Question

all the I-f bond's in IF7 are not equivalent give reason​

Answer 1

Answer:

it can help u.

Explanation:

Well, you got a seven-coordinate iodine centre, whose coordination geometry is pentagonal bipyramidal. You will have to consult your text, but I think at room temperature, this structure is conformationally mobile, i.e. there is a SOFT energy of exchange between axial and equatorial positions, and thus say the [math] ^{19}F[/math] NMR spectrum would display a SINGLET. Below room temperature (quite possibly well below) the spectrum would resolve into two absorptions (a triplet, and a sextet, why?)…corresponding to axial and equatorial positions…

Answer 2

Answer:

All the IF bonds in IF7 are not equivalent because of Jahn-Teller effect.

Explanation:

• The Jahn-Teller effect is a non-linear chemical system's geometric distortion that reduces its symmetry and energy.
• In octahedral complexes, where the two axial bonds are either shorter or longer than the equatorial bonds, this distortion is most common.
• Tetrahedral compounds exhibit this behaviour as well. This effect is determined by the system's electronic state.
• When the degeneracy is disrupted by the stabilisation (dropping in energy) of the d orbital s with a z component, while the orbitals without a z component are destabilised, elongation Jahn-Teller distortions result (higher in energy).
• To attend stability the axial bonds which are on z axis in pentagonal bipyramidal so they try to stretch or elongate or sometimes compression takes place in IF7.