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Give the name and formula of each of the binary compounds formed from the following elements. a. \(\mathrm{Li}\) and \(\mathrm{N}\) b. \(\mathrm{Na}\) and \(\mathrm{Br}\) c. \(\mathrm{K}\) and \(\mathrm{S}\)

Short Answer

Expert verified
a. The compound formed from Li and N is lithium nitride, with the formula \(Li_3N\). b. The compound formed from Na and Br is sodium bromide, with the formula \(NaBr\). c. The compound formed from K and S is potassium sulfide, with the formula \(K_2S\).

Step by step solution

01

a. Li and N

Lithium (Li) is an alkali metal with a valency of +1. Nitrogen (N) is an element in group 15 with a valency of -3. To find the formula of the binary compound formed between Li and N, we can use the cross-over rule. 1. Write the symbols of both elements with their valencies as superscripts: Li^+1 N^-3 2. Swap the valencies: Li^3 N^1 or Li3N Hence, the formula of the compound is Li3N. The name of this compound is lithium nitride.
02

b. Na and Br

Sodium (Na) is an alkali metal with a valency of +1. Bromine (Br) is a halogen with a valency of -1. To find the formula of the binary compound formed between Na and Br, we can use the cross-over rule. 1. Write the symbols of both elements with their valencies as superscripts: Na^+1 Br^-1 2. Swap the valencies: Na^1 Br^1 or NaBr Hence, the formula of the compound is NaBr. The name of this compound is sodium bromide.
03

c. K and S

Potassium (K) is an alkali metal with a valency of +1. Sulfur (S) is an element in group 16 with a valency of -2. To find the formula of the binary compound formed between K and S, we can use the cross-over rule. 1. Write the symbols of both elements with their valencies as superscripts: K^+1 S^-2 2. Swap the valencies: K^2 S^1 or K2S Hence, the formula of the compound is K2S. The name of this compound is potassium sulfide.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Valency
Valency is a fundamental concept in chemistry that indicates the ability of an atom to bond with other atoms. It is determined by the number of electrons an atom can lose, gain, or share.
For instance, when combining elements like lithium (Li) and nitrogen (N), understanding their valencies is crucial. Lithium is an alkali metal and has a valency of +1. This means it can easily lose one electron to achieve a stable configuration. On the other hand, nitrogen, located in group 15 of the periodic table, has a valency of -3, indicating it tends to gain three electrons to become stable.
Valency helps in predicting how elements will interact and what kind of binary compounds they form. It's essential to grasp this concept to write correct chemical formulas.
Cross-Over Rule
The cross-over rule is a handy method for determining the chemical formula of compounds. It involves crossing over the valencies of the elements involved to balance the charges and form a neutral compound.
Let's see how this applies to sodium (Na) and bromine (Br). Sodium, like other alkali metals, has a valency of +1. Bromine, a member of the halogen family, has a valency of -1. By applying the cross-over rule, the valency of sodium becomes the subscript for bromine, and the valency of bromine becomes the subscript for sodium.
Thus, the formula becomes NaBr, which is sodium bromide. This simple method ensures the compound formed is chemically stable. It's an essential skill for writing formulas for binary compounds.
Alkali Metals
Alkali metals are the elements found in Group 1 of the periodic table. They include lithium (Li), sodium (Na), potassium (K), and others. These metals are characterized by having a single electron in their outermost shell, giving them a valency of +1.
Due to their single valency electron, alkali metals are highly reactive, especially with nonmetals like halogens. For example, potassium (K) reacts with sulfur (S), which has a valency of -2. By using the cross-over rule, we form the compound K extsubscript{2}S, known as potassium sulfide.
Learning about the reactivity and properties of alkali metals helps in understanding their role in forming binary compounds.
Halogens
Halogens are found in Group 17 of the periodic table and include fluorine, chlorine, bromine (Br), and others. These elements are highly electronegative, meaning they have a strong tendency to attract electrons. They typically have a valency of -1, showing they usually gain one electron to achieve a complete outer shell.
Their nature makes them highly reactive, especially with alkali metals. For example, bromine reacts with sodium (Na) to form sodium bromide (NaBr), through the interaction of their respective valencies.
Understanding halogens and their chemical behavior is key in predicting the formation of compounds and their resulting properties.

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Most popular questions from this chapter

One bit of evidence that the quantum mechanical model is "correct" lies in the magnetic properties of matter. Atoms with unpaired electrons are attracted by magnetic fields and thus are said to exhibit paramagnetism. The degree to which this effect is observed is directly related to the number of unpaired electrons present in the atom. Consider the ground-state electron configurations for \(\mathrm{Li}, \mathrm{N}, \mathrm{Ni}, \mathrm{Te}, \mathrm{Ba}\), and \(\mathrm{Hg} .\) Which of these atoms would be expected to be paramagnetic, and how many unpaired electrons are present in each paramagnetic atom?

An electron is excited from the \(n=1\) ground state to the \(n=3\) state in a hydrogen atom. Which of the following statements are true? Correct the false statements to make them true. a. It takes more energy to ionize (completely remove) the electron from \(n=3\) than from the ground state. b. The electron is farther from the nucleus on average in the \(n=3\) state than in the \(n=1\) state. c. The wavelength of light emitted if the electron drops from \(n=3\) to \(n=2\) will be shorter than the wavelength of light emitted if the electron falls from \(n=3\) to \(n=1\). d. The wavelength of light emitted when the electron returns to the ground state from \(n=3\) will be the same as the wavelength of light absorbed to go from \(n=1\) to \(n=3\). e. For \(n=3\), the electron is in the first excited state.

Assume that a hydrogen atom's electron has been excited to the \(n=5\) level. How many different wavelengths of light can be emitted as this excited atom loses energy?

Write the expected electron configurations for each of the following atoms: \(\mathrm{Cl}, \mathrm{Sb}, \mathrm{Sr}, \mathrm{W}, \mathrm{Pb}, \mathrm{Cf}\).

It takes \(7.21 \times 10^{-19} \mathrm{~J}\) of energy to remove an electron from an iron atom. What is the maximum wavelength of light that can do this?

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