
Find the value of X in the tremolite asbestos
\[C{{a}_{2}}M{{g}_{x}}{{(S{{i}_{4}}{{O}_{11}})}_{2}}{{(OH)}_{2}}\]
Answer
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Hint: In this question we should find the number of Mg atoms in the tremolite asbestos. We can find X by solving the oxidation state of the molecule. Follow the rules assigned to calculate oxidation number and find the X.
Complete Solution :
There are certain rules that can be used to calculate the oxidation number of a particular atom:
1. The oxidation number of an element in its free state is zero. The oxidation number of ${{H}_{2}}$, $C{{l}_{2}}$,${{P}_{4}}$ are zero.
2. Oxidation number of an atom in a monatomic ion is equal to the number of positive or negative charges on the ion.
Example: Oxidation number of Al atoms in $A{{l}^{+3}}$ ion is +3. The Cl atom in $C{{l}^{-}}$ ion is -1.
3. The oxidation number of H in a compound is assigned as +1, except in metallic hydrides where its oxidation number is -1.
Example: The oxidation number of hydride is -1 in $Ca{{H}_{2}}$.
4. The oxidation number of O in compounds is assigned as -2.
Exceptions:
- In peroxides (${{H}_{2}}{{O}_{2}}$,$N{{a}_{2}}{{O}_{2}}$) where the oxidation number of O is -1.
-In superoxides ($K{{O}_{2}}$) the oxidation number of O is $-\frac{1}{2}$.
5. The oxidation number halogen in halides or any compound is -1.
-The oxidation number of sulphur in sulphides is -2.
-The alkali metals that are the elements of group 1 possess oxidation number +1.
-The elements of group 2 that is alkaline metals possess oxidation number +2.
6. The algebraic sum of oxidation numbers of all atoms in a compound or ion is equal to the zero.
7. In coordination compounds,
- The oxidation number of neutral ligands like$NO$, $CO$,$N{{H}_{3}}$ and ${{H}_{2}}O$ is zero.
-The oxidation number of $C{{H}_{3}}$,${{C}_{6}}{{H}_{5}}$are +1.
-The oxidation number of CN, OH, Cl are -1.
\[C{{a}_{2}}M{{g}_{x}}{{(S{{i}_{4}}{{O}_{11}})}_{2}}{{(OH)}_{2}}\]
Ca and Mg has oxidation number + 2 as these two elements belongs to group 2, (With reference to 5)
The oxidation number of oxygen is - 2. (With reference to rule 4)
The oxidation number of hydrogen is + 1. (With reference to rule 3)
The valence electron in silicon is + 4. Oxidation number of atoms gives the number of valence electrons, visa -versa.
- Let us consider O.S is the short form of oxidation number,
O.S of Ca $\times $ 2 + O.S of Mg $\times $ X + O.S of Si $\times $ 8 + O.S of O $\times $ 22 + O.S of O $\times $2 + O.S of H $\times $ 2 = 0
(With reference to rule 7)
- Now apply all the values in above equation,
2 $\times $ 2 + 2x + 4 $\times $ 8 + (-2) $\times $ 22 + (-2) $\times $ 2 + 1 $\times $2 = 0
4 + 2x + 32 - 44 - 4 + 2 = 0
2x - 10 = 0
x = 5
- Thus, the formula of tremolite asbestos
\[C{{a}_{2}}M{{g}_{5}}{{(S{{i}_{4}}{{O}_{11}})}_{2}}{{(OH)}_{2}}\]
Note: Firstly, to solve these patterns of question the rules to calculate oxidation number should be in our tip, this might help us to answer the question within no time. During calculation be careful with the sign because that may lead to wrong answers. Oxidation number of an atom gives the number of valence electrons.
Complete Solution :
There are certain rules that can be used to calculate the oxidation number of a particular atom:
1. The oxidation number of an element in its free state is zero. The oxidation number of ${{H}_{2}}$, $C{{l}_{2}}$,${{P}_{4}}$ are zero.
2. Oxidation number of an atom in a monatomic ion is equal to the number of positive or negative charges on the ion.
Example: Oxidation number of Al atoms in $A{{l}^{+3}}$ ion is +3. The Cl atom in $C{{l}^{-}}$ ion is -1.
3. The oxidation number of H in a compound is assigned as +1, except in metallic hydrides where its oxidation number is -1.
Example: The oxidation number of hydride is -1 in $Ca{{H}_{2}}$.
4. The oxidation number of O in compounds is assigned as -2.
Exceptions:
- In peroxides (${{H}_{2}}{{O}_{2}}$,$N{{a}_{2}}{{O}_{2}}$) where the oxidation number of O is -1.
-In superoxides ($K{{O}_{2}}$) the oxidation number of O is $-\frac{1}{2}$.
5. The oxidation number halogen in halides or any compound is -1.
-The oxidation number of sulphur in sulphides is -2.
-The alkali metals that are the elements of group 1 possess oxidation number +1.
-The elements of group 2 that is alkaline metals possess oxidation number +2.
6. The algebraic sum of oxidation numbers of all atoms in a compound or ion is equal to the zero.
7. In coordination compounds,
- The oxidation number of neutral ligands like$NO$, $CO$,$N{{H}_{3}}$ and ${{H}_{2}}O$ is zero.
-The oxidation number of $C{{H}_{3}}$,${{C}_{6}}{{H}_{5}}$are +1.
-The oxidation number of CN, OH, Cl are -1.
\[C{{a}_{2}}M{{g}_{x}}{{(S{{i}_{4}}{{O}_{11}})}_{2}}{{(OH)}_{2}}\]
Ca and Mg has oxidation number + 2 as these two elements belongs to group 2, (With reference to 5)
The oxidation number of oxygen is - 2. (With reference to rule 4)
The oxidation number of hydrogen is + 1. (With reference to rule 3)
The valence electron in silicon is + 4. Oxidation number of atoms gives the number of valence electrons, visa -versa.
- Let us consider O.S is the short form of oxidation number,
O.S of Ca $\times $ 2 + O.S of Mg $\times $ X + O.S of Si $\times $ 8 + O.S of O $\times $ 22 + O.S of O $\times $2 + O.S of H $\times $ 2 = 0
(With reference to rule 7)
- Now apply all the values in above equation,
2 $\times $ 2 + 2x + 4 $\times $ 8 + (-2) $\times $ 22 + (-2) $\times $ 2 + 1 $\times $2 = 0
4 + 2x + 32 - 44 - 4 + 2 = 0
2x - 10 = 0
x = 5
- Thus, the formula of tremolite asbestos
\[C{{a}_{2}}M{{g}_{5}}{{(S{{i}_{4}}{{O}_{11}})}_{2}}{{(OH)}_{2}}\]
Note: Firstly, to solve these patterns of question the rules to calculate oxidation number should be in our tip, this might help us to answer the question within no time. During calculation be careful with the sign because that may lead to wrong answers. Oxidation number of an atom gives the number of valence electrons.
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