A $[C r (H_{2} O)_{6}]^{3 -}$ complex typically absorbs at around $574 n m$ . It is allowed to react with ammonia to form a new complex $[C r (N H_{3})_{6}]^{3 +}$ that should have absorption at :

800 n m

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580 n m

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620 n m

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320 n m

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Solution

The correct option is C $580 n m$
Ammonia is stronger ligand than water. Thus the value of the CFSE of $[M (N H_{3})_{5}]^{2 +}$ is greater than the value of the CFSE of $[M (H_{2} O)_{6}]^{2 +}$ .
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 $[M (N H_{3})_{6}]^{2 +}$ should have absorption at 580 nm.

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Similar questions

The absorption maxima of some complexions are as follows:

$[C r C l_{6}]^{3 -} - 760 n m$

$[C r (N H_{3})_{6}]^{3 +} - 470 n m$

$[C r (H_{2} O)_{6}]^{3 +} - 695 n m$

The order of $Δ_{o}$ for these ions is

C r^{3 +}

[C r (C l)_{6}]^{3 -}, [C r (H_{2} O)_{6}]^{3 +}, [C r (N H_{3})_{6}]^{3 +}

[C r (C N)_{6}]^{3 -}

(Δ_{o})

{[T i {(H_{2} O)}_{6}]}_{6}^{3 +}

{[T i {(N H_{3})}_{6}]}_{6}^{3 +}

Which of the following complex ions is expected to absorb visible light?

(At number Zn = 30, Sc = 21, Ti = 22, Cr = 24)

[C r (H_{2} O)_{6}] C l_{3}

(A)

[C r (N H_{3})_{6}] C l_{3}

(B)

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