Single Choice

A $$[Cr(H_2O)_6]^{3-}$$ complex typically absorbs at around $$574 \ nm$$. It is allowed to react with ammonia to form a new complex $$[Cr(NH_3)_6]^{3+}$$ that should have absorption at :

A$$800 \ nm$$
B$$580 \ nm$$
Correct Answer
C$$620 \ nm$$
D$$320 \ nm$$

Solution

Ammonia is stronger ligand than water. Thus the value of the CFSE of $$[M(NH_3)_5]^{2+}$$ is greater than the value of the CFSE of $$[M(H_2O)_6]^{2+}$$.
Higher energy corresponds to shorter wavelength.
Thus the absorption shifts to shorter wavelength.
The shift in the wavelength is small as the difference between the strengths of ammonia and water ligands is small.
Hence, the new complex $$[M(NH_3)_6]^{2+}$$ should have absorption at 580 nm.

SIMILAR QUESTIONS

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Which of the following is diamagnetic?

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Given complexes are low spin complexes, [V(CN)6]4− [Cr(NH3)6]2+ [Ru(NH3)6]3+ [Fe(CN)6]4− then order of magnetic moment (μ) for V2+,Fe2+,Cr+3,Ru3+ is:

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