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physical science

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not exactly physical science, but I'm confused about how you tell the charge of elements in the B groups on the periodic table

  • physical science -

    B groups? Your periodic table is out of date by 30 years. Let me go find one with B groups.

    OK, there are two outdated periodic tables, each of which use B for something different. I suspect from your question you are asking about the valence of the transition elements, in which the outer electrons occupy the d level energy. There is no way to know the valence the elements will form except memory, most can have several valence states, due to the interaction of the d and s levels. Manganese is a favorite of mine, examine its common valence states.

    Is is possible your instructor can start teaching the new periodic table, something the rest of the world has used since 1980, having eighteen families. See

  • physical science -

    And your question is a little confusing, too. ALL elements in the periodic table have a charge of zero when they are in their elemental state. You may be asking how to tell the charge on the ions. For the B group of elements, ONE of the possible ionic charges (valence) is always the number of the group (same as the A families). There may be other possible valences in their compounds, too, and those you need to either memorize OR look at the electron configuration. SOMETIMES, the electron configuration will help you decide if another valence is possible.
    Example 1. Cu is in group IB and ONE of its valences is + 1 and there are a number of compounds in which Cu displays a valence of +1. Cu also has a valence of +2 and that is its more common valence. Au in group IB also has a valence of +1 but it also has a valence of +3. I don't know how to do those except by memory.
    Example 2: Mn in group VIIB has a possible valence of +7. It ALSO has common valences of +2, +4, less common is +6 and some others. For Mn (element #25) the electron configuration is
    1s2 2s2 2p6 3s2 3p6 3d5 4s2.
    The +2 valence can be seen easily by simply removing the 2 4s2 electrons. The +7 is easily seen by removing all 3d and 4s electrons (7 total). The other valences aren't so easily predicted from this venue. I hope this helps. Let me know if this isn't close to what you were looking for or if you need further explanation.

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