The increasing order of energies of various molecular orbitals of \[{N_2}\] is given below:
\[\sigma 1s < \sigma *1s < \sigma 2s < \sigma *2s < \pi 2{p_x} = \pi 2{p_y} < \sigma 2{p_z} < \pi *2{p_x} = \pi *2{p_y} < \sigma *2{p_z}\]
The above sequence is not true for the molecule:
A.\[{C_2}\]
B.\[{B_2}\]
C.\[{O_2}\]
D.\[B{e_2}\]
Answer
618.6k+ views
Hint:We know that for all elements which have atomic number more than 7, that is beyond nitrogen (\[{N_2}\]), then the energy of \[\sigma 2{p_z}\] is lower than \[\pi 2{p_x}\] and \[\pi 2{p_y}\] orbitals. So, the sequence given in the question is true for \[{N_2}\] and lower molecules.
Complete answer:
There are basically two different ways of filling the electrons, the first one is for the elements that have atomic number less than or equal to 7 i.e. till nitrogen, and the second one is for the elements that have atomic number more than 7.
For the homonuclear diatomic molecules of second row elements like \[L{i_2}\], \[B{e_2}\], \[{B_2}\], \[{C_2}\], \[{N_2}\], the \[\sigma 2{p_z}\]molecular orbitals is higher in energy than \[\pi \;2{p_x}\] and \[\pi {\text{ }}2{p_y}\] molecular orbitals
For these atoms, i.e., \[{N_2}\]and lower molecules, the order is:
\[\sigma 1s\], \[\sigma *1s\],\[\;\sigma 2s\] , \[\sigma *2s\], \[\left[ {\pi 2{p_x}\; = \pi 2{p_y}} \right]\],\[\sigma 2{p_z}\], \[\left[ {\pi *2{p_x} = \pi *2{p_y}} \right]\], \[\sigma *2{p_z}\]
Here in the options, only \[{O_2}\] is of atomic number more than 7,
The above sequence is applicable for atoms those have atomic no. less than or equal to 7, So, oxygen is correct. The correct order of energy of various molecular orbitals is as follows:
For\[{O_2}\] and the higher molecules:
\[\sigma 1s\], \[\sigma *1s\],\[\;\sigma 2s\] , \[\sigma *2s\], \[\sigma 2{p_z}\], \[\left[ {\pi 2{p_x}\; = \pi 2{p_y}} \right]\], \[\left[ {\pi *2{p_x} = \pi *2{p_y}} \right]\], \[\sigma *2{p_z}\]
Therefore, the correct answer is option (C).
Note:
The factors upon which relative energies of molecular orbitals depend are the energies of the atomic orbitals combining to form molecular orbitals and the extent of overlapping between the atomic orbitals. The greater the overlap, the more the bonding orbital is lowered and the antibonding orbital is raised in energy relative to atomic orbitals.
Complete answer:
There are basically two different ways of filling the electrons, the first one is for the elements that have atomic number less than or equal to 7 i.e. till nitrogen, and the second one is for the elements that have atomic number more than 7.
For the homonuclear diatomic molecules of second row elements like \[L{i_2}\], \[B{e_2}\], \[{B_2}\], \[{C_2}\], \[{N_2}\], the \[\sigma 2{p_z}\]molecular orbitals is higher in energy than \[\pi \;2{p_x}\] and \[\pi {\text{ }}2{p_y}\] molecular orbitals
For these atoms, i.e., \[{N_2}\]and lower molecules, the order is:
\[\sigma 1s\], \[\sigma *1s\],\[\;\sigma 2s\] , \[\sigma *2s\], \[\left[ {\pi 2{p_x}\; = \pi 2{p_y}} \right]\],\[\sigma 2{p_z}\], \[\left[ {\pi *2{p_x} = \pi *2{p_y}} \right]\], \[\sigma *2{p_z}\]
Here in the options, only \[{O_2}\] is of atomic number more than 7,
The above sequence is applicable for atoms those have atomic no. less than or equal to 7, So, oxygen is correct. The correct order of energy of various molecular orbitals is as follows:
For\[{O_2}\] and the higher molecules:
\[\sigma 1s\], \[\sigma *1s\],\[\;\sigma 2s\] , \[\sigma *2s\], \[\sigma 2{p_z}\], \[\left[ {\pi 2{p_x}\; = \pi 2{p_y}} \right]\], \[\left[ {\pi *2{p_x} = \pi *2{p_y}} \right]\], \[\sigma *2{p_z}\]
Therefore, the correct answer is option (C).
Note:
The factors upon which relative energies of molecular orbitals depend are the energies of the atomic orbitals combining to form molecular orbitals and the extent of overlapping between the atomic orbitals. The greater the overlap, the more the bonding orbital is lowered and the antibonding orbital is raised in energy relative to atomic orbitals.
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