Arrange the following ions in the order of decreasing $X-O$ bond length, where $X$ is the central atom

Arrange the following ions in the order of decreasing $X-O$ bond length, where $X$ is the central atom
  1. $\mathrm{ClO}_{4}^{-}, \mathrm{SO}_{4}^{2-}, \mathrm{PO}_{4}^{3-}, \mathrm{SiO}_{4}^{-}$
  2. $\mathrm{SiO}_{4}^{4-}, \mathrm{PO}_{4}^{3-}, \mathrm{SO}_{4}^{2-}, \mathrm{ClO}_{4}^{-}$
  3. $\mathrm{SiO}_{4}^{4-}, \mathrm{PO}_{4}^{3-}, \mathrm{ClO}_{4}^{-}, \mathrm{SO}_{4}^{2-}$
  4. $\mathrm{SiO}_{4}^{4-}, \mathrm{SO}_{4}^{2-}, \mathrm{PO}_{4}^{3-}, \mathrm{ClO}_{4}^{-}$

Solution

Arranged by decreasing bond length: 1. \(\mathrm{SiO}_4^{4-}\) : Silicon has the largest atomic radius and the lowest oxidation state, leading to the longest bond lengths. 2. \(\mathrm{PO}_4^{3-}\) : Phosphorus is smaller than silicon but has a higher oxidation state. 3. \(\mathrm{SO}_4^{2-}\) : Sulfur is smaller than phosphorus and has a higher oxidation state. 4. \(\mathrm{ClO}_4^{-}\): Chlorine has the smallest atomic radius and the highest oxidation state, resulting in the shortest bond lengths. Thus, the correct order of decreasing \(\mathrm{X}-\mathrm{O}\) bond length is: (2) \(\mathrm{SiO}_4^{4-}, \mathrm{PO}_4^{3-}, \mathrm{SO}_4^{2-}, \mathrm{ClO}_4^{-}\) /

Asked in: JEE-TOPICTESTS-CHEMISTRY

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