To determine the geometry and hybridisation of the central atom in ${\mathrm{BF}}_3$, we first need to find the number of valence electrons of the central atom, Boron (B).

Step 1: Determine the number of valence electrons for the central atom.
Boron (B) is in Group 13, so it has 3 valence electrons.

Step 2: Determine the number of bond pairs and lone pairs.
Boron forms 3 single bonds with 3 Fluorine (F) atoms. Since Boron has 3 valence electrons and forms 3 bonds, there are no lone pairs on the central Boron atom.
Number of bond pairs = 3
Number of lone pairs = 0

Step 3: Determine the steric number.
Steric number = (Number of bond pairs) + (Number of lone pairs) = 3 + 0 = 3.

Step 4: Determine the hybridisation and geometry.
A steric number of 3 corresponds to $s{p}^2$ hybridisation. For a molecule with 3 bond pairs and 0 lone pairs, the electron geometry and molecular geometry are both trigonal planar.

Option Analysis:

• A) linear, $sp$: This corresponds to a steric number of 2 (e.g., ${\mathrm{BeCl}}_2$). Incorrect.
• B) trigonal planar, $s{p}^2$: This corresponds to a steric number of 3 with no lone pairs, which matches ${\mathrm{BF}}_3$. Correct.
• C) tetrahedral, $s{p}^3$: This corresponds to a steric number of 4 (e.g., ${\mathrm{CH}}_4$). Incorrect.
• D) pyramidal, $s{p}^3$: This corresponds to a steric number of 4 with one lone pair (e.g., ${\mathrm{NH}}_3$). Incorrect.

Correct Answer: (B)