Problem 5. Provide answers to the following questions. A) Oxygen is more electro

Question

Problem 5. Provide answers to the following questions. A) Oxygen is more electronegative than nitrogen, fluorine is more electronegative than the other halogens. Fluoride is a stronger field ligand than the other halides, but ammonia is a stronger field ligand than water. Why? B) Except in cases where ligand geometry requires it, square-planar geometry occurs most commonly with d^7, d^8, and d^9 ions and with strong-field pi acceptor ligands. Suggest why these conditions support square-planar geometry. C) [Co(H2O)6]^3+ is a strong oxidizing agent that will oxidize water, but [Co(NH3)6]^3+ is stable in aqueous solution. Explain this difference. D) Why are phosphine ligands (PR3), especially those with electronegative substituents, pi acceptors?

Explanation / Answer

A) Strength of ammonia and water ligands can be explained by pi bonding. Water has two lone pairs of electrons on oxygen so after donating one lone pair to metal for sigma bond it has one more lone pair of electrons. This pair can be donated to pi bonding system which leads to more number of electrons in antibonding pi orbitals. Then energy gap between HOMO (Highest Occupied Molecular Orbital) to LUMO(Lowest unoccupied molecular orbital) decreases this results lower crystal field splitting energy. Hence water is weak ligand. But ammonia has no lone pair of electrons after sigma bond is formed.

B) Square planar geometry needs dsp2 hybridization. It is only possible when electrons paired in d subshell. In d7 and d8 systems electrons filled in four d orbitals which leave an empty lower d orbital. This dx2-y2 orbital is available for bonding with lignads. But it is possible in presence of strong field ligands such as pi acceptor ligands.

C) [Co(H2O)6]3+ is weak field complex and it has lower crystal fieled stabilization energy. But [Co(NH3)6]3+ is strong field complex (low spin) complex. It has greater crystal field stabilization energy as all the six electrons filled in t2g set.

D) In phosphine P has empty d orbital which can be used accomidate the electrons accepted from metal. When PR3 has electronegative substituents electron density on P decreases which increase the pi acceptor capacity of PR3 ligand.

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