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