Is a compound ionic or covalent?
-
Does the compound contain a metal?
-
Yes. The compound is ionic (a salt).
-
No. The compound is covalent.
Is a compound binary or ternary?
-
Does the compound contain two elements?
-
Yes. The compound is binary.
-
No. The compound is not binary.
-
Does the compound contain three elements?
-
Yes. The compound is ternary.
-
No. The compound is not ternary.
Is a compound an acid or a base?
-
Yes. Compound is an acid.
-
Yes. Compound is a base.
How to name a polyatomic ion from its formula.
If the ion is
-
binary and
-
a basic formula then
-
(name of ion) = (basic name)
-
has one less $\ce{O}$ atom than the basic formula then
-
(name of ion) = (basic name) – “ate” + “ite”
-
has two less $\ce{O}$ atoms than the basic formula then
-
(name of ion) = “hypo” + (basic name) – “ate” + “ite”
-
has one more $\ce{O}$ atoms than the basic formula then
-
(name of ion) = “per” + (basic name)
-
ternary and
-
$\ce{H}_n$ followed by a binary ion then
-
(name of ion) = “hydrogen” + (binary ion name)
How to find the oxidation state
If formula is
-
an element then
-
(element oxidation number) = 0.
-
an ion then
-
(ion oxidation number) = (magnitude of charge).
-
a compound and
-
ionic (a salt) and
-
first element is a metallic monatomic ion and
-
compound is a peroxide (has an $\ce{O—O}$ single bond) then
-
($\ce{O}$ oxidation state) = –1.
-
(metal oxidation state) = +1.
-
compound is a metallic hydride
($\ce{H}$ combined with a metal of lower
electronegativity than $\ce{H}$.)
-
($\ce{H}$ oxidation state) = –1.
-
(metal oxidation state) = +1.
-
a formula component is a polyatomic ion
-
(polyatomic ion oxidation state) = (magnitude of charge).
-
(metal oxidation state) =
– (polyatomic ion oxidation state).
-
other than above and
-
nonmetal is $\ce{H}$ then
-
($\ce{H}$ oxidation state) = +1.
-
(metal oxidation state) = –1.
-
nonmetal is not $\ce{H}$ then
-
(nonmetal oxidation state) = (magnitude of charge).
-
(metal oxidation state) = – (nonmetal oxidation state).
-
first element is a metallic polyatomic ion.
-
(polyatomic ion oxidation state) = (magnitude of charge).
-
(nonmetal oxidation state) = – (polyatomic ion oxidation state).
-
covalent and
-
a peroxide (has an $\ce{O—O}$ single bond) then
-
($\ce{O}$ oxidation state) = –1.
-
(remaining formula component oxidation state) = +1.
-
a formula component is a polyatomic ion then
-
(polyatomic ion oxidation state) = (magnitude of charge).
-
(remaining formula component oxidation state) =
– (polyatomic ion oxidation state).
-
something else, then for (electronegativity of nonmetal $X$)
< (electronegativity of nonmetal $Y$)
-
(nonmetal $X$ oxidation state) = +.
-
(nonmetal $Y$ oxidation state) = –.
