To determine the geometry of a species, we can use the VSEPR theory, which is based on the number of electron domains (bond pairs and lone pairs) around the central atom. The steric number (SN) is the sum of the number of atoms bonded to the central atom and the number of lone pairs on the central atom.

Step 1: Analyze option A) ${\mathrm{BH}}_4^{-}$

The central atom is Boron (B). Boron has 3 valence electrons. The charge is -1, so we add 1 electron. Total valence electrons = 3 + 1 (from charge) + 4 (from 4 H atoms) = 8 electrons. Each H forms a single bond, so there are 4 bond pairs. Number of lone pairs = (8 - 4*2)/2 = 0. Steric number = 4 (bond pairs) + 0 (lone pairs) = 4. A steric number of 4 with no lone pairs corresponds to a tetrahedral geometry.

Step 2: Analyze option B) ${\mathrm{NH}}_2^{-}$

The central atom is Nitrogen (N). Nitrogen has 5 valence electrons. The charge is -1, so we add 1 electron. Total valence electrons = 5 + 1 (from charge) + 2 (from 2 H atoms) = 8 electrons. There are 2 bond pairs. Number of lone pairs = (8 - 2*2)/2 = 2. Steric number = 2 (bond pairs) + 2 (lone pairs) = 4. A steric number of 4 with two lone pairs corresponds to a bent or V-shaped geometry.

Step 3: Analyze option C) ${\mathrm{CO}}_3{}^{2-}$

The central atom is Carbon (C). Carbon has 4 valence electrons. The charge is -2, so we add 2 electrons. Total valence electrons = 4 + 2 (from charge) + 3*6 (from 3 O atoms) = 24 electrons. Carbon forms one double bond and two single bonds with oxygen atoms (resonance structures exist, but for VSEPR, we consider the number of electron domains). There are 3 bond pairs (considering each C-O bond as one domain). Number of lone pairs on carbon = (24 - 3*2)/2 = 9. This calculation is incorrect for the central atom. For the central atom, Carbon, it forms 3 bonds (one double, two single, or three equivalent bonds due to resonance). So, there are 3 electron domains around carbon. Number of lone pairs on carbon = 0. Steric number = 3 (bond pairs) + 0 (lone pairs) = 3. A steric number of 3 with no lone pairs corresponds to a trigonal planar geometry.

Step 4: Analyze option D) ${\mathrm{H}}_3{\mathrm{O}}^{+}$

The central atom is Oxygen (O). Oxygen has 6 valence electrons. The charge is +1, so we subtract 1 electron. Total valence electrons = 6 - 1 (from charge) + 3 (from 3 H atoms) = 8 electrons. There are 3 bond pairs. Number of lone pairs = (8 - 3*2)/2 = 1. Steric number = 3 (bond pairs) + 1 (lone pair) = 4. A steric number of 4 with one lone pair corresponds to a trigonal pyramidal geometry.

Based on the analysis, only ${\mathrm{BH}}_4^{-}$ has a tetrahedral geometry.

Correct Answer: (A)