The [CrCl6]3 ion has a maximum in its absorption spectrum at 735 nm. Calculate the crystal field splitting energy (in kJ/mol) for this ion.

Question

Carl Swisher · Accepted Answer

So here, we have the maximum absorbance of [CrCl6]3− is 735 nm or 735×10−7m. And we are required to find its crystal-field splitting energy in kJmol. If we calculate the energy of the photon that is absorbed(E) we are also calculating the crystal-field splitting energy (△0) since they are equal. So. we have: E=△0h⋅ν And ν=cν; △0=h⋅cν; Where: △0 -crystal-field splitting energy h-Plane&#039;s constant (6.626×10−34J−s) c-Speed of light (3.0×108ms) λ -wavelength (7.35×107m) △0=6.626×10−34J⋅s(3.0×108ms7.35×10−7m) △0=2.70×10−19J 2.70×10−19J is the energy of only one ion. So, to get the crystal-field splitting energy in kJ/mol. We need to convert J to kJ(I000)=IkJ) and multiply the energy to Avogadro&#039;s number which is 6.023×1023. So, we have △0=2.70×10−19J×1kJ1000J(×6.023×10231mol) △0=162.621kJmol

Thomas White · Accepted Answer

△0=hcλ=(6.626×10−34J⋅s)(2.99×108ms)735×10−9m=2.70×10−19J  This is the energy per photon change to per mol of photons  (2.70×10−19Jphoton)(6.02214×1023photons1mol)(1kJ1000J)=162kJmol

The [C_{r}Cl_{6}]^{3} ion has a maximum in its absorption spectrum

Answered question

Answer & Explanation

New Questions in Other