The ion that has sp³d² hybridization for the central atom is:

1. [ICl₄]^–

2. [ICl₂]^–

3. [BrF₂]^–

4. [IF₆]^–

1. $N_2^{2-}$

2. O₂

3. $C_2^{2-}$

4. $O_2^{2-}$

The correct statement about ICl₅ and  ICl₄^- is:

During the change of O₂ to $O_2^{-}$, the incoming electron that goes to the orbital is:

Among the following, the molecule expected to be stabilized by anion formation is : C₂, O₂, NO, F₂

1. NO
2. O₂
3. C₂
4. F₂

The shape/structure of [XeF₅]^– and XeO₃F₂, respectively, are :

1. Pentagonal planar and trigonal bipyramidal

2. Trigonal bipyramidal and pentagonal planar

3. Octahedral and square pyramidal

4. Trigonal bipyramidal and trigonal bipyramidal

Of the species, NO , NO⁺, NO²⁺ and NO^– , the one with minimum bond strength is: 

1.  NO²⁺

2.  NO⁺

3.  NO 

4.  NO^–

The molecule in which hybrid MOs involve only one d-orbital of the central atom is:

1. \(X e F_{4}\)

2. \(\left[ N i (CN)_{4} \right]^{2 -}\)

3. \(B r F_{5}\)

4. \(\left[ C r F_{6} \right]^{3 -}\)

The structure of PCl₅ in the solid state is: 

1. Square pyramidal.
2. Tetrahedral [PCl₄]⁺ and octahedral [PCl₆]^–
3. Square planar [PCl₄] and octahedral [PCl₆]^–
4. Trigonal bipyramidal.

The compound that has the largest H–M–H bond angle (M=N, O, S, C), is: