Hybridisation & Structure of Carbon Compounds Past Year (2006 - 2015) MCQs Organic Chemistry - Some Basic Principles And Techniques Chemistry NEET Practice Questions, MCQs, Past Year Questions (PYQs), NCERT Questions, Question Bank, Class 11 and Class 12 Questions, and PDF solved with answers

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 |

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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 |

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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 |

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The hybridisation of benzyl carbonium ion

is:

1.	sp²	2.	spd²
3.	sp²d	4.	sp³

Subtopic: Hybridisation & Structure of Carbon Compounds |

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The outer orbitals of C in an ethene molecule can be considered to be hybridized to given three equivalent sp² orbitals. The total number of sigma ($\sigma$) and pi($\pi$) bonds in ethene molecule are:

1. 3 sigma ($\sigma$) and 2 pi($\pi$) bonds
2. 4 sigma ($\sigma$) and 1 pi($\pi$) bonds
3. 5 sigma ($\sigma$) and 1 pi($\pi$) bonds
4. 1 sigma ($\sigma$) and 2 pi($\pi$) bonds

Subtopic: Hybridisation & Structure of Carbon Compounds |

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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 |

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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 |

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The correct order regarding the electronegativity of hybrid orbitals of carbon is:

1. $s p > s p^{2} < s p^{3}$

2. $s p > s p^{2} > s p^{3}$

3. $s p < s p^{2} > s p^{3}$

4. $s p < s p^{2} < s p^{3}$

Subtopic: Hybridisation & Structure of Carbon Compounds |

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