Molecular orbital
source : chemed.chem.purdue.edu
Relationship between Electronic configuration and Molecular behaviour :
1) Bond order :
It is defined as the number of covalent bonds between the two combining atoms of a molecule.
Bond order = 0.5 (Nb – Na)
Nb = number of bonding electrons or number of electrons in bonding M.O’s
Na = number of antibonding electrons
2) Stability of molecule :
It is determined by bond order.Higher is the bond order greater is the stability of molecule.
i) If Nb > Na, then molecule will be stable.
ii) If Nb=Na, then formation of the molecule will not take place.
iii) If Nb< Na, then molecule will be unstable.
3) Magnetic property :
A molecule in which all the electrons are paired, is called diamagnetic while molecule which has one or more unpaired electron is called paramagnetic.
Molecular orbital diagram of H2 (Hydrogen molecule) :
Number of electrons in H2 = 2
Electronic configuration of H2 = σ1s2
Bond order = 0.5 (Nb-Na)
Nb=2
Na=0
B.O = 0.5 (2-0)
B.O= 1
M.O. diagram of He2 molecule :
2He = 1s2
Electronic configuration of He2 = σ1s2, σ*1s2
Nb =2, Na=2
B.O =0.5(0-0)
B.O=0 (Zero)
Molecules having zero bond order do not exist.
Molecular orbital diagram of H2+ (Hydrogen molecule ion) :
H2+ = σ1s1
B.O =0.5 (Nb- Na)
Nb =1 , Na = 0
B.O = 0.5(1-0)
B.O =0.5
Molecular orbital diagram of Li2 & Be2 :
Number of electrons in Li2 molecule =6
Li2 = σ1s2,σ*1s2,σ2s2
Nb=4, Na=2
B.O =0.5(Nb- Na)
B.O = 0.5 (4-2)
B.O = 1
Number of electrons in Be2 molecule = 8
Be2 = σ1s2,σ*1s2,σ2s2,σ*2s2
Nb=4, Na=4
B.O =0.5(Nb- Na)
B.O = 0.5 (4 – 4)
B.O = 0
source : isite.lps.org
Molecular orbital diagram of C2 molecule :
Number of electrons in C2 molecule = 12
C2 = σ1s2,σ*1s2,σ2s2,σ*2s2, π2py2= π2pz2 ,σ2px0
Nb=8, Na=4
B.O =0.5(Nb- Na)
B.O = 0.5 (8-4)
B.O =2
Molecular orbital diagram of N2 molecule :
7N = 1s2,2s2,2p3
Number of electrons in N2 molecule = 14
N2 = σ1s2,σ*1s2,σ2s2,σ*2s2,π2py2 = π2pz2,σ2px2,
Nb=10, Na=4
B.O =0.5(Nb- Na)
B.O = 0.5 (10-4)
B.O =3
Molecular orbital diagram of O2 molecule :
8O – 1s2 ,2s2,2p4
Number of electrons in O2 molecule = 16
Electronic configuration of O2:
σ1s2,σ*1s2,σ2s2,σ*2s2,σ2px2,π2py2= π2pz2,π*2py1=π*2pz1,σ*2px0.
Nb= 10
Na =6
B.O = 0.5 (10-6)=2
O2 molecule is paramagnetic because two unpaired electrons are present. One unpaired in π*2py1 and one in π*2pz1.
Molecular Orbital diagram of O2– ion :
This is superoxide ion. It has 8+8+1=17 electrons. The M.O configuration is O2– is
σ1s2,σ*1s2,σ2s2,σ*2s2,σ2px2,π*2py2= π2pz2,π*2py2=π*2pz1,σ*2px0.
Nb=10
Na=7
B.O =0.5 (Nb-Na)
B.O=0.5(10-7)
B.O=1.5
Molecular Orbital diagram of O2— ion :
This is peroxide ion. It has 8+8+2 =18 electrons. The M.O configuration is:
O2– –— = σ1s2,σ*1s2,σ2s2,σ*2s2,σ2px2, π2py2= π2pz2,π*2py2=π*2pz2,σ*2px0.
Nb=10
Na=8
B.O =0.5 (Nb-Na)
B.O=0.5(10-8)
B.O= 1
Molecular Orbital diagram of Hetero nuclear diatomic molecule :
Molecular Orbital diagram of Carbon monoxide molecule (CO):
Total electrons:6+8 =14
CO =σ1s2,σ*1s2,σ2s2,σ*2s2, σ2px2, π2py2= π2pz2.
Nb=10
Na=4
B.O =0.5 (Nb-Na)
B.O=0.5(10-4)
B.O= 3
Molecular Orbital diagram of NO(nitric oxide)molecule :
Electronic configuration of NO: σ1s2,σ*1s2,σ2s2,σ*2s2,σ2px2,π2py2= π2pz2, π*2py1=π*2pz0,
Number of electrons =7+8 =15
Nb=10
Na=5
B.O =0.5 (Nb-Na)
B.O=0.5(10-5)
B.O =2.5
