
Which one among the following possesses an sp hybridised carbon in its structure
A) \[{\rm{C}}{{\rm{H}}_{\rm{2}}} = {\rm{C}}{\rm{.Cl}} - {\rm{CH}} = {\rm{C}}{{\rm{H}}_{\rm{2}}}\]
B) \[{\rm{C}}{\rm{.C}}{{\rm{l}}_{\rm{2}}} = {\rm{C}}{\rm{.C}}{{\rm{l}}_{\rm{2}}}\]
C) \[{\rm{C}}{{\rm{H}}_{\rm{2}}} = {\rm{C}} = {\rm{C}}{{\rm{H}}_{\rm{2}}}\]
D) \[{\rm{C}}{{\rm{H}}_{\rm{2}}} = {\rm{CH}} - {\rm{CH}} = {\rm{C}}{{\rm{H}}_{\rm{2}}}\]
Answer
163.2k+ views
Hint: An hp hybridised carbon atom is one in which the carbon atom is surrounded by two groups. The surrounding atom can be the atoms that are covalently bonded with the atom and the lone pair of the atom.
Complete Step by Step Answer:
Let’s discuss all the options one by one.
Option A is\[\mathop {{\rm{C}}{{\rm{H}}_{\rm{2}}}}\limits_{s{p^2}} = \mathop {\rm{C}}\limits_{s{p^2}} {\rm{.Cl}} - \mathop {\rm{C}}\limits_{s{p^2}} {\rm{H}} = \mathop {\rm{C}}\limits_{s{p^2}} {{\rm{H}}_{\rm{2}}}\]. Here, we see that all the carbon atoms are surrounded by three groups, so, all the carbon atoms are \[s{p^2}\] hybridised. Therefore, no sp hybridised carbon atom is present in the compound.
Option B is \[\mathop {{\rm{C}}{\rm{.}}}\limits_{s{p^2}} {\rm{C}}{{\rm{l}}_{\rm{2}}} = \mathop {\rm{C}}\limits_{s{p^2}} {\rm{.C}}{{\rm{l}}_{\rm{2}}}\]. Here, both the carbon atoms are \[s{p^2}\]hybridised because they are surrounded by three groups. So, no sp hybridised carbon atom is present in the compound.
Option C is \[\mathop {\rm{C}}\limits_{s{p^2}} {{\rm{H}}_{\rm{2}}} = \mathop {\rm{C}}\limits_{sp} = \mathop {\rm{C}}\limits_{s{p^2}} {{\rm{H}}_{\rm{2}}}\]. Here, C1 is \[s{p^2}\]hybridised because it is surrounded by three groups. C2 is sp hybridised as it is surrounded by two atoms and C3 is \[s{p^2}\]hybridised because it is surrounded by three groups. So, one sp hybridised C atom is present in the compound.
Option D is \[{\rm{C}}{{\rm{H}}_{\rm{2}}} = {\rm{CH}} - {\rm{CH}} = {\rm{C}}{{\rm{H}}_{\rm{2}}}\]. Here, C1 is \[s{p^2}\]hybridised because it is surrounded by two groups. C2 is \[s{p^2}\] hybridised as it is surrounded by three atoms and C3 is \[s{p^2}\]hybridised because it is surrounded by three groups and C4 is \[s{p^2}\] hybridised. So, no sp hybridised C atom is present.
Therefore, option C is right.
Note: It is to be noted that, while counting the surrounding groups of an atom to find its hybridization, the double bond or triple-bonded atom is to be counted as one group and not as two groups.
Complete Step by Step Answer:
Let’s discuss all the options one by one.
Option A is\[\mathop {{\rm{C}}{{\rm{H}}_{\rm{2}}}}\limits_{s{p^2}} = \mathop {\rm{C}}\limits_{s{p^2}} {\rm{.Cl}} - \mathop {\rm{C}}\limits_{s{p^2}} {\rm{H}} = \mathop {\rm{C}}\limits_{s{p^2}} {{\rm{H}}_{\rm{2}}}\]. Here, we see that all the carbon atoms are surrounded by three groups, so, all the carbon atoms are \[s{p^2}\] hybridised. Therefore, no sp hybridised carbon atom is present in the compound.
Option B is \[\mathop {{\rm{C}}{\rm{.}}}\limits_{s{p^2}} {\rm{C}}{{\rm{l}}_{\rm{2}}} = \mathop {\rm{C}}\limits_{s{p^2}} {\rm{.C}}{{\rm{l}}_{\rm{2}}}\]. Here, both the carbon atoms are \[s{p^2}\]hybridised because they are surrounded by three groups. So, no sp hybridised carbon atom is present in the compound.
Option C is \[\mathop {\rm{C}}\limits_{s{p^2}} {{\rm{H}}_{\rm{2}}} = \mathop {\rm{C}}\limits_{sp} = \mathop {\rm{C}}\limits_{s{p^2}} {{\rm{H}}_{\rm{2}}}\]. Here, C1 is \[s{p^2}\]hybridised because it is surrounded by three groups. C2 is sp hybridised as it is surrounded by two atoms and C3 is \[s{p^2}\]hybridised because it is surrounded by three groups. So, one sp hybridised C atom is present in the compound.
Option D is \[{\rm{C}}{{\rm{H}}_{\rm{2}}} = {\rm{CH}} - {\rm{CH}} = {\rm{C}}{{\rm{H}}_{\rm{2}}}\]. Here, C1 is \[s{p^2}\]hybridised because it is surrounded by two groups. C2 is \[s{p^2}\] hybridised as it is surrounded by three atoms and C3 is \[s{p^2}\]hybridised because it is surrounded by three groups and C4 is \[s{p^2}\] hybridised. So, no sp hybridised C atom is present.
Therefore, option C is right.
Note: It is to be noted that, while counting the surrounding groups of an atom to find its hybridization, the double bond or triple-bonded atom is to be counted as one group and not as two groups.
Recently Updated Pages
JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

JEE Energetics Important Concepts and Tips for Exam Preparation

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Types of Solutions

Degree of Dissociation and Its Formula With Solved Example for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Other Pages
NCERT Solutions for Class 12 Chemistry Chapter 1 Solutions

JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Solutions Class 12 Notes: CBSE Chemistry Chapter 1

NCERT Solutions for Class 12 Chemistry Chapter 6 Haloalkanes and Haloarenes

NCERT Solutions for Class 12 Chemistry Chapter 2 Electrochemistry

Electrochemistry Class 12 Notes: CBSE Chemistry Chapter 2
