The correct hybridization states of carbon atoms in C₆H₆ is/are:

1.  sp² 

2. sp

3. sp³

4.  All of the above

During estimation of nitrogen present in an organic compound, the ammonia evolved from 0.5 g of the compound in Kjeldahl’s estimation of nitrogen, neutralized 10 mL of 1 M H₂SO₄. The percentage of nitrogen in the compound is-

1. 46.0 %

2. 51.0%

3. 56.0%

4. 49.0%

In sulphur estimation, 0.157 g of an organic compound gave 0.4813 g of barium sulphate. The percentage of sulphur in the compound is-

1. 39.10 %

2. 48.13 %

3. 42.10 %

4. 52. 43 %

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

$\stackrel{1}{C}{H}_2=\stackrel{2}{C}H-\stackrel{3}{C}\equiv N$

1. C¹= sp , C²= sp³ , C³= sp²

2. C¹= sp² , C²= sp³ , ${\mathrm{C}}^3$= sp³

3. C¹= sp² , C²= sp² , ${\mathrm{C}}^3$= sp

4. C¹= sp³ , C²= sp³ , ${\mathrm{C}}^3$ = sp³

The correct hybridization states of carbon atoms marked as 1,2,3 in the following compound are -

$\stackrel{1}{C}{H}_3-\stackrel{}{C}$ $2H=\stackrel{3}{C}{H}_2$

1. C₁= sp , C₂= sp³ , C₃= sp²

2. C₁= sp² , C₂= sp³ , C₃= sp³

3. C₁= sp³ , C₂= sp² , C₃= sp²

4. C₁= sp³ , C₂= sp³ , C₃= sp³

$\stackrel{1}{C}{H}_2$ $=\stackrel{2}{C}=O$

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

1. C₁= sp , C₂= sp³

2. C₁= sp² , C₂= sp

3. C₁= sp³ , C₂= sp

4. C₁= sp , C₂= sp²

Chlorine atom can be classified as -

1. Carbocation

2. Nucleophile

3. Electrophile

4. Carbanion

The number of σ and π bonds in the molecule ${\mathrm{CH}}_2{\mathrm{Cl}}_2$ are -

1.  2 = (${\sigma }_{C-Cl})$ , 1= $\left({\sigma }_{C-H}\right)$ , and 1= π

2. 2 = (${\sigma }_{C-Cl})$ , 2= $\left({\sigma }_{C-H}\right)$ , and 0= π

3. 2 = (${\sigma }_{C-Cl})$ , 1= $\left({\sigma }_{C-H}\right)$ , and 1= π

4. 2 = (${\sigma }_{C-Cl})$ , 1= $\left({\sigma }_{C-H}\right)$ , and 1= π

The number of σ and π bonds in the molecule $C_6{H}_{12}$ are-

1. 7 = (${\sigma }_{C-C})$ , 11 =  $\left({\sigma }_{C-H}\right)$, and 0= π 

2. 6 = (${\sigma }_{C-C})$ , 12 =  $\left({\sigma }_{C-H}\right)$, and 0= π 

3. 12 = (${\sigma }_{C-C})$ , 6 =  $\left({\sigma }_{C-H}\right)$, and 1= π 

4. 5 = (${\sigma }_{C-C})$ , 13 =  $\left({\sigma }_{C-H}\right)$, and 1= π 

The number of σ and π bonds in the molecule $C_6{H}_6$ are -