Question
Question: The percentage s-character of the hybrid orbitals in methane, ethene and ethyne are respectively: ...
The percentage s-character of the hybrid orbitals in methane, ethene and ethyne are respectively:
(A) 25, 33, 50
(B) 25, 50, 75
(C) 50, 75, 100
(D) 10, 20, 40
Solution
Methane has 1 carbon atom attached to 4 H atoms giving it a tetrahedral shape. While ethene has a double bond between both C atoms and ethyne has a triple bond. Now find their hybridisation. The type of hybridisation will itself give you the percentage s-character.
Formula used: Formula for finding out the hybridization of any molecule:
H=1/2[V+M−C+A]
Where, V = number of valence electrons;
M = monovalent atoms;
C = positive charge;
A = negative charge.
Complete step by step answer:
First we need to find out the hybridisation of all the 3 compounds.
-Let’s begin with methane. Methane has the molecular formula of: CH4. In this the carbon atom has 4 electrons that are available for bonding and all 4 of them are bonded to a H atom each. So, the hybridisation will be sp3. Even though by imagining the structure we get the hybridisation we can find it out using the formula also.
Using above given formula we can find the hybridisation: H=1/2[V+M−C+A]
For CH4: V = 4, M = 4, C = 0, A = 0.
H = 21[4+4+0+0]
= 4
Since the hybridisation number is 4 so the hybridisation will be sp3. The hybridisation type sp3 has 25 % s character.
-For ethene: It’s molecular formula is: C2H4. The 2 C atoms are connected to each other by a double bond (1 sigma bond and 1 pi bond) and each carbon is also attached to 2 H atoms. This means that 1 atom of C has 3 bonding domains and hence its hybridisation number will be 3. So, it has sp2 type hybridisation. For this the percentage character is 33%.
-For ethyne: It’s molecular formula is C2H2. In this both the carbons are bonded to each other by a triple bond (1 sigma bond and 2 pi bonds) and each carbon is also bonded with 1 H atom each through a sigma bond. This means that each carbon atom has 2 bonding domains and hence its hybridisation is sp. For this the percentage s- character is 50%.
So, the correct option is: (A) 25, 33, 50
Note: Remember that sp has 50% s character because we are talking about (1s+ 1p) orbitals, sp2 has 33% s-character due to (1s+2p) orbitals and in the same way sp3has 25% s-character. The same way it goes on for others.