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Chemistry Question on Molecular Orbital Theory

Number of molecules having bond order 2 from the following molecule is ...................
C2,O2,Be2,Li2,Ne2,N2,He2C_2, O_2, Be_2, Li_2, Ne_2, N_2, He_2

Answer

The bond order (B.O.) is calculated using the formula:
B.O.=Number of bonding electrons−Number of antibonding electrons2.\text{B.O.} = \frac{\text{Number of bonding electrons} - \text{Number of antibonding electrons}}{2}.
{C2_2:}
(12e−):σ1s2,σ1s∗2,σ2s2,σ2s∗2,[π2px=π2py]4.(12e^-) : \sigma_{1s}^2, \sigma_{1s}^{{\ast}2}, \sigma_{2s}^2, \sigma_{2s}^{{\ast}2}, [\pi_{2px} = \pi_{2py}]^4.
B.O.=8−42=2.\text{B.O.} = \frac{8 - 4}{2} = 2.
{O2_2:} (16e−):σ1s2,σ1s∗2,σ2s2,σ2s∗2,σ2pz2,[π2px=π2py]4,[π2px∗=π2py∗]2.(16e^-) : \sigma_{1s}^2, \sigma_{1s}^{{\ast}2}, \sigma_{2s}^2, \sigma_{2s}^{{\ast}2}, \sigma_{2p_z}^2, [\pi_{2px} = \pi_{2py}]^4, [\pi_{2px}^{{\ast}} = \pi_{2py}^{{\ast}}]^2.
B.O.=10−62=2.\text{B.O.} = \frac{10 - 6}{2} = 2.
{Be2_2:} (8e−):σ1s2,σ1s∗2,σ2s2,σ2s∗2.(8e^-) : \sigma_{1s}^2, \sigma_{1s}^{{\ast}2}, \sigma_{2s}^2, \sigma_{2s}^{{\ast}2}.
B.O.=4−42=0.\text{B.O.} = \frac{4 - 4}{2} = 0.
{Li2_2:} (6e−):σ1s2,σ1s∗2,σ2s2.(6e^-) : \sigma_{1s}^2, \sigma_{1s}^{{\ast}2}, \sigma_{2s}^2.
B.O.=4−22=1.\text{B.O.} = \frac{4 - 2}{2} = 1.
{Ne2_2:} (20e−):σ1s2,σ1s∗2,σ2s2,σ2s∗2,σ2pz2,[π2px=π2py]4,[π2px∗=π2py∗]4,σ2pz∗2.(20e^-) : \sigma_{1s}^2, \sigma_{1s}^{{\ast}2}, \sigma_{2s}^2, \sigma_{2s}^{{\ast}2}, \sigma_{2p_z}^2, [\pi_{2px} = \pi_{2py}]^4, [\pi_{2px}^{{\ast}} = \pi_{2py}^{{\ast}}]^4, \sigma_{2p_z}^{{\ast}2}.
B.O.=10−102=0.\text{B.O.} = \frac{10 - 10}{2} = 0.
{N2_2:} (14e−):σ1s2,σ1s∗2,σ2s2,σ2s∗2,π2px2,π2py2,σ2pz2.(14e^-) : \sigma_{1s}^2, \sigma_{1s}^{{\ast}2}, \sigma_{2s}^2, \sigma_{2s}^{{\ast}2}, \pi_{2px}^2, \pi_{2py}^2, \sigma_{2p_z}^2.
B.O.=10−42=3.\text{B.O.} = \frac{10 - 4}{2} = 3.
{H2_2:} (2e−):σ1s2.(2e^-) : \sigma_{1s}^2.
B.O.=2−02=1.\text{B.O.} = \frac{2 - 0}{2} = 1.
Molecules with bond order 2: C2_2 and O2_2.

Explanation

Solution

The bond order (B.O.) is calculated using the formula:
B.O.=Number of bonding electrons−Number of antibonding electrons2.\text{B.O.} = \frac{\text{Number of bonding electrons} - \text{Number of antibonding electrons}}{2}.
{C2_2:}
(12e−):σ1s2,σ1s∗2,σ2s2,σ2s∗2,[π2px=π2py]4.(12e^-) : \sigma_{1s}^2, \sigma_{1s}^{{\ast}2}, \sigma_{2s}^2, \sigma_{2s}^{{\ast}2}, [\pi_{2px} = \pi_{2py}]^4.
B.O.=8−42=2.\text{B.O.} = \frac{8 - 4}{2} = 2.
{O2_2:} (16e−):σ1s2,σ1s∗2,σ2s2,σ2s∗2,σ2pz2,[π2px=π2py]4,[π2px∗=π2py∗]2.(16e^-) : \sigma_{1s}^2, \sigma_{1s}^{{\ast}2}, \sigma_{2s}^2, \sigma_{2s}^{{\ast}2}, \sigma_{2p_z}^2, [\pi_{2px} = \pi_{2py}]^4, [\pi_{2px}^{{\ast}} = \pi_{2py}^{{\ast}}]^2.
B.O.=10−62=2.\text{B.O.} = \frac{10 - 6}{2} = 2.
{Be2_2:} (8e−):σ1s2,σ1s∗2,σ2s2,σ2s∗2.(8e^-) : \sigma_{1s}^2, \sigma_{1s}^{{\ast}2}, \sigma_{2s}^2, \sigma_{2s}^{{\ast}2}.
B.O.=4−42=0.\text{B.O.} = \frac{4 - 4}{2} = 0.
{Li2_2:} (6e−):σ1s2,σ1s∗2,σ2s2.(6e^-) : \sigma_{1s}^2, \sigma_{1s}^{{\ast}2}, \sigma_{2s}^2.
B.O.=4−22=1.\text{B.O.} = \frac{4 - 2}{2} = 1.
{Ne2_2:} (20e−):σ1s2,σ1s∗2,σ2s2,σ2s∗2,σ2pz2,[π2px=π2py]4,[π2px∗=π2py∗]4,σ2pz∗2.(20e^-) : \sigma_{1s}^2, \sigma_{1s}^{{\ast}2}, \sigma_{2s}^2, \sigma_{2s}^{{\ast}2}, \sigma_{2p_z}^2, [\pi_{2px} = \pi_{2py}]^4, [\pi_{2px}^{{\ast}} = \pi_{2py}^{{\ast}}]^4, \sigma_{2p_z}^{{\ast}2}.
B.O.=10−102=0.\text{B.O.} = \frac{10 - 10}{2} = 0.
{N2_2:} (14e−):σ1s2,σ1s∗2,σ2s2,σ2s∗2,π2px2,π2py2,σ2pz2.(14e^-) : \sigma_{1s}^2, \sigma_{1s}^{{\ast}2}, \sigma_{2s}^2, \sigma_{2s}^{{\ast}2}, \pi_{2px}^2, \pi_{2py}^2, \sigma_{2p_z}^2.
B.O.=10−42=3.\text{B.O.} = \frac{10 - 4}{2} = 3.
{H2_2:} (2e−):σ1s2.(2e^-) : \sigma_{1s}^2.
B.O.=2−02=1.\text{B.O.} = \frac{2 - 0}{2} = 1.
Molecules with bond order 2: C2_2 and O2_2.