Important Formulae Summary for NEET UG Chemistry Hydrocarbons Preparation

1. Stoichiometry and Chemical Reactions

Molecular Formula for Alkanes ( $C n H 2 n + 2$ ):
- The general formula for alkanes is $C n H 2 n + 2$ , where $n$ represents the number of carbon atoms. This formula is essential for calculating the molecular formula of any alkane.
General Combustion Equation for Alkanes:
- The general combustion reaction for alkanes is: $C n H 2 n + 2 + \frac{3 n + 1}{2} O 2 \to n CO 2 + (n + 1) H 2 O$ This formula is crucial for determining the amount of oxygen required and the products formed during the complete combustion of an alkane.
Halogenation of Methane:
- Methane undergoes halogenation through a free radical substitution mechanism: $C H 4 + X 2 \to C H 3 X + H X$ where XXX can be a halogen like Cl or Br.

2. Thermodynamics

Enthalpy of Combustion:
- The enthalpy change during the combustion of alkanes, for example: $C H 4 + 2 O 2 \to CO 2 + 2 H 2 O;, Δ H = - 890, kJ/mol$ This value is used to calculate the energy released during combustion reactions.
Bond Enthalpy for C-H and C-C Bonds:
- The bond enthalpies are essential for calculating the energy changes during chemical reactions:
- - C-H bond: $413 kJ/mol$
  - C-C bond: $348 kJ/mol$

3. Physical Chemistry (Chemical Bonding and Molecular Structure)

Bond Lengths in Hydrocarbons:
- C-H bond length: 109 pm
- C-C bond length (in alkanes): 154 pm
- C=C bond length (in alkenes): 134 pm
- C≡C bond length (in alkynes): 120 pm

4. Derivations and Key Concepts

Derivation of Alkane General Formula:
- Starting from the basic principle that each carbon atom forms four bonds and each hydrogen forms one, the general formula for alkanes can be derived as $C n H 2 n + 2$ .
Bonding in Alkenes:
- The carbon-carbon double bond in alkenes consists of one sigma (σ) bond and one pi (π) bond, which can be represented as: $C = C$
- The sigma bond is formed by the overlap of sp2 hybridized orbitals, and the pi bond is formed by the lateral overlap of unhybridized p orbitals.

5. Example Applications

Combustion Example:
- For the combustion of ethane ( $C 2 H 6$ ): $C 2 H 6 + \frac{7}{2} O 2 \to 2 CO 2 + 3 H 2 O$
- Calculate the amount of oxygen needed and the energy released using the enthalpy of combustion.
Stoichiometric Calculations:
- To determine the amount of product formed, for instance, in the complete combustion of propane ( $C 3 H 8$ ), use the stoichiometric coefficients from the balanced equation.

6. Common Mistakes

Incorrect Use of General Formula:
- Students often misapply the general formula for alkanes, forgetting to adjust for functional groups like double or triple bonds. Always confirm whether the compound is an alkane, alkene, or alkyne before using the general formula.
Neglecting Bond Angles in Geometry:
- When determining molecular geometry, students might ignore bond angles. For example, in methane ( $C H 4$ ), the bond angle is 109.5°, not 90° as sometimes incorrectly assumed.

Final Review and Tips for NEET

Focus on mastering the stoichiometry related to hydrocarbon reactions as these frequently appear in NEET exams.
Understand the derivations of key formulae, such as those for calculating the number of hydrogen atoms in hydrocarbons and energy released during combustion.
Practice identifying common mistakes, such as incorrect use of molecular formulas or misinterpreting the conditions for specific reactions.

This summary provides a concise yet comprehensive review of the critical formulae and concepts from the chapter on hydrocarbons, tailored for NEET preparation.