A valid Lewis structure of IF3 cannot be drawn without violating the octet rule.
Answer: IF3 (Iodine Trifluoride)
This is because, I (Iodine) and F (Fluorine) both have odd number of valence electrons (7) which also means that there are too many valence electrons in the valence shell.
Answer:
Ge: [Ar] 3d10 4s2 4p2 => 6 electrons in the outer shell
Br: [Ar] 3d10 4s2 4p5 => 7 electrons in the outer shell
Kr: [Ar] 3d10 4s2 4p6 => 8 electrons in the outer shell
Explanation:
The electron affinity or propension to attract electrons is given by the electronic configuration. Remember that the most stable configuration is that were the last shell is full, i.e. it has 8 electrons.
The closer an atom is to reach the 8 electrons in the outer shell the bigger the electron affinity.
Of the three elements, Br needs only 1 electron to have 8 electrons in the outer shell, so it has the biggest electron affinity (the least negative).
Ge: needs 2 electrons to have 8 electrons in the outer shell, so it has a smaller (more negative) electron affinity than Br.
Kr, which is a noble gas, has 8 electrons and is not willing to attract more electrons at all, the it has the lowest (more negative) electron affinity of all three to the extension that really the ion is so unstable that it does not make sense to talk about a number for the electron affinity of this atom.
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