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MDCAT
Chemistry
2017

Chemistry · Fundamental Principles of Organic Chemistry

Work through this past-paper style MCQ, then read the full explanation. Practice more chemistry questions on mMCQ with adaptive practice and topic analytics.

Question

What type of hybridization is/are present in Hex-4-ene 1-yne:

Options
  • A

    Sp2

  • B

    Sp

  • C

    Sp and Sp2

  • D

    Sp. Sp2. Sp3

Explanation

The first carbon atom has an sp hybridization, as it is involved in a triple bond with another carbon atom.
The second and third carbon atoms have sp2 hybridization, as they are involved in a double bond with a carbon atom and a single bond with two other carbon atoms.
The fourth carbon atom has sp hybridization, as it is involved in a triple bond with a carbon atom.
The fifth carbon atom has sp2 hybridization, as it is involved in a double bond with a carbon atom and a single bond with two other carbon atoms.
The sixth carbon atom has sp3 hybridization, as it is bonded to two other carbon atoms and two hydrogen atoms, resulting in a tetrahedral geometry.
Therefore, the correct answer is D) sp, sp2, sp, sp2, sp, sp3.

Explanation is given below.

It is an incorrect option.The first carbon atom has an sp hybridization, as it is involved in a triple bond with another carbon atom.

The second and third carbon atoms have sp2 hybridization, as they are involved in a double bond with a carbon atom and a single bond with two other carbon atoms.

The fourth carbon atom has sp hybridization, as it is involved in a triple bond with a carbon atom.

The fifth carbon atom has sp2 hybridization, as it is involved in a double bond with a carbon atom and a single bond with two other carbon atoms.

The sixth carbon atom has sp3 hybridization, as it is bonded to two other carbon atoms and two hydrogen atoms, resulting in a tetrahedral geometry.

Therefore, the correct answer is D) sp, sp2, sp, sp2, sp, sp3.

It is an incorrect option. The first carbon atom has an sp hybridization, as it is involved in a triple bond with another carbon atom.

The second and third carbon atoms have sp2 hybridization, as they are involved in a double bond with a carbon atom and a single bond with two other carbon atoms.

The fourth carbon atom has sp hybridization, as it is involved in a triple bond with a carbon atom.

The fifth carbon atom has sp2 hybridization, as it is involved in a double bond with a carbon atom and a single bond with two other carbon atoms.

The sixth carbon atom has sp3 hybridization, as it is bonded to two other carbon atoms and two hydrogen atoms, resulting in a tetrahedral geometry.

Therefore, the correct answer is D) sp, sp2, sp, sp2, sp, sp3.

It is an incorrect option. The first carbon atom has an sp hybridization, as it is involved in a triple bond with another carbon atom.


The second and third carbon atoms have sp2 hybridization, as they are involved in a double bond with a carbon atom and a single bond with two other carbon atoms.

The fourth carbon atom has sp hybridization, as it is involved in a Triple bond with a carbon atom.

The fifth carbon atom has sp2 hybridization, as it is involved in a double bond with a carbon atom and a single bond with two other carbon atoms.

The sixth carbon atom has sp3 hybridization, as it is bonded to two other carbon atoms and two hydrogen atoms, resulting in a tetrahedral geometry.

Therefore, the correct answer is D) sp, sp2, sp, sp2, sp, sp3.

The first carbon atom has an sp hybridization, as it is involved in a triple bond with another carbon atom.

The second and third carbon atoms have sp2 hybridization, as they are involved in a double bond with a carbon atom and a single bond with two other carbon atoms.

The fourth carbon atom has sp hybridization, as it is involved in a triple bond with a carbon atom.

The fifth carbon atom has sp2 hybridization, as it is involved in a double bond with a carbon atom and a single bond with two other carbon atoms.

The sixth carbon atom has sp3 hybridization, as it is bonded to two other carbon atoms and two hydrogen atoms, resulting in a tetrahedral geometry.

Therefore, the correct answer is D) sp, sp2, sp, sp2, sp, sp3.

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Tagged under Chemistry · Fundamental Principles of Organic Chemistry · 2017