Question

pls tell the answer .......

Answer 1

BORON molecule:     B atom: 1s2 2s2  2p1

  2s-2p bonding is absent  means that there is no 2s-2p hybridization to result in sp, sp2,sp3 orbitals.  Consider the atoms linked as two balls connected to the two ends of a rod (interatomic axis).   We look at the molecular orbitals MO, which are bonding orbitals and antibonding orbitals.  First bonding orbitals are filled and then antibonding orbitals are filled.

  When we write the electronic configuration with 12, 2s, 2p bonding and antibonding orbitals combining the electrons in both atoms, we find that the unpaired electrons in BORON  molecules give rise to magnetism.
 
  In other molecules the electrons have spins which cancel out the magnetic effect.

$Be_2:\sigma_{1s}^2 \ \sigma_{1s}^{*\ 2}\sigma_{2s}^2 \ \sigma_{2s}^{*\ 2}\\\\B_2:\sigma_{1s}^2 \ \sigma_{1s}^{*\ 2}\sigma_{2s}^2 \ \sigma_{2s}^{*\ 2}\ \ \pi_{2p_y}^{\ \ 1}\ \ \pi_{2p_z}^{\ \ 1}\\\\C_2:\sigma_{1s}^2 \ \sigma_{1s}^{*\ 2}\sigma_{2s}^2 \ \sigma_{2s}^{*\ 2}\ \ \pi_{2p_y}^{\ \ 2}\ \ \pi_{2p_z}^{\ \ 2}\\\\N_2:\sigma_{1s}^2 \ \sigma_{1s}^{*\ 2}\sigma_{2s}^2 \ \sigma_{2s}^{*\ 2}\ \ \pi_{2p_y}^{\ \ 2}\ \ \pi_{2p_z}^{\ \ 2}\ \ \sigma_{2p}^{\ \ 2}$

We find that B2 has two bonding orbitals in two directions and unpaired electrons.  So it is paramagnetic.  Others are not.

Similarly, O2 molecule has two electrons in antibonding orbitals of 2p.  So O2 is paramagnetic  too.