Given below are two statements : One is labeled as Assertion A  and the other is labeled as Reason R

Assertion A:  The reduction of a metal oxide is easier if the metal formed is in liquid state than solid state.                               

Reason R:  The value of $\Delta G^{\Theta }$  becomes more on negative side as entropy is higher in liquid state than solid state.

In the light of the above statements, choose the most appropriate answer from the options given below

<p>&nbsp;<math><mrow><mi>Δ</mi><mi>G</mi><mo>=</mo><mi>Δ</mi><mi>H</mi><mo>−</mo><mi>T</mi><mi>Δ</mi><mi>S</mi></mrow></math></p><p><math><mrow><mo>∴</mo><mi>Δ</mi></mrow></math>&nbsp;S increases then (Ve) value of&nbsp;<math><mrow><mi>Δ</mi><mi>G</mi></mrow></math>&nbsp;increases and it is easy to reduce. Entropy of liquid is greater than solid. On melting, entropy increases and&nbsp;<math><mrow><mi>Δ</mi><mi>G</mi></mrow></math>&nbsp;becomes more negative.</p>

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<p>&nbsp;<math><mrow><mi>? </mi><mi>G</mi><mo>=</mo><mi>? </mi><mi>H</mi><mo>? </mo><mi>T</mi><mi>? </mi><mi>S</mi></mrow></math></p><p><math><mrow><mo>? </mo><mi>? </mi></mrow></math>&nbsp;S increases then (Ve) value of&nbsp;<math><mrow><mi>? </mi><mi>G</mi></mrow></math>&nbsp;increases and it is easy to reduce. Entropy of liquid is greater than solid. On melting, entropy increases and&nbsp;<math><mrow><mi>? </mi><mi>G</mi></mrow></math>&nbsp;becomes more negative.</p>

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