H₂O is not linear because oxygen is sp³ hybridised in H₂O. For linear shape, sp hybridisation is required.

As hydrogen molecule does not have any lone pair of electron, hence it can not form a coordinate bond but Cl₂ can.

Central atom is sp² hybridized in BF₃ and 

Hybridisation = number of sigma bonds + no of lone pairs. 

From the structure of molecules, hybridization can be known.

Two compounds are said to be isostructural, if number of bond pairs and lone pairs are same for the given pairs. 

NO^-₂ = 2σ + 1 lp = 3, ie, sp² hybridization 
NO₃^- = 3σ, i.e. sp² hybridization 
NH₂^- = 2σ + 2lp = 4, ie, sp³ hybridization 
NH⁺₄ = 4σ =4, ie, sp³ hybridization

SCN- : C has 2 sigma bonds and 2 pi bonds and hence has sp hybridisation.
Therefore, out of the given options, only NO^-₂ and NO₃^-have the same hybridization.

For sp2 hybridization, there must be 3σ-bonds or 2σ-bonds along with a lone pair of electrons. 

(i) NO^-₂ = 2σ + 1 lp = 3, ie, sp² hybridization 

(ii) NH₃ = 3σ + 1 lp = 4, ie, sp³ hybridization 

(iii) BF₃ = 3σ + 0 lp = 3, ie, sp² hybridization 

(iv) NH^-₂ = 2σ + 2 lp = 4, ie, sp³ hybridization 

(v) H₂O = 2σ + 2 lp = 4, ie, sp³ 

3σ-bonds or 2σ-bonds and a lone pair must be present in the sp2 hybridization. Therefore, BF₃ and NO^-₂ have sp² hybridization.

more the % s character more the electronegativity. So the order of electronegativity of carbon in sp, sp2 and sp3 hybridized states is sp>sp2>sp3

Note that a multiple bond (double or triple bond) is shorter than the corresponding single bond. 

The C-atom in is sp² hybridized as shown below

The C-atom in CO₂ is sp hybridized with bond length of carbon-oxygen bond as 122 pm. 
O _{=C =O} ↔ ⁺O≡C -_O^- ↔ ^-O—C ≡ 

So the correct order of bond length is CO < CO₂<