Among the following species which two have trigonal pyramidal shape? (I) $$NI_3$$ (II) $$I_3^-$$ (III) $$SO_3^{2-}$$ (IV) $$NO_3^-$$

A molecule or ion exhibits a trigonal pyramidal shape when its central atom is $$sp^3$$ hybridized with a steric number of 4, comprising exactly 3 bonding pairs and 1 lone pair ($$\text{AB}_3\text{E}$$ type).

• (I) $$\text{NI}_3$$ (Nitrogen Triiodide)

  • The central nitrogen ($$\text{N}$$) atom belongs to Group 15 and has 5 valence electrons. It uses 3 electrons to form 3 single covalent bonds with iodine atoms, leaving 2 non-bonding electrons.
  • Steric Number: $$3 \text{ bonding pairs} + 1 \text{ lone pair} = 4$$ ($$sp^3$$ hybridization).
  • Shape: Trigonal Pyramidal.



• (II) $$\text{I}_3^-$$ (Triiodide Ion)

  • The central iodine ($$\text{I}$$) atom has 7 valence electrons, plus 1 extra electron from the negative charge, totaling 8. It bonds with 2 surrounding iodine atoms, leaving 6 non-bonding electrons.
  • Steric Number: $$2 \text{ bonding pairs} + 3 \text{ lone pairs} = 5$$ ($$sp^3d$$ hybridization).
  • Shape: Linear.



• (III) $$\text{SO}_3^{2-}$$ (Sulfite Ion)

  • The central sulfur ($$\text{S}$$) atom belongs to Group 16 and has 6 valence electrons, plus 2 extra electrons from the $$2-$$ charge, totaling 8. It uses 6 electrons to form bonds with 3 oxygen atoms, leaving 2 non-bonding electrons.
  • Steric Number: $$3 \text{ bonding pairs} + 1 \text{ lone pair} = 4$$ ($$sp^3$$ hybridization).
  • Shape: Trigonal Pyramidal.



• (IV) $$\text{NO}_3^-$$ (Nitrate Ion)

  • The central nitrogen ($$\text{N}$$) atom has 5 valence electrons, plus 1 from the charge, totaling 6. It shares these to form bonds with 3 oxygen atoms, leaving no lone pairs.
  • Steric Number: $$3 \text{ bonding pairs} + 0 \text{ lone pairs} = 3$$ ($$sp^2$$ hybridization).
  • Shape: Trigonal Planar.

Conclusion:

Both $$\text{NI}_3$$ (I) and $$\text{SO}_3^{2-}$$ (III) have a steric number of 4 with 1 lone pair, giving them identical trigonal pyramidal geometries.

Answer: Option A — I and III