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Question: Amongst \[N{O_3}^ - ,{\text{ }}As{O_3}^{3 - },{\text{ }}C{O_3}^{2 - },{\text{ }}Cl{O_3}^ - ,{\text{ ...

Amongst NO3, AsO33, CO32, ClO3, SO32  and  BO33N{O_3}^ - ,{\text{ }}As{O_3}^{3 - },{\text{ }}C{O_3}^{2 - },{\text{ }}Cl{O_3}^ - ,{\text{ }}S{O_3}^{2 - }\;and\;B{O_3}^{3 - }​, the non-planar species are :

{A:N{O_3}^ - ,{\text{ }}C{O_3}^{2 - },{\text{ }}and{\text{ }}B{O_3}^{3 - }} \\\ {B:{\text{ }}S{O_3}^{2 - },{\text{ }}Cl{O_3}^ - ,{\text{ }}and{\text{ }}As{O_3}^{3 - }} \\\ {C:{\text{ }}C{O_2}^{3 - },{\text{ }}S{O_3}^{2 - }{,^{}}andB{O_3}^{3 - }} \\\ {D:{\text{ }}As{O_3}^{3 - },{\text{ }}C{O_3}^{2 - },{\text{ }}and{\text{ }}S{O_3}^{2 - }} \end{array}$$
Explanation

Solution

A planar molecule refers to a molecule with all its atoms lying in one two-dimensional plane. Molecules having linear, square or triangular geometries are said to be planar molecules. In other cases, molecules may form three-dimensional shapes, like tetrahedrons, bipyramids or octahedrons.

Complete step by step answer:
First of all you have to find the hybridization of molecules. You can remember following rules to decide whether a molecule is planar or non-planar:

  1. The molecule is not considered to be planar If there exists a sp3sp_3 hybridized carbon (or nitrogen).
  2. The molecule is considered to be planar if there is no sp3sp_3hybridized carbon (or nitrogen), but there is one sp2sp_2 h hybridized carbon or nitrogen atom.
  3. The molecule is not considered to be planar if there is no sp3 hybridized atom but there are two sp2sp_2 hybridized atoms which are separated by an even number of double bonds.
    We can say that the molecule is not planar if there exists either a sp3sp_3 hybridized atom or two sp2sp_2 hybridized atoms that are being separated by even numbers of double bonds. In other cases, its structure is planar. For the present scenario, consider the following table:
SpeciesHybridizationGeometryNo. of lone pairs
NO3-sp20
AsO33-sp31
CO32-sp20
ClO3-sp31
SO32-sp31
BO33-sp20

NO3, CO32and BO33N{O_3}^ - ,{\text{ }}C{O_3}^{2 - } and {\text{ }}B{O_3}^{3 - } possess sp2sp_2 hybridisation and thus, are planar species.

Hence, the correct answer is Option B.

Note:
You can remember that in order to find the hybridization of a molecule, count the valence electrons of each atom in a compound (fact to be noted is that count valence electron of hydrogen = 7). Add them all and divide the number by 8. Quotient represents bond pair while remainder2\dfrac{{remainder}}{2} represents lone pair. Add quotient and remainder2\dfrac{{remainder}}{2} and assume it as x. If (i) x is 3, molecule is planar with sp2 hybridisation, (ii) x is 2, molecule is linear with sp hybridization and x is 4, molecule is non-planar with sp3 hybridization.