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Because iodine is a homonuclear diatomic element, an unbonded atom of iodine will have the subscript of 2. If we draw the Lewis structures for both iodine atoms, we will find that each has 7 dots around it. This is because iodine is in group 7A on the periodic table. Then, by putting the two structures side by side, with the lone electrons (or dots) are facing each other. This will create a single bond. The entire molecule should look similar to this:

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:I-I:

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13y ago
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5mo ago

The Lewis structure of B2 is a linear molecule with a triple bond between the two boron atoms, each boron having three lone pairs. This arrangement follows the octet rule for boron, with each boron atom having a full outer shell of electrons.

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10y ago

This is an exception to the rule of 8 valence electrons. Two boron atoms connect with a triple bond, and there are only 6 electrons present in each atoms outer shell.

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15y ago

I - I - I with 3 pairs of electrons on each each Iodine atom for a total of 22 valence electrons. The middle I has 3 pairs of electrons and that is why it disagrees with the octet rule

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13y ago

same as fluorine just with an I instead of F. 7 valence electrons in group 17

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13y ago

There is no Lewis structure for sodium iodide it is an ionic bond so therefore requires an electron dot diagram

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12y ago

I-S-I, with 16 nonbonded electrons (4 around S and 6 around each I)

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11y ago

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:o-i=o:

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Q: What is the Lewis structure of B2?
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