1.

Compare the relative stability of the following species and indicate their magnetic properties: `O_(2)`,`O_(2)^(o+)`,`O_(2)^(ө)`(super oxide),`O_(2)^(-2)`(peroxide).

Answer» There are 16 electrons in a molecule of dioxygen,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 number of bonding electrons=8=`N_(b)` and the number of anti-bonding orbitals=4=`N_(a)`
Bond order `=(1)/(2)(N_(b)-N_(*a))`
`=(1)/(2)(8-4)`
`=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 `
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`
Electronic configuration of `O_(2)^(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))]^(2)`
`N_(b)=8`
` N_(a)=6`
Bond order of `O_(2)^(2-)=(1)/(2)(8-6)=1`
Bond dissociation energy is directly proportional to bond order. Thus, the higher the bond order, the greater will be the stability. On this basis, the order of stability is `O_(2)^(+) gt O_(2) gt O_(2)^(-) gt O_(2)^(2-)`.


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