Given below are two statements :
Statement I : According to Ellingham diagram, any metal oxide with higher $$\Delta$$G$$^\circ$$ is more stable than the one with lower $$\Delta$$G$$^\circ$$.
Statement II : The metal involved in the formation of oxide placed lower in the Ellingham diagram can reduce the oxide of a metal placed higher in the diagram.
In the light of the above statements, choose the most appropriate answer from the options given below :
Consider the following reaction:
$$2HSO_4^ - (aq)\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{(2)\,Hydrolysis}^{(1)\,Electrolysis}} 2HSO_4^ - + 2{H^ + } + A$$
The dihedral angle in product A in its solid phase at 110 K is :
The correct order of melting point is :
The correct order of melting points of hydrides of group 16 elements is :