Draw the molecular orbital diagram for:
(i) Be₂
(ii) B₂ and predict bond order and magnetic properties.

(i) Be2 molecule: The electronic configuration of Be(Z = 4) is:
₄ Be 1s² 2s¹
Be₂ molecule is formed by the overlap of atomic orbitals of both beryllium atoms.
Number of valence electrons in Be atom = 2
Thus in the formation of Be₂ molecule, two outer electrons of each Be atom i.e. 4 in all, have to be accommodated in various molecular orbitals in the increasing order of their energies.
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The molecular orbital electronic configuration,

Magnetic property: Since bond order is zero, Be₂ molecule does not exist. It is diamagnetic due to the absence of any unpaired electron.
B₂ molecule: The electronic configuration of B atom (Z = 5) is

B₂ molecule is formed by the overlap of atomic orbitals of both boron atoms. A number of valence electrons of each boron atom = 3.
In the formation of B₂ molecule, three valence electrons of each boron atom i.e. 6 in all, have to be accommodated in various molecular orbitals in the increasing order of their energies.

MO electronic configuration:

Bond order:  Here Nb = 4,   Na = 2
Bond order  = 
The two boron atom is B2 molecules are linked by one covalent bond.
Magnetic properties: Since each 2p_x and 2p_y MO contains unpaired electron, therefore B₂ molecule is paramagnetic.