Question

why is B2 paramagnetic while C2 is diamagnetic?

Answer 1

B2 has two unpaired electron so it is paramagnetic whereas C2 has only paired electrons so it is diamagnetic.

Answer 2

Answer : $B_2$ is paramagnetic while $C_2$ is diamagnetic because $B_2$ has two number of unpaired electrons.

Explanation :

According to the molecular orbital theory, the general molecular orbital configuration of boron and carbon 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 5 electrons present in boron and 6 electrons present in carbon.

(i) The number of electrons present in $B_2$ molecule = 2(5) = 10

The molecular orbital configuration of $B_2$ molecule will be,

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

The number of unpaired electron in $B_2$ molecule are, 2. So, this is paramagnetic. That means, more the number of unpaired electrons, more paramagnetic

(ii) The number of electrons present in $C_2$ molecule = 2(6) = 12

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 number of unpaired electron in $C_2$ molecule is zero. So, this is diamagnetic.

$B_2$ is paramagnetic while $C_2$ is diamagnetic because $B_2$ has two number of unpaired electrons.