BF₃ is a planar and an electron deficient compound. Hybridization and number of electrons around the central atom, respectively are: 

Which of the compounds given below will exhibit a linear structure?

Match the coordination number and type of hybridization with the distribution of hybrid orbitals in space based on Valence bond theory.

Consider the given compound:


In the given options, the following for the above compound are, respectively:
(i) sp² hybridised carbon atoms
(ii) $\mathrm{\pi }$ bonds 

The hybridisations of atomic orbitals of nitrogen in \(\mathrm{NO}^{+}\), \(N O_3^{-}\) and NH₃ respectively are:

1. $\mathrm{sp},$ ${\mathrm{sp}}^3$ $\mathrm{and}$ ${\mathrm{sp}}^2$

2. ${\mathrm{sp}}^2,$ ${\mathrm{sp}}^3$ $\mathrm{and}$ $\mathrm{sp}$

3. $\mathrm{sp},$ ${\mathrm{sp}}^2$ $\mathrm{and}$ ${\mathrm{sp}}^3$

4. ${\mathrm{sp}}^2,$ $\mathrm{sp}$ $\mathrm{and}$ ${\mathrm{sp}}^3$

Which of the following pairs are isoelectronic and isostructural?

1. ${\mathrm{CO}}_3^{2-},$ ${\mathrm{NO}}_3^{-}$

2. ${\mathrm{ClO}}_3^{-},$ ${\mathrm{CO}}_3^{2-}$

3. ${\mathrm{SO}}_3^{2-},$ ${\mathrm{NO}}_3^{-}$

4. ${\mathrm{ClO}}_2^{-},$ ${\mathrm{SO}}_3^{2-}$

The correct shape and hybridization for XeF₄ are:

1. Octahedral, sp³d²

2. Trigonal bipyramidal, sp³d³

3. Planar triangle, sp³d³

4. Square planar, sp³d²

In which orbital is the pair of electrons located in the provided carbanion, CH₃C≡C^– ?

1. sp³
2. sp²
3. sp
4. 2p