CHAPTER XI. VALENCE.

55. The Symbols NaCl and MgCl2 differ in two ways.--What are
they? Let us see why the atom of Mg unites with two Cl atoms,
while that of Na takes but one. If the atoms of two elements
attract each other, there must be either a general attraction all
over their surfaces, or else some one or more points of
attraction. Suppose the latter to be true, each atom must have
one or more poles or bonds of attraction, like the poles of a
magnet. Different elements differ in their number of bonds. Na
has one, which may be written graphically Na-; Cl has one, -Cl.
When Na unites with Cl, the bonds of each element balance, as
follows: Na-Cl. The element Mg, however, has two such bonds, as
Mg= or -Mg-. When Mg unites with Cl, in order to balance, or
saturate, the bonds, it is evident that two atoms of Cl must be
used, as Cl-Mg-Cl, or MgCl2.

A compound or an element, in order to exist, must have no free
bonds. In organic chemistry the exceptions to this rule are very
numerous, and, in fact, we do not know that atoms have bonds at
all; but we can best explain the phenomena by supposing them, and
for a general statement we may say that there must be no free
bonds. In binaries the bonds of each element must balance.

56. The Valence, Quantivalence, of an Element is its Combining
Power Measured by Bonds.--H, having the least number of bonds,
one, is taken as the unit. Valence has always to be taken into
account in writing the symbol of a compound. It is often written
above and after the elements [i.e. written like an exponent], as
K^I, Mg^II.

An element having a valence of one is a monad; of two, a dyad;
three, a triad; four, tetrad; five, pentad; six, hexad, etc. It
is also said to be monovalent, di- or bivalent, etc. This theory
of bonds shows why an atom cannot exist alone. It would have free
or unused bonds, and hence must combine with its fellow to form a
molecule, in case of an element as well as in that of a compound.
This is illustrated by these graphic symbols in which there are
no free bonds: H-H, O=O, N[3-bond symbol]N, C[4-bond symbol]C. A
graphic symbol shows apparent molecular structure.

After all, how do we know that there are twice as many Cl atoms
in the chloride of magnesium as in that of sodium? The compounds
have been analyzed over and over again, and have been found to
correspond to the symbols MgCl2 and NaCl. This will be better
understood after studying the chapter on atomic weights. In
writing the symbol for the union of H with O, if we take an atom
of each, the bonds do not balance, H-=O, the former having one;
the latter, two. Evidently two atoms of H are needed, as H-O-H,
or

H
= O , or H2O. In the union of Zn and O, each has two bonds;
H

hence they unite atom with atom, Zn = O, or ZnO.

Write the grapbic and the common symbols for the union of H^I and
Cl^I; of K^I and Br^I; Ag^I and O^II; Na^I and S^II; H^I and
P^III. Study valences. It will be seen that some elements have a
variable quantivalence. Sn has either 2 or 4; P has 3 or 5. It
usually varies by two for a given element, as though a pair of
bonds sometimes saturated each other;. e.g. =Sn=, a quantivalence
of 4, and |Sn=, a quantivalence of 2. There are, therefore, two
oxides of tin, SnO and SnO2, or Sn=O and O=Sn=O. Write symbols
for the two chlorides of tin; two oxides of P; two oxides of
arsenic.

The chlorides of iron are FeCl2 and Fe2Cl6. In the latter, it
might be supposed that the quantivalence of Fe is 3, but the
graphic symbol shows it to be 4. It is called a pseudo-triad, or
false triad. Cr and Al are also pseudo-triads.

Cl Cl | | Cl--Fe--Fe--Cl | | Cl Cl

Write formulae for two oxides of iron; the oxide of Al.

57. A Radical is a Group of Elements which has no separate
existence, but enters into combination like a single atom; e.g.
(NO3) in the compounds HNO3 or KNO3; (SO4) in H2SO4. In HNO3 the
radical has a valence of 1, to balance that of H, H-NO3). In
H2SO4, what is the valence of (SO4)? Give it in each of these
radicals, noting first that of the first element: K(NO3),
Na2(SO4), Na2(CO3), K(ClO3), H3(PO4), Ca3(PO4)2, Na4(SiO4).

Suppose we wish to know the symbol for calcium phosphate. Ca and
PO4 are the two parts. In H3(PO4) the radical is a triad, to
balance H3. Ca is a dyad, Ca==(P04). The least common multiple of
the bonds (2 and 3) is 6, which, divided by 2 (no. Ca bonds),
gives 3 (no. Ca atoms to be taken). 6 / 3 (no. (PO4) bonds) gives
2 (no. PO4 radicals to be taken). Hence the symbol Ca3(P04)2.
Verify this by writing graphically.

Write symbols for the union of Mg and (SO4), Na and (PO4), Zn and
(NO3), K and (NO3), K and (SO4), Mg and (PO4), Fe and (SO4) (both
valences of Fe), Fe and (NO3), taking the valences of the
radicals from HNO3, H2SO4, H3PO4.