The oxidation number of Co in the complex ion | vedclass

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(B) Let the oxidation state of $Co$ be $x$.$en$ (ethylenediamine) is a neutral ligand,so its charge is $0$.$NH_2^-$ (amido) has a charge of $-1$.$OH^-

Vedclass · Accepted Answer

$+3$

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(B) Let the oxidation state of $Co$ be $x$.$en$ (ethylenediamine) is a neutral ligand,so its charge is $0$.$NH_2^-$ (amido) has a charge of $-1$.$OH^-$ (hydroxo) has a charge of $-1$.The total charge on the complex ion is $+3$.Therefore,the equation is: $2x + 4(0) + (-1) + (-1) = +3$.$2x - 2 = +3$.$2x = +5$.$x = +2.5$.However,in such bridged complexes,the oxidation state is often represented as an average. If we assume the question implies the average oxidation state,the answer is $+2.5$. Given the options provided,there might be a discrepancy. If the complex was $[Co(NH_3)_6]^{3+}$,the oxidation state would be $+3$. Based on the provided image,the calculation leads to $+2.5$. If we must choose the closest standard integer value for such complexes,$+3$ is often cited for $Co$ in similar coordination environments,but mathematically it is $+2.5$.

The oxidation number of $Co$ in the complex ion $[(en)_2Co(\mu-NH_2)(\mu-OH)Co(en)_2]^{3+}$ is:

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