be calculated by the number of bonds between the two atoms in a molecule. That means if an atom is bonded to another atom via single bond, then its bond order is 1, if it is bonded via double bond the bond order is 2 and so on. In case of cyanide ion, the carbon atom is bonded to nitrogen ion via triple bond, So the bond order of carbon is 3 in $ HCN $ molecule.
Complete Step By Step Answer:
Bond order is defined as the number of covalent bonds between two atoms in a molecule which is equal to half the difference between the number of electrons in bonding molecular orbital and antibonding molecular orbitals. Mathematically, the expression for determination of bond order can be calculated as follows:
$ B.O=\dfrac{{{N}_{b}}-{{N}_{a}}}{2} $
Where, $ {{N}_{b}} $ represents the number of electrons in bonding molecular orbitals while $ {{N}_{a}} $ represents the number of electrons in antibonding molecular orbitals.
We know that molecular orbital diagram of a triatomic linear molecule is very complicated and as the given molecule i.e., $ HCN $ has a tendency to exist in its ionic form which is $ {{H}^{+}} $ and $ C{{N}^{-}} $ ions. So, we can calculate the bond order of carbon atoms by drawing the molecular orbital diagram for cyanide ions.
The molecular orbital diagram for cyanide ion will be as follows:
As per the given molecular orbital diagram, the number of electrons in bonding molecular orbital $ =10 $ and number of electrons in antibonding molecular orbitals $ =4 $ . So, the bond order will be as follows:
$ B.O=\dfrac{10-4}{2}\Rightarrow 3 $
Thus, we can conclude that the bond order of $ C $ in $ HCN $ is 3.
Note:
It is important to note that the bond order can also be calculated by the number of bonds between the two atoms in a molecule. That means if an atom is bonded to another atom via single bond, then its bond order is 1, if it is bonded via double bond the bond order is 2 and so on. In case of cyanide ion, the carbon atom is bonded to nitrogen ion via triple bond, So the bond order of carbon is 3 in $ HCN $ molecule.
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