Question

The mineral colemanite is finely powdered and boiled with sodium carbonate solution. The precipitate of $CaC{{O}_{3}}$ and the other two products X and Y are formed. On passing a current of $C{{O}_{2}}$, Y produces X. What is the oxidation state of boron in compound Y?
(a)- +1
(b)- +3
(c)- +5
(d)- 0

Answer 1

Hint: For calculating the oxidation state of boron in the compounds, take the oxidation state of oxygen -2, the oxidation state of sodium +1, and the oxidation state of calcium +2.

Complete answer:
So the question says, the mineral colemanite is finely powdered and boiled with sodium carbonate. The formula of colemanite is $C{{a}_{2}}{{B}_{6}}{{O}_{11}}$ and the formula of sodium carbonate is $N{{a}_{2}}C{{O}_{3}}$. Hence the products will be calcium carbonate, sodium tetraborate, and sodium metaborate. The reaction is given below:
$C{{a}_{2}}{{B}_{6}}{{O}_{11}}+2N{{a}_{2}}C{{O}_{3}}\xrightarrow[{{H}_{2}}O]{fuse}2CaC{{O}_{3}}+N{{a}_{2}}{{B}_{4}}{{O}_{7}}+2NaB{{O}_{2}}$
So the compound X will be sodium tetraborate whose formula is $N{{a}_{2}}{{B}_{4}}{{O}_{7}}$ and the compound Y will be sodium metaborate $NaB{{O}_{2}}$
When the Y is subjected to a current of carbon dioxide, then there is the formation of X. so when the sodium metaborate is treated with carbon dioxide, sodium tetraborate is formed. The reaction is given below:
$4NaB{{O}_{2}}+C{{O}_{2}}\to N{{a}_{2}}{{B}_{4}}{{O}_{7}}+N{{a}_{2}}C{{O}_{3}}$
So the compound Y is $NaB{{O}_{2}}$and for calculating the oxidation of boron, the oxidation state of sodium will be +1 and the oxidation state of oxygen will be -2.
$+1+x+2(-2)=0$
$+1+x-4=0$
$x=+3$
So the oxidation of the boron will be +3.
Therefore, the correct answer is an option (b)- +3.

Note:
The compound X sodium tetraborate whose formula is $N{{a}_{2}}{{B}_{4}}{{O}_{7}}$ is used for Borax bead test. When the sodium tetraborate is heated there is a formation of sodium metaborate and glassy bed. The reaction is given below:
$N{{a}_{2}}{{B}_{4}}{{O}_{7}}\to 2NaB{{O}_{2}}+{{B}_{2}}{{O}_{3}}$