Thermal Decomposition of Group 2 Carbonates and Nitrates: PECTAA Class 12 Chemistry, Punjab Board

Heat a carbonate and it breaks apart. Heat it more and more as you go down Group 2, because the bigger the metal ion, the harder the break. One idea, polarizing power, explains the whole trend.

Section 17.4 of the PECTAA Chemistry 12 book (Group 2 Elements) covers thermal decomposition of carbonates and nitrates. This note is for FSc Part 2 (2nd year, Class 12) students in Lahore, Faisalabad, Sargodha, Rawalpindi, Multan, Gujranwala and the other Punjab boards. Gujrat district comes under the Gujranwala board. PECTAA writes the books. Your own board sets your paper.

What to know

Definition. Thermal decomposition is a chemical reaction where a single compound breaks down into two or more simpler substances when heated. It usually requires energy (endothermic) to break the chemical bonds in the compound.

Why it is useful. The book says it helps in understanding and analyzing Group 2 carbonates and nitrates, because their thermal stability follows a clear trend down the group. By noting the temperature at which a compound decomposes, chemists can identify unknown compounds and confirm the presence of specific elements.

Carbonates. They decompose on heating to a metal oxide (MO) and carbon dioxide gas.

• MgCO₃(s) → MgO(s) + CO₂(g)
• CaCO₃(s) → CaO(s) + CO₂(g)

Thermal stability of carbonates increases down the group. BeCO₃ decomposes at a lower temperature than MgCO₃, which decomposes at a lower temperature than CaCO₃, and so on. BaCO₃ needs the highest temperature.

Nitrates. They decompose on heating to a metal oxide (MO), brown nitrogen dioxide gas and oxygen gas.

• 2Ca(NO₃)₂(s) → 2CaO(s) + 4NO₂(g) + O₂(g)

Their stability also increases down the group. Be(NO₃)₂ is the least stable and Ba(NO₃)₂ is the most stable.

The reason: polarizing power. The trend depends on the size of the M²⁺ ion.

• A small, highly charged cation such as Be²⁺ has a greater polarizing power.
• It distorts the large carbonate ion (CO₃²⁻) more, which weakens its bonds. The carbonate then decomposes more easily with heat.
• Down the group the metal ion gets larger and its polarizing power decreases. The carbonates become more stable and need more heat.
• The same argument applies to the nitrate ion (NO₃⁻).

Key points

Table 17.3 in the book gives the temperature of decomposition of the Group 2 carbonates at 1 atm.

The last column is worked out from the book table. Each temperature is higher than the one above it. This is the trend "stability increases down the group".

Gas clues. Carbonates give CO₂. Nitrates give brown NO₂ and also O₂.

Where the oxide fits. Table 17.2 in the book compares the two groups. Group 1 forms oxides of the type M₂O, for example Na₂O, which are highly basic. Group 2 forms oxides of the type MO, for example CaO, which are less basic than Group 1 oxides. The solid left after heating a Group 2 carbonate or nitrate is this MO oxide.

Two ways to read the trend. Read the table down the page to see the temperature rise. Read the diagram in the book across: the small Mg²⁺ ion distorts the carbonate ion strongly, while the much larger Ba²⁺ ion distorts it far less.

Worked examples

Example 1. Write the equation for the thermal decomposition of SrCO₃. (Book Quick Check 17.2)

1. A Group 2 carbonate gives the oxide and CO₂.
2. SrCO₃(s) → SrO(s) + CO₂(g).
3. Atoms: Sr 1 = 1, C 1 = 1, O 3 = 1 + 2. It is balanced.

Example 2. Why is MgCO₃ less stable than CaCO₃?

1. Mg²⁺ is smaller than Ca²⁺.
2. So Mg²⁺ has the greater polarizing power.
3. It distorts the CO₃²⁻ ion more, so MgCO₃ decomposes at a lower temperature. Table 17.3 gives 540 °C for MgCO₃ and 900 °C for CaCO₃.

Example 3. Write the equation for the thermal decomposition of magnesium nitrate, using the same pattern as the book calcium equation.

1. 2Mg(NO₃)₂(s) → 2MgO(s) + 4NO₂(g) + O₂(g).
2. Count atoms on each side. Mg: 2 = 2. N: 4 = 4. O: 12 = 2 + 8 + 2 = 12.
3. It is balanced.

Example 4. By how much does the decomposition temperature rise from CaCO₃ to SrCO₃?

1. SrCO₃ is 1289 °C and CaCO₃ is 900 °C.
2. 1289 - 900 = 389 °C.

Example 5. Predict which decomposes at the lower temperature, magnesium nitrate or barium nitrate. (Book Quick Check 17.2, part e)

1. Mg²⁺ is smaller than Ba²⁺, so it has the greater polarizing power.
2. It distorts the nitrate ion more and weakens it.
3. So Mg(NO₃)₂ decomposes at the lower temperature. Ba(NO₃)₂ is the most stable of the group.
4. Barium nitrate follows the same pattern: 2Ba(NO₃)₂(s) → 2BaO(s) + 4NO₂(g) + O₂(g).

Common mistakes in the exam

1. Saying stability decreases down the group. The book says it increases for both carbonates and nitrates.
2. Forgetting oxygen in the nitrate equation. The products are MO, NO₂ and O₂.
3. Writing CO₂ for nitrates. Nitrates give nitrogen dioxide, not carbon dioxide.
4. Not balancing. For nitrates, 2 mol of the nitrate give 4 mol of NO₂ and 1 mol of O₂.
5. Giving only "ion size". Say that smaller ions have greater polarizing power and distort the anion.
6. Mixing the order. BeCO₃ is the least stable and BaCO₃ the most stable.

Exam technique

• Write the general equation first: carbonate gives MO + CO₂. Nitrate gives MO + NO₂ + O₂.
• For a "reason" question, use these words: small cation, high polarizing power, distorts the anion, weakens bonds.
• Quote two or three values from Table 17.3 to support your answer.
• Check your board's notice for the sections in your paper.

Practice questions

1. Define thermal decomposition. Answer: A reaction in which a single compound breaks down into two or more simpler substances when heated.
2. Write the equation for heating CaCO₃. Answer: CaCO₃(s) → CaO(s) + CO₂(g).
3. Which carbonate in Table 17.3 needs the highest temperature? Answer: BaCO₃, at 1360 °C.
4. Name the three products of heating a Group 2 nitrate. Answer: A metal oxide, brown nitrogen dioxide and oxygen.
5. Which has the greater polarizing power, Be²⁺ or Ba²⁺? Answer: Be²⁺, because it is smaller and highly charged.

Ready to practise this? Continue in 2nd Year Chemistry.