Explanation
Which one of the following molecular geometries (i.e. shapes) is not possible for the $$sp^{3}d^{2}$$ hybridization?
Which of the following are isostructural ?
(I)$$NO_{3}^{-}$$
(II)$$CO_{3}^{2-}$$
(III)$$CIO_{3}^{-}$$
(IV)$$SO_{3}$$
(V)$$ XeO_{3}$$
$$NO_2$$
The total number of valence electrons is $$5+2(6)=17$$
There is the persistent radical character on nitrogen because it has an unpaired electron.
$$ClO_2$$
A Lewis structure shows $$ClO_2$$ has a total of $$2\times 6+7=19$$ valence electrons, or 9 electron pairs and a lone electron.
Cl is the central atom.
The molecule $$ClO_2$$ has an odd number of valence electrons.
$$NO$$
14 valence electrons in your trial structure.
The valence electrons you have available are: $$1 N + 1 O = 1\times 5 + 1\times 6 = 11$$.
The molecule $$NO$$, $$NO_2$$ and $$ClO_2$$ has an odd number of valence electrons.
Hint:
Mathematically, the product of the magnitude of the charge and the distance between the nucleus of both the charge is termed as dipole moment.
Explanation:
Final Answer:
Thus, from the resonant structures we conclude that the overall dipole moment is maximum in compound C
So, the correct answer is option C.
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