LiCoO_2 crystallizes in a rhombohedral struct | vedclass

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(B) In the initial compound $LiCoO_2$, the oxidation state of $Li$ is $+1$ and oxygen is $-2$.Let the oxidation state of $Co$ be $x$.$1(1) + x + 2(-2)

Vedclass · Accepted Answer

$16.66\%$ increase

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(B) In the initial compound $LiCoO_2$, the oxidation state of $Li$ is $+1$ and oxygen is $-2$.Let the oxidation state of $Co$ be $x$.$1(1) + x + 2(-2) = 0$$x - 3 = 0 \implies x = +3$.After $50\%$ lithium extraction, the formula becomes $Li_{0.5}CoO_2$.Let the new oxidation state of $Co$ be $x'$.$0.5(1) + x' + 2(-2) = 0$$x' + 0.5 - 4 = 0 \implies x' = +3.5$.The change in oxidation state of $Co$ is $3.5 - 3.0 = +0.5$.The percentage increase in the oxidation state is $\frac{0.5}{3} 	imes 100 = 16.66\%$.

$LiCoO_2$ crystallizes in a rhombohedral structure. Consider a situation where $50\%$ lithium $(Li)$ is extracted from the lattice. To keep the crystal electrically neutral, the change in the average oxidation state of $Co$ is

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