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Barium is a group 2 element and iodine is a group 17 element, which means barium (Ba) is metal and iodine is a halogen. Metals and halogens react to form metal halide compounds known as salts, such as barium iodide.
The chemical formula of barium iodide is BaI2. Commonly, it is found as minerals witherite (BaCo3) and barite (BaSO4). Mainly we produce it by electrolysis of barium chloride (BaCl2). We also receive it by the decomposition of barium by metallic sodium.
Key Terms: Barium Iodide, Anhydrous Barium Iodide, Dihydrate Barium Iodide, Iodine, Properties of Barium Iodide, Structure of Barium Iodide
What is Barium Iodide?
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Barium (Ba) is an element of the periodic table with atomic number 56 and is a soft, silvery, alkaline metal. It is a group II element. And, it is highly reactive so barium can never be found in the free state of nature. It is an s-block element having an atomic weight of 137.32amu.
- Barium iodide (Bal2) is obtained by the decomposition of barium by metallic sodium.
- Barium iodide crystallizes in small, colorless needles, which deform slightly and are very soluble in water.
- Barium (Gk barys, heavy) is a heavy metal, first isolated in England by Sir Humphrey Davy in 1808 by electrolysis of molten barium salts.
- It is prepared by the action of hydroiodic acid on barium carbonate solution or barium hydroxide.
Barium iodide is an inorganic compound composed of barium and iodide ions and is soluble in water and has a white crystalline structure. It is useful as an X-ray contrast media component. When barium iodide is exposed to X-rays, it creates a bright image on the film that allows doctors to see inside the body.
How To Prepare Barium Iodide?
We can prepare barium iodide by dehydrating it using ammonium carbonate and barium hydroxide in the presence of a hydrogen periodate reagent. By treating Ba metal with 1, 2-diiodoethane in ether we can prepare anhydrous barium iodide. It reacts with alkyl potassium compounds for the formation of organobarium compounds. Additionally, we can reduce it with lithium biphenyl which gives a highly active form of barium metal.
Barium Iodide Structure
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Barium iodide formula is BaI2. It contains two forms - an anhydrous form and a dihydrate form. The dihydrate form of barium iodide is when each barium iodide molecule is bonded to 2 other water molecules.
- The chemical formula of dihydrate barium iodide is written as BaI2.2H2O.
- Anhydrous barium iodide has no water molecules attached to it.
- From dehydrated barium iodide, we can easily obtain anhydrous barium iodide by heating the latter and allowing the water molecules to evaporate. They are crystalline solids.
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Chemical Properties of Barium Iodide
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Chemical properties of Barium Iodide are tabulated below:
| Chemical Properties of Barium Iodide | |
|---|---|
| BaI2 | Barium iodide |
| Density | 5.15 g/cm³ |
| Molecular Weight/ Molar Mass | 391.136 g/mol |
| Boiling Point | 2,027 °C |
| Melting Point | 711 °C |
| Chemical Formula | BaI2 |
The following are the chemical properties of Barium iodide (BaI):
- The chemical properties of barium iodide can vary slightly between anhydrous barium iodide and dihydrate barium iodide. This is because of the existence of water molecules attached to the barium iodide molecules. This makes the molecule soluble in ethanol, water, and acetone. However in the case of anhydrous barium iodide, despite being a stable salt, it is sensitive to air, light, and moisture.
- It is also hygroscopic; which means that it will readily absorb water to crystallize in the dihydrate form of barium iodide. Both are sensitive to oxides and toxic for consumption. As they are metal halide compounds they are also ionic in nature.
- Barium iodide reacts with potassium bromide to form a yellow precipitate of barium bromide and potassium iodide.
