The hybridization of orbitals of N atom in $N{O}_3^{-},<mspace\; width="1em"></mspace>N{O}_2^{+,\hspace{0.17em}}and<mspace\; width="1em"></mspace>N{H}_4^{+}$ are respectively:

1. $sp,<mspace\; width="1em"></mspace>s{p}^2,<mspace\; width="1em"></mspace>s{p}^3$

2. $s{p}^2,<mspace\; width="1em"></mspace>sp,<mspace\; width="1em"></mspace>s{p}^3$

3. $sp,<mspace\; width="1em"></mspace>s{p}^3,<mspace\; width="1em"></mspace>s{p}^2$

4. $s{p}^2,<mspace\; width="1em"></mspace>s{p}^3,<mspace\; width="1em"></mspace>sp$

Xe F${}_2$ is isostructural with : 

1. ${\mathrm{SbCl}}_3$

2. ${\mathrm{BCl}}_3$

3. ${\mathrm{TeF}}_2$

4. ${\mathrm{ICl}}_2^{-}$

Given the following hydrocarbon:

The state of hybridisation of carbons marked 2, 3, 5 and 6 in the following hydrocarbon are, respectively :

1. sp, sp³, sp² and sp³

2. sp³, sp², sp² and sp

3. sp, sp², sp² and sp³

4. sp, sp² sp³ and sp²

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

1. sp > sp² < sp³             

2. sp > sp² > sp³

3. sp < sp² > sp³             

4. sp < sp² < sp³

Which of the following species has a bond angle of \(120^\circ\)?
1. \(\mathrm{{SF}_6}\)
2. \(\mathrm{NCl_3}\)
3. \(\mathrm{BCl_3}\)
4. \(\mathrm{PH_3}\)

Match the compounds of Xe in column I with the molecular structure in column II.

Match the compounds given in Column I with the hybridization and shape given in Column II and mark the correct option. 

Mg₂C₃ reacts with water forming propyne ${\mathrm{C}}_3^{4-}$ . The number of sigma and pie bonds in \(C_{3}^{4-}\)is-  

1. Two sigma and two pi bonds.             

2. Three sigma and one pi bonds.           

3. Two sigma and one pi bonds.          

4. Two sigma and three pi bonds.