Saved Bookmarks
| 1. |
Use the molecular orbital energy level diagram to show that N_(2) would be expected to have atriple bond, F_(2) a single bond and Ne_(2), no bond. |
|
Answer» Solution :Formation of `N_(2)` molecule : Electronic configuration of N-atom `""_(7)N =1s^(2), 2s^(2) , 2p_(x)^(1) 2p_(y)^(1) , 2p_(z)^(1)` `N_(2)` molecule = `sigma 1s^(2), sigma^(**) 1s^(2), sigma 2s^(2) , sigma^(**) 2s^(2), PI 2sp_(x)^(2) = pi 2p_(y)^(2), sigma 2p_(z)^(2)`  Bond ORDER value of 3 means that `N_(2)` CONTAINS a triple bond Formation of `F_(2)` molecule : `""_(9)F = 1s^(2), 2s^(2, 2p_(x)^(2), 2p_(y)^(2), 2p_(z)^(1)` `F_(2) " molecule " = 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)^(2) = pi 2p_(y)^(2)`  Bond order = `(1)/(2) [N_(b) - N_(a) ]= (1)/(2) (10 - 8) =1 ` Bond order value 1 means that `F_(2)` contains SINGLE bond. Formation of `Ne_(2)` molecule : `""_(10)Ne = 1s^(2), 2s^(2) , 2p_(x)^(2) , 2p_(y)^(2) , 2p_(z)^(2)` `Ne_(2) "molecule" = sigma 1s^(2), sigma^(**) 2s^(2), 2 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) = sigma 2p_(z)^(2)` Bond order = `(1)/(2) [ N_(b) - N_(a) ]= (1)/(2) (10 - 10) =0 `  Bond order value zero means that there is no formation of bond between two Ne-atoms. Hence, `Ne_(2)` molecuule does not exist, |
|