Q2. (a) describe the differences in chemical bonding that can be used to distinguish a semi-bridging h2-co ligand from a symmetrical 12-co ligand? (5 marks) (b) explain why the v- o stretching frequency changes when bis(acetylacetonato)oxovanadium(iv) is dissolved in pyridine. (4 marks) (c) the interaction of na fe(co), with (ch3)2nch i gives the carbene complex (co) fechn(ch3)2. write a plausible reaction sequence. (8 marks)

Must be negative.substance increases

B"hf and f-"  is the correct answer took the test on edgenuity and got it right     

Answerthis solution will not form a precipitateexplanationsolubility equilbrium:         pbf₂(s)  ⇄ pb²⁺(aq) + 2 f⁻(aq)the solubility product expression is    (we exclude solids from the equilibrium expression).now suppose we have the trial ion product q, which we can define as    then, we can interpret  q as the product of the concentration of the ions that we have and ksp as the product of the ion concentrations that are needed to form a saturated solution.if q > ksp, then that implies that there will be a surplus of ions that exceed the amount that is required to form a saturated solution. a precipitate will then form (the precipitate continually  forms until the surplus ions are removed and q is decreased to ksp's value).if q = ksp, the saturated solution that forms is barely saturated and the precipitate will be minimal.if q < ksp, then the ion concentration will not be enough to form a saturated solution (so no precipitate).using our information, we calculate the trial ion product:     since q = 2.7  × 10⁻⁸ < ksp =  4  ×  10⁻⁸, a precipitate will not form.