1.

Among H_(2),He_(2)^(+) ,Li_(2),Be_(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»


Solution :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 .


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