Animated Solution for Chemistry - Chemical Bonding and Molecular Structure: Which one of the following has the regular tetrahedral structure ? (Atomic number : $\text{B}=5,\text{S}=16,\text{Ni}=28,\text{Xe}=54$)

When we talk about a regular tetrahedral structure in chemistry, we are talking about perfect, undisturbed symmetry. For a molecule to achieve this beautiful geometry, its central atom must meet two strict conditions. First, it must undergo $s{p}^3$ hybridization, which provides four equivalent hybrid orbitals pointing towards the corners of a tetrahedron. Second, and most importantly, it must possess zero lone pairs.

Why are lone pairs such a dealbreaker? According to VSEPR (Valence Shell Electron Pair Repulsion) theory, lone pairs exert a stronger repulsive force than bond pairs. Even a single lone pair will push the surrounding bonds closer together, distorting the perfect $109.5^{\circ }$ bond angles and ruining the regular tetrahedral shape. Let's investigate our four candidates to see which one survives this test.

Xenon is a noble gas, meaning it has a full octet of 8 valence electrons. In ${\text{XeF}}_4$, it forms 4 single bonds with fluorine atoms. This uses up 4 electrons, leaving 4 electrons behind, which pair up to form 2 lone pairs.

With 4 bond pairs and 2 lone pairs, the steric number is $4+2=6$. This corresponds to $s{p}^3{d}^2$ hybridization, which has an octahedral electron geometry. To minimize repulsion, the two lone pairs position themselves exactly opposite to each other (axially). The four fluorine atoms are left in the equatorial plane, resulting in a flat, square planar shape. Definitely not tetrahedral!

Sulfur, sitting in group 16, has 6 valence electrons. It uses 4 of these to bond with fluorine, leaving 2 electrons, which constitutes exactly 1 lone pair.

Adding the 4 bond pairs and 1 lone pair gives a steric number of $4+1=5$, leading to $s{p}^{3}d$ hybridization. The base geometry is trigonal bipyramidal. The lone pair, seeking maximum space to minimize repulsion, occupies an equatorial position. This pushes the axial fluorine atoms slightly away, creating a distorted shape known as a see-saw. So, ${\text{SF}}_4$ is also out of the running.

This species is a coordination complex rather than a simple covalent molecule. Nickel is in a $+2$ oxidation state, giving it a $3d^8$ electron configuration. The cyanide ion (${\text{CN}}^{-}$) is a strong field ligand.

Strong field ligands force the unpaired d-electrons of the central metal to pair up. This pairing frees up one of the inner $3d$ orbitals. The nickel ion then uses this empty $d$ orbital, along with one $s$ and two $p$ orbitals, to hybridize as $ds{p}^2$. In coordination chemistry, $ds{p}^2$ hybridization always corresponds to a square planar geometry. Thus, no tetrahedron here either.

Finally, let's examine the tetrafluoroborate ion. Boron is in group 13 and normally has 3 valence electrons. However, the overall negative charge on the ion indicates the presence of an extra electron, bringing Boron's total valence electrons to 4.

Boron uses all 4 of these electrons to form single bonds with four fluorine atoms. This leaves absolutely zero lone pairs! With 4 bond pairs and 0 lone pairs, the steric number is exactly 4, which means perfect $s{p}^3$ hybridization.

Without any lone pairs to cause unequal repulsion, the four $\text{B-F}$ bonds spread out symmetrically in 3D space, separated by perfect $109.5^{\circ }$ angles. We have found our winner: ${\text{BF}}_4^{-}$ boasts a flawless, regular tetrahedral structure.