Question

Orthoboric acid when heated to red hot gives-
A) Metaboric acid
B) Pyroboric acid
C) Boron and water
D) Boric anhydride

Answer 1

Hint: As we know that in chemistry, periodic tables play a vital role. In the periodic table there are totally \[118\] elements. In the periodic table there are totally \[18\] columns and \[7\] rows. The columns are called groups. Hence, \[18\] groups in the periodic table. The rows are called periods. Hence, totally \[7\] period in the table. Boron is one of the elements in the periodic table. The atomic number of boron is \[5\]. The mass number of boron is\[10.8\]. Boron is present in the second period and \[13th\] group in the periodic table.

Complete answer:
We need to know that the molecular formula of Orthoboric is \[{H_3}B{O_3}\].
Orthoboric acid when heated to give the product of \[HB{O_2}\] and the side product of water in one mole.
The chemical equation for the above discussion given below,
 \[{H_3}B{O_3}\xrightarrow{\Delta }HB{O_2} + {H_2}O\]
Orthoboric acid when heated to red hot gives the product of \[{B_2}{O_3}\] and the side product of water in three moles.
\[2{H_3}B{O_3}\xrightarrow{{redhot}}{B_2}{O_3} + 3{H_2}O\]
The molecular formula of boric anhydride is \[{B_2}{O_3}\].
According to the above discussion, we conclude orthoboric acid when heated to red hot gives boric anhydride.
Hence, the option D is the correct answer, because orthoboric acid when heated to red hot gives boric anhydride.
The option A is not the correct answer, because orthoboric acid when heated to red hot does not give metaboric acid.
The option B is not the correct answer, because orthoboric acid when heated to red hot does not give pyroboric acid.
The option C is not the correct answer, because orthoboric acid when heated to red hot does not give boron and water.

Hence, the option D is the correct answer, because orthoboric acid when heated to red hot gives boric anhydride.

Note:
We have to know that the isotopes mean two atoms having the same atomic number but different mass number and also having the same atomic symbol. Here \[{}^{{\text{10}}}{\text{B and }}{}^{{\text{11}}}{\text{B}}\] having same atomic number of five and same symbol of B but having different mass number of \[{\text{10 and 11}}\]. The atomic number of the element is nothing but the number of electrons or number of protons. The mass number of the atom is nothing but the sum of the number of protons and the number of neutrons.