In the organic compound${\mathrm{CH}}_2=\mathrm{CH}–{\mathrm{CH}}_2–{\mathrm{CH}}_2–\mathrm{C}\equiv \mathrm{CH}$, the pair of hydridized orbitals involved in the formation of C₂ – C₃ bond is-
1. sp – ${\mathrm{sp}}^2$ 

2. sp – ${\mathrm{sp}}^3$

3. ${\mathrm{sp}}^2$ – ${\mathrm{sp}}^3$

4. ${\mathrm{sp}}^3$ – ${\mathrm{sp}}^3$ 

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

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 

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

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

The correct hybridization states of carbon atoms marked as 1,2,3 in the following compound are -$\stackrel{1}{\mathrm{ \; C}}{H}_3-\stackrel{}{C}$ $2H=\stackrel{3}{C}{H}_2$

$O$
$\left|\right|$
$\underset{1}{C}{H}_3—\underset{2}{C}—\underset{3}{C}{H}_3$

The correct hybridization  of carbon atoms are-

1. C₁= sp , C₂= sp³ , C₂= sp²

2. C₁= sp² , C₂= sp³ , C₂= sp³

3. C₁= sp³ , C₂= sp² , C₂= sp³

4. C₁= sp³ , C₂= sp³ , C₂= sp³

The correct hybridization states of carbon atoms in the following compound are - $\stackrel{1}{\mathrm{ \; \; C}}{H}_2=\stackrel{2}{C}H-\stackrel{3}{C}\equiv N$
 

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

1.  sp² 

2. sp

3. sp³

4.  All of the above