Estimate the ligand fi eld splitting for (a) [CrCl6]3 (max 740 nm), (b) [Cr(NH3)6]3 (max 460 nm), and (c) [Cr(OH2)6]3 (max 575 n
m), where max is the wavelength of the most intensely absorbed light. (d) Arrange the ligands in order of increasing ligand fi eld strength.
1 answer:
Answer:
NH3>H2O>Cl-
Explanation:
The given wavelengths of maximum absorption for each complex can be used to estimate the magnitude of field splitting of the respective ligands as shown in the image attached. The field splitting is reported in the unit kilojoule per mole (KJmol-1).
It can be seen from the calculation in the image attached that ammonia shows the highest crystal field splitting followed by water and lastly the chloride anion. This corresponds to the respective positions of these species in the spectrochemical series. Water and the chloride ion are weak field ligands.
You might be interested in
We have to know the molarity of solution obtained when 5.71 g of Na₂CO₃.10 H₂O is dissolved in water and made up to 250 cm³ solution.
The molarity of solution obtained when 5.71 g of sodium carbonate-10-water (Na₂CO₃.10 H₂O) is dissolved in water and made up to 250.0 cm^3 solutionis: (A) 0.08 mol dm⁻³
The molarit y of solution means the number of moles of solute present in one litre of solution. Here solute is Na₂CO₃.10 H₂O and solvent is water. Volume of solution is 250 cm³.
Molar mass of Na₂CO₃.10 H₂O is 286 grams which means mass of one mole of Na₂CO₃.10 H₂O is 286 grams.
5.71 grams of Na₂CO₃.10 H₂O is equal to = 0.0199 moles of Na₂CO₃.10 H₂O. So, 0.0199 moles of Na₂CO₃.10 H₂O present in 250 cm³ volume of solution.
Hence, number of moles of Na₂CO₃.10 H₂O present in one litre (equal to 1000 cm³) of solution is = 0.0796 moles. So, the molarity of the solution is 0.0796 mol/dm³ ≅ 0.08 mol/dm³