Correct Answer - D
MOEC of
`B_(2)(10)= sigma 1s^(2), overset(**)sigma1s^(2), sigma 2s^(2), overset(**)sigma3s^(2), pi 2p_(x)^(1), ~~ pi 2p_(y)^(1)`
MOEC of `O_(2)(16)= sigma1s^(2), overset(**)sigma 1s^(2), overset(**)sigma 2s^(2), sigma 2p_(z)^(2)`
`pi 2p_(x)^(2) ~~ 2pi_(z)^(2), overset(** ) sigma 2p_(x)^(1)~~ overset(**) sigma 2p_(y)^(1)`
MOEC of `NO(15)= sigma 1s^(2), overset(**) sigma 1s^(2), sigma 2s^(2), overset(**) sigma 2s^(2), sigma p_(z)^(2)`
`pi 2 p_(x)^(2)~~ pi 2p_(y)^(2),overset(**) pi 2p_(x)^(1)~~ overset(**) pi 2p_(y)`
Due to the presence of unpaired electron, `B_(2), O_(2) and NO` all are paramgnetic.