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Which of the following statement is not correct from the view point of molecular orbital theory ? |
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Answer» `Be_(2)` is not a stable molecule Existence of molecule, bonding NATURE and energy order of molecular orbitals can be explained on the basis of molecular orbital theory as follows : (i) if molecules having ZERO bond order than it does not exists if molecular having non-z.ero bond order it exists or expected to exist. (ii) Higher the value of bond order, higher will be its bond strength. The electrons present in bonding molecular orbital are KNOWN as bonding electrons (Nb) and electrons present on anti-bonding molecular orbital are known as anti-bonding electrons `(N_(a))`and half of their difference is known as bond order. i.e., `Be_(2) (4 + 4) = 8 ) = sigma 1s^(2) , sigma^(**) 1s^(2) , sigma 1s^(2), sigma^(**) 2s^(2) ` Bond order (BO) =` (1)/(2)` [ Number of bonding electrons `(N_(B))` - NUmber of anti-bonding electrons `N_(a) = (4 - 4)/(2) = 0 ` Here, bond order of `Be_(2)` is zero. Thus, it does not exist. `He_(2) (2 + 2 = 4) = sigma 1 s^(2) , sigma^(**) 1s^(2) ` BO = `(2- 2)/(2)= 0 ` Bond order of `Be_(2)` is zero . So it does not exist. `He_(2)^(+) (2 + 2 - 1 = 3) = sigma 1 s^(2), sigma^(**) 1s^(1)` BO = `(2- 1)/(2) = 0.5 ` As, the bond order is not zero, this molecule is expected to exist. `N_(2) (7 + 7 = 14 )=sigma 1s^(2), sigma^(**) 1s^(2), sigma 2s^(2) , sigma^(**) 2s^(2), pi 2p_(x)^(2)= pi 2p_(y)^(2) sigma^(**) 2p_(x)^(2) ` BO = `(10 - 4)/(2) = 3` So, dinitrogen`(N_(2))` molecule contain triple bond and no any molecule of second period have more than double bond. Hence, bondstrength of `N_(2)` is maximum amongst the homonuclear diatomic molecules belonging to the second period. (D)It is incorrect. The correct order of energies of molecular orbitals in `N_(2)` molecule is `sigma 2s lt sigma^(**) 2s lt (pi 2p_(x)= pi 2p_(y)) lt pi 2p_(x) lt pi^(**) 2p_(x) = pi 2p_(y) lt sigma^(**) 2p_(z)` |
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