KBr + BaI2 \(\to\) KI + BaBr2
- Barium sulfate is created when barium iodide is treated with sodium sulfate. The chemical formula is mentioned below:
BaI2 + Na2SO4 \(\to\) BaSO4 + NaI
Physical Properties of Barium Iodide – BaI2
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The following table shows the physical properties of Barium Iodide (BaI2).
| Physical Properties of Barium Iodide | |
|---|---|
| Odour | Odourless |
| Appearance | White solid |
| Covalently-Bonded Unit | 3 |
| Complexity | 2.8 |
| pH | 7 |
| Solubility | Readily soluble in alcohol |
| Chemical formula | BaI2 |
| Molecular weight | 391.136 g/mol (anhydrous) 427.167 g/mol (dihydrate) |
| Density | 5.15 g/cm3 (anhydrous) 4.916 g/cm3 (dihydrate) |
| Solubility in water | 166.7 g/100 mL (0 °C) |
| Melting point | 711 °C |
Important Reactions Involving Barium Iodide
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Barium iodide can be made by reacting barium and iodine molecules because metals and halogens are highly reactive. Since group 17 contains less iodine, we need to provide heat for the reaction to take place.
| Ba + I2 \(\to\) BaI2 |
The reaction of potassium iodide with barium nitrate in the presence of water will give the dihydrate barium iodide.
| KI(aq) + Ba(NO3)2 (aq) \(\to\) KNO3 (aq) + BaI2(aq) |
This type of reaction is known as a double displacement reaction and is reversible.
Since this reaction takes place in the presence of water, after the formation of barium iodide, water molecules form ligands of barium iodide, and so the product we get is the dihydrate barium hydroxide. Potassium nitrate is soluble in water and hence on heating, we get crystallized anhydrous barium iodide.
Uses of Barium Iodide
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Barium iodide is toxic like other soluble salts containing barium. It has very limited use compared to many other halide forms. Some uses of barium iodide include:
- Useful for preparing barium dioxide, but has no medicinal usage.
- Useful for manufacturing other iodide compounds.
- Useful for densifying the copper castings defined in a paper prior to the Electrochemical Society.
- The most ordinary use of barium iodide is in photography. It is used as a photographic developer. It makes the image more visible and helps to improve the contrast of the image.
- In metallurgy, barium iodide is also useful as a flux. It is used to help lower the melting point of metals so that they can be melted and cast more easily.
Things to Remember
- Barium iodide is an inorganic compound composed of barium and iodide ions. It contains a white crystalline structure and is soluble in water.
- Barium iodide (BaI2) is a salt of iodide and barium ions. It is a white crystalline compound that is soluble in water.
- Barium iodide (BaI2) contains two forms - an anhydrous form and a dihydrate form. The chemical formula of dihydrate barium iodide is written as BaI2.2H2O. Anhydrous barium iodide exists in 2 different cell structures: a PbCl2 structure and a Fe2P-type structure.
- Anhydrous barium iodide’s molar mass is 391.136 g/mol and its density is 5.15 g/cm3. Dihydrate barium iodide’s molar mass is 427.167 g/mol and its density is 5.0 g/cm3.
- Barium iodide (BaI2) is useful for preparing barium dioxide, manufacturing other iodide compounds, and densifying the copper castings defined in a paper prior to the Electrochemical Society. It is also used in photography.
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Sample Questions
Ques. Answer the following Questions: (3 Marks)
(1) State the uses of barium iodide.
(2) Is barium iodide ionic or covalent?
(3) State the structure of barium iodide.
Ans: (1) Barium iodide is widely useful to manufacture other important compounds of iodine. This compound is also useful in preparing barium dioxide.
(2) An ionic compound is a solid composed of oppositely charged ions. When dissolved in water, such compounds make a solution, allowing the conduction of electrical current. Therefore, barium iodide is an ionic compound.
(3) The structure of the anhydrous form resembles that of lead(II) chloride (PbCl2) where each barium center is bound to nine iodide ligands. The crystalline packing structure is similar to that of BaCl2.
Ques. State the primary distinction between anhydrous barium iodide and dihydrate barium iodide. (3 Marks)
Ans: The following table shows the primary differences between anhydrous barium iodide and dihydrate barium iodide:
| Anhydrous Barium Iodide | Dihydrate Barium Iodide |
|---|---|
| Anhydrous barium nitrate is a white solid. | Dihydrate barium nitrate is a colorless crystal. |
| Its molar weight is 391.136g/mol. | Its molar weight is 427.267g/mol. |
| Its density is 5.15g/cm3. | Its density is 5g/cm3. |
| Its melting point is 711°C. | Its melting point is 740°C. |
Ques. State the complete Balanced Reaction: (3 Marks)
Potassium bromide(aq) + Barium iodide(aq) \(\to\) Potassium iodide(aq) + Barium bromide(s)
Ans: BaI2(aq) + 2KBr(aq) \(\to\) BaBr2(S) + 2KI(aq).
