How do you count valence electrons in compounds?

Answer 1

The number of valence electrons in the transition metals is somewhat different than main group elements. As you go from left to right across the Periodic Table, the electrons added to the transition metals go into the d-orbitals. However, because the energy of the 4s orbital is lower than the 3d orbital (and the 5s is lower than the 4d, etc.), the 4s orbital fills first. Therefore the electron configuration of iron for instance is [Ar]4s23d6.

Because the valence electrons are defined as the electrons in the outermost or highest energy shell, for iron, that would be the 4th shell. So the 6 electrons in the 3d orbital don't count. Only the 2 electrons in the 4s orbital count since they are in the 4th shell. Most transition metals thus have 2 valence electrons (although some, such as chromium, only have one because of exceptions to the filling rules -- the configuration of chromium is [Ar]4s13d5).

However, despite this, when drawing Lewis dot structures or drawing molecular orbital diagrams for transition metals, which are the main reasons to count valence electrons, all of them count. In fact, while the main group elements follow the "octet rule" (for a complete valence shell of 8 electrons), the transition metals follow the "18 electron rule" since the 10 electrons in the d-orbitals are now included. When counting electrons for Lewis dot structures, they all go into the count for the structure and transition metals are most stable when they have 18 electrons in the structure in the same way that main group elements are most stable when they have 8.

See the Related Questions to the left for more information on counting valence electrons. See the Web Links for an excellent periodic table with the electron configuration of each element.

Answer 2

Valence goes up to the group number for the first half of the d-block (2–7), then climbs back down (6–2) for the end. The bottom rows approach the extremes of bond number (8) and are more consistent and smooth in this transition. The top valences for the last half of the d-block, those atoms with paired suborbitals, are saturated in their solid state and show up as much fewer in all but coordinative or vapid states. Thus, osmium tetroxide must be a vapor. (Valence and oxidation state, however, aren't the same—the former is the fluoridation state. Because of the greatter room the one-bond fluorine atoms take up, valence tends to be fewer than oxidation at the top. However, the weakker bonding of oxygen means that oxidation tends to be fewer than valence at the bot, such as for the end of the d-block at silver and other nobil metals.) The hither-bottom atoms even reach negative valences, I guess to fill the platinum or rhenium octet: osmium(−II), rhenium(−III); tungsten(−I).

Answer 3

1) Decipher which elements are present in your compound

2) Determine the number of valence (oxidation number) electrons of each element by looking at the periodical table

3) Count the number of bonds necessary to get that zero sum value.

Ammonia NH3

One nitrogen has 5 valence electrons

Three hydrogen, each with one valence electron, totals 3

This means there are 8 valence electrons, making 4 pairs

Answer 4

Transition metals have 1 to 2 valence electrons

Answer 5

1 or 2 valence electrons.