Question

# Four diatomic species are listed, identify the correct order in which the bond order is increasing in them.

A
NO<O2<C22<He+2
B
O2<NO<C22<He+2
C
C22<He+2<O2<NO
D
He+2<O2<NO<C22

Solution

## The correct option is D $$He^+_2 < O^-_2 < NO < C^{2-}_2$$A/c to Molecular orbital energy diagram, bond order is calculated from the formula, $$\dfrac { { N }_{ b }-{ N }_{ a } }{ 2 }$$$${ N }_{ b }=$$ no. of electrons is bonding $$M.O.$$.$${ N }_{ a }=$$ no. of electrons in antibonding $$MO$$.For $${ He }_{ 2 }^{ + }$$ :- $${ \left( \sigma 1S \right) }^{ 2 }{ \left( \sigma \ast 1S \right) }^{ 1 }$$                  $$\Rightarrow \dfrac { 2-1 }{ 2 } =0.5$$For $$NO$$ :- $${ \left( \sigma 2S \right) }^{ 2 }{ \left( \sigma \ast 2S \right) }^{ 2 }{ \left( \pi 2px \right) }^{ 2 }{ \left( \pi 2py \right) }^{ 2 }{ \left( \sigma 2pz \right) }^{ 2 }{ \left( { \pi }^{ \ast }2px \right) }^{ 1 }$$                 $$\dfrac { 8-3 }{ 2 } =2.5$$For $${ O }_{ 2 }^{ - }$$ :- $${ \left( \sigma 2S \right) }^{ 2 }{ \left( { \sigma }^{ \ast }2S \right) }^{ 2 }{ \left( \sigma 2pz \right) }^{ 2 }{ \left( \pi 2px \right) }^{ 2 }{ \left( \pi 2py \right) }^{ 2 }{ \left( { \pi }^{ \ast }2px \right) }^{ 2 }{ \left( { \pi }^{ \ast }2py \right) }^{ 1 }$$                $$\dfrac { 8-5 }{ 2 } =1.5$$For $${ C }_{ 2 }^{ 2- }$$ :- $$\dfrac { 6-0 }{ 2 } =3$$So, the correct option is 'D'.Chemistry

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