How to name a compound from its formula
If the compound is
-
a polyatomic ion then
-
(name of compound) = (name of polyatomic ion)
-
ionic (a salt) and
-
first element is a metallic monatomic ion and
-
compound is binary and
-
the metal has a fixed oxidation state then
-
(name of compound)
= (metal name)
+ (space)
+ (nonmetal name)
– (nonmetal name ending)
+ “ide”
-
the metal has a variable oxidation state then
-
(name of compound)
= (metal name)
+ (metal oxidation state in roman numerals)
+ (nonmetal name)
– (ending)
+ “ide”
-
compound is ternary and second formula component is a polyatomic ion;
or compound is quaternary and second formula component is a ternary polyatomic ion, and
-
the metal has a fixed oxidation state then
-
(name of compound)
= (metal name)
+ (space)
+ (polyatomic ion name)
-
the metal has a variable oxidation state then
-
(name of compound)
= (metal name)
+ (metal oxidation state in roman numerals)
+ (space)
+ (polyatomic ion name)
-
first element is a metallic polyatomic ion and compound is ternary then
-
(name of compound)
= (polyatomic ion name)
+ (space)
+ (nonmetal name)
– (nonmetal name ending)
+ “ide”
-
covalent and
-
an acid in aqueous solution and
-
binary then
-
(name of compound)
= “hydro”
+ (name of second nonmetal)
– (ending of second nonmetal)
+ “ic acid”
-
ternary and second formula component is a polyatomic ion and
-
compound is an oxyacid and
-
name of polyatomic ion ends in “-ate” then
-
(name of compound)
= (name of polyatomic ion)
– “ate”
+ “ic acid”
-
name of polyatomic ion ends in “ite” then
-
(name of compound)
= (name of polyatomic ion)
– “ate”
+ “ous acid”
-
else ?????
-
quaternary and second formula component is a ternary polyatomic ion then
-
a nonacid or acid out of aqueous solution and
-
binary and
-
(subscript on nonmetal with positive oxidation state) = 1.
-
(name of compound)
= (name of nonmetal with positive oxidation state)
+ (space)
+ (prefix for number of nonmetals in negative oxidation state)
+ (name of nonmetal with negative oxidation state)
– (ending of nonmetal with negative oxidation state)
+ “ide”
-
(subscript on nonmetal with positive oxidation state) > 1.
-
(name of compound)
= (prefix for number of nonmetals in positive oxidation state)
+ (space)
+ (name of nonmetal with positive oxidation state)
+ (space)
+ (prefix for number of nonmetals in negative oxidation state)
+ (name of nonmetal with negative oxidation state)
– (ending of nonmetal with negative oxidation state)
+ “ide”
-
ternary and second formula component is a polyatomic ion then
-
quaternary and second formula component is a ternary polyatomic ion then
-
compound is a hydrate then
-
(name of compound)
= (name of salt)
+ (prefix for number of molecules of water)
+ “hydrate”
How to draw the Lewis structure of a compound
-
Determine the central atom or atoms and write the
surrounding atoms around the central atoms, joining
each surrounding atom to the central atom by a single
bonding pair of electrons.
-
Calculate the total number of valence electrons, $v,$
contributed to the compound by each atom. This is the
number of electrons initially available for completing
the noble gas configuration of each atom.
-
Subtract from this number the number of electrons, $p,$
which have already been written into the structure so far
from the first step. This gives the number of electrons
left over, $l,$ to complete the noble gas configurations.
The operation is $v-pk=l.$
-
For each atom, subtract the number of valence electrons, $pn,$
which have already been written into the structure so far from
the first step for that atom, from the number of valence electrons
in a noble gas, and add these up. This sum is the total number, $r,$
of electrons still required by the structure for each atom in the
structure to have a noble gas configuration. The operation is
$r=(8-p1)+(8-p2)+(8-p3)+\ldots+(8-pk).$
-
Subtract $r$ from $l$ to find $d,$ the deficit (shortage or excess)
between available electrons and required electrons.
The operation is $l-r=d.$
-
If
-
$r=0$ then
-
Complete the noble gas configuration of each atom
by writing as many lone pairs around each atom as
are needed so that the total number of electrons
around each atom is 8.
-
$r\lt0$ and
-
$r=-2$ then
-
$r=-4$ then
-
One Triple bond or two double bonds.
-
$r=-6$
-
One triple and one double bond, or three double bonds.
-
$r\gt0$ then
-
Complete the noble gas configuration of each atom
by writing as many lone pairs around each atom as
are needed so that the total number of electrons
around each atom is 8.
-
Add the excess $d$ electrons to the central atom.
Alex Notes