Given below are two statements:
Assertion A: In an Ellingham diagram, the oxidation of carbon to carbon monoxide shows a negative slope with respect to temperature.
Reason R: CO tends to get decomposed at higher temperature.

In the Ellingham diagram, the reaction $$2C(s) + O_2(g) \to 2CO(g)$$ has $$\Delta n_g = 2 - 1 = +1$$ (an increase in gas moles). By the relation $$\Delta G = \Delta H - T\Delta S$$, an increase in gaseous moles means $$\Delta S > 0$$, so the $$-T\Delta S$$ term becomes more negative with increasing temperature. This causes $$\Delta G$$ to decrease (become more negative) as temperature rises, giving the line a negative slope. Therefore, Assertion A is correct.

Reason R claims that CO tends to get decomposed at higher temperature. This is incorrect — in fact, the negative slope of the Ellingham line for CO formation means that CO becomes more thermodynamically stable at higher temperatures, not less. CO does not decompose at higher temperatures; rather, it is favoured as a reducing agent at elevated temperatures precisely because of this trend.

Since A is correct but R is not correct, the answer is Option D.