write the balanced equation. write the ionic equation and net ionic equations for the following reaction. (6marks)
part a
a. balance equation (predict the products and their state of matter)
bi(no₃)₃(aq) + naoh(aq) -------

b. ionic equation

c. net ionic equation

part b
using lewis diagrams illustrate the formations of bond(s) between sodium na and phosphorous p. name this compound (4 marks)

Question

write the balanced equation. write the ionic equation and net ionic equations for the following reaction. (6marks)
part a
 a. balance equation (predict the products and their state of matter)
bi(no₃)₃(aq) + naoh(aq) ------->

b. ionic equation

c. net ionic equation

part b
using lewis diagrams illustrate the formations of bond(s) between sodium na and phosphorous p. name this compound (4 marks)

Accepted Answer

### Part A
#### a. Balance equation (predict the products and their state of matter)
# Answer:
$\ce{Bi(NO_{3})_{3}(aq) + 3NaOH(aq) -> Bi(OH)_{3}(s) + 3NaNO_{3}(aq)}$
# Explanation:
## Step 1: Predict products
Bismuth(III) nitrate ($\ce{Bi(NO_{3})_{3}}$) reacts with sodium hydroxide ($\ce{NaOH}$) in a double - displacement reaction. The products are bismuth(III) hydroxide ($\ce{Bi(OH)_{3}}$) and sodium nitrate ($\ce{NaNO_{3}}$). Bismuth(III) hydroxide is insoluble in water (so it is a solid, $s$), and sodium nitrate is soluble (so it is in aqueous state, $aq$).
## Step 2: Balance the equation
- For the nitrate ions ($\ce{NO_{3}^-}$): There are 3 $\ce{NO_{3}^-}$ in $\ce{Bi(NO_{3})_{3}}$, so we need 3 $\ce{NaNO_{3}}$ on the product side. This means we need 3 $\ce{NaOH}$ on the reactant side to balance the sodium ions ($\ce{Na^+}$).
- For the hydroxide ions ($\ce{OH^-}$) and bismuth ions ($\ce{Bi^{3+}}$): 3 $\ce{OH^-}$ from 3 $\ce{NaOH}$ react with 1 $\ce{Bi^{3+}}$ from $\ce{Bi(NO_{3})_{3}}$ to form 1 $\ce{Bi(OH)_{3}}$. So the balanced equation is $\ce{Bi(NO_{3})_{3}(aq) + 3NaOH(aq) -> Bi(OH)_{3}(s) + 3NaNO_{3}(aq)}$.

#### b. Ionic equation
# Answer:
$\ce{Bi^{3+}(aq) + 3NO_{3}^-(aq) + 3Na^+(aq) + 3OH^-(aq) -> Bi(OH)_{3}(s) + 3Na^+(aq) + 3NO_{3}^-(aq)}$
# Explanation:
## Step 1: Dissociate soluble compounds
- $\ce{Bi(NO_{3})_{3}(aq)}$ dissociates into $\ce{Bi^{3+}(aq)}$ and $\ce{3NO_{3}^-(aq)}$ because it is soluble in water.
- $\ce{NaOH(aq)}$ dissociates into $\ce{Na^+(aq)}$ and $\ce{OH^-(aq)}$, and since we have 3 moles of $\ce{NaOH}$, we have $\ce{3Na^+(aq)}$ and $\ce{3OH^-(aq)}$.
- $\ce{NaNO_{3}(aq)}$ dissociates into $\ce{Na^+(aq)}$ and $\ce{NO_{3}^-(aq)}$, and with 3 moles of $\ce{NaNO_{3}}$, we have $\ce{3Na^+(aq)}$ and $\ce{3NO_{3}^-(aq)}$.
- $\ce{Bi(OH)_{3}(s)}$ does not dissociate as it is a solid.
## Step 2: Write the ionic equation
Substitute the dissociated ions into the balanced molecular equation: $\ce{Bi^{3+}(aq) + 3NO_{3}^-(aq) + 3Na^+(aq) + 3OH^-(aq) -> Bi(OH)_{3}(s) + 3Na^+(aq) + 3NO_{3}^-(aq)}$.

