Question

The hybridization of carbon atoms in ethane doesn’t change during:
A.Pyrolysis
B.Combustion
C.Halogenation
D.Both halogenation and combustion

Answer 1

Hint: To answer this question, you should recall the concept of hybridization. Check the hybridization states of the reactants and products to find the answer to this question. The hybridization of an atom changes when the number of monovalent atoms attached to it changes. The presence of a single, double or triple bond also changes the hybridization of the central atom.

Complete Step by step solution:
Analysing each option systematically:
Pyrolysis can be defined as heating of a natural material, for example, biomass, without oxygen. Since no oxygen is available the material doesn't combust yet the compounds that form that material thermally deteriorate into flammable gases and charcoal. This option is incorrect and can be eliminated.
Combustion is defined as the chemical process in which a substance responds quickly with oxygen and emits heat. The first substance is known as the fuel, and the source that provides oxygen is known as the oxidizer. Combustion of ethane leads to the evolution of carbon dioxide. As hybridization of product and reactant C atoms is different. This option is incorrect and can be eliminated.
Halogenation is a chemical reaction in which one or more halogen to a compound. The pathway and stoichiometry of halogenation rely upon the basic highlights and useful gatherings of the natural substrate, just as on the particular halogen.
From the above definition, it is evident that the correct response is halogenation.

Hence, the correct option is option C.

Note: Even if you are not able to calculate the hybridisation using the above-mentioned you can find the hybridization $(X)$ using the formula: \[\dfrac{1}{2}(V+H-C+A)\] where
$V$= Number of valence electrons in the central atom
$H$= Number of surrounding monovalent atoms
$C$= Cationic charge
$A$= Anionic charge. The value of X will determine the hybridisation of the molecule. If $X$=2 then $sp$ ; if 3 then $s{{p}^{2}}$ ; if 4 then $s{{p}^{3}}$ ; if 5 then $s{{p}^{3}}d$ ; if 6 then $s{{p}^{3}}{{d}^{2}}$ ; if 7 then $s{{p}^{3}}{{d}^{3}}$hybridization.