Among `H_(2), He_(2)^(+), Li_(2), Be_(2), B_(2), C_(2), N_(2), O_(2)^(-)` and `F_(2)`, the number of diamagnetic species is
(Atomic numbers: `H=1, He=2, Li=3, Be=4, B=5, C=6, N=7, O=8, F=9`)

Answer

6

Explanation

M.O. Configuration are
`H_(2)(2) = sigma (1s)^(2)` (Diamagnetic)
`He_(2)^(+) = sigma (1s)^(2), sigma^(**) (1s)^(1)` (Paramagnetic)
`Li (2) (6) = sigma (1s)^(2) , sigma^(**) (1s)^(2) , sigma(2s)^(2) ` (Diamgnetic)
` Be_(2) (8) = sigma (1s)^(2) , sigma^(**) (1s)^(2) , sigma(2s)^(2) , sigma^(**) (2s)^(2)` (Diamagnetic)
` Be_(2) (10) = sigma (1s)^(2) , sigma^(**) (1s)^(2) , sigma(2s)^(2) , sigma^(**) (2s)^(2), `
`pi(2p_(x))^(1) = pi(2p_(y))^(1)` (Paramagnetic)
` C_(2) (12) = sigma (1s)^(2) , sigma^(**) (1s)^(2) , sigma(2s)^(2) , sigma^(**) (2s)^(2), `
`pi(2p_(x))^(2) = pi(2p_(y))^(2)` (Diamagnetic)
` N_(2) (14) = sigma (1s)^(2) , sigma^(**) (1s)^(2) , sigma(2s)^(2) , sigma^(**) (2s)^(2), `
`pi(2p_(x))^(2) = pi(2p_(y))^(2),sigma(2p_(z))^(2)` (Diamagnetic)
` O_(2)^(-) (17) = sigma (1s)^(2) , sigma^(**) (1s)^(2) , sigma(2s)^(2) , sigma^(**) (2s)^(2), `
`sigma(2p_(z))^(2) = pi(2p_(y))^(2),sigma(2p_(z))^(2)`
` pi^(**) (2p_(x))^(2) = pi^(**) (2p_(y))^(1) ` (paramagnetic)
` F_(2) (18) = sigma (1s)^(2) , sigma^(**) (1s)^(2) , sigma(2s)^(2) , sigma^(**) (2s)^(2), `
`sigma(2p_(z))^(2) = pi(2p_(x))^(2),sigma(2p_(y))^(2)`
`pi^(**) (2p_(x))^(2) = pi^(**) (2p_(y))^(2)` (Diamagnetic)
Therefore , the total number of diamagnetic species
is six .