Given below are two statements: Assertion A: [CoCl(NH3)5]2+ absorbs at lower wavelength of light with respect to…

🧠 Compare Δ → Compare λ Absorbed

Larger Δ → higher-energy d–d transition → shorter absorbed wavelength.

Both complexes are Co(III). Difference is in the sixth ligand: Cl⁻ in $[\mathrm{CoCl(NH_3)_5}]^{2+}$ vs H₂O in $[\mathrm{Co(NH_3)_5(H_2O)}]^{3+}$.

Spectrochemical series: $\mathrm{Cl^-} < \mathrm{H_2O}$. So replacing H₂O with Cl⁻ lowers $\Delta_o$ → longer wavelength absorbed.

So $[\mathrm{CoCl(NH_3)_5}]^{2+}$ absorbs at higher (longer) wavelength than $[\mathrm{Co(NH_3)_5(H_2O)}]^{3+}$.

The assertion A says the opposite ("absorbs at lower wavelength"). A is FALSE by the chemistry.

🗺️ Reason R Examined

R says wavelength absorbed depends on the oxidation state of the metal.

This is a generally true statement (higher OS → larger Δ → smaller λ, holding ligand and geometry constant). For example, comparing $[\mathrm{Cr(H_2O)_6}]^{2+}$ vs $[\mathrm{Cr(H_2O)_6}]^{3+}$: the +3 has larger Δ.

But here, both complexes have Co(III) — the same OS. So R is true as a general principle, but not the relevant explanation for this comparison (where the difference is ligand identity, not OS).

⚡ Charge of Complex Reinforces the Conclusion

The $[\mathrm{CoCl(NH_3)_5}]^{2+}$ ion (charge +2) has a smaller overall positive charge than $[\mathrm{Co(NH_3)_5(H_2O)}]^{3+}$ (charge +3). Lower complex charge = weaker electrostatic attraction = smaller Δ — which also pushes [CoCl(NH₃)₅]²⁺ toward longer wavelength absorbed. Both effects (ligand and overall charge) act in the same direction.

⚠️ Watch the Direction of "Higher" vs "Lower" Wavelength

• Higher wavelength = longer wavelength = lower energy = smaller Δ.
• Lower wavelength = shorter wavelength = higher energy = larger Δ.

Mixing this up is the easiest way to mis-answer this kind of question.

$\boxed{\text{Answer: (1) A is false but R is true}}$