This is the fully balanced chemical equation of the reaction. Here we witness a double displacement chemical reaction. Here both the reactants exchange ions between themselves to form new substances. The properties of the products are completely different from those of the corresponding reactants.
By balancing the equation we confirm that the mass conservation law is sustained. In chemistry, unbalanced chemical equations do not support conservation laws.
Ques. Write the balanced chemical equation for the following and identify the type of reaction in each case: (5 Marks)
(a) Potassium bromide(aq) + Barium iodide(aq) \(\to\) Potassium iodide(aq) + Barium bromide(s)
(b) Hydrogen(g) + Chlorine(g) \(\to\) Hydrogen chloride(g)
(c) Magnesium(s) + Hydrochloric acid(aq) \(\to\) Magnesium chloride(aq) + Hydrogen(g)
(d) Zinc carbonate(s) \(\to\) Zinc oxide(s) + Carbon dioxide(g)
Ans:
(a) Potassium bromide(aq) + Barium iodide(aq) \(\to\) Potassium iodide(aq) + Barium bromide(s)
2KBr (aq) +BaI2 (aq) \(\to\) 2KI (aq) +BaBr2(s)
In this reaction, both reactants exchange ions to form two new products. Therefore, it is a double displacement reaction.
(b) Hydrogen(g) + Chlorine(g) \(\to\) Hydrogen chloride(g)
H2(g)+Cl2(g) \(\to\) 2HCl (g)
Two separate reactants react with each other to create a single product in this reaction. Hence, it is known as a combination reaction.
(c) Magnesium(s) + Hydrochloric acid(aq) \(\to\) Magnesium chloride(aq) + Hydrogen(g)
Mg(s) +2HCl(aq) \(\to\) MgCl2(aq) + H2(g)
In this reaction, the more reactive Mg replaces the less reactive H. Therefore, it is a displacement reaction.
(d) Zinc carbonate(s) \(\to\) Zinc oxide(s) + Carbon dioxide(g)
ZnCO3 (s) \(\to\)ZnO(s) +CO2(g)
In this reaction, one substance produces two different products. So, it is a decomposition reaction.
Ques: Convert the following information into chemical formulas and then balance the formulas: (5 Marks)
(a) Hydrogen sulfide gas burns in the air to give water and sulfur dioxide.
(b) Carbon disulfide burns in the air to give carbon dioxide and sulfur dioxide.
(c) Aluminum metal replaces iron from ferric oxide, Fe2O3, giving aluminum oxide and iron.
(d) Phosphorus burns in oxygen to give phosphorus pentoxide.
(e) Barium chloride reacts with zinc sulfate to give zinc chloride and barium sulfate.
Ans:
(a) 2H2S + 3O2 \(\to\) 2H2O + 2SO2
(b) CS2 + 3O2 \(\to\) CO2 + 2SO2
(c) 2Al + Fe2O3 \(\to\) Al2O3 + 2Fe
(d) P4 + 5O2\(\to\) 2P2O5
(e) BaCl2 + ZnSO4 \(\to\)ZnCl2 + BaSO4
Ques: Answer the following questions: (5 Marks)
(a) State the various characteristics of chemical reactions.
(b) State a characteristic of every chemical reaction that occurs when:
1. Dilute hydrochloric acid is added to sodium carbonate.
2. Lemon juice is slowly added to the solution of potassium permanganate.
3. Dilute sulfuric acid is added to the solution of barium chloride.
4. Quicklime is treated with water.
5. Wax is roasted in the form of a candle.
Ans: (a) The various characteristics of chemical reactions are:
- Evolution of a gas
- Formation of a precipitate
- Change in color
- Change in temperature
- Change in the state.
(b) The characteristic of chemical reaction:
- Evolution of carbon dioxide gas
- Change in color from purple to colorless
- Formation of a white precipitate of barium sulfate
- Change in temperature
- Conversion from solid to liquid and gas.
