Question #05211

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Question #05211

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Answer 1 · 2015-11-05T20:34:50

Here's what I got.

Explanation:

I'll show you how to do the first four, and leave the last two for you as practice.

Your tool of choice here will be the periodic table.

Let's sart with the first two ions. You that in order to become an ion, an atom must either gain electrons, in which case it becomes an anion, or negatively charged ion, or lose electrons, in which case it becomes a cation, or positively charged ion.

So right from the start you know that're looking for electron configurations of atoms that would become isoelectronic with argon, $Ar$ $"Ar"$ , if electrons were to either be added or taken away from them.

The electron configuration of argon is

$Ar: [Ne] 3 s^{2} 3 p^{6}$ $"Ar: " ["Ne"] 3s^2 3p^6$

So, what are the elements that are closest to argon in the periodic table?

Well, the halogen that shares a period with argon is chlorine, $Cl$ $"Cl"$ . If chlorine gains an electron, its electron configuration would go from

$Cl: [Ne] 3 s^{2} 3 p^{5}$ $"Cl: " ["Ne"] 3s^2 3p^5$

Notice that it's on eelectron shy of reaching argon's electron configuration. When chlorine gains an electron it forms the chloride anion, ${Cl}^{-}$ $"Cl"^(-)$

${Cl}^{-} : [Ne] 3 s^{2} 3 p^{6}$ $"Cl"^(-): ["Ne"] 3s^2 3p^6$

Another element that's very close to argon in the periodic table is potassium, $K$ $"K"$ , which has the electron configuration

$K: [Ar] 4 s^{1}$ $"K: " ["Ar"] 4s^1$

When potassium loses an electron, it forms the potassium cation, $K^{+}$ $"K"^(+)$ , which has the electron configuration

$K^{+} : [Ar]$ $"K"^(+): ["Ar"]$

Now for the next two, which this time must have the electron configuration $[Ar] 3 d^{5}$ $["Ar"] 3d^5$ .

This one is a little tricky because you're dealing with transition metals, but the same approach can be used here as well.

Notice that this electron configuration is missing the 4s-orbital electrons, which means that you must look for elements that have an electron configuration that features at least on electron in the 4s-orbital.

Manganese, $Mn$ $"Mn"$ , is one option. A neutral manganese atom has the electron configuration

$Mn: [Ar] 3 d^{5} 4 s^{2}$ $"Mn: " ["Ar"] 3d^5 4s^2$

This means that if manganese were to lose two electrons from its 4s-orbital, the electron configuration of the resulting manganese(II) cation, ${Mn}^{+}$ $"Mn"^(+)$ , would be

${Mn}^{2 +} : [Ar] 3 d^{5}$ $"Mn"^(2+): ["Ar"] 3d^5$

Chromium fits here as well, since its electron configuration is

$Cr: [Ar] 3 d^{5} 4 s^{1}$ $"Cr: " ["Ar"] 3d^5 4s^1$

This time, only the one electron that resides in the 4s-orbital must be lost in order for the chromium(I) cation, ${Cr}^{+}$ $"Cr"^(+)$ , to be formed

${Cr}^{+} : [Ar] 3 d^{5}$ $"Cr"^(+): ["Ar"] 3d^5$

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Question #05211

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