The group having triangular planar structure is:

First, let us recall the basic idea that the shape of a molecule or ion is decided by the steric number of the central atom and the presence or absence of lone-pairs. For main-group species, we use the Valence Shell Electron Pair Repulsion (VSEPR) concept.

The rule says: if the steric number is $$3$$ and there are no lone-pairs on the central atom, the hybridisation is $$sp^2$$ and the geometry is trigonal (triangular) planar. Whenever one of the three positions is occupied by a lone-pair, the shape becomes trigonal pyramidal instead of planar.

Now we examine every species that appears in the four option sets.

1. $$BF_3$$  The central boron has three valence electrons and forms three $$\sigma$$ bonds with three fluorine atoms. The steric number is $$3$$ and there is no lone-pair on boron, so $$BF_3$$ is $$sp^2$$ hybridised and trigonal planar.

2. $$NF_3$$  Nitrogen has five valence electrons. Three are used for three $$\sigma$$ bonds with fluorine; the remaining two constitute one lone-pair. Thus steric number $$= 3 \text{ bonds } + 1 \text{ lone-pair } = 4$$, giving $$sp^3$$ hybridisation and a trigonal pyramidal shape, not planar.

3. $$CO_3^{2-}$$ (carbonate ion)  Carbon forms three $$\sigma$$ bonds with three oxygen atoms. No lone-pair remains on carbon. Steric number $$=3$$, hybridisation $$sp^2$$, shape trigonal planar.

4. $$NO_3^-$$ (nitrate ion)  Nitrogen is bonded to three oxygen atoms with no lone-pair on nitrogen (the extra electron is delocalised in $$\pi$$ bonding). Steric number $$=3$$, hybridisation $$sp^2$$, geometry trigonal planar.

5. $$SO_3$$  Sulphur uses three $$\sigma$$ bonds for three oxygen atoms; there is no lone-pair on sulphur in this structure. Steric number $$=3$$, hybridisation $$sp^2$$, and the molecule is trigonal planar.

6. $$NH_3$$  Three $$\sigma$$ bonds + one lone-pair on nitrogen give steric number $$4$$, $$sp^3$$ hybridisation, and trigonal pyramidal shape.

7. $$NCl_3$$  Exactly analogous to $$NH_3$$: three $$\sigma$$ bonds + one lone-pair, hence trigonal pyramidal.

8. $$BCl_3$$  Like $$BF_3$$, boron has no lone-pair; steric number $$3$$, so $$sp^2$$ and trigonal planar.

Having classified every species, we list those that are truly triangular planar:

$$BF_3,\; BCl_3,\; CO_3^{2-},\; NO_3^- ,\; SO_3$$

We now compare each option:

A. $$BF_3,\; NF_3,\; CO_3^{2-}$$  contains $$NF_3$$ (pyramidal) – reject.

B. $$CO_3^{2-},\; NO_3^-,\; SO_3$$  every member is trigonal planar – accept.

C. $$NH_3,\; SO_3,\; CO_3^{2-}$$  contains $$NH_3$$ (pyramidal) – reject.

D. $$NCl_3,\; BCl_3,\; SO_3$$  contains $$NCl_3$$ (pyramidal) – reject.

Only Option B satisfies the requirement that all species in the group possess a triangular planar structure.

Hence, the correct answer is Option 2.