Ques: There are two metals, X and Y, which form the salts XSO4 and Y2SO4, respectively. The solution of salt XSO4 is blue in color whereas that of Y2SO4 is colorless. When barium chloride solution is added to the XSO4 solution, then a white precipitate Z is formed along with a salt which turns the solution green. And when barium chloride solution is added to the Y2SO4 solution, then the same white precipitate Z is formed along with the colorless common salt solution. (3 Marks)
(a) Which metals could the X and Y be?
(b) State the name and formula of salt XSO4.
(c) State the name and formula of salt Y2SO4.
(d) State the name and formula of white precipitate Z
(e) State the name and formula of the salt which turns the solution green in the first case.
Ans:
(a) Metal X: Copper; Metal Y: Sodium
(b) Copper sulfate, CuSO4
(c) Sodium sulfate, Na2SO4
(d) Barium sulfate, BaSO4
(e) Copper chloride, CuCl2
Ques. What holds a sample of Barium iodide, BaI2 together: (3 Marks)
1. Hydrogen bonding
2. Ionic bonding
3. Metallic bonding
4. Nonpolar covalent bonding
5. Polar covalent bonding
Ans: The correct option is 2.
- We know that the greater the difference in electronegativity of two atoms, the greater the chances of ionic bond formation. Barium is metal and iodine is non-metal. Barium loses two electrons to gain an electronic configuration similar to a stable noble gas, forming a bivalent cation, and iodine gains one electron, forming a monovalent ion.
- Two iodide ions are required to satisfy the valency and neutralize the charge of the barium cation, and thus barium iodide is formed by the formation of an ionic bond between the barium cation and the two iodide ions.
- An ionic bond is a chemical bond that forms because of the electrostatic attraction between a cation and anions. In barium iodide, ionic bonding is the attractive force that holds the sample together. Thus, the correct option is B.
Ques. The atomic number of the element barium is 56. See its position in the periodic table and answer the questions mentioned below: (3 Marks)
(i) Is it a metal or a non-metal?
(ii) How reactive is it compared to calcium?
Ans: (i) The metallic character indicates the level of reactivity of the metal. Metals lose electrons in chemical reactions due to their low ionization energy. The metallic character raises, as we move the group down. Going with periods in the periodic table is irrelevant. Therefore, barium (Ba) is a metal.
(ii) Reactivity indicates how vigorously or likely an atom reacts with other substances. This is usually determined by how easily electrons can be removed and how badly they want to take electrons from other atoms (electronegativity) because it is the transfer of electrons that is the basis of chemical reactions. For metals reactivity decreases as you move from left to right through the period and reactivity increases as you move down the group. Therefore, Ba is less reactive than Cs.
Ques. The element Barium has atomic number 56. See its position in the Periodic Table and answer the questions mentioned below: (3 Marks)
(i) What is its valency?
(ii) State the formula of its phosphate.
(iii) Is it larger or smaller than cesium (Cs) in size?
Ans: (i) Barium (Ba) is an alkaline earth metal with 2 outer shells. Like other metals in this group, it readily loses this electron to form the Ba2+ ion. So, the valency of barium is 2 which is the same if we go down the group with some exceptions.
(ii) Barium phosphate is an inorganic salt consisting of 3 cations and 2 anions. The a cation is of barium (Ba2+) and the anion is phosphate (PO42−). Barium phosphate’s molecular or chemical formula is Ba3(PO4)2.
(iii) The size of neutral atoms is drawn from the atomic radius, which is half the distance between two atoms touching each other. The size of the atoms increases, as you move an element group (column) down. The reason behind this is that each atom further down the column has more neutrons and protons and also gains an extra electron energy shell. Thus, cesium (Cs) is bigger in size than barium (Ba).
Ques. What happens if barium iodide is reacted with sodium sulfate and potassium bromide? (2 Marks)
Ans: When barium iodide reacts with barium bromide, potassium bromide, and potassium iodide:
KBr + BaI2 \(\to\) Kl + BaBr2
When barium iodide reacts with sodium sulfate, barium sulfate, and sodium iodide are formed:
BaI2 + Na2SO4 \(\to\) BaSO4 + NaI
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