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  • 1(current)
Q. No.
Match List-I with List-II.
List-I
(Molecule)
List-II
(Number and types of bond/s between two carbon atoms)
A.  Ethane I.  \(\small{\mathrm{One ~\sigma-bond~ and~ Two~ \pi-bonds}}\)
B.  Ethene II.  \(\text { Two } \pi \text {-bonds }\)
C.  Carbon
molecule, \(\mathrm{C} _2\)		 III.  \(\text { One } \sigma \text {-bond }\)
D.  Ethyne IV.  \(\small{\mathrm{One ~\sigma-bond ~and ~One~ \pi-bond}}\)
Choose the correct answer from the options given below:
1. A-IV, B-III, C-II, D-I 2. A-III, B-IV, C-II, D-I
3. A-III, B-IV, C-I, D-II 4. A-I, B-IV, C-II, D-III
Subtopic:  Hybridisation & Structure of Carbon Compounds |
 75%
Level 2: 60%+
NEET - 2024
Hints
The number of \(\sigma\)  bonds, \(\pi\) bonds and lone pair of electrons in pyridine, respectively, are: 
1. 12, 2, 1
2. 11, 2, 0 
3. 12, 3, 0 
4. 11, 3, 1
Subtopic:  Hybridisation & Structure of Carbon Compounds |
 69%
Level 2: 60%+
NEET - 2023
Hints
What is the hybridization shown by C1 and C2 carbons, respectively in the given compound?
   
\(\mathrm{ OHC-CH=CH-{^2}CH_2-{^1}COOCH_3}\)
 
1. sp2 and sp3 2. sp2 and sp2
3. sp3 and sp2 4. sp3 and sp3
Subtopic:  Hybridisation & Structure of Carbon Compounds |
 66%
Level 2: 60%+
NEET - 2022
Hints
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The number of sigma (σ) and pi (π) bonds in pent-3-en-1-yne are:

1.  13 σ bonds and no π bond

2.  10 σ bonds and 3 π bonds

3.  8 σ bonds and 5 π bonds

4.  11 σ bonds and 2 π bonds

Subtopic:  Hybridisation & Structure of Carbon Compounds |
 87%
Level 1: 80%+
NEET - 2019
Hints

The enolic form of ethyl acetoacetate is given below:

 

The number of σ and π bonds in the enolic form are respectively:

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 |
 89%
Level 1: 80%+
NEET - 2015
Hints
The total number of \(\pi\)-bond electrons in the given below structure are:
1. Four (4) 2. Eight (8)
3. Twelve (12) 4. Sixteen (16)
Subtopic:  Hybridisation & Structure of Carbon Compounds |
 69%
Level 2: 60%+
NEET - 2015
Hints
Links
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The radical,  is aromatic because it has:
1. 7 p-orbitals and 6 \(\pi\) electrons
2. 7 p-orbitals and 7 \(\pi\) electrons
3. 6 p-orbitals and 7 \(\pi\) electrons
4. 6 p-orbitals and 6 \(\pi\) electrons
Subtopic:  Hybridisation & Structure of Carbon Compounds |
Level 3: 35%-60%
AIPMT - 2013
Hints

The correct order of increasing bond length of  C-H, C-O, C-C and C=C is:

1.  C-C<C=C<C-O<C-H

2.  C-O<C-H<C-C<C=C

3.  C-H<C-O<C-C<C=C

4.  C-H<C=C<C-O<C-C

Subtopic:  Hybridisation & Structure of Carbon Compounds |
 64%
Level 2: 60%+
AIPMT - 2011
Hints
Links

The following compounds are given for reference
(1) CH3-CH2–
(2) H2C=CH–
(3) HC≡C– 

The increasing order of basic strength is:
1. (2)>(1)>(3)
2. (3) >(2) >(1)
3. (1) >(3) >(2)
4. (1) >(2)>(3)

Subtopic:  Hybridisation & Structure of Carbon Compounds |
 70%
Level 2: 60%+
AIPMT - 2008
Hints
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The correct order regarding the electronegativity of hybrid orbitals of carbon is:

1. sp>sp2<sp3

2. sp>sp2>sp3

3. sp<sp2>sp3

4. sp<sp2<sp3

Subtopic:  Hybridisation & Structure of Carbon Compounds |
 78%
Level 2: 60%+
AIPMT - 2006
Hints
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