Question

The correct order of hybridization of the central atom in the following species \(\mathrm{NH}_{3}, \mathrm{PtCl}_{4}^{-2}, \mathrm{PCl}_{5}\) and \(\mathrm{BCl}_{3}\) is: (a) \(\mathrm{dsp}^{2}, \mathrm{sp}^{3} \mathrm{~d}, \mathrm{sp}^{2}\) and \(\mathrm{sp}^{3}\) (b) \(\mathrm{sp}^{3}, \mathrm{dsp}^{2}, \mathrm{sp}^{3} \mathrm{~d}, \mathrm{sp}^{2}\) (c) \(\mathrm{dsp}^{2}, \mathrm{sp}^{2}, \mathrm{sp}^{3}, \mathrm{sp}^{3} \mathrm{~d}\) (d) \(\mathrm{dsp}^{2}, \mathrm{sp}^{3}, \mathrm{sp}^{2}, \mathrm{sp}^{3} \mathrm{~d}\)

Short answer

The correct order is (b): sp^3, dsp^2, sp^3d, sp^2.

Step-by-step solution

Determine Hybridization of NH3

The molecule NH3 has a central atom N (nitrogen) which forms three sigma bonds with hydrogen atoms and has one lone pair. This results in a total of 4 electron pairs around nitrogen. According to the VSEPR theory, a total of 4 electron pairs corresponds to sp^3 hybridization. Therefore, NH3 has sp^3 hybridization.

Determine Hybridization of PtCl4^(-2)

PtCl4^(-2) consists of a central platinum (Pt) atom bonded to four chloride (Cl) ions. Since it is a negatively charged ion (anion) with a charge of -2, we expect that the central platinum atom will undergo dsp^2 hybridization to accommodate the four ligands in a square planar geometry, which is typical for such complexes.

Determine Hybridization of PCl5

In PCl5, the central atom is phosphorus (P), which forms five sigma bonds with chlorine atoms. To accommodate the 5 bonds, phosphorus uses sp^3d hybridization, resulting in a trigonal bipyramidal shape. Therefore, PCl5 exhibits sp^3d hybridization.

Determine Hybridization of BCl3

In BCl3, the central atom boron (B) forms three sigma bonds with chlorine atoms, and it does not have lone pairs. With 3 electron domains (all bonding), the hybridization of boron in BCl3 is sp^2, resulting in a trigonal planar geometry.

Assemble the Correct Order

Now, based on the hybridizations calculated: NH3 (sp^3), PtCl4^(-2) (dsp^2), PCl5 (sp^3d), and BCl3 (sp^2), the correct order of hybridization should be sp^3 for NH3, dsp^2 for PtCl4^(-2), sp^3d for PCl5, and sp^2 for BCl3. This corresponds with option (b): sp^3, dsp^2, sp^3d, sp^2.

Key concepts

VSEPR Theory

The Valence Shell Electron Pair Repulsion (VSEPR) theory is an essential principle when predicting the geometry of molecules. It's based on the idea that electron pairs surrounding an atom will repel each other and thus tend to stay as far apart as possible. This gives molecules their three-dimensional shapes, dictated by the number and types of electron pairs around the central atom. For example, in \(\text{NH}_3\), nitrogen forms three sigma bonds with hydrogen atoms and has one lone pair of electrons. These four electron pairs arrange themselves in a tetrahedral shape, to minimize repulsions, which results in the molecule's trigonal pyramidal geometry. By considering these arrangements, VSEPR theory helps us determine the hybridization, such as \(\text{sp}^3\) for \(\text{NH}_3\). Understanding VSEPR allows you to visualize how bonds and lone pairs interact to determine molecular shapes, forming the basis of hybridization explanations.

sp3 Hybridization

Hybridization is a concept used to describe the mixing of atomic orbitals to form new hybrid orbitals, which influences the geometry and bonding characteristics of molecules. In \(\text{sp}^3\) hybridization, one s orbital combines with three p orbitals to form four \(\text{sp}^3\) hybrid orbitals. These orbitals are equivalent in energy and shape, and they form a tetrahedral geometry around the central atom. The angle between any two \(\text{sp}^3\) orbitals is about 109.5 degrees.A prime example of an \(\text{sp}^3\) hybridized molecule is \(\text{NH}_3\). The nitrogen atom in \(\text{NH}_3\) is surrounded by four areas of electron density: three from bonding with hydrogens and one from its lone pair. This leads to the formation of \(\text{sp}^3\) hybrid orbitals, giving the molecule a trigonal pyramidal shape.

dsp2 Hybridization

\(\text{dsp}^2\) hybridization is crucial for understanding the geometries of specific transition metal complexes. Here, one d orbital, one s orbital, and two p orbitals combine to create four equivalent \(\text{dsp}^2\) hybrid orbitals. These orbitals arrange themselves in a square planar shape, often seen in transition metal complexes. For instance, in \(\text{PtCl}_4^{-2}\), the central platinum atom bonds with four chloride ligands. The \(\text{dsp}^2\) hybridization allows these ligands to assume a planar configuration with 90-degree angles between them. This hybridization type accommodates the specific spatial coordination needed for a stable square planar arrangement in such metal complexes.

sp3d Hybridization

When an atom forms five bonds, \(\text{sp}^3d\) hybridization is at play. This involves the mixing of one s orbital, three p orbitals, and one d orbital to create five \(\text{sp}^3d\) hybrid orbitals. These orbitals are aligned in a trigonal bipyramidal geometry, where three are placed equatorially (in a plane) and two axially (above and below the plane). Take \(\text{PCl}_5\) as an example. Phosphorus is the central atom, bonding with five chlorine atoms using these hybrid orbitals. The resulting shape is trigonal bipyramidal, with three chlorine atoms in a plane at 120-degree angles and two in a linear alignment, making \(\text{sp}^3d\) housing flexible for larger molecular geometries. This hybridization explains how molecules can expand beyond the simple geometries covered by \(\text{sp}^3\), accommodating five pairs of bonds or electron domains.