#### c. Net ionic equation
# Answer:
$\ce{Bi^{3+}(aq) + 3OH^-(aq) -> Bi(OH)_{3}(s)}$
# Explanation:
## Step 1: Identify spectator ions
Spectator ions are ions that appear on both sides of the ionic equation and do not participate in the chemical reaction. In the ionic equation $\ce{Bi^{3+}(aq) + 3NO_{3}^-(aq) + 3Na^+(aq) + 3OH^-(aq) -> Bi(OH)_{3}(s) + 3Na^+(aq) + 3NO_{3}^-(aq)}$, the $\ce{Na^+(aq)}$ and $\ce{NO_{3}^-(aq)}$ ions are spectator ions.
## Step 2: Remove spectator ions
After removing the spectator ions ($\ce{Na^+}$ and $\ce{NO_{3}^-}$) from the ionic equation, we get the net ionic equation: $\ce{Bi^{3+}(aq) + 3OH^-(aq) -> Bi(OH)_{3}(s)}$.

### Part B
# Answer:
The compound formed is sodium phosphide, formula $\ce{Na_{3}P}$. The Lewis diagram formation:
- **Sodium (Na)**: Sodium has an electron configuration of $1s^{2}2s^{2}2p^{6}3s^{1}$. It has one valence electron. In order to achieve a stable octet (like the noble gas neon, $1s^{2}2s^{2}2p^{6}$), sodium tends to lose its single valence electron to form a $\ce{Na^+}$ ion. The Lewis symbol for Na is $\ce{Na\cdot}$ (with the single dot representing the valence electron), and after losing the electron, it becomes $\ce{Na^+}$.
- **Phosphorus (P)**: Phosphorus has an electron configuration of $1s^{2}2s^{2}2p^{6}3s^{2}3p^{3}$. It has 5 valence electrons. To achieve a stable octet, it needs to gain 3 electrons. The Lewis symbol for P is $\ce{\cdot \overset{\cdot \cdot}{P} \cdot}$ (with 5 valence electrons shown as dots).
- **Ionic bond formation**: Three sodium atoms each lose one electron (total of 3 electrons lost) and these 3 electrons are gained by one phosphorus atom. The Lewis diagram for the formation can be represented as:

$\ce{3Na\cdot -> 3Na^+ + 3e^-}$

$\ce{\cdot \overset{\cdot \cdot}{P} \cdot + 3e^- -> [:\overset{\cdot \cdot}{P}:]^{3 - }}$

Then, the electrostatic attraction between the $\ce{Na^+}$ ions and the $\ce{P^{3 - }}$ ion forms the ionic compound $\ce{Na_{3}P}$, and the Lewis structure of the compound is $\ce{[Na^+]_{3}[:\overset{\cdot \cdot}{P}:]^{3 - }}$ (or simply represented as $\ce{Na_{3}P}$ with the understanding of the ionic bonding between $\ce{Na^+}$ and $\ce{P^{3 - }}$).
# Brief Explanations:
- **Sodium's electron behavior**: Sodium is a Group 1 metal, so it has one valence electron. It loses this electron to form a cation ($\ce{Na^+}$) to achieve a stable electron configuration (noble gas - like).
- **Phosphorus's electron behavior**: Phosphorus is a Group 15 non - metal. It has 5 valence electrons and needs 3 more electrons to complete its octet. So it gains 3 electrons to form an anion ($\ce{P^{3 - }}$).
- **Compound formation**: The ratio of sodium ions to phosphide ions is 3:1 because the charge of $\ce{Na^+}$ is + 1 and the charge of $\ce{P^{3 - }}$ is - 3. To balance the charges in the ionic compound, we need 3 $\ce{Na^+}$ ions for every 1 $\ce{P^{3 - }}$ ion. The Lewis diagrams show the transfer of electrons from sodium atoms to the phosphorus atom and the resulