Observe the following molecules : `C_(2), N_(2), O_(2), F_(2)` Which one of the following statement is correct for the above molecules ?

Options

A. They exhibit same magnetic property
B. The have same number of bonding molecular orbitals and same number of antibonding molecular orbitals
C. The sequence of molecular orbitals is as follows `sigma 2 p_(z) lt (pi 2p_(x) = pi 2 p_(y)) lt (pi 2 p_(x)) = pi 2 p_(y)) lt sigma 2 p_(z)`
D. They have same bond order

Answer

C

Explanation

The electronic configuration, magnetic property and bond order of given molecules are as follows :
(i) `C_(2) = 12`
`= sigma 1s^(2) lt sigma^(**) 1s^(2) lt sigma 2s^(2) lt sigma^(**) 2s^(2) lt pi 2p_(x)^(2) = pi 2p_(y)^(2)`
it is diamagnetic in nature.
`BO = (1)/(2)[N_(b) - N_(a)]`
`= (1)/(2)[8-4] = 2`
(ii) `N_(2)= 14`
`= sigma 1s^(2) lt sigma^(**) 1s^(2) lt sigma 2s^(2) lt sigma^(**) 2s^(2) lt pi 2p_(x)^(2) = pi 2p_(y)^(2) lt sigma 2pi_(x)^(2)`
It is diamagnetic in nature.
`BO = (1)/(2)[10 - 4] = 3`
(iii) `O_(2) = 16 = sigma 1s^(2) lt sigma^(**) 1s^(2) lt sigma 2s^(2) lt sigma^(**) 2s^(2) lt sigma 2p_(z)^(2) lt sigma 2p_(x)^(2) lt pi 2p_(y)^(2) lt pi^(**) 2p_(x)^(1) = pi^(**) 2p_(y)^(1)`
It is paramagnetic in nature.
`BO = (1)/(2)[10 - 6] = 2`
(iv) `F_(2) = 18 = sigma 1s^(2) lt sigma^(**)1s^(2) lt sigma 2s^(2) lt sigma^(**) 2s^(2) lt sigma 2pz^(2)`
`pi 2p_(x)^(2) = pi 2p_(y)^(2) lt pi^(**) 2p_(x)^(2) = pi^(**) 2p_(y)^(2)`
it is diamagnetic in nature.
`BO = (1)/(2)[10 - 8] = 1`
Here, the filling of electrons in bonding and antibonding orbital is different but they all have same number of molecular and antimolecular orbital (even are empty).
Thus, statement (b) is correct.