Choose incorrect stability order.

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$[Cr {({NH}_{3})}_{6}]^{1 +} < [Cr {({NH}_{3})}_{6}]^{2 +} < [Cr {({NH}_{3})}_{6}]^{3 +}$

$[Cu {({NH}_{3})}_{4}]^{2 +} < [Cu {(en)}_{2}]^{2 +} < {[Cu (trien)]}^{2 +}$

$[Fe {(H_{2} O)}_{6}]^{3 +} < [Fe {({NO}_{2})}_{6^{}}]^{3 +} < [Fe {({NH}_{3})}_{6}]^{3 +}$

$[Co {(H_{2} O)}_{6}]^{3 +} < [Rh {(H_{2} O)}_{6}]^{3 +} < [Ir {(H_{2} O)}_{6}]^{3 +}$

answer is B.

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Detailed Solution

Increasing stability order:
$[Cu {({NH}_{3})}_{4}]^{2 +} < [Cu {(en)}_{2}]^{2 +} < {[Cu (trien)]}^{2 +}$
Their formation of entropy increases in the same order, because denticity of ligand increases $[Fe {(H_{2} O)}_{6}]^{3 +} < [Fe {({NO}_{2})}_{6^{}}]^{3 +} < [Fe {({NH}_{3})}_{6}]^{3 +}$
${NO}_{2}$ is stronger ligand than ${NH}_{3}$
$[Co {(H_{2} O)}_{6}]^{3 +} < [Rh {(H_{2} O)}_{6}]^{3 +} < [Ir {(H_{2} O)}_{6}]^{3 +}$
Z_eff value increases from ${Co}^{3 +}$ to ${Ir}^{3 +}$
$[Cr {({NH}_{3})}_{6}]^{1 +} < [Cr {({NH}_{3})}_{6}]^{2 +} < [Cr {({NH}_{3})}_{6}]^{3 +}$
Oxidation state of Cr atom increases from +1 to +3