QBManager

Hydrocarbons

230690 One mole of alkene $X$ on ozonolysis gave one mole of acetaldehyde and one mole of acetone. The IUPAC name of $X$ is

1 2-methyl-2-butene

2 2-methyl-1-butene

3 2-butene

4 1-butene

Explanation:

: To determine the structure of alkene that undergoes ozonolysis, add the products and replace $O$ by double $(=)$ bond. Thus,
original image

VITEEE- 2009

Hydrocarbons

230691 What is the hybridization of oxygen atom in an alcohol molecule?

1

{sp}^{3}

2

sp

3

{sp}^{2}

4

{sp}^{2} d

Explanation:

: Oxygen atom in alcohol molecule is ${sp}^{3}$ hybridised.

VITEEE- 2006

Hydrocarbons

230693 Which of the d-orbital is used in $s p^{3} d$ hybridization?

1

d_{x y}

2

d_{x^{2} - y^{2}}

3

d_{z^{2}}

4

d_{y z}

Explanation:

: Hybridisation is defined as concept of mixing two atomic orbital's having the same energy levels to give a new degenerate orbital's. In ${sp}^{3} d$ hybridisation five orbitals are degenerate that is their energies are the same but configurations are different. In this ${sp}^{3} d$ hybridisation there are a total 5 orbital's which are $s, px$ , py, pz and ${dz}^{2}$ orbital take part in hybridisation.

AMU-2011

Hydrocarbons

230694 In the compound
${CH}_{2} = CH - {CH}_{2} - {CH}_{2} - C \equiv CH$ , the pair of hybridized which involved in the formation of $C_{2} - C_{3}$ bond is

1

sp - {sp}^{2}

2

s p - sp p^{3}

3

s p^{2} - s p^{3}

4

{sp}^{3} - {sp}^{3}

Explanation:

: ${CH}_{3} = CH - {CH}_{2} - {CH}_{2} - C \equiv CH$
Hybridisation at $C_{2} = {sp}^{2}$ and $C_{3} = {sp}^{3}$

Assam CEE-2020

Hydrocarbons

230690 One mole of alkene $X$ on ozonolysis gave one mole of acetaldehyde and one mole of acetone. The IUPAC name of $X$ is

1 2-methyl-2-butene

2 2-methyl-1-butene

3 2-butene

4 1-butene

Explanation:

: To determine the structure of alkene that undergoes ozonolysis, add the products and replace $O$ by double $(=)$ bond. Thus,
original image

VITEEE- 2009

Hydrocarbons

230691 What is the hybridization of oxygen atom in an alcohol molecule?

1

{sp}^{3}

2

sp

3

{sp}^{2}

4

{sp}^{2} d

Explanation:

: Oxygen atom in alcohol molecule is ${sp}^{3}$ hybridised.

VITEEE- 2006

Hydrocarbons

230693 Which of the d-orbital is used in $s p^{3} d$ hybridization?

1

d_{x y}

2

d_{x^{2} - y^{2}}

3

d_{z^{2}}

4

d_{y z}

Explanation:

: Hybridisation is defined as concept of mixing two atomic orbital's having the same energy levels to give a new degenerate orbital's. In ${sp}^{3} d$ hybridisation five orbitals are degenerate that is their energies are the same but configurations are different. In this ${sp}^{3} d$ hybridisation there are a total 5 orbital's which are $s, px$ , py, pz and ${dz}^{2}$ orbital take part in hybridisation.

AMU-2011

Hydrocarbons

230694 In the compound
${CH}_{2} = CH - {CH}_{2} - {CH}_{2} - C \equiv CH$ , the pair of hybridized which involved in the formation of $C_{2} - C_{3}$ bond is

1

sp - {sp}^{2}

2

s p - sp p^{3}

3

s p^{2} - s p^{3}

4

{sp}^{3} - {sp}^{3}

Explanation:

: ${CH}_{3} = CH - {CH}_{2} - {CH}_{2} - C \equiv CH$
Hybridisation at $C_{2} = {sp}^{2}$ and $C_{3} = {sp}^{3}$

