Naming compounds with metals

Write the names for the following compounds.
  1. LiF
  2. CaCl₂
  3. Na₂O
  4. KNO₃
hint
For compounds containing metals, we name the metal first. Then we name the nonmetal but replace the suffix with "-ide."

If the nonmetals are a polyatomic ion, then you can just give the name of the polyatomic ion (no "-ide" suffix modification needed).

Remember that the compounds need to be charge-neutral overall, so work out the charges of the metals and nonmetals using the periodic table or polyatomic ions.
solutions 1 & 2
The names for the compounds are:
  1. lithium fluoride
  2. calcium chloride
SOLUTIONs 3 & 4
The names for these compounds are:
  1. sodium oxide
  2. potassium nitrate

Nitrate is a polyatomic ion with the formula: NO₃⁻.
Examples: KCl, MgBr₂, and CaSO₄

Naming compounds with metals & Roman numerals

Write the names for the following compounds.
  1. TiO₂
  2. Ni₂S₃
  3. PbSO₄
  4. AgCl and ZnCl₂
hint
For compounds containing transition metals, we name the transition metal first, but we include Roman numerals to denote the charge of the metal. The reason for this is that transition metals can adopt many different charges unlike the case of, for example, group 1 and group 2 metals.

Then we name the nonmetal but replace the suffix with "-ide." If the nonmetals are a polyatomic ion, then you can just give the name of the polyatomic ion (no "-ide" suffix modification needed).

Remember that the compounds need to be charge-neutral overall, so you can work out the charge of the transition metal by knowing the charge of the nonmetals first.
solution 1
When oxygen forms a compound with a metal, it will take on a -2 charge. There are two oxygen anions, giving a total charge of -4.

Therefore, the titanium must have a +4 charge to cancel out the charges from both oxygen anions. The name of the compound would be: titanium(IV) oxide
SOLUTION 2
When sulfur forms a compound with a metal, it will take on a -2 charge. There are three sulfur anions, giving a total charge of -6.

Therefore, the two nickel cations must each have a +3 charge to cancel out the charges from the three sulfur anions. The name of the compound would be: nickel(III) sulfide
solution 3
Sulfate is a polyatomic ion with the formula and charge: SO₄²⁻

Therefore, the lead cation must  have a +2 charge to cancel out the charges from the sulfate anion. The name of the compound would be: lead(II) sulfate.
solution 4
There are two transition metals that only adopt one kind of charge.
  • Silver will always form +1 cations: Ag⁺
  • Zinc will always form +2 cations: Zn²⁺
In compounds containing Ag and Zn, we do not include Roman numerals.

So the names of the compounds are silver chloride and zinc chloride.
Examples: FeO, Fe₂O₃, Fe(NO₃)₃

Writing electron configuration for transition metals

Write the electron configuration for ...
  1. Ni and Ni²⁺
  2. Fe and Fe³⁺
  3. Mn and Mn⁴⁺
Solution 1
Electron configuration for Ni is
\[ [\text{Ar}] 4s^2 3d^8 \]
If we lose two electrons from neutral Ni, then the electron configuration for Ni²⁺ is
\[ [\text{Ar}] 3d^8 \]
For transition metal cations, the electrons removed are usually from the s shell first.
solution 2
Electron configuration for Fe is
\[ [\text{Ar}] 4s^2 3d^6 \]
If we lose three electrons from neutral Fe, then the electron configuration for Fe³⁺ is
\[ [\text{Ar}] 3d^5 \]
For transition metal cations, the electrons removed are usually from the s shell first.
SOLUTION 3
Electron configuration for Mn is
\[ [\text{Ar}] 4s^2 3d^5 \]
If we lose four electrons from neutral Mn, then the electron configuration for Mn⁴⁺ is
\[ [\text{Ar}] 3d^3 \]
For transition metal cations, the electrons removed are usually from the s shell first.
Watch me work out examples for vanadium: V, V²⁺, and V⁴⁺.

Exceptions for transition metal electron configurations

Write the electron configuration for ...
  1. Cr
  2. Cu
  3. Ag
Solution 1
At first glance, it may seem like the electron configuration for Cr should be
\[ [\text{Ar}] 4s^2 3d^4 \]
However, what is actually the case is
\[ [\text{Ar}] 4s^1 3d^5 \]
where the 4s orbital and the 3d orbitals are both half-filled. This anomalous electron configuration would also be observed for Mo and W (both below Cr on the periodic table).
solution 2
At first glance, it may seem like the electron configuration for Cu should be
\[ [\text{Ar}] 4s^2 3d^9 \]
However, what is actually the case is
\[ [\text{Ar}] 4s^1 3d^{10} \]
where the 4s orbital is half-filled and the 3d orbitals are fully-filled. This anomalous electron configuration would also be observed for Ag and Au (both below Cu on the periodic table).
SOLUTION 3
Just like copper (above silver), the electron configuration for Ag would be
\[ [\text{Kr}] 5s^1 4d^{10} \]
Watch me work out examples for Cr and Cu.

Naming compounds with ONLY NONmetals

Write the names for the following compounds.
  1. N₂O
  2. P₂O₅
  3. P₄S₆
  4. S₂F₄
  5. ClO₃
hint
For compounds containing all nonmetals, we first name the element listed first. Then we name the second element but replace the suffix with "-ide."

Remember that for these covalent compounds, we need to use prefixes to denote the number of each element in the formula.
solution 1
dinitrogen monoxide
SOLUTION 2
diphosphorus pentoxide
solution 3
tetraphophorus hexasulfide
solution 4
disulfur tetrafluoride
solution 5
chlorine trioxide
Examples: N₂O₄, CO, SO₃, and PCl₅

Naming ACIDS

Write the names for the following oxyacids.
  1. HNO₂
  2. CH₃COOH
  3. H₂CrO₄
  4. H₂SO₃
  5. HBr
  6. HCN
hint
Acidic compounds will always contain one (or more) H⁺ cations. Once you isolate the H⁺ cation, the rest of the compound is the anion.

The following conventions are used for the different kinds of acids.

For oxyacids:
  • If the anion name ends in -ate, substitute with -ic acid.
  • If the anion name ends in -ite, substitute with -ous acid.

For binary acids:
  • Begin with prefix hydro-
  • Then name the anion, but substitute the suffix to -ic acid.

solution 1
H⁺ and NO₂⁻ (nitrite)

Nitrous acid
SOLUTION 2
H⁺ and CH₃COO⁻ (acetate)

Acetic acid
solution 3
H⁺, H⁺, and CrO₄²⁻ (chromate)

Chromic acid
solution 4
H⁺, H⁺, and SO₃²⁻ (sulfite)

Sulfurous acid
solution 5
H⁺ and Br⁻ (bromide)

Hydrobromic acid
solution 6
H⁺ and CN⁻ (cyanide)

Hydrocyanic acid
Naming oxyacids: HClO, HClO₂, HClO₃, HClO₄
Oxyacids with P or S: H₂SO₄ and H₃PO₃
Binary acids: HCl and H₂S