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Write the Lewis symbols for each of the following ions:(a) As3–(b) I–(c) Be2+(d) O2–(e) Ga3+(f) Li+(g) N3–
a) $[: \ddot{\mathrm{As}}:]^{3-}$$\\$b)$$\text { [:?: }$$$\\$c) $[\mathrm{Be}]^{2+}$$\\$d) $[: \ddot{0}:]^{2-}$$\\$e) $[\mathrm{Ga}]^{3+}$$\\$f) $[\mathrm{Li}]^{+}$$\\$g) $[: \ddot{\mathrm{N}}:]^{3-}$
03:11
Linhan Y.
Chemistry 101
Chapter 7
Chemical Bonding and Molecular Geometry
Chemical Bonding
Molecular Geometry
Heben B.
October 18, 2020
University of Central Florida
Rice University
University of Toronto
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when drawing the louis symbols of ions, they are typically found in brackets. If they are an ions we show the valence electrons as dots surrounding the element. If they are cat ions, typically the valence electrons have been removed and you will not see the dots. So if we look at A. S. A. S. Has five valence electrons. The three minus means there's three more giving it eight. So we'll see all eight dots around A. S. In brackets. With the three minus charge on the outside, iodine has seven valence electrons. The one minus charge gives it eight. So we've got eight dots with brackets and then a one minus charge. Beryllium has two valence electrons, brilliant. two plus we'll have those two valence electrons removed. So we just have B. E. With no dots brackets and a two plus charge. Oxygen has six valence electrons. The tu minus charge gives it to more. So there's eight valence electrons. Eight thoughts around oxygen brackets and a tu minus charge gallium has three valence electrons. The three are removed with three plus charge. So we just have gallium no dots in brackets. With the three plus charge. lithium has one valence electron with the plus charge. The one is removed so it's just lithium no dots brackets and a plus one charge. Nitrogen has five valence electrons with the three minus charge. There are three more, giving it eight. So it'll be nitrogen, eight dots brackets and a three minus charge.
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