`H_2` molecule : Total no. of electrons `= 2`
Arrangement: `sigma_(1s)^2`
Bond order `: 1/2 (2 - 0) = 1`
`H_2^(+)` molecule: Total no. of electrons `= 1`
Arrangement `: sigma_(1s)^(1)`
Bond order `: 1/2 (1 - 0) = 1 /2`
`He_2^` molecule: Total no. of elect rons = 4
Arrangement `:sigma_(1s)^2 , sigma_(1s)^(ast 2)`
Bond order : `1/2 (2 - 2) = 0`
`He_2` molecule does not exist.
`He_2^(+)` molecule : Total no. of electrons = 3
Arrangement : `sigma_(1s)^2 , sigma_(1s)^(ast1)`
Bond order `: 1/2 (2 - 1) = 1 /2`
So `He_2^(+)` exists and has been detected in discharge tubes.
`Li_2` molecule :Total no. of electrons = 6
Arrangement : `sigma_(1s)^2 , sigma_(1s)^(ast 2) , sigma_(2s)^2`
Bond order: `1/2 (4- 2) = 1`
No unpaired e's so diamagnetic
`Be_2` molecule : Total no. of electrons = 8
Arrangement : `sigma_(1s)^2 , sigma_(1s)^(ast2) , sigma_(2s)^2 , sigma_(2s)^(ast 2)`
Bond order: `1/2 (4- 4) = 0`
No unpaired `e^(-)`'s so diamagnetic.
`B_2` molecule: Total no. of electrons = 10
Arrangemen t : `sigma_(1s)^2 , sigma_(1s)^(ast2) , sigma_(2s)^2 , sigma_(2s)^(ast 2) pi_(2p_y)^(1) pi_(2p_z)^(1)`
Bond order `: 1/2 (6 - 4) = 1`
It is paramagnetic
`C_2` molecule : Total no. of electrons = 12
Arrangement: `sigma_(1s)^2 , sigma_(1s)^(ast2) , sigma_(2s)^2 , sigma_(2s)^(ast2) pi_(2p_y)^(2) pi_(2p_y)^(2)`
Bond order : `1/2 (4 - 0) = 2`
It is diamagnetic
`N_2` molecule : Total no. of electrons = 14
Arrangement: `sigma_(1s)^2 , sigma_(1s)^(ast2) , sigma_(2s)^2 , sigma_(2s)^(ast2) pi_(2p_y)^(2) pi_(2p_z)^(2) sigma_(2p_x)^2`
Bond order `: 1/2 (6 - 0) = 2`
It is diamagnetic
`O_2` molecule: Total no. of electrons = 16
Arrangement : `sigma_(1s)^2 , sigma_(1s)^(ast 2), sigma_(2s)^2 , sigma_(2s)^(ast2), sigma_(2p_x)^2 pi_(2p_y)^(2) pi_(2p_z)^(2) pi_(2p_y)^(ast2) pi_(2p_z)^(ast2)`
Bond order `: 1/2 (6 - 2) = 2`
It is paramagnetic
`F_2` molecule : Total no. of electrons = 18
Arrangement : `sigma_(1s)^2 , sigma_(1s)^(ast2) , sigma_(2s)^2 , sigma_(2s)^(ast2), sigma_(2p_x)^2 pi_(2p_y)^(2) pi_(2p_z)^(2) pi_(2p_y)^(ast2) pi_(2p_z)^(ast2) sigma_(2p_x)^(ast)`
Bond order `: 1/2 (6 - 4) = 1`
`H_2` molecule : Total no. of electrons `= 2`
Arrangement: `sigma_(1s)^2`
Bond order `: 1/2 (2 - 0) = 1`
`H_2^(+)` molecule: Total no. of electrons `= 1`
Arrangement `: sigma_(1s)^(1)`
Bond order `: 1/2 (1 - 0) = 1 /2`
`He_2^` molecule: Total no. of elect rons = 4
Arrangement `:sigma_(1s)^2 , sigma_(1s)^(ast 2)`
Bond order : `1/2 (2 - 2) = 0`
`He_2` molecule does not exist.
`He_2^(+)` molecule : Total no. of electrons = 3
Arrangement : `sigma_(1s)^2 , sigma_(1s)^(ast1)`
Bond order `: 1/2 (2 - 1) = 1 /2`
So `He_2^(+)` exists and has been detected in discharge tubes.
`Li_2` molecule :Total no. of electrons = 6
Arrangement : `sigma_(1s)^2 , sigma_(1s)^(ast 2) , sigma_(2s)^2`
Bond order: `1/2 (4- 2) = 1`
No unpaired e's so diamagnetic
`Be_2` molecule : Total no. of electrons = 8
Arrangement : `sigma_(1s)^2 , sigma_(1s)^(ast2) , sigma_(2s)^2 , sigma_(2s)^(ast 2)`
Bond order: `1/2 (4- 4) = 0`
No unpaired `e^(-)`'s so diamagnetic.
`B_2` molecule: Total no. of electrons = 10
Arrangemen t : `sigma_(1s)^2 , sigma_(1s)^(ast2) , sigma_(2s)^2 , sigma_(2s)^(ast 2) pi_(2p_y)^(1) pi_(2p_z)^(1)`
Bond order `: 1/2 (6 - 4) = 1`
It is paramagnetic
`C_2` molecule : Total no. of electrons = 12
Arrangement: `sigma_(1s)^2 , sigma_(1s)^(ast2) , sigma_(2s)^2 , sigma_(2s)^(ast2) pi_(2p_y)^(2) pi_(2p_y)^(2)`
Bond order : `1/2 (4 - 0) = 2`
It is diamagnetic
`N_2` molecule : Total no. of electrons = 14
Arrangement: `sigma_(1s)^2 , sigma_(1s)^(ast2) , sigma_(2s)^2 , sigma_(2s)^(ast2) pi_(2p_y)^(2) pi_(2p_z)^(2) sigma_(2p_x)^2`
Bond order `: 1/2 (6 - 0) = 2`
It is diamagnetic
`O_2` molecule: Total no. of electrons = 16
Arrangement : `sigma_(1s)^2 , sigma_(1s)^(ast 2), sigma_(2s)^2 , sigma_(2s)^(ast2), sigma_(2p_x)^2 pi_(2p_y)^(2) pi_(2p_z)^(2) pi_(2p_y)^(ast2) pi_(2p_z)^(ast2)`
Bond order `: 1/2 (6 - 2) = 2`
It is paramagnetic
`F_2` molecule : Total no. of electrons = 18
Arrangement : `sigma_(1s)^2 , sigma_(1s)^(ast2) , sigma_(2s)^2 , sigma_(2s)^(ast2), sigma_(2p_x)^2 pi_(2p_y)^(2) pi_(2p_z)^(2) pi_(2p_y)^(ast2) pi_(2p_z)^(ast2) sigma_(2p_x)^(ast)`
Bond order `: 1/2 (6 - 4) = 1`