Q. In which of the following,bond order increases and magnetic behaviour changes? and why? a) `N_2 -> N_2^+` b)`C_2->C_2^+` c)`NO->NO^+` d) `O_2->O_2^+`

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llltkl | College Teacher | (Level 3) Valedictorian

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The bond order and magnetic behaviour of small molecules and molecular ions can be elegantly traced with the help of molecular orbital (M.O.) diagrams.

a) For `N_2` (14),

The M.O. picture is `sigma1s^2,sigma'1s^2,sigma2s^2,sigma'2s^2,pi2p_y^2,pi2p_z^2,sigma2p_x^2,pi'2p_y,pi'2p_z,sigma'2p_x`Number of unpaired electrons =0, magnetic character: diamagnetic

Bond order=(Number of electrons in bonding MO-Number of electrons in antibonding MO)/2

=(10-4)/2=3

For `N_2^+` (13),

The M.O. picture is` sigma1s^2,sigma'1s^2,sigma2s^2,sigma'2s^2,pi2p_y^2,pi2p_z^2,sigma2p_x^1,pi'2p_y,pi'2p_z,sigma'2p_x`

Number of unpaired electrons =1, magnetic character: 1e-paramagnetic

Bond order=(Number of electrons in bonding MO-Number of electrons in antibonding MO)/2

=(9-4)/2=2.5

Hence, bond order increases and magnetic behaviour changes.

b) For `C_2` (12),

The M.O. picture is `sigma1s^2,sigma'1s^2,sigma2s^2,sigma'2s^2,pi2p_y^2,pi2p_z^2,sigma2p_x,pi'2p_y,pi'2p_z,sigma'2p_x`

Number of unpaired electrons =0, magnetic character: diamagnetic

Bond order=(Number of electrons in bonding MO-Number of electrons in antibonding MO)/2

=(8-4)/2=2

For `C_2^+` (11),

The M.O. picture is `sigma1s^2,sigma'1s^2,sigma2s^2,sigma'2s^2,pi2p_y^2,pi2p_z^1,sigma2p_x,pi'2p_y,pi'2p_z,sigma'2p_x`

Number of unpaired electrons =1, magnetic character: 1e-paramagnetic

Bond order=(Number of electrons in bonding MO-Number of electrons in antibonding MO)/2

=(7-4)/2=1.5

Hence, bond order decreases and magnetic behaviour changes.

c) For NO (15),

The M.O. picture is `sigma1s^2,sigma'1s^2,sigma2s^2,sigma'2s^2,pi2p_y^2,pi2p_z^2,sigma2p_x^2,pi'2p_y^1,pi'2p_z,sigma'2p_x`

Number of unpaired electrons =1, magnetic character: 1e-paramagnetic

Bond order=(Number of electrons in bonding MO-Number of electrons in antibonding MO)/2

=(10-5)/2=2.5

For `NO^+` (14),

The M.O. picture is `sigma1s^2,sigma'1s^2,sigma2s^2,sigma'2s^2,pi2p_y^2,pi2p_z^2,sigma2p_x^2,pi'2p_y,pi'2p_z,sigma'2p_x`

Number of unpaired electrons =0, magnetic character: diamagnetic

Bond order=(Number of electrons in bonding MO-Number of electrons in antibonding MO)/2

=(10-4)/2=3

Hence, bond order increases and magnetic behaviour changes.

d) For `O_2` (16),

The M.O. picture is `sigma1s^2,sigma'1s^2,sigma2s^2,sigma'2s^2,sigma2p_x^2,pi2p_y^2,pi2p_z^2,pi'2p_y^1,pi'2p_z^1,sigma'2p_x`

Number of unpaired electrons =2, magnetic character: 2e-paramagnetic

Bond order=(Number of electrons in bonding MO-Number of electrons in antibonding MO)/2

=(10-6)/2=2

For `O_2^+` (15),

The M.O. picture is `sigma1s^2,sigma'1s^2,sigma2s^2,sigma'2s^2,sigma2p_x^2,pi2p_y^2,pi2p_z^2,pi'2p_y^1,pi'2p_z,sigma'2p_x`

Number of unpaired electrons =1, magnetic character: 1e-paramagnetic

Bond order=(Number of electrons in bonding MO-Number of electrons in antibonding MO)/2

=(10-5)/2=2.5

Hence, bond order increases and magnetic behaviour changes.

Therefore, correct answers are options a), c) and d).

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llltkl | College Teacher | (Level 3) Valedictorian

Posted on

I'm sorry!

In a), the bond order actually decreases from 3 to 2.5.

So, exclude this option.

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