Hybridisation Structure Of Carbon Compounds

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

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

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

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

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

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)

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

1. $\mathrm{C}-\mathrm{C}<\mathrm{C}=\mathrm{C}<\mathrm{C}-\mathrm{O}<\mathrm{C}-\mathrm{H}$

2. $\mathrm{C}-\mathrm{O}<\mathrm{C}-\mathrm{H}<\mathrm{C}-\mathrm{C}<\mathrm{C}=\mathrm{C}$

3. $\mathrm{C}-\mathrm{H}<\mathrm{C}-\mathrm{O}<\mathrm{C}-\mathrm{C}<\mathrm{C}=\mathrm{C}$

4. $\mathrm{C}-\mathrm{H}<\mathrm{C}=\mathrm{C}<\mathrm{C}-\mathrm{O}<\mathrm{C}-\mathrm{C}$

The radical,  is aromatic because it has:
1. 7 p-orbitals and 6 electrons
2. 7 p-orbitals and 7 electrons
3. 6 p-orbitals and 7 electrons
4. 6 p-orbitals and 6 electrons

The number of sigma ($\mathrm{\backslash sigma}$) and pi ($\mathrm{\backslash pi}$) bonds in pent-3-en-1-yne are:

1. 13 $\mathrm{\backslash sigma}$ bonds and no $\mathrm{\backslash pi}$ bond

2. 10 $\mathrm{\backslash sigma}$ bonds and 3 $\mathrm{\backslash pi}$ bonds

3. 8 $\mathrm{\backslash sigma}$ bonds and 5 $\mathrm{\backslash pi}$ bonds

4. 11 $\mathrm{\backslash sigma}$ bonds and 2 $\mathrm{\backslash pi}$ bonds

The number of bonds, 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

Match List-I with List-II.

	List-I (Molecule)		List-II (Number and types of bond/s between two carbon atoms)
A.	Ethane	I.
B.	Ethene	II.
C.	Carbon molecule,	III.
D.	Ethyne	IV.

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