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 |
 64%
From NCERT
NEET - 2023
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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 |
 55%
From NCERT
NEET - 2022
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The number of sigma (σ) and pi (π) bonds in pent-2-en-4-yne is:

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 |
 85%
From NCERT
NEET - 2019
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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 |
 88%
From NCERT
NEET - 2015
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The total number of pi-bond electrons in the following structure are:
1. 4 2. 8
3. 12 4. 16
Subtopic:  Hybridisation & Structure of Carbon Compounds |
 66%
From NCERT
NEET - 2015
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The radical,  is aromatic because it has:
1. 7 p-orbitals and 6 unpaired electrons
2. 7 p-orbitals and 7 unpaired electrons
3. 6 p-orbitals and 7 unpaired electrons
4. 6 p-orbitals and 6 unpaired electrons
Subtopic:  Hybridisation & Structure of Carbon Compounds |
From NCERT
AIPMT - 2013
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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 |
 65%
From NCERT
AIPMT - 2011
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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 |
 69%
From NCERT
AIPMT - 2008
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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 |
 76%
From NCERT
AIPMT - 2006
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