Given the order of density: Diamond $>$ Graphite$>$ Fullerene ${\mathrm{C}}_{60}$; choose the correct order for $\mathrm{C}-\mathrm{C}$ bond length.

(Consider the larger bond length if there are two different bond lengths)

An element $\mathrm{E}=\{\mathrm{B}, \mathrm{C}, \mathrm{Si}, \mathrm{Ge}\}$, predict $\mathrm{E}$ on the basic of given conditions:

I : Powdered $\mathrm{E}$ reacts with ${\mathrm{O}}_2$ to form an oxide.

II: Oxide formed in $\mathrm{I}$ reacts with $\mathrm{NaOH}$

III: E reacts with steam on red heat forming two gaseous products, which can be used as a fuel. $\mathrm{E}$ can be :

$2\mathrm{E}+{\mathrm{N}}_2⟶2\mathrm{EN}$ (very hard substance)

$\mathrm{EN}+{\mathrm{H}}_2\mathrm{O}⟶$ Acid $+$ The pungent smelling gas

The acid  in the above set of reactions is:

$\mathrm{B}(\mathrm{OH}{)}_3+\mathrm{NaOH}⟶\mathrm{Na}\left[\mathrm{B}(\mathrm{OH}{)}_4\right](\mathrm{aq})$.

Then addition of which of the following shifts the reaction in the backward direction.

${\mathrm{H}}_3{\mathrm{BO}}_3\left(\mathrm{S}\right)+\text{aqueous} \mathrm{NaOH}\stackrel{\mathrm{\Delta }}{⟶}\left(\mathrm{X}\right)$

${\mathrm{H}}_3{\mathrm{BO}}_3\left(\mathrm{S}\right)+\mathrm{molten} \mathrm{NaOH}\stackrel{∆}{⟶}\left(\mathrm{Y}\right)$

Compound $\left(\mathrm{X}\right)$ and $\left(\mathrm{Y}\right)$ are respectively,