Understanding Ionic Compounds
Ionic compounds form between metals and nonmetals where metals lose electrons to become cations (positive ions) and nonmetals gain electrons to become anions (negative ions).
Step-by-Step Process to Write Ionic Compound Formulas
Step 1: Identify Symbols and Charges
- Write the symbol of the metal with its charge.
- Write the symbol of the nonmetal with its charge.
Example:
- Magnesium (Mg) is a metal in group 2A, so it forms Mg2+.
- Oxygen (O) is a nonmetal in group 6A, so it forms O2−.
Step 2: Use Magnitude of Charges as Subscripts
- Take the magnitude (absolute value) of each ion's charge.
- Use the magnitude of the metal's charge as the subscript for the nonmetal.
- Use the magnitude of the nonmetal's charge as the subscript for the metal.
Example:
- Mg2+ and O2− become Mg2O2.
Step 3: Simplify Subscripts to the Smallest Whole Numbers
- Reduce the subscripts to the smallest whole number ratio.
Example:
- Mg2O2 simplifies to MgO.
Step 4: Verify Charge Neutrality
- Confirm that the total positive charge equals the total negative charge.
Example:
- Mg2+ (1 ion) and O2− (1 ion) charges cancel out, confirming MgO is correct.
Additional Examples
Aluminum and Oxygen
- Aluminum: Al3+
- Oxygen: O2−
- Exchange magnitudes: Al2O3
- Check charges: 2 × (+3) = +6 and 3 × (-2) = -6, charges balance.
Potassium and Nitride
- Potassium: K+
- Nitride: N3−
- Formula: K3N
- Charges: 3 × (+1) = +3 and 1 × (-3) = -3, balanced.
Calcium and Bromide
- Calcium: Ca2+
- Bromide: Br−
- Formula: CaBr2
- Charges: 1 × (+2) = +2 and 2 × (-1) = -2, balanced.
Aluminum and Sulfide
- Aluminum: Al3+
- Sulfide: S2−
- Formula: Al2S3
- Charges: 2 × (+3) = +6 and 3 × (-2) = -6, balanced.
Key Takeaways
- Always write metal first, then nonmetal.
- Use ion charges to determine subscripts by swapping magnitudes.
- Simplify subscripts to smallest whole numbers.
- Confirm total charges cancel to ensure neutrality.
This method ensures accurate and neutral ionic compound formulas essential for chemistry studies and applications. For a deeper understanding of the concepts involved, check out Understanding Atomic, Molecular Elements and Ionic vs Molecular Compounds and Understanding Ions and Ionic Bonds in Chemistry. Additionally, for a comprehensive review of related topics, see Comprehensive Chemistry Review: Isotopes, Ionic Compounds, and Molar Mass. If you're interested in mastering the balancing of redox reactions, you might find Mastering the Half Reaction Method to Balance Redox Reactions helpful.
Now let's try to write some formulas of ionic compounds. Now we know when you see ionic compounds that means we have
to consider a metal and non-metal. And also you should know that metals lose electrons and become cations. Nonmetals
gain electrons and become annions. Let's try to see magnesium and oxygen. Magnesium is a metal. Oxygen is a
nonmetal. It should be ionic. What type of ions they make? Magnesium always make two positive because it is in 2A. Oxygen
always make two negative because it's in 6 A. It will take two electrons and become negative two charge ion. There
are certain steps. Let's go over the steps. How to write a formula of ionic compound. First step is write the symbol
of the metal and its charge. We just found out symbol of the metal is mg and it charge is 2 plus followed by the
symbol of the nonmetal and its charge. Nonmetal is oxygen. It charge is -2. Let's go to the second step. Make the
magnitude of the charge of each ion. So magnitude of charge of each ion mean the magnitude of magnesium is two two
positive the magnitude is two. Similarly oxygen also in oxide ion magnitude of charge is two. Let's see what they say.
Make the magnitude of the charge of each ion without the sign. Yeah magnitude is not the sign become the subscript of the
other ion. What does that mean? is asking us to make the magnitude of the charge to become the subscript of the
other ion. You know subscript is the this is a subscript that is a subscript. Let's try to do what it has to do. Make
the magnitude of the charge the subscript of the other. Basically mg2 this two please understand this two is
coming from here. That's why it says make the magnity of the charge of each ion become the subscript of the other
ion. So magnesium mg2 because it's a substrate of the other ion. Similarly oxygen I will that two here the two in
oxygen means actually this one to be interchanged. Let's go to the next step. If possible if possible reduce the
subscript we have subset now two and two reduce the substitute give the ratio with the smallest whole number. So it's
asking look at this subscript and see if we can give smallest whole numbers. We have two and two ratios. We can make it
even smaller. We can make it 1 and one. Still small and whole ratio go because I divide both sides by two. The next one
is to check to make sure the sum of charges of the cations exactly cancels the sum of charges of the annions. I
just need to emphasize that when you are writing a formula unit in a Cl or any ionic compounds, this is the smallest
and neutral. There cannot be a charge. It's asking check to make sure sum of the charges of kions exactly cancels the
sum of the charges of annions. Let's see magnesium is positive2 oxygen is -2. There's one of each. Therefore, it
cancels out each other. Therefore, this MGO should be correct. These are the steps and that's how you write a
formula. Let's try to do one more. Write the formulas with ionic compounds that form from aluminum and oxygen. Magnesium
oxygen. We just did aluminum and oxygen. We have the steps. Write the symbol for metal and it charge. Metal is aluminum
L3+ followed by the nonmetal and it charge. Nonmetal is oxygen and it charge is -2
O2 minus. Make the magnitude of the charge of each ion becomes subscript. That means I'm going to exchange this
magnitude and then I will get Al23. This two is coming from oxygen. This three is coming from aluminum. The third
step is if possible reduce the subscript to a smallest whole numbers. We this is the smallest we can have two and three.
There's no way we can simplify this without having fractions. But we cannot have fractions. We have to have whole
numbers. Finally check to make sure the sum of the charges of cations exactly cancels the sum of the charges of
annions. Here we have two aluminum 2 * each aluminum is positive3 + 3 and other one is 3 * because there are three
oxygen that's that's what this subscript shows three times each oxygen is -2 this one is pos6 this one is -6 that cancels
out therefore Al23 is correct what are the formulas for compounds made out of following ions
potassium with nitride potassium ium is K plus K plus nitride is N minus these are the steps I will go very fast we
exchange them first then K3 is from this nitrogen and N1 but in chemistry we don't write one here similarly we did
not write K + one similarly we don't write here this is the chemical formula K3N and if you see we cannot simplify 3
and 1 and you can also check if canceling out because we have three potassium atoms each one is positive one
is going to be three positive and And this one is -3. Positive and negative cancels out. That is correct. K3 N.
Third one is calcium ion with bromide ion. Calcium is this one. Ca2 plus and bro mean bromide ion is this one. Br
minus. Let's interchange that then you will get C A this Br has negative one. We do not write one here. We do not
write one. And I will write B R 2 two from this calcium ch. Then we cannot simplify this one and two ratios. These
1 and two ratios we cannot simplify. And finally let's try to see the charges are zero. One calcium that is positive2 from
here we can see and there are two broines. Each one is -1 from here -1 positive2 and -2 they cancels out each
other. That is correct. I'm going to erase everything and C A B are 2. Finally aluminum and sulfide. Aluminum
is 3 plus a L3+ sulfide S2 minus we'll exchange L2 S3
and if you just see if it is going to cancel out. All right.
Heads up!
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