
How do you determine the correct subscripts in a chemical formula?
Answer
561k+ views
Hint: First you should know about the charge which each atom carries in the compound and then balance the charge by the addition of the atoms, so that both the atoms which are present in the molecule carries the equal amount of charge on them. Now solve it.
Complete answer:
- We can determine the subscripts in a chemical formula by balancing the charge on the atoms.
- In other words, we can say that we have to calculate actually how many atoms will be needed in order to balance the charge.
- For example;- It can be explained by taking the example of calcium fluoride’. We know that, calcium has the electronic configuration as ${{[Ar]}^{18}}4{{s}^{2}}$ i.e. it has strong tendency to lose two electrons from its outermost valence shell and thus, carries +2 charge in its most of the compounds.
- On the other hand, fluorine has the electronic configuration as $1{{s}^{2}}2{{s}^{2}}{{p}^{5}}$ i.e. it has a strong tendency to gain one electron to complete its octet and thus, carries -1 charge in its most of the compounds.
- Now we will balance the charge by adding the number of atoms.
- Since, calcium carries $+2$ charge, so in order to balance the charge, there should be two fluorine atoms each of them carrying $-1$ charge.
- So, thus, the formula of calcium fluoride comes out to be $Ca{{F}_{2}}$.
Hence, in this way, we can find the subscripts in a chemical formula.
Note: Calcium is a metal which comes in the category of the alkaline earth metals and belongs to the s-block elements. On the other hand, fluorine is a non-metal which comes in the category of the halogens and belongs to the p-block elements.
Complete answer:
- We can determine the subscripts in a chemical formula by balancing the charge on the atoms.
- In other words, we can say that we have to calculate actually how many atoms will be needed in order to balance the charge.
- For example;- It can be explained by taking the example of calcium fluoride’. We know that, calcium has the electronic configuration as ${{[Ar]}^{18}}4{{s}^{2}}$ i.e. it has strong tendency to lose two electrons from its outermost valence shell and thus, carries +2 charge in its most of the compounds.
- On the other hand, fluorine has the electronic configuration as $1{{s}^{2}}2{{s}^{2}}{{p}^{5}}$ i.e. it has a strong tendency to gain one electron to complete its octet and thus, carries -1 charge in its most of the compounds.
- Now we will balance the charge by adding the number of atoms.
- Since, calcium carries $+2$ charge, so in order to balance the charge, there should be two fluorine atoms each of them carrying $-1$ charge.
- So, thus, the formula of calcium fluoride comes out to be $Ca{{F}_{2}}$.
Hence, in this way, we can find the subscripts in a chemical formula.
Note: Calcium is a metal which comes in the category of the alkaline earth metals and belongs to the s-block elements. On the other hand, fluorine is a non-metal which comes in the category of the halogens and belongs to the p-block elements.
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