Assam CEE-2020

Hydrocarbons

230690 One mole of alkene $X$ on ozonolysis gave one mole of acetaldehyde and one mole of acetone. The IUPAC name of $X$ is

1 2-methyl-2-butene

2 2-methyl-1-butene

3 2-butene

4 1-butene

Explanation:

: To determine the structure of alkene that undergoes ozonolysis, add the products and replace $O$ by double $(=)$ bond. Thus,
original image

VITEEE- 2009

Hydrocarbons

230691 What is the hybridization of oxygen atom in an alcohol molecule?

1

{sp}^{3}

2

sp

3

{sp}^{2}

4

{sp}^{2} d

Explanation:

: Oxygen atom in alcohol molecule is ${sp}^{3}$ hybridised.

VITEEE- 2006

Hydrocarbons

230693 Which of the d-orbital is used in $s p^{3} d$ hybridization?

1

d_{x y}

2

d_{x^{2} - y^{2}}

3

d_{z^{2}}

4

d_{y z}

Explanation:

: Hybridisation is defined as concept of mixing two atomic orbital's having the same energy levels to give a new degenerate orbital's. In ${sp}^{3} d$ hybridisation five orbitals are degenerate that is their energies are the same but configurations are different. In this ${sp}^{3} d$ hybridisation there are a total 5 orbital's which are $s, px$ , py, pz and ${dz}^{2}$ orbital take part in hybridisation.

AMU-2011

Hydrocarbons

230694 In the compound
${CH}_{2} = CH - {CH}_{2} - {CH}_{2} - C \equiv CH$ , the pair of hybridized which involved in the formation of $C_{2} - C_{3}$ bond is

1

sp - {sp}^{2}

2

s p - sp p^{3}

3

s p^{2} - s p^{3}

4

{sp}^{3} - {sp}^{3}

Explanation:

: ${CH}_{3} = CH - {CH}_{2} - {CH}_{2} - C \equiv CH$
Hybridisation at $C_{2} = {sp}^{2}$ and $C_{3} = {sp}^{3}$

Assam CEE-2020

Hydrocarbons

230690 One mole of alkene $X$ on ozonolysis gave one mole of acetaldehyde and one mole of acetone. The IUPAC name of $X$ is

1 2-methyl-2-butene

2 2-methyl-1-butene

3 2-butene

4 1-butene

Explanation:

: To determine the structure of alkene that undergoes ozonolysis, add the products and replace $O$ by double $(=)$ bond. Thus,
original image

VITEEE- 2009

Hydrocarbons

230691 What is the hybridization of oxygen atom in an alcohol molecule?

1

{sp}^{3}

2

sp

3

{sp}^{2}

4

{sp}^{2} d

Explanation:

: Oxygen atom in alcohol molecule is ${sp}^{3}$ hybridised.

VITEEE- 2006

Hydrocarbons

230693 Which of the d-orbital is used in $s p^{3} d$ hybridization?

1

d_{x y}

2

d_{x^{2} - y^{2}}

3

d_{z^{2}}

4

d_{y z}

Explanation:

: Hybridisation is defined as concept of mixing two atomic orbital's having the same energy levels to give a new degenerate orbital's. In ${sp}^{3} d$ hybridisation five orbitals are degenerate that is their energies are the same but configurations are different. In this ${sp}^{3} d$ hybridisation there are a total 5 orbital's which are $s, px$ , py, pz and ${dz}^{2}$ orbital take part in hybridisation.

AMU-2011

Hydrocarbons

230694 In the compound
${CH}_{2} = CH - {CH}_{2} - {CH}_{2} - C \equiv CH$ , the pair of hybridized which involved in the formation of $C_{2} - C_{3}$ bond is

1

sp - {sp}^{2}

2

s p - sp p^{3}

3

s p^{2} - s p^{3}

4

{sp}^{3} - {sp}^{3}

Explanation:

: ${CH}_{3} = CH - {CH}_{2} - {CH}_{2} - C \equiv CH$
Hybridisation at $C_{2} = {sp}^{2}$ and $C_{3} = {sp}^{3}$

Assam CEE-2020