Long Answer TypeWhich of the following combinations of orbtials are allowed in LCAO method (considering z-axis to be molecular axis) and sketch the shapes of molecular orbitals formed by their addition and subtraction:
(i) s and pz
(ii) px and px
(iii) pz and py
(iv) s and px.
Short Answer TypeGive the number of electrons which occupy the bonding molecular orbital in H2+ H2 and He2.
What do you understand by bond order? How is it related with bond length and bond energy? Explain on the basis of bond order that He2 molecule does not exist.
Long Answer TypeUsing LCAO method for the formation of molecular orbitals in case of homonuclear diatomic hydrogen molecule.
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.
.
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 
2px and 
2py MO contains unpaired electron, therefore B2 molecule is paramagnetic.
Short Answer Type
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