Question

${{H}_{3}}B{{O}_{3}}\xrightarrow{{{T}_{1}}}X\xrightarrow{{{T}_{2}}}Y\xrightarrow{red\,hot}{{B}_{2}}{{O}_{2}}$, if ${{T}_{1}}<{{T}_{2}}$, then X and Y respectively are:
A.X = Metaboric acid and Y = Tetraboric acid
B. X = Tetraboric acid and Y=Metaboric acid
C. X = Borax and Y= Metaboric acid
D. X = Tetraboric acid and Y = Borax

Answer 1

Hint: The concept behind this question is about the properties of boron and its compounds in the p-block group of elements. Temperature plays a role in the formation of compounds when reagents are heated.

Complete step by step solution:
In order to answer our question, we need to learn about the properties of Boron and it’s group. Group III A contains six elements: boron, aluminium, gallium, indium, thallium and ununtrium. The penultimate shell (next to the outermost) contains $1{{s}^{2}}$ in boron, $2{{s}^{2}}2{{p}^{6}}$ (8 electrons) in aluminium and $(n-1){{s}^{2}}(n-1){{p}^{6}}(n-1){{d}^{10}}$ (18 electrons) in other elements. This shows why boron differs from aluminium and both boron and aluminium having noble gas kernels differ from other four elements. Boron is non-metal and always forms covalent bonds. The Boron family is known as the most heterogeneous family as there is no regular trend in all properties, as it comes after d-block, lanthanoid contraction, poor shielding of d-orbital, they have large deviations in properties. The heavier members of this group show inert pair effects. Boron compounds, especially the hydrides are electron deficient compounds which can accept a lone pair of electrons hence behave as Lewis acids. Now let us come to the question. We can represent the reactions as:
\[{{H}_{3}}B{{O}_{3}}\xrightarrow{{{100}^{0}}C}HB{{O}_{2}}\xrightarrow{{{160}^{0}}C}{{H}_{2}}{{B}_{4}}{{O}_{7}}\xrightarrow{red\,hot}{{B}_{2}}{{O}_{3}}\]
Here we can see that X comes out to be metaboric acid and Y is tetraboric acid.

Hence, the correct answer for this question is option A.

NOTE: Boron compounds have three covalent bonds hence, require two electrons to complete octet hence, electron deficient compounds. They have a +3 group oxidation state. Boron is not found freely in nature, but in boric acid.