Question

Write the molecular orbital electronic configuration for carbon molecule. Calculate its bond order and comment on its magnetic property

Answer 1

Bond order is 2 and molecule is diamagnetic

Explanation:

According to the molecular orbital theory, the general molecular orbital configuration will be,

$(\sigma_{1s}),(\sigma_{1s}^*),(\sigma_{2s}),(\sigma_{2s}^*),[(\pi_{2p_x})=(\pi_{2p_y})],(\sigma_{2p_z}),[(\pi_{2p_x}^*)=(\pi_{2p_y}^*)],(\sigma_{2p_z}^*)$

As there are 12 electrons present in carbon molecule.

The molecular orbital configuration of $C_2$ molecule will be,

$(\sigma_{1s})^2,(\sigma_{1s}^*)^2,(\sigma_{2s})^2,(\sigma_{2s}^*)^2,[(\pi_{2p_x})^2=(\pi_{2p_y})^2],(\sigma_{2p_z})^0,[(\pi_{2p_x}^*)^0=(\pi_{2p_y}^*)^0],(\sigma_{2p_z}^*)^0$

The formula of bond order = $\frac{1}{2}\times (\text{Number of bonding electrons}-\text{Number of anti-bonding electrons})$

Bond order of $C_2=\frac{1}{2}\times(8-4)=2$

As all the electrons are paired, the molecule is diamagnetic.

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Answer 2

Answer:

Write the molecular orbital electronic configuration for carbon molecule. Calculate its bond order and comment on its magnetic property

Explanation:

Write the molecular orbital electronic configuration for carbon molecule. Calculate its bond order and comment on its magnetic property