MCQ
Choose incorrect stability order
  • A
    $[Cu(NH_3)_4]^{2+} < [Cu(en)_2]^{2+} < [Cu(trien)]^{2+}$
  • $[Fe(H_2O)_6]^{3+} < [Fe(NO_2)_6]^{3-} < [Fe(NH_3)_6]^{3+}$
  • C
    $[Co(H_2O)_6]^{3+} < [Rh(H_2O)_6]^{3+} < [Ir(H_2O)_6]^{3+}$
  • D
    $[Cr(NH_3)_6]^{1+} < [Cr(NH_3)_6]^{2+} < [Cr(NH_3)_6]^{3+}$

Answer

Correct option: B.
$[Fe(H_2O)_6]^{3+} < [Fe(NO_2)_6]^{3-} < [Fe(NH_3)_6]^{3+}$
b
$(b)$ Increasing stability order :

$(a)\, [Cu(NH_3)_4]^{2+} <  [Cu(en)_2]^{2+} <  [Cu(trien)]^{2+}$

Their formation entropy increases in the same order, because denticity of   ligand increases

$(b)\, [Fe(H_2O)_6]^{3+} <  [Fe(NO_2)_6]^{3-} < [Fe(NH_3)_6]^{3+}$

$NO_2^-$ is stronger ligand than $NH_3$

$(c)\, [Co(H_2O)_6]^{3+} < [Rh(H_2O)_6]^{3+} <  [Ir(H_2O)_6]^{3+}$

$Z_{eff}$ value increases from $Co^{3+}$ to $Ir^{3+}$

$(d)\, [Cr(NH_3)_6]^{1+} < [Cr(NH_3)_6]^{2+} <  [Cr(NH_3)_6]^{3+}$

Oxidation state of $Cr$ atom increases from $+ 1$ to $+3$.

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