Explanation
Hint: Bond order can be calculated by using molecular orbital theory.
Step 1: Determination of Bond Order :
Total number of electrons in $$O_2=16$$ but due to a positive charge one electron is deducted and thus,
the total number of electrons in $${O_2}^+ = 15$$
Molecular orbital configuration can be written as,
$$(\sigma 1s)^2 <(\sigma\ast 1s)^2< (\sigma 2s)^2< (\sigma\ast 2s)^2< (\sigma 2pz)^2< (\pi 2px)^2 \approx (\pi 2py)^2< (\pi\ast 2px)^1 \approx (\pi\ast 2py)^0 $$
$$\ast \ -$$ mark represent the electron is present in anti-bonding molecular orbital.
Bond order can be defined as:
$$Bond\ Order = \dfrac{Number\ of\ electrons\ in\ B.M.O\ –\ Number\ of\ electrons\ in\ A.B.M.O.}{2} $$
( where, B.M.O. - Bonding Molecular Orbital, A.B.M.O - Anti-bonding Molecular Orbital )
$${O_2}^+$$ has $$10$$ bonding electrons and $$5$$ antibonding electrons.
So, the bond order is $$\dfrac{10-5}{2}=2.5$$.
Final Step: Correct Answer - (C) $$2.5$$.
A) Para-dimethyl benzene is non polar because, he individual bond dipoles cancel each other.
On the other hand, para-dihydroxy benzene is polar because the dipoles of two O-H bonds do not cancel each other.
Hence, option A is correct.
(B) Ammonia has higher dipole moment than NF₃ because in ammonia, the bond dipoles and the dipole due to lone pair reinforce each other whereas in NF₃ they oppose each other.
Hence, option B is correct.
(C) The direction of the dipole moment in CO is from oxygen to carbon as oxygen is more electronegative.
Hence, option C is incorrect.
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