Predict which will be more stable, [CrO4 ] 2− or [WO4 ] 2−, and explain

Predict which will be more stable, [CrO4 ] 2− or [WO4 ] 2−, and explain.

The correct answer and explanation is :

The tungstate ion, ([WO_4]^{2-}), will be more stable than the chromate ion, ([CrO_4]^{2-}).

Explanation:

The stability of oxyanions like ([CrO_4]^{2-}) and ([WO_4]^{2-}) depends on several factors, including the size and electronegativity of the central metal atom, bond strength, and oxidation state stability.

  1. Atomic Size and Charge Distribution:
  • Tungsten (W) is in Group 6 and belongs to Period 6, while Chromium (Cr) is in Period 4.
  • Being a heavier element, tungsten has a larger atomic radius than chromium. This larger size allows the negative charge in ([WO_4]^{2-}) to be more delocalized, reducing electrostatic repulsion and increasing stability.
  • In contrast, Cr has a smaller size, leading to a higher charge density, making ([CrO_4]^{2-}) less stable.
  1. Bond Strength and Covalency:
  • W–O bonds in ([WO_4]^{2-}) are stronger than Cr–O bonds in ([CrO_4]^{2-}) because of better orbital overlap and stronger metal-ligand bonding interactions.
  • Tungsten exhibits stronger d-orbital participation, leading to greater covalency in ([WO_4]^{2-}), which contributes to stability.
  1. Oxidation State Stability:
  • Both Cr and W exhibit the +6 oxidation state in their oxyanions, but W(+6) is more stable than Cr(+6) due to the inert pair effect in heavier elements.
  • Cr(+6) is a strong oxidizing agent and tends to get reduced easily, making ([CrO_4]^{2-}) less stable in aqueous environments.

Conclusion:

Due to better charge delocalization, stronger W–O bonds, and greater stability of the +6 oxidation state, ([WO_4]^{2-}) is more thermodynamically and kinetically stable than ([CrO_4]^{2-}).

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