Q21 / 120 · VITEEE 2015 · Chemistry

Point out incorrect stability order

  • A
    $\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+} < \left[\mathrm{Cu}(\mathrm{en})_{2}\right]^{2+} < [\mathrm{Cu}$ $($ trien $)]^{2+}$
  • Correct answer
    B
    $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+} < \left[\mathrm{Fe}\left(\mathrm{NO}_{2}\right)_{6}\right]^{3-} < \left[\mathrm{Fe}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}$
  • C
    $\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+} < \left[\mathrm{Rh}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+} < \left[\operatorname{Ir}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}$
  • D
    $\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}\right]^{+} < \left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+} < \left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}$

Solution

Increasing order of stability (a) $\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+} < \left[\mathrm{Cu}(\mathrm{en})_{2}\right]^{2+}$ $ < \left[\mathrm{Cu}(\text { trein })^{2+}\right.$ Their formation of entropy increases in the same order. Ligand denticity is increased. (b) $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+} < \left[\mathrm{Fe}\left(\mathrm{NO}_{2}\right)_{6}\right]^{3-}$ $ < \left[\mathrm{Fe}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}$ $\because \quad \mathrm{NH}_{3}$ is weaker ligand than $\mathrm{NO}_{2}^{-}$. $\therefore \quad$ The correct stability order is $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)^{3+} < \left[\mathrm{Fe}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+} < \left[\mathrm{Fe}\left(\mathrm{NO}_{2}\right)_{6}\right]^{3-}\right.$ (c) $\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+} < \left[\mathrm{Rh}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}$ $\mathrm{Z}_{\text {of }}$ value increases from $\mathrm{Co}^{3+}$ to $\mathrm{Ir}^{3+}$, (d) $\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}\right]^{1+} < \left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}$ O.S of Cr atom increases from $+1$ to $+3$.

Hence, the correct answer is option: 'B'.

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