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Chemical Bonding and Molecular Structure - Hybridization

Grade 11CBSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Hybridization is defined as the process of intermixing of the orbitals of slightly different energies of the same atom so as to redistribute their energies, resulting in the formation of a new set of orbitals of equivalent energies and shape.

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The number of hybrid orbitals formed is always equal to the number of atomic orbitals that get hybridized.

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Hybrid orbitals are always equivalent in energy and shape. They are more effective in forming stable bonds than pure atomic orbitals.

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spsp Hybridization: Involves the mixing of one ss and one pp orbital, resulting in two spsp hybrid orbitals arranged linearly at an angle of 180∘180^\circ. Example: BeCl2BeCl_2, C2H2C_2H_2.

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sp2sp^2 Hybridization: Involves the mixing of one ss and two pp orbitals, resulting in three sp2sp^2 hybrid orbitals directed towards the corners of an equilateral triangle at 120∘120^\circ. Example: BCl3BCl_3, C2H4C_2H_4.

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sp3sp^3 Hybridization: Involves the mixing of one ss and three pp orbitals, resulting in four sp3sp^3 hybrid orbitals directed towards the four corners of a regular tetrahedron at 109.5∘109.5^\circ. Example: CH4CH_4, NH3NH_3, H2OH_2O.

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sp3dsp^3d Hybridization: Involves one ss, three pp, and one dd orbital, resulting in a Trigonal Bipyramidal geometry. Example: PCl5PCl_5.

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sp3d2sp^3d^2 Hybridization: Involves one ss, three pp, and two dd orbitals, resulting in an Octahedral geometry. Example: SF6SF_6.

📐Formulae

H=12[V+M−C+A]H = \frac{1}{2} [V + M - C + A]

Where: V=Number of valence electrons of the central atom\text{Where: } V = \text{Number of valence electrons of the central atom}

M=Number of monovalent atoms (H, F, Cl, Br, I) surrounding the central atomM = \text{Number of monovalent atoms (H, F, Cl, Br, I) surrounding the central atom}

C=Charge on the cationC = \text{Charge on the cation}

A=Charge on the anionA = \text{Charge on the anion}

If H=2→sp,H=3→sp2,H=4→sp3,H=5→sp3d,H=6→sp3d2\text{If } H=2 \rightarrow sp, H=3 \rightarrow sp^2, H=4 \rightarrow sp^3, H=5 \rightarrow sp^3d, H=6 \rightarrow sp^3d^2

💡Examples

Problem 1:

Predict the hybridization and geometry of the PCl5PCl_5 molecule.

Solution:

For Phosphorus (PP), valence electrons V=5V = 5. Number of monovalent atoms (ClCl) M=5M = 5. C=0,A=0C = 0, A = 0. H=12[5+5−0+0]=5H = \frac{1}{2}[5 + 5 - 0 + 0] = 5.

Explanation:

Since H=5H = 5, the hybridization is sp3dsp^3d. The geometry of the PCl5PCl_5 molecule is Trigonal Bipyramidal with bond angles of 120∘120^\circ (equatorial) and 90∘90^\circ (axial).

Problem 2:

Determine the hybridization of the central atom in H2OH_2O.

Solution:

For Oxygen (OO), V=6V = 6. Number of monovalent atoms (HH) M=2M = 2. H=12[6+2]=4H = \frac{1}{2}[6 + 2] = 4.

Explanation:

Since H=4H = 4, the hybridization is sp3sp^3. Although the electron geometry is tetrahedral, the presence of two lone pairs on the oxygen atom results in a 'Bent' or 'V-shape' molecular geometry with a bond angle of approximately 104.5∘104.5^\circ.

Problem 3:

What is the hybridization of Carbon in Ethene (C2H4C_2H_4)?

Solution:

Each Carbon atom is bonded to two Hydrogen atoms (monovalent) and one other Carbon atom via a double bond.

Explanation:

In C2H4C_2H_4, each Carbon atom forms 3 σ\sigma bonds and 1 π\pi bond. Since hybridization only involves σ\sigma bonds and lone pairs, the steric number is 3. Thus, the hybridization is sp2sp^2 and the geometry around each carbon is Trigonal Planar.