Magnesium (Mg) reacts with chlorine \(\left(\mathrm{Cl}_{2}\right)\) to form the compound \(\mathrm{MgCl}_{2}\). (a) Predict the formulas for the compounds formed when all the other alkaline earth metals react with chlorine. (b) Predict the formulas for the compounds formed when all the alkaline earth metals react with bromine. (c) Explain what principle you used to make your predictions.

Short Answer

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Compounds formed when alkaline earth metals react with chlorine are: BeCl₂, CaCl₂, SrCl₂, and BaCl₂. Compounds formed when alkaline earth metals react with bromine are: BeBr₂, MgBr₂, CaBr₂, SrBr₂, and BaBr₂. The octet rule was used to make these predictions, as it states that atoms tend to combine in such a way that they each have eight valence electrons, giving them the same electronic configuration as a noble gas.

Step by step solution

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(a) Predicting formulas for compounds formed with chlorine

To predict the formulas for the compounds formed when all other alkaline earth metals react with chlorine, we need to know the valence electrons of both these metals and chlorine. Alkaline earth metals are found in group 2 of the periodic table, and therefore have 2 valence electrons. Chlorine, in group 17, has 7 valence electrons. According to the octet rule, the alkaline earth metals will lose their 2 valence electrons and the chlorine will gain 1 electron to complete its octet. This means that 1 alkaline earth metal atom will form a bond with 2 chlorine atoms, resulting in compounds with a 1:2 ratio of metal to chlorine. So, the formulas for these compounds are: - BeCl₂ (Beryllium chloride) - CaCl₂ (Calcium chloride) - SrCl₂ (Strontium chloride) - BaCl₂ (Barium chloride)
02

(b) Predicting formulas for compounds formed with bromine

Similar to part (a), we need to know the valence electrons for both the alkaline earth metals and bromine. Bromine is also in group 17, one element below chlorine, and therefore has 7 valence electrons. Using the octet rule, 1 alkaline earth metal atom will form a bond with 2 bromine atoms, resulting in compounds with a 1:2 ratio of metal to bromine. So, the formulas for these compounds are: - BeBr₂ (Beryllium bromide) - MgBr₂ (Magnesium bromide) - CaBr₂ (Calcium bromide) - SrBr₂ (Strontium bromide) - BaBr₂ (Barium bromide)
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(c) Explaining the principle used to make the predictions

The principle used to make the predictions in this exercise is the octet rule. The octet rule states that atoms tend to combine in such a way that they each have eight valence electrons, giving them the same electronic configuration as a noble gas. In this case, the alkaline earth metals lose their 2 valence electrons, and the non-metal atoms (chlorine or bromine) gain 1 or more electrons to complete their octet, forming ionic compounds with a 1:2 ratio. This principle allows us to predict the formulas for the compounds formed when alkaline earth metals react with both chlorine and bromine.

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Most popular questions from this chapter

A \(110.99-g\) sample of a compound containing only calcium and chlorine is found to contain \(40.98 \mathrm{~g}\) of calcium. (a) How many grams of chlorine are in the compound? (b) What is the percent by mass of calcium in the compound? (c) What is the percent by mass of chlorine in the compound?

Calcium reacts with fluorine to form the compound \(\mathrm{CaF}_{2}\). In the reaction, each atom of one of the elements loses two electrons, and each atom of the other element gains one electron. (a) Atoms of which element gain electrons? (b) Atoms of which element lose electrons? (c) What is the charge on the atoms that gain one electron? (d) What is the charge on the atoms that lose two electrons? (e) Write the full atomic symbol, including charges, for each element. Give Ca 20 neutrons and F 10 neutrons.

Metals are excellent conductors of electricity. Nonmetals are poor conductors of electricity and are called insulators. Some elements have the ability to conduct electricity roughly between that of metals and nonmetals. Give some examples of these elements, and name the class they belong to.

Use Dalton's hook atomic models to sketch a molecule of \(\mathrm{H}_{2} \mathrm{~S}\).

Two different compounds, both consisting of sodium (Na) and oxygen (O), were analyzed. The data are given below: $$ \begin{array}{cccc} & \begin{array}{l} \text { Mass of } \\ \text { sample } \\ \text { Compound } \end{array} & \begin{array}{l} \text { Mass of } \\ \text { analyzed } \end{array} & \text { O present } & \begin{array}{l} \text { Mass of } \\ \text { Na present } \end{array} \\ \hline \mathrm{A} & 19.50 \mathrm{~g} & 8.00 \mathrm{~g} & ? \\ \mathrm{~B} & 61.98 \mathrm{~g} & 16.00 \mathrm{~g} & ? \\ \hline \end{array} $$ (a) Fill in the last column of the table. (b) Calculate the \(\% \mathrm{Na}\) and \(\% \mathrm{O}\) for both compounds.

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