The correct hybridization states of carbon atoms in the following compound are -     C1H2=C2H-C3N
 

1. C1= sp , C2= sp, C3= sp2 2. C1= sp2 , C2= sp,  C 3 = sp3
3. C1= sp2 , C2= sp,  C 3 = sp 4. C1= sp3 , C2= sp,  C 3 = sp3
Subtopic:  Hybridisation & Structure of Carbon Compounds |
 91%
From NCERT

To unlock all the explanations of this course, you need to be enrolled.

Hints

To unlock all the explanations of this course, you need to be enrolled.


The number of σ and π bonds in the molecule CH2Cl2 are -

1.  2 = (σC-Cl) , 1= (σC-H) , and 1= π

2. 2 = (σC-Cl) , 2= (σC-H) , and 0= π

3. 2 = (σC-Cl) , 1= (σC-H) , and 1= π

4. 2 = (σC-Cl) , 1= (σC-H) , and 1= π

Subtopic:  Hybridisation & Structure of Carbon Compounds |
 93%
From NCERT

To unlock all the explanations of this course, you need to be enrolled.

Hints

To unlock all the explanations of this course, you need to be enrolled.


C1H2 =C2=O

The correct hybridization states of carbon atoms in the above compound are - 

1. C1= sp , C2= sp3 2. C1= sp, C2= sp
3. C1= sp, C2= sp 4. C1= sp , C2= sp2
Subtopic:  Hybridisation & Structure of Carbon Compounds |
 92%

To unlock all the explanations of this course, you need to be enrolled.

Hints

To unlock all the explanations of this course, you need to be enrolled.


advertisementadvertisement

The correct hybridization states of carbon atoms marked as 1,2,3 in the following compound are:
\({ }^1 \mathrm{CH}_3-\mathrm{ }^2{C} \mathrm{H}={}^3{\mathrm{C}} \mathrm{H}_2 \)

1. C1= sp , C2= sp, C3= sp2 2. C1= sp2 , C2= sp, C3= sp3
3. C1= sp3 , C2= sp, C3= sp2 4. C1= sp3 , C2= sp, C3= sp3
Subtopic:  Hybridisation & Structure of Carbon Compounds |
 93%
From NCERT

To unlock all the explanations of this course, you need to be enrolled.

Hints

To unlock all the explanations of this course, you need to be enrolled.


The number of σ and π bonds in the molecule C6H6 are -

1. 6 C – C sigma ( σ C - C )  bonds, 5 C–H sigma ( ( σ C - H )  bonds, and 3 C=C pi ( π C - C ) 
2. 6 C – C sigma ( σ C - C )  bonds, 5 C–H sigma ( ( σ C - H )  bonds, and 2 C=C pi ( π C - C ) 
3. 6 C – C sigma ( σ C - C )  bonds, 6 C–H sigma ( ( σ C - H )  bonds, and 3 C=C pi ( π C - C ) 
4. 6 C – C sigma ( σ C - C )  bonds, 6 C–H sigma ( ( σ C - H )  bonds, and 2 C=C pi ( π C - C ) 

Subtopic:  Hybridisation & Structure of Carbon Compounds |
 88%
From NCERT

To unlock all the explanations of this course, you need to be enrolled.

Hints
Links

To unlock all the explanations of this course, you need to be enrolled.


The number of primary carbon atoms in the following compound are:

1. 6 2. 2
3. 4 4. 3
Subtopic:  Hybridisation & Structure of Carbon Compounds |
 87%
From NCERT

To unlock all the explanations of this course, you need to be enrolled.

Hints
Links

To unlock all the explanations of this course, you need to be enrolled.


advertisementadvertisement

The enolic form of ethyl acetoacetate is given below. The number of sigma and pi bonds in the enolic form of ethyl acetoacetate are -

1. 18 sigma bonds and 2 pi-bonds

2. 16 sigma bonds and 1 pi-bond

3. 9 sigma bonds and 2 pi-bonds

4. 9 sigma bonds and 1 pi-bond

Subtopic:  Hybridisation & Structure of Carbon Compounds |
 86%
From NCERT
NEET - 2015

To unlock all the explanations of this course, you need to be enrolled.

Hints
Links

To unlock all the explanations of this course, you need to be enrolled.


The correct hybridization states of carbon atoms in C6His/are:

1.  sp

2. sp

3. sp3

4.  All of the above

Subtopic:  Hybridisation & Structure of Carbon Compounds |
 84%
From NCERT

To unlock all the explanations of this course, you need to be enrolled.

Hints

To unlock all the explanations of this course, you need to be enrolled.


The cylindrical shape of an alkyne is due to the presence of- 

1. Three sigma C-C bonds

2. Three π C-C bonds

3. Two sigma C-C and one π C-C bonds

4. One sigma C-C and two π C-C bonds

Subtopic:  Hybridisation & Structure of Carbon Compounds |
 80%
From NCERT

To unlock all the explanations of this course, you need to be enrolled.

Hints

To unlock all the explanations of this course, you need to be enrolled.


advertisementadvertisement

The \(C - H\) bond distance is longer in - 

1. \(C_2H_2\) 2. \(C_2H_4\)
3. \(C_2H_6\) 4. \(C_2H_2Br_2\)
Subtopic:  Hybridisation & Structure of Carbon Compounds |
 77%
From NCERT

To unlock all the explanations of this course, you need to be enrolled.

Hints

To unlock all the explanations of this course, you need to be enrolled.