Types of Hybridization and Geometry of Molecules

What this figure shows. In $sp^3$ hybridization, one 's' and three 'p' orbitals of comparable energy mix and recast into four equivalent $sp^3$ hybrid orbitals, each with one large and one small lobe, oriented at 109°28' to one another and pointing toward the four corners of a regular tetrahedron. For methane specifically: carbon's ground state $1s^2\,2s^2\,2p_x^1\,2p_y^1\,2p_z^0$ is excited (one 2s electron promoted to the empty $2p_z$), giving four singly-occupied orbitals (2s, $2p_x$, $2p_y$, $2p_z$) which then mix into four equivalent $sp^3$ hybrids, each holding one unpaired electron. Each of these overlaps axially with a hydrogen 1s orbital to form one C–H sigma bond, giving CH4 its four identical, tetrahedrally-arranged C–H σ bonds. NH3 and H2O are cited as further examples where the central atom is $sp^3$ hybridized (with, respectively, one and two of the four hybrid orbitals occupied by lone pairs …

What this figure shows. In $sp^2$ hybridization, one 's' and two 'p' orbitals mix into three equivalent $sp^2$ hybrid orbitals, all lying in one plane, maximally apart at 120° to each other; the third, unhybridized p orbital is left perpendicular to that plane. For BF3: boron's ground state $1s^2\,2s^2\,2p_x^1\,2p_y^0\,2p_z^0$ is excited (one 2s electron promoted to the empty $2p_y$), giving three half-filled orbitals (2s, $2p_x$, $2p_y$) that mix into three $sp^2$ hybrids oriented at the corners of an equilateral triangle (120° apart); each overlaps axially with a half-filled fluorine $2p_z$ orbital to form a B–F sigma bond by $sp^2$-p overlap, giving BF3 its observed trigonal-planar shape with all F–B–F angles equal to 120°. For C2H4 (ethene): each carbon undergoes $sp^2$ hybridization (one 's' and two 'p' orbitals forming three $sp^2$ hybrids per carbon), leaving one unhybridized p orbital per carbon standing perpendicular to the molecular plane. Two of each carbon's $sp^2$ hybrids overlap axially with a hydrogen 1s orbital (giving four C–H σ bonds total), and the remaining $sp^2$ hybrid on each carbon overlaps axially with its counterpart on the other carbon (one C–C σ bond, $sp^2$-$sp^2$); the two unhybridized p orbitals (one per carbon) then overlap sideways to form one C–C π bo …

What this figure shows. In sp hybridization, one 's' and one 'p' orbital mix into two equivalent sp hybrid orbitals placed at 180° to each other (linear); the remaining two, unhybridized p orbitals stay at right angles both to each other and to the sp hybrids. For BeCl2 (worked as Problem 5.4 in the source): beryllium's ground state $1s^2\,2s^2\,2p_z^0$ is excited (one 2s electron promoted to $2p_z$), and the resulting 2s/$2p_z$ pair mixes into two sp hybrids at 180°; each overlaps with a half-filled $2p_z$ orbital on a chlorine atom to form a Be–Cl sigma bond by sp-p overlap, giving BeCl2 its observed linear, 180° Cl–Be–Cl geometry. For acetylene, C2H2: each carbon undergoes sp hybridization (giving two sp hybrids and two unhybridized p orbitals per carbon); one sp hybrid on each carbon overlaps axially with a hydrogen 1s orbital (C–H σ bond, sp-s overlap), and the other sp hybrid on each carbon overlaps axially with its counterpart on the other carbon (one C–C σ bond, sp-sp overlap); the two pairs of remaining unhybridized p orbitals (one pair per axis) then overlap sideways to give two mutually perpendicular C–C π bonds. So acetylene has, in total, three bonds between the two carbons — one C–C σ (sp-sp) plus two C–C π (p-p) — completing carbon's fourth valency via the sp-s C–H σ bond, and the whole H–C≡C–H molecule is linear (180°). NOTE: the source PDF labels this figure 'Fig 5.5', duplicating the figure number used later, in section 5.5.3, for the entirely different molecular-orbital bo …