
The bond order of \[O_{2}^{+}\] is:
A. 1
B. 1.5
C. 2.5
D. 3
Answer
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Hint: To discover the bond request of \[O_{2}^{+}\] we can utilize the idea of Molecular Orbital Theory. In this strategy we need to include the quantity of atoms in the Bonding Molecular Orbital and the Antibonding Molecular Orbital.
Complete step-by-step answer:
During the 1s shell there are 2 electrons in both BMO and ABMO. Likewise, during the 2s shell there are two electrons in both the BMO and ABMO. In any case, in the 2p shell 6 electrons in the BMO and 1 electron in the ABMO. In this procedure we should utilize this recipe:
(no. of electrons in BMO - no. of electrons in ABMO)/2
There are 10 electrons in BMO and 5 electrons in ABMO.
\[\dfrac{\left( 10-5 \right)}{2}=2.5\]
So, the bond request of \[O_{2}^{+}\] is \[2.5\].
you can elude the photos underneath. In any case, before the survey, ensure that you fill the BMO first since it has less vitality contrasted with ABMO. Yet, in 2p it invests in light of the fact that there are in excess of 14 electrons.
The higher the security request, the more grounded the draw between the two particles and the shorter the security length. The nonpartisan oxygen is paramagnetic as indicated by MO hypothesis since it winds up with two unpaired electrons in two ruffian pi antibonding sub-atomic orbitals.
The correct answer is C.
Note: The other two are paramagnetic on the grounds that they have an odd number of electrons so it doesn't make a difference what sort of holding they are engaged with, the electrons can't be totally matched up.
Complete step-by-step answer:
During the 1s shell there are 2 electrons in both BMO and ABMO. Likewise, during the 2s shell there are two electrons in both the BMO and ABMO. In any case, in the 2p shell 6 electrons in the BMO and 1 electron in the ABMO. In this procedure we should utilize this recipe:
(no. of electrons in BMO - no. of electrons in ABMO)/2
There are 10 electrons in BMO and 5 electrons in ABMO.
\[\dfrac{\left( 10-5 \right)}{2}=2.5\]
So, the bond request of \[O_{2}^{+}\] is \[2.5\].
you can elude the photos underneath. In any case, before the survey, ensure that you fill the BMO first since it has less vitality contrasted with ABMO. Yet, in 2p it invests in light of the fact that there are in excess of 14 electrons.
The higher the security request, the more grounded the draw between the two particles and the shorter the security length. The nonpartisan oxygen is paramagnetic as indicated by MO hypothesis since it winds up with two unpaired electrons in two ruffian pi antibonding sub-atomic orbitals.
The correct answer is C.
Note: The other two are paramagnetic on the grounds that they have an odd number of electrons so it doesn't make a difference what sort of holding they are engaged with, the electrons can't be totally matched up.
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