To determine which species contains no lone pair of electrons, we need to draw their Lewis structures and count the lone pairs on the central or constituent atoms.

The hydrogen molecule (H₂) consists of two hydrogen atoms, each with one valence electron. They share these two electrons to form a single covalent bond. Each hydrogen atom achieves a duet configuration.

H — H

There are no lone pairs on either hydrogen atom.

The oxygen molecule (O₂) consists of two oxygen atoms, each with six valence electrons. They form a double bond and each oxygen atom has two lone pairs.

:O=O:

Each oxygen atom has two lone pairs, so O₂ contains lone pairs.

The carbonate ion (CO₃²⁻) has carbon as the central atom. Carbon has 4 valence electrons, and each oxygen has 6 valence electrons. The ⁻2 charge adds 2 electrons. Total valence electrons = $4+3\times 6+2=24$.

In its resonance structures, the central carbon atom forms one double bond and two single bonds with oxygen atoms. The carbon atom has no lone pairs. However, each oxygen atom in the carbonate ion has lone pairs (the double-bonded oxygen has two lone pairs, and the single-bonded oxygens each have three lone pairs).

Thus, CO₃²⁻ contains lone pairs on the oxygen atoms.

In nitrogen trifluoride (NF₃), nitrogen is the central atom with 5 valence electrons, and each fluorine atom has 7 valence electrons. Total valence electrons = $5+3\times 7=26$.

Nitrogen forms three single bonds with three fluorine atoms and has one lone pair. Each fluorine atom has three lone pairs.

Thus, NF₃ contains lone pairs on both nitrogen and fluorine atoms.

Based on the analysis, only H₂ contains no lone pair of electrons.

Correct Answer: (A)