To determine the shape of a molecule or ion, we need to find the hybridization of the central atom and the number of lone pairs and bond pairs around it. The VSEPR theory helps predict the geometry.

The central atom in H₃O⁺ is oxygen. Oxygen has 6 valence electrons. It forms three single bonds with three hydrogen atoms and carries a +1 charge. The number of electrons involved in bonding is 3. The number of non-bonding electrons (lone pairs) is $(6-3-1)/2=1$. So, there are 3 bond pairs and 1 lone pair around the central oxygen atom. This corresponds to a steric number of 4 (3 bond pairs + 1 lone pair), leading to sp³ hybridization. Due to the presence of one lone pair, the molecular geometry is distorted from tetrahedral to pyramidal.

• A) BF₃: The central atom is boron. Boron has 3 valence electrons and forms three single bonds with fluorine atoms. There are no lone pairs on boron. This corresponds to a steric number of 3 (3 bond pairs + 0 lone pairs), leading to sp² hybridization and a trigonal planar shape.
• B) H₃O⁺: As explained above, it has 3 bond pairs and 1 lone pair, resulting in a pyramidal shape.
• C) NO₃^-: The central atom is nitrogen. Nitrogen has 5 valence electrons. It forms three bonds with oxygen atoms (one double bond and two single bonds, which resonate). The overall charge is -1. The number of non-bonding electrons (lone pairs) is $(5-3-(-1))/2=0$. So, there are 3 bond pairs and 0 lone pairs around the central nitrogen atom. This corresponds to a steric number of 3, leading to sp² hybridization and a trigonal planar shape.
• D) CO₃^2-: The central atom is carbon. Carbon has 4 valence electrons. It forms three bonds with oxygen atoms (one double bond and two single bonds, which resonate). The overall charge is -2. The number of non-bonding electrons (lone pairs) is $(4-3-(-2))/2=0$. So, there are 3 bond pairs and 0 lone pairs around the central carbon atom. This corresponds to a steric number of 3, leading to sp² hybridization and a trigonal planar shape.

Correct Answer: (B)