The correct statements among the following is/are

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Most five-coordinate compounds adopt either the Trigonal bipyramidal (TBP) or square pyramidal(SP structures).

When the central atom is a main group element, the axial bonds are longer than the equatorial bonds in TBP structure but reverse is the case for the SP structure

When the central atom is a transition element in TBP $(e x . C u C l_{5}^{3 -})$ the axial $C u - C l$ bonds are shorter than equatorial ones but on the other hand in sp $(e x . N i {(C N)}_{5}^{3 -})$ the axial $N i - C$ bonds are longer than the equatorial ones.

In $B r F_{5}$ the bromine atom is below the basal plane of square pyramid but in $N i {(C N)}_{5}^{3 -}$ the $N i$ atom is above the basal plane

answer is A, B, C, D.

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Detailed Solution

The situation in B is well known and may be explained in terms of $s p^{3} d$ hybridization or the VSEPR theory.
For $N i {(C N)}_{5}^{3 -}$ in (C) the highest d-orbital $d_{x^{2} - y^{2}}$ is empty, thus allowing the basal lingads to move in closer to the metal. For $C u C l_{5}^{3 -}$ the highest d-orbital $d_{z^{2}}^{}$ is only half filled leading to shorter axial bonds.
The lone pair in $B r F_{5}^{}$ is below the basal plane and repel the in-plane ligands upward. On the other hand no such effect is found in $N i {(C N)}_{5}^{3 -}$ . In addition ligands repulsions will force the basal ligands downward.