Ncert Exercises & Solutions

The colour of the coordination compounds depends on the crystal field splitting. What will be the correct order of absorption of the wavelength of light in the visible region for the complexes?

$1. \ \left[\mathrm{Co}(\mathrm{CN})_{6}\right]^{3-}>\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}>\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+} \\ 2. \ \left[\mathrm{Co}(\mathrm{CN})_{6}\right]^{3-}>\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}>\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+} \\ 3. \ \left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}>\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}>\left[\mathrm{Co}(\mathrm{CN})_{6}\right]^{3-} \\ 4. \ \left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}>\left[\mathrm{Co}(\mathrm{CN})_{6}\right]^{3-}>\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}$

Hint: Crystal field splitting depends on the nature of ligand

The splitting of d orbital depends on the interaction between ligand and metal ion. Strong field ligand interacts, strongly with metal as compare to weak field ligand.

Hence, $∆ E = \frac{h c}{λ} \propto$
$∆ E \propto \frac{1}{λ}$
More is the splitting energy, lesser is the wavelength use. Splitting energy depends on the strength of ligand. Thus, strength of ligand increases, ∆ E increases, CFSE increases and λ absorbed decreases.

Thus, the correct order is

${[C o {(H_{2} O)}_{6}]}^{3 +} > {[C o {(N H_{3})}_{6}]}^{3 +} > {[C o {(C N)}_{6}]}^{3 -}$

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