vi. A covalent bond between two elements is formed by sharing two “valence
electrons.” “Valence electrons” are those that can be implied in bond
formation.
vii. Elements can be classified according to their “attitude to attract electrons” or
“electronegativity, E”. For the elements we shall deal with, it is E H < E C < E N
< E O < E F .
viii. Between two elements (of the same or different species) can be formed
multiple bonds: double bonds such as in O=O, triple bonds such as NN or
even four bonds.
ix. When a covalent bond is formed, for the calculation of the formal oxidation
state of an element, by convention, the bonding pair(s) is(are) attributed to
the element that has a higher electronegativity. So, for example, in C=O both
electron pairs are attributed to O that is more electronegative than C.
x. The oxidation state of an element is obtained by calculating the difference
between the number of valence electrons (H = 1; C = 4; N = 5; O = 6) and
the number of electrons after attributing the bonding electrons to the more
electronegative atom. In covalent compounds, the oxidation state does not
represent a net charge on the atoms but is a formal number useful to
understand whether the element has been oxidized or reduced in making that
specific compound. In general, the highest oxidation state of an element is
given by the number of valence electrons, and the lowest oxidation state by
the number of lacking electrons for completing the level. In H-H, having the
two atoms the same value of E, the two electrons (represented by “-”) are
attributed one to each atom: therefore each atom has 1 electron as in the atom
and the n ox of H in H 2 is zero. In H-F, the electron pair is attributed to F, and
the oxidation state of H, is 1 − 0 = +1. In C = O the four electrons of the
double bond (coming two from C and 2 from O) are attributed to oxygen so
that C is left with only 2 of the original 4 valence electrons and has an
oxidation state of n ox = 4 − 2 = +2, while oxygen has 2e
− in addition to the
original 6 and has an oxidation state of n ox = 6 − 8 = −2. In CH 4 , H is +1, C
is −4; in CH 2 O, n oxH = +1, n oxO = −2, n oxC = 0. It is clear that the n ox are just
conventional values that may help to understand the electron flow in a
reaction.
Note that in a species, the sum of the oxidation states of the elements multiplied
each for the relevant stoichiometric coefficient must be equal to the charge of the
species. So, in CH 4 , being E C < E H , each H has n ox = +1, and C has n ox = −4,
therefore: 4x(+1) H − 4 C = 0 (methane is a neutral species). In CO 3
2− , C has n ox =
+4, O has n ox = −2, in total: +4+[3x(−2)] = −2 that is the charge of the ion.
Appendix A
231
electrons.” “Valence electrons” are those that can be implied in bond
formation.
vii. Elements can be classified according to their “attitude to attract electrons” or
“electronegativity, E”. For the elements we shall deal with, it is E H < E C < E N
< E O < E F .
viii. Between two elements (of the same or different species) can be formed
multiple bonds: double bonds such as in O=O, triple bonds such as NN or
even four bonds.
ix. When a covalent bond is formed, for the calculation of the formal oxidation
state of an element, by convention, the bonding pair(s) is(are) attributed to
the element that has a higher electronegativity. So, for example, in C=O both
electron pairs are attributed to O that is more electronegative than C.
x. The oxidation state of an element is obtained by calculating the difference
between the number of valence electrons (H = 1; C = 4; N = 5; O = 6) and
the number of electrons after attributing the bonding electrons to the more
electronegative atom. In covalent compounds, the oxidation state does not
represent a net charge on the atoms but is a formal number useful to
understand whether the element has been oxidized or reduced in making that
specific compound. In general, the highest oxidation state of an element is
given by the number of valence electrons, and the lowest oxidation state by
the number of lacking electrons for completing the level. In H-H, having the
two atoms the same value of E, the two electrons (represented by “-”) are
attributed one to each atom: therefore each atom has 1 electron as in the atom
and the n ox of H in H 2 is zero. In H-F, the electron pair is attributed to F, and
the oxidation state of H, is 1 − 0 = +1. In C = O the four electrons of the
double bond (coming two from C and 2 from O) are attributed to oxygen so
that C is left with only 2 of the original 4 valence electrons and has an
oxidation state of n ox = 4 − 2 = +2, while oxygen has 2e
− in addition to the
original 6 and has an oxidation state of n ox = 6 − 8 = −2. In CH 4 , H is +1, C
is −4; in CH 2 O, n oxH = +1, n oxO = −2, n oxC = 0. It is clear that the n ox are just
conventional values that may help to understand the electron flow in a
reaction.
Note that in a species, the sum of the oxidation states of the elements multiplied
each for the relevant stoichiometric coefficient must be equal to the charge of the
species. So, in CH 4 , being E C < E H , each H has n ox = +1, and C has n ox = −4,
therefore: 4x(+1) H − 4 C = 0 (methane is a neutral species). In CO 3
2− , C has n ox =
+4, O has n ox = −2, in total: +4+[3x(−2)] = −2 that is the charge of the ion.
Appendix A
231