1.

Four diatomic species are listed in different sequence .Which of these represent the correct order of their increasing bond order?A. `O_(2)^(-) lt NO lt C_(2)^(2-) lt He_(@)^(+)`B. `C_(2)^(2-) lt He_(2)^(+) lt O_(2)^(-) lt NO`C. `He_(2)^(+) lt O_(2)^(-) lt NO lt C_(2)^(2-)`D. `NO lt O_(2)^(-) lt C_(2)^(2-) lt He_(2)^(+)`

Answer» Correct Answer - C
`C_(2)^(20) rArr "total electron = 14"`
configuration
`KK sigma2s^(2) sigma^(**)2s^(2)pi2px^(2) pi2py^(2) simga2pz^(2)`
`"Bouderder"=(Nb-Na)/(2)=(8-2)/(2)=3`
`"NO "rArr" total electron"=5+6=11`
`KK sigma2s^(2)sigma^(**)2s^(2) sigma2pz^(2) pi2px^(2) pi2py^(2) pipx^(1)`
`"Bond order"=((8-3))/(2)=2.5`
`O_(2)^(-)rArr KK sigma(2s)^(2)sigma^(**)(2s)^(2)sigma(2pz)^(2)pi(2px)^(2)pi(2py)^(2) pi^(**)(2px)^(2)pi^(**)(2py)^(1)`
`B.O=(8-5)/(2)=1.5`
`He^(2+)rArr sigma1s^(2)sigma^(**)1s^(1)`
`B.O.=(2-1)/(2)=0.5`


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