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Organic Chemistry – Some Basic Principles and Techniques - Tetravalence of Carbon: Shapes of Organic Compounds

Grade 11CBSEChemistry

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

🔑Concepts

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Carbon has an atomic number of 6 with the electronic configuration [He]2s22p2[He] 2s^2 2p^2. To form four bonds, it undergoes excitation to [He]2s12px12py12pz1[He] 2s^1 2p_x^1 2p_y^1 2p_z^1, demonstrating its tetravalence.

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Hybridization is the process of mixing atomic orbitals to form new hybrid orbitals. In organic compounds, carbon undergoes three types of hybridization: sp3sp^3, sp2sp^2, and spsp.

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sp3sp^3 Hybridization: Resulting in a tetrahedral shape with bond angles of 109.5∘109.5^\circ. It occurs when carbon is bonded to four other atoms via single bonds (e.g., CH4CH_4).

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sp2sp^2 Hybridization: Resulting in a trigonal planar shape with bond angles of 120∘120^\circ. It occurs when carbon is bonded to three other atoms (e.g., CH2=CH2CH_2=CH_2), involving one π\pi bond.

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spsp Hybridization: Resulting in a linear shape with a bond angle of 180∘180^\circ. It occurs when carbon is bonded to two other atoms (e.g., HC≡CHHC \equiv CH), involving two π\pi bonds.

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Bond Length and Strength: spsp hybridized carbons have the shortest bond length and highest bond strength due to higher s-character, whereas sp3sp^3 has the longest bond length.

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Electronegativity: The electronegativity of carbon increases with the increase in s-character. Therefore, the order is sp>sp2>sp3sp > sp^2 > sp^3.

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Sigma (σ\sigma) bonds are formed by the axial (head-on) overlap of orbitals, while Pi (π\pi) bonds are formed by the lateral (sideways) overlap of unhybridized p-orbitals.

📐Formulae

s-character in spn=11+n×100%\text{s-character in } sp^n = \frac{1}{1+n} \times 100\%

p-character in spn=n1+n×100%\text{p-character in } sp^n = \frac{n}{1+n} \times 100\%

Total σ bonds in open chain CnHm=(n+m−1)\text{Total } \sigma \text{ bonds in open chain } C_n H_m = (n+m-1)

Hybridization Index (H)=(Number of σ bonds)+(Number of lone pairs)\text{Hybridization Index } (H) = (\text{Number of } \sigma \text{ bonds}) + (\text{Number of lone pairs})

💡Examples

Problem 1:

Identify the hybridization of each carbon atom in the molecule: CH3−CH=CH−C≡NCH_3-CH=CH-C \equiv N.

Solution:

  1. C1C_1 in CH3CH_3 is bonded to 4 atoms via σ\sigma bonds: sp3sp^3.
  2. C2C_2 in −CH=-CH= is bonded to 3 atoms (one π\pi bond): sp2sp^2.
  3. C3C_3 in =CH−=CH- is bonded to 3 atoms (one π\pi bond): sp2sp^2.
  4. C4C_4 in −C≡N-C \equiv N is bonded to 2 atoms (two π\pi bonds): spsp.

Explanation:

Hybridization is determined by counting the number of σ\sigma bonds. 4σ=sp34 \sigma = sp^3, 3σ=sp23 \sigma = sp^2, and 2σ=sp2 \sigma = sp.

Problem 2:

How many σ\sigma and π\pi bonds are present in the molecule ethyne (HC≡CHHC \equiv CH)?

Solution:

Ethyne contains 3σ3 \sigma bonds and 2π2 \pi bonds.

Explanation:

In ethyne, there are two C−HC-H single bonds (σ\sigma) and one C≡CC \equiv C triple bond. A triple bond consists of 1σ1 \sigma and 2π2 \pi bonds. Total σ=1+1+1=3\sigma = 1+1+1=3; Total π=2\pi = 2.