Types of Hybridisation

Figure 4.10 is the visual anchor for the concept of sp hybridisation — the simplest kind of orbital mixing. The figure has two parts, (a) and (b), and each teaches a different layer of the same idea.

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Part (a): The formation of two sp hybrid orbitals

This panel shows the before and after of orbital mixing. On the left, you see two separate atomic orbitals: one s orbital (spherical) and one p orbital (dumbbell-shaped, specifically the $p_z$ orbital, which points along the z-axis). The s orbital has no directional preference; the p orbital has a positive lobe (say, along $+z$) and a negative lobe (along $-z$).

The arrow or process symbol between them indicates that these two orbitals combine mathematically — they are not physically glued together but rather their wavefunctions are added and subtracted. The result, shown on the right, is a pair of equivalent sp hybrid orbitals. Each hybrid looks like a lopsided dumbbell: one large lobe (the positive lobe) and a very small back lobe (the negative lobe). Crucially, the two hybrids point in exactly opposite directions, $180^\circ$ apart, along the z-axis.

The small negative lobes are usually drawn but are often omitted in simplified diagrams; they are a consequence of the subtraction step in the hybridisation process. The key takeaway: one s + one p → two sp hybrids, linear, opposite.

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Part (b): The linear BeCl₂ molecule

Here the figure shows how those two sp hybrids on the central beryllium atom actually form bonds. Beryllium in its ground state has the configuration $1s^2 2s^2$ — no unpaired electrons. To form two bonds, one of the $2s$ electrons is promoted to an empty $2p$ orbital (the $2p_z$), giving the configuration $1s^2 2s^1 2p_z^1$. These two half-filled orbitals (one s, one p) then hybridise into the two sp hybrids from part (a).

Each sp hybrid on Be now overlaps end-on (axially) with a chlorine 3p orbital (specifically, the $3p_z$ orbital of each Cl atom). This head-on overlap creates a sigma ($\sigma$) bond. Because the two sp hybrids point in opposite directions, the two Cl atoms end up on opposite sides of Be, giving the linear geometry Cl–Be–Cl with a bond angle of exactly $180^\circ$.

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The physical idea this figure teaches

Hybridisation is a mathematical model that explains why certain molecules have the shapes they do. Without hybridisation, you would expect Be to form bonds using its $2s$ orbital (which is spherical) and its $2p$ orbital (which is directional but at $90^\circ$ to each other in a pure p set). That would predict a bent molecule — which is wrong. By mixing the s and p orbitals, you get two equivalent orbitals that are optimally directed for bonding: $180^\circ$ apart. This is why BeCl₂ is linear, not bent.

The figure also teaches that the number of hybrid orbitals equals the number of atomic orbitals mixed (here, 1 s + 1 p = 2 sp hybrids), and that the geometry is determined by the repulsion between these hybrid orbitals (they point as far apart as possible — $180^\circ$ for two).

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The key formula the textbook develops with this figure

The central quantitative idea is the percentage s-character in each hybrid. For sp hybridisation:

More generally, for any hybridisation involving $m$ s orbitals and $n$ p orbitals (and possibly d orbitals), the s-character of each equivalent hybrid is:

For sp: $m=1$, $n=1$ → 50% s, 50% p. …

Fig. 4.11 is a two-part schematic that shows how atomic orbitals on a boron atom reorganise to form three equivalent hybrid orbitals, and how those hybrids then bond with chlorine atoms to give the BCl₃ molecule.

The left panel depicts the mixing process. Three atomic orbitals from boron’s valence shell — one 2s orbital (spherical) and two 2p orbitals (dumbbell-shaped, lying in the same plane) — are shown combining. The result is three identical sp² hybrid orbitals, each drawn as a lopsided lobe: a large positive lobe and a much smaller negative lobe. The three hybrids are arranged in a plane, pointing toward the corners of an equilateral triangle, with 120° angles between them. The right panel shows the complete BCl₃ molecule. Each sp² hybrid on boron overlaps end‑to‑end (axially) with a 2p orbital from a chlorine atom, forming a sigma bond. The three chlorine atoms sit at the vertices of the triangle, and the entire molecule is flat — trigonal planar — with all Cl–B–Cl bond angles equal to 120°.

The physical idea is that mixing one s and two p orbitals produces three equivalent orbitals that are more directional than the original s orbital and more concentrated along the bonding directions than the original p orbitals. This directional character allows stronger overlap with the chlorine orbitals, giving a stable, symmetric molecule. The 120° angle is a direct consequence of the three hybrids repelling each other equally in a plane — the geometry that minimises electron‑pair repulsion for three equivalent bonds.

The textbook uses this figure to introduce the concept that hybridisation determines molecular shape. For sp² hybridisation, the key result is that the three hybrid orbitals lie in a plane at 120° to each other, and the molecule formed (like BCl₃) is trigonal planar. There is no separate formula beyond the geometric fact that the bond angle is fixed at 120° for a perfect sp²‑hybridised centre with no lone pairs. …

Fig. 4.12 is a two-part diagram that shows the entire journey from atomic orbitals to a finished methane molecule. The left side illustrates the mixing process — how one s orbital and three p orbitals (pₓ, p_y, p_z) of carbon combine to form four identical sp³ hybrid orbitals. The right side shows the final molecule: these four hybrids overlapping with the 1s orbitals of four hydrogen atoms to give tetrahedral CH₄.

The left panel is essentially a "before and after" of the carbon atom's valence orbitals. Before hybridisation, you see four separate orbitals: a spherical 2s orbital and three dumbbell-shaped 2p orbitals oriented along the x, y, and z axes. The figure then shows these four orbitals merging into four equivalent sp³ hybrids. Each hybrid has a large lobe (the part that points outward) and a much smaller lobe on the opposite side. The four large lobes are directed toward the four corners of a regular tetrahedron, with the carbon nucleus at the centre. The angle between any two hybrid orbitals is exactly 109.5°.

The right panel of Fig. 4.12 shows the complete CH₄ molecule. Each sp³ hybrid orbital on carbon overlaps end‑to‑end (axially) with the 1s orbital of a hydrogen atom, forming a sigma (σ) bond. Because all four hybrids are identical and point to tetrahedral corners, all four C–H bonds are equivalent in length and strength, and the molecule has a perfect tetrahedral shape with H–C–H bond angles of 109.5°.

The figure anchors the concept of sp³ hybridisation; its quantitative companion is the s/p composition of each hybrid:

Here, the 1 comes from the single s orbital and the 3 from the three p orbitals that participate in mixing. Each of the four sp³ hybrids therefore has 25% s‑character and 75% p‑character. This ratio directly influences bond properties: greater s‑character makes a bond shorter and stronger (as in sp hybrids with 50% s‑character), while greater p‑character makes bonds longer and weaker.

The figure also implicitly teaches the relationship between hybridisation and geometry. For a central atom with four electron domains (bond pairs or lone pairs), the domains arrange tetrahedrally. When all four domains are bond pairs, as in CH₄, the molecule is tetrahedral with 109.5° angles. When one domain is a lone pair (NH₃) or two are lone pairs (H₂O), the geometry distorts — the bond angles shrink to 107° and 104.5° respectively — but the underlying hybridisation remains sp³. …

Fig. 4.13 is a schematic diagram of the ammonia molecule, NH₃, built from the idea of sp³ hybridisation. The figure does not show a graph or plot; it is a structural drawing that illustrates how the atomic orbitals of nitrogen mix and then overlap with hydrogen 1s orbitals to form the molecule.

The central nitrogen atom is shown at the apex of a trigonal pyramid. Three sp³ hybrid orbitals, each containing one electron, point toward the three corners of the base of the pyramid. Each of these three hybrids overlaps end‑to‑end (sigma overlap) with the 1s orbital of a hydrogen atom, forming three N–H sigma bonds. The fourth sp³ hybrid orbital, which contains a lone pair of electrons, points toward the fourth corner of a tetrahedron — but since no hydrogen is attached there, that corner is empty. The lone pair occupies more space than a bonding pair, so it pushes the three N–H bonds closer together. The result is a bond angle of 107° instead of the full tetrahedral angle of 109.5°, and the overall shape is pyramidal, not tetrahedral.

The physical idea the figure teaches is that the geometry of a molecule is determined not only by the number of sigma bonds but also by the number of lone pairs on the central atom. The lone pair repels the bonding pairs more strongly, distorting the ideal tetrahedral angle. This is the core of the VSEPR (Valence Shell Electron Pair Repulsion) theory, which the textbook uses alongside hybridisation to explain molecular shapes.

The key formula that emerges from this figure is the relationship between the number of hybrid orbitals and the number of atoms plus lone pairs. For NH₃:

Here, the central nitrogen atom has 5 valence electrons. In its ground state, the configuration is $1s^2 2s^2 2p_x^1 2p_y^1 2p_z^1$. Under sp³ hybridisation, one 2s orbital and three 2p orbitals mix to form four equivalent sp³ hybrid orbitals. Three of these hybrids each contain one electron and form sigma bonds with hydrogen. The fourth hybrid contains two electrons — the lone pair. The bond angle is given by:

This is less than the ideal tetrahedral angle of $109.5^\circ$ because the lone pair exerts a stronger repulsive force on the bonding pairs than the bonding pairs exert on each other. …