Which of the bicarbonates does not exist in solid state? (t) \(\mathrm{NaHCO}_{3}\) (b) \(\mathrm{KHCO}_{3}\) \(\mathrm{Ca}\left(\mathrm{HCO}_{3}\right)_{2}\) (d) \(\mathrm{RbHCO}_{3}\)

Short Answer

Expert verified
Calcium bicarbonate \(\mathrm{Ca}\left(\mathrm{HCO}_{3}\right)_{2}\) does not exist in solid state.

Step by step solution

01

Understanding Bicarbonates in Solid State

Certain bicarbonates can exist in solid state, whereas others can decompose into carbon dioxide, water, and a carbonate of that metal. Bicarbonates of Group 1 metals typically form stable solids.
02

Analyzing the Given Bicarbonates

Examine each given bicarbonate to determine if it belongs to Group 1 of the periodic table or if it has properties that would prevent it from existing in a solid state.
03

Identifying the Bicarbonate that Does Not Exist in Solid State

Because sodium, potassium, and rubidium are Group 1 metals, their bicarbonates can exist in solid state. However, calcium is a Group 2 metal, and its bicarbonate is not stable in solid form and thus cannot be isolated; it decomposes into solid calcium carbonate, water, and carbon dioxide gas.

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Key Concepts

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

Group 1 Metals
Group 1 metals, also known as alkali metals, reside on the far left side of the periodic table and include elements like lithium, sodium, potassium, rubidium, and cesium. These elements are characterized by a single electron in their outermost shell, which makes them highly reactive, particularly with water. Group 1 metals are also known for forming stable compounds with nonmetals. For instance, when combined with the hydrogen carbonate ion (HCO3-), they form bicarbonates.

Because of their high reactivity, these metals are never found in nature as free elements but always as parts of compounds. For students, a key concept to grasp is the similarity in chemical properties among Group 1 metals, due to their single valence electron which governs their bonding behaviors. While these metals bond readily, their compounds such as bicarbonates often have distinct stabilities and decomposition patterns,which are crucial to understand when studying their chemistry.
Stability of Bicarbonates
Bicarbonates, or hydrogen carbonates, are interesting compounds that have both a hydrogen and a carbonate ion. When discussing the stability of bicarbonates, especially of those from Group 1 metals, it is important to consider their ability to stay intact as solid compounds. The stability of a bicarbonate is influenced by the size of the metal ion it is associated with; generally, the larger the metal ion, the less stable the bicarbonate.

In the case of Group 1 metals, the bicarbonates tend to be quite stable in solid form due to the relatively smaller ionic sizes of these metals and their propensity to form ionic bonds. So, when diving into the stability of these compounds, it's instrumental for students to link the properties of Group 1 metals—such as ionic size and charge density—to the resulting stability of their bicarbonates. This understanding allows for a better grasp of why some bicarbonates are readily isolable in the solid state while others cannot be obtained this way.
Decomposition of Bicarbonates

Decomposition Reactions

The decomposition of bicarbonates is a chemical process where these compounds break down upon heating to produce carbon dioxide (CO2), water (H2O), and a corresponding carbonate. In the context of Group 1 metal bicarbonates, decomposition usually requires high temperatures and results in the formation of stable carbonates.

Predicting Decomposition

Factors such as the position of the metal in the periodic table can influence the temperature at which bicarbonates decompose. Group 1 metal bicarbonates are more stable and require higher temperatures for decomposition than those of heavier group members or other groups.

It's essential for students to understand that since the bicarbonates are an intermediate compound, the conditions of their stability and the nature of their decomposition are perfect illustrations of the dynamic changes compounds can undergo. Exercises that deal with such transformations not only test their comprehension of chemical stability but also the application of heat and reaction prediction in inorganic chemistry.

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Most popular questions from this chapter

4\. Complete the following equations: (i) \(\left.\mathrm{Na}_{2} \mathrm{O}_{2}+2 \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{CW}\right)+\mathrm{H}_{2} \mathrm{O}_{2}\) (ii) \(2 \mathrm{KO}_{2}+2 \mathrm{H}_{2} \mathrm{O} \rightarrow \underline{(X)}+\underline{(Y)}+\mathrm{O}_{2}\) (iii) \(\mathrm{Na}_{2} \mathrm{O}+\mathrm{CO}_{2} \rightarrow \underline{(Z)}\) \(\begin{array}{llll}W & X & Y & Z\end{array}\) \(\begin{array}{lllll}\text { (a) } 4 \mathrm{Na} & \mathrm{K}_{2} \mathrm{O} & \mathrm{H}_{2} \mathrm{O} & \mathrm{Na}_{2} \mathrm{O}_{2}\end{array}\) \(\begin{array}{lllll}\text { (b) } 4 \mathrm{Na} & \mathrm{K}_{2} \mathrm{O} & \mathrm{H}_{2} \mathrm{O}_{2} & \mathrm{Na}_{2} \mathrm{CO}_{3}\end{array}\) (c) \(4 \mathrm{NaOH} \quad 2 \mathrm{KOH} \quad \mathrm{H}_{2} \mathrm{O} \quad \mathrm{Na}_{2} \mathrm{O}_{2}\) \(\begin{array}{lllll}\text { (d) } 2 \mathrm{NaOH} & 2 \mathrm{KOH} & \mathrm{H}_{2} \mathrm{O}_{2} & \mathrm{Na}_{2} \mathrm{CO}_{3}\end{array}\)

Beryllium shows diagonal relationship with aluminium. Which of the following similarity is incorrect? (a) \(\mathrm{Be}_{2} \mathrm{C}\) like \(\mathrm{Al}_{4} \mathrm{C}_{3}\) yields methane on hydrolysis. (b) Be like \(\mathrm{Al}\) is rendered passive by \(\mathrm{HNO}_{3} .\) (c) \(\mathrm{Be}(\mathrm{OH})_{2}\) like \(\mathrm{Al}(\mathrm{OH})_{3}\) is basic. (d) Be forms beryllates and \(\mathrm{Al}\) forms aluminates.

Which of the following statements is correct? (a) Sodium carbonate decomposes on heating (b) Sodium bicarbonate is more soluble in water than potassium bicarbonate. (c) Sodium when heated with excess of \(\mathrm{O}_{2}\) gives peroxide. (d) Lithium halides are highly ionic in nature.

Which nitrate will decompose to give \(\mathrm{NO}_{2}\) on heating? (a) \(\mathrm{NaNO}_{3}\) (b) \(\mathrm{KNO}_{3}\) (c) \(\mathrm{RbNO}_{3}\) (d) \(\mathrm{LiNO}_{3}\)

When kept open in air, the crystals of washing soda lose 9 molecules of water to form a monohydrate. \(\mathrm{Na}_{2} \mathrm{CO}_{3} \cdot 10 \mathrm{H}_{2} \mathrm{O} \frac{\text { exposed }}{\text { to air }}>\mathrm{Na}_{2} \mathrm{CO}_{3} \mathrm{H}_{2} \mathrm{O}+9 \mathrm{H}_{2} \mathrm{O}\) This process is called (a) efflorescence (b) deliquescence (c) dehydration (d) hydration.

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