Observe the following molecules `//` ions
`H_(2) ,N_(2) ,O_(2) ,N_(2)^(+) , O_(2)^(+) , O_(2)^(-), F_(2).` Identify correct statement .

Options

A. `H_(2),N_(2),O_(2)F_(2)` show diamagnetic property
B. `O_(2),O_(2),O_(2)^(-),N_(2)^(+)` show paramagnetic property
C. `N_(2),F_(2),O_(2)^(+),O_(2)^-)` show diamagnetic property
D. `H_(2) N_(2),O_(2)^(+),O_(2)^(-)` show paramagnetic property

Answer

B

Explanation

`H_(2) -(ols)^(2)`
`H_(2)` does not have any unpaired electron hence diamagnetic in nature
`N_(2) to KK (Sigma 2s)^(2) (sigma 2s)^(2) (pi 2p)^(2) (pi 2p_(y))^(2) (sigma p_(z))^(2)`
`N_(2)` does not have any unpaired electron hence diamagnetic in nature
`O_(2) to KK^(1) (Sigma ) 2 s)^(2) (Sigma 2s)^(2) (pi 2pi_(k))^(2) (pi 2p_(y))^(2)`
`O_(2)` have two unpaired electrons hence
paramagnetic in nature
`N_(2)^(+) to KK^(1) (sigma 2s)^(2)(sigma 2s)^(2) (pi 2p_(y))^(2) (pi 2p_(y))^(2) (pi 2p_(z))^(1)`
`N_(2)` have one unpaired electron so it paramagnetic in nature
`O_(2)^(+) to KK(sigma 2 s)^(2) (sigma 2s)^(2) (pi 2p_(x))^(2) (pi 2p_(y))^(2)`
`F_(2)` does not have unpaired electron hence diamagnetic in nature
Hence the molecules `N_(2)^(+) ,O_(2) ,O_(2)^(+) , O_(2)^(-)` and `H_(2) ,N_(2) ,F_(2)` are paramagnetic and diamagnetic respectively due to presence of unpaired and paired electrons Hence option (b ) is the correct answer .