1.

Consider the following species : CN^(+), CN^(-), NO and CN Which one of these will have the highestbond order ?

Answer»

`CN^(+)`
`CN^(-)`
NO
CN

Solution :`NO (15E^(-)) : sigma (1s)^(2) , sigma^(**) (1s^(2)), sigma^(**) (1s)^(2) , sigma (2s)^(2)`,
` sigma^(**) (2s)^(2), sigma (2p_(z))^(2) , pi(2p_(x))^(2)`
`pi (2p_(y))^(2) . Pi^(**) (2p_(x))^(1) = pi^(**) (2p_(y))^(0)`
B.O. ` = (10 - 5)/(2) = 2.5 `
`CN^(-) (14 e^(-)) : sigma(1s)^(2), sigma^(**) (1s)^(2), sigma (2s)^(2)` ,
` sigma^(**) (2s)^(2) , pi (2p_(x))^(2) = pi (2p_(y))^(2)`
`sigma (2p_(z))^(2)`
B.O. ` = (10- 4)/(2) = 3 `
` CN (13 e^(-)) : sigma(1s)^(2), sigma^(**) (1s)^(2), sigma (2s)^(2)` ,
` sigma^(**) (2s)^(2) , pi (2p_(x))^(2) = pi (2p_(y))^(2)`
`sigma (2p_(z))^(1)`
B.O.`= (9-4)/(2) = 2.5 `
`CN (12 e^(-)) : sigma(1s)^(2), sigma^(**) (1s)^(2), sigma (2s)^(2)` ,
`sigma^(**) (2s)^(2) , pi (2p_(x))^(2) = pi (2p_(y))^(2)`
B.O. ` = (8 - 4)/(2) = 2 `
Hence , `CN^(-)` has the highest bond order .


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