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InterviewSolution
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What is meant by the term bond order? Calculate the bond order of `N_(2)`,`O_(2)`,`O_(2)^(o+)` and `O_(2)^(ө)`. |
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Answer» Bond order is defined as one half of the difference between the number of electrons present in the bonding and anti-bonding orbitals of a molecule. If `N_(a)` is equal to the number of electrons in an anti-bonding orbital, then `N_(b)` is equal to the number of electrons in a bonding orbital. Bond order `=(1)/(2)(N_(b)-N_(a))` If `N_(b) gt N_(a)`, then the molecule is said be stable. However , if `N_(b) le N_(a)`, then the molecule is considered to be unstable. Bond order of `N_(2)` can be calculated form its electronic configuration as : `[sigma(1s)]^(2)[sigma^(**)(1s)]^(2)[sigma(2s)]^(2)[sigma^(**)(2s)]^(2)[pi(2p_(x))]^(2)[pi(2p_(y))]^(2)[sigma(2p_(z))]^(2)` Number of bonding electron , `N_(b)=10` Number of anti-bonding electron , `N_(a)=4` Bond order of nitrogen molecule =`(1)/(2)(10-4)=3` There are 16 electrons in a dioxygen molecule, 8 from each oxygen atom. The electronic configuration of oxygen molecule can be written as : `[sigma -(1s)]^(2)[sigma^(**)(1s)]^(2)[sigma(2s)]^(2)[sigma^(**)(2s)]^(2)[sigma(1p_(z))]^(2)[pi(2p_(x))]^(2)[pi(2p_(y))]^(2)[pi^(**)(2p_(x))]^(1)[pi^(**)(2p_(y))]^(1)` Since the 1s orbital of each oxygen atom is not involved in bonding, the numebr of bonding electrons=8=`N_(b)` and the number of anti-bonding electrons =4=`N_(a)` Bond order `=(1)/(2)(N_(b)-N_(a))` `=(1)/(2)(8-4)=2` Hence, the bond order of oxygen molecule is 2. Similarly, the electronic configuration of `O_(2)^(+)` can be written as : `KK[sigma(2s)]^(2)[sigma^(**)(2s)]^(2)[sigma(2p_(z))]^(2)[pi(2p_(x))]^(2)[pi(2p_(y))]^(2)[pi^(**)(2p_(x))]^(1)` `N_(b)=8` `N_(a)=3` Bond order of `O_(2)^(+)=(1)/(2)(8-3)=2.5` Thus, the bond order of `O_(2)^(+)` is 2.5 The electronic configuration of `O_(2)^(-)` ion will be : `KK[sigma(2s)]^(2)[sigma^(**)(2s)]^(2)[sigma(2p_(z))]^(2)[pi(2p_(x))]^(2)[pi(2p_(y))]^(2)[pi^(**)(2p_(x))]^(2)[pi^(**)(2p_(y))]^(1)` `N_(b)=8` `N_(a)=5` Bond order of `O_(2)^(-)=(1)/(2)(8-5)=1.5` Thus, the bond order of `O_(2)^(-)` ion is 1.5. |
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