What would be the correct formula of the ionic compound formed if a certain metal M with valence electron configuration Ns2 reacted with a nonmetal valence electron Ns2 Np4?

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What would be the correct formula of the ionic compound formed if a certain metal M with valence electron configuration Ns2 reacted with a nonmetal valence electron Ns2 Np4?

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Answer 1 · 2018-07-05T06:35:28

MX

Explanation:

Just some basic background information, when forming ionic compound, the metal will give away electron(s) to become cation (positively charged ion) while the nonmetal will accept electron(s) to become anion (negatively charged ion).

The metal, M, with valence electron configuration of $n s^{2}$ $ns^2$ means it has 2 valence electrons. This metal belongs to Group IIA or Group 2. That means, it will want to donate 2 of its valence electrons in order to achieve octet. It will therefore become $M^{2+}$ $M^"2+"$ .

As for the nonmetal (I'll just call it X) with valence electron configuration of $n s^{2} n p^{4}$ $ns^2 np^4$ means it has ** 6 valence electrons ** (2 + 4 = 6). This nonmetal belongs to Group VIA or Group 16. That means, it will want to accept 2 electrons in order to achieve octet, and it will become $X^{2-}$ $X^"2-"$ .

Combining the two ions, $M^{2+}$ $M^"2+"$ and $X^{2-}$ $X^"2-"$ will produce $M_{2} X_{2}$ $M_2X_2$ , which should be simplified to be $M X$ $MX$ .

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What would be the correct formula of the ionic compound formed if a certain metal M with valence electron configuration Ns2 reacted with a nonmetal valence electron Ns2 Np4?

1 Answer

Explanation:

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