Question

Match the species in Column I with the geometry/shape in Column II.

Column I	Column II
(i) \[\ce{H3O+}\]	(a) Linear
(ii) \[\ce{HC ≡ CH}\]	(b) Angular
(iii) \[\ce{ClO^{-}2}\]	(c) Tetrahedral
(iv) \[\ce{NH^{+}4}\]	(d) Trigonal bipyramidal
	(e) Pyramidal

Answer 1

Column I	Column II
(i) \[\ce{H3O+}\]	(e) Pyramidal
(ii) \[\ce{HC ≡ CH}\]	(a) Linear
(iii) \[\ce{ClO^{-}2}\]	(b) Angular
(iv) \[\ce{NH^{+}4}\]	(c) Tetrahedral

Explanation:

(i) The shape of \[\ce{H3O+}\] is pyramidal this structure has the total 8 valence electrons, six in oxygen atom, three in each hydrogen atoms and one electron is lost due to positive charge. Out of five valence electrons in oxygen three form a sigma bond with the hydrogen atom and a lone pair is left on the oxygen atom. Although the four electron pairs make a tetrahedral geometry but to the lone pair of electrons, lone-pair bond-pair repulsion occurs which distorts the shape and makes it pyramidal.

(ii) The ethyne molecule has a linear geometry, because the two carbon atoms make two pi bonds with each other and one sigma bond with the hydrogen atom. The carbon and hydrogen atoms are spsp hybridized.

(iii) \[\ce{ClO^{-}2}\] has 20 valence electrons in its structure. Chlorine atoms make one pi-bond and two sigma bonds with the oxygen atoms. The pi-bond is in resonance and hence \[\ce{ClO^{-}2}\] has two resonating structures. Two lone pairs of electrons are left on the oxygen atom and hence, due to lone pair-lone pair repulsion the bond angle reduces and the shape becomes linear.

(iv) \[\ce{NH^{+}4}\] has the total of 8 valence electrons. Due to the presence of positive charge, nitrogen has 4 valence electrons and hydrogen has one in every four hydrogen atoms. The four valence electrons in nitrogen make sigma bonds with the four hydrogen atoms, thereby making a tetrahedral geometry and having sp³ hybridized orbitals.