In the Lewis structures listed below, M and X represent various elements in the third period of the periodic table. Write the formula of each compound using the chemical symbols of each element:





What are the easiest ways to answer these questions?

I am working on these questions,meanwhile if any member of CSE knows the answers to these questions may answer these questions.

Solution: Now, in the third period, following elements are included. Na, Mg, Al, Si, P,S, Cl, Ar.

For question (a), NaCl is the answer$\big[Na\big]\big[:\ddot{Cl}:\big]^{2-}$. Note:-Cl has lone pair of $e^{-}$ below it.

In all the four questions X has 8 valence electrons. I want to know the meaning of $\pm$ charge inside the bracket containing M and $\pm$ charge as a superscript of the bracket containing X.

I don't know how to draw lewis strucures of element, compounds using LaTeX.

If any member knows it may reply to me.

  • 2
    $\begingroup$ No. Absolutely not. For (a), the idea is that you should combine an ion with charge +2 and an ion with charge -2. You should consult your notes on which elements are likely to give those ions. And for (b) for example, you are looking for an ion at +3 and an ion at -1 (of which there are 3 to balance the charge). And so forth... $\endgroup$
    – Zhe
    Jul 24, 2020 at 14:56
  • 1
    $\begingroup$ I would recommend to insert the upper index ($1$- or $2$-) inside the brackets. In this case, a) is $\ce{MgO}$, and d) is $\ce{Al_2S_3}$. Try to find yourself b and c ! $\endgroup$
    – Maurice
    Jul 24, 2020 at 16:04
  • $\begingroup$ @Maurice, How to draw Lewis dot structures on Chemistry stack exchange? M and X represents various elements from the third period of periodic table. $\endgroup$ Jul 24, 2020 at 16:43
  • $\begingroup$ @Maurice, O belongs to the second period of periodic table. $\endgroup$ Jul 24, 2020 at 16:53
  • 1
    $\begingroup$ OK. Replace the oxygen atom by the sulfur atom in MgO that becomes MgS. $\endgroup$
    – Maurice
    Jul 24, 2020 at 17:06

1 Answer 1


The elements in the third period of the periodic table has following electron configurations and they achieve their most stable ionic forms by gaining or losing indicated electrons (final electron configurations of each is also shown):

  1. $\ce{Na}$: $\mathrm{[Ne]3s^1} \quad \ce{ ->[-e^-] Na+}: \mathrm{[Ne]}$
  2. $\ce{Mg}$: $\mathrm{[Ne]3s^2} \quad \ce{ ->[-2e^-] Mg^2+}: \mathrm{[Ne]}$
  3. $\ce{Al}$: $\mathrm{[Ne]3s^23p^1} \quad \ce{ ->[-3e^-] Al^3+}: \mathrm{[Ne]}$
  4. $\ce{Si}$: $\mathrm{[Ne]3s^23p^2}$
  5. $\ce{P}$: $\mathrm{[Ne]3s^23p^3} \quad \ce{ ->[+3e^-] P^3-}: \mathrm{[Ne]3s^23p^6}\ce{#[Ar]}$
  6. $\ce{S}$: $\mathrm{[Ne]3s^23p^4} \quad \ce{ ->[+2e^-] S^2-}: \mathrm{[Ne]3s^23p^6}\ce{#[Ar]}$
  7. $\ce{Cl}$: $\mathrm{[Ne]3s^23p^5} \quad \ce{ ->[+e^-] Cl-}: \mathrm{[Ne]3s^23p^6}\ce{#[Ar]}$
  8. $\ce{Ar}$: $\mathrm{[Ne]3s^23p^6}\ce{#[Ar]}$

Briefly, the first element of 3rd period, $\ce{Na}$ can loose one electron to gain neon electron configuration (octet) than gaining 7 electrons to achieve argon electron configuration as shown in entry (1) above. Consequently, loosing one electron make it $\ce{Na+}$. Similarly, the seventh element of 3rd period, $\ce{Cl}$ can gain one electron to achieve argon electron configuration (also an octet) than loosing 7 electrons to achieve neon electron configuration as shown in entry (7) above. As a result of gaining one electron, $\ce{Cl}$ becomes $\ce{Cl-}$.

For $\ce{Si}$, it can gain 4 electrons or loose 4 electrons. However, either is very difficult to achieve so that its nature (along with that of carbon) is explained well using molecular orbital theory.

Your list contains compounds with three cations: $\ce{M+, M^2+,}$ and $\ce{M^3+}$. Those are $\ce{Na+, Mg^2+,}$ and $\ce{Al^3+}$, respectively. Even though it wasn't shown, all of these have octets in the last cell as explained above.

Also, your list contains compounds with only two anions: $\ce{X-}$ and $\ce{X^2-}$. Those are $\ce{Cl-}$ and $\ce{S^2-}$, respectively. Those also have octets in the last cell as explained above.

Since you know the cations and anions in the list, it's now easy to make the compounds.

  • $\begingroup$ So answer to the a, b, c, d questions are MgS(Magnesium Sulfide), $AlCl_3$Aluminum Chloride, $Na_2S$ Sodium Sulfide,$Al_2S_3$Aluminum Sulfide respectively. $\endgroup$ Jul 26, 2020 at 6:15
  • $\begingroup$ Your answers are correct. $\endgroup$ Jul 26, 2020 at 15:49

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