Thermal decomposition of Sodium Hydrogen Carbonate (NaHCO₃)

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20. Thermal decomposition of Sodium Hydrogen Carbonate (NaHCO $_3$ )

a) Sodium Hydrogen Carbonate (NaHCO $_3$ ) is an ionic compound composed of a sodium ion and a hydrogen carbonate ion.
The two radicals (polyatomic ions) present or closely related in this context are:

Hydrogen Carbonate radical (also known as Bicarbonate radical): $\boxed{\text{HCO}_3^-}$
Carbonate radical: $\boxed{\text{CO}_3^{2-}}$ (This radical is formed during the decomposition of sodium hydrogen carbonate).

b) The thermal decomposition of Sodium Hydrogen Carbonate produces Sodium Carbonate, Water, and Carbon (IV) oxide.

Write the unbalanced chemical equation with state symbols.
$\text{NaHCO}_3(s) \xrightarrow{\Delta} \text{Na}_2\text{CO}_3(s) + \text{H}_2\text{O}(g) + \text{CO}_2(g)$
Balance the sodium atoms.
There is 1 Na on the left and 2 Na on the right. Place a coefficient of 2 in front of NaHCO $_3$ .
$2\text{NaHCO}_3(s) \xrightarrow{\Delta} \text{Na}_2\text{CO}_3(s) + \text{H}_2\text{O}(g) + \text{CO}_2(g)$
Verify the balance of all other atoms
- Sodium (Na): Left = $2 \times 1 = 2$ , Right = $1 \times 2 = 2$ (Balanced)
- Hydrogen (H): Left = $2 \times 1 = 2$ , Right = $1 \times 2 = 2$ (Balanced)
- Carbon (C): Left = $2 \times 1 = 2$ , Right = $1 \times 1 (\text{in Na}_2\text{CO}_3) + 1 \times 1 (\text{in CO}_2) = 2$ (Balanced)
- Oxygen (O): Left = $2 \times 3 = 6$ , Right = $1 \times 3 (\text{in Na}_2\text{CO}_3) + 1 \times 1 (\text{in H}_2\text{O}) + 1 \times 2 (\text{in CO}_2) = 3 + 1 + 2 = 6$ (Balanced) The balanced chemical equation for this thermal decomposition is:
$\boxed{2\text{NaHCO}_3(s) \rightarrow \text{Na}_2\text{CO}_3(s) + \text{H}_2\text{O}(g) + \text{CO}_2(g)}$
21. Doping Silicon

a) Silicon (Si) has an atomic number Z=14, meaning a neutral Silicon atom has 14 electrons.
Fill the atomic orbitals in order of increasing energy.
- 1s orbital: 2 electrons
- 2s orbital: 2 electrons
- 2p orbital: 6 electrons
- 3s orbital: 2 electrons
- 3p orbital: The remaining $14 - (2+2+6+2) = 2$ electrons. The s, p configuration for a neutral Silicon atom is:
$\boxed{1s^2 2s^2 2p^6 3s^2 3p^2}$
b) The element used for doping has "exactly one more electron in its p-orbital than Silicon."
Silicon's outermost p-orbital configuration is $3p^2$ .
An element with one more electron in its p-orbital would have a $3p^3$ configuration.
This means it is in the same period (Period 3) but one group to the right of Silicon (Group 14).
The element in Period 3, Group 15 is Phosphorus (P).
Verify the electron configuration for Phosphorus (P, Z=15).
- 1s orbital: 2 electrons
- 2s orbital: 2 electrons
- 2p orbital: 6 electrons
- 3s orbital: 2 electrons
- 3p orbital: The remaining $15 - (2+2+6+2) = 3$ electrons. The element used for doping is Phosphorus (P). Its s, p configuration is:
$\boxed{1s^2 2s^2 2p^6 3s^2 3p^3}$

Thermal decomposition of Sodium Hydrogen Carbonate (NaHCO₃)

Step-by-step solution

Write the unbalanced chemical equation with state symbols.

Balance the sodium atoms.

Verify the balance of all other atoms

Fill the atomic orbitals in order of increasing energy.

Silicon's outermost p-orbital configuration is $3p^2$ .

An element with one more electron in its p-orbital would have a $3p^3$ configuration.

The element in Period 3, Group 15 is Phosphorus (P).

Verify the electron configuration for Phosphorus (P, Z=15).

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Thermal decomposition of Sodium Hydrogen Carbonate (NaHCO₃)

Step-by-step solution

Write the unbalanced chemical equation with state symbols.

Balance the sodium atoms.

Verify the balance of all other atoms

Fill the atomic orbitals in order of increasing energy.

Silicon's outermost p-orbital configuration is 3p23p^23p2.

An element with one more electron in its p-orbital would have a 3p33p^33p3 configuration.

The element in Period 3, Group 15 is Phosphorus (P).

Verify the electron configuration for Phosphorus (P, Z=15).

Still stuck on this one?

More Physics Questions

Silicon's outermost p-orbital configuration is $3p^2$ .

An element with one more electron in its p-orbital would have a $3p^3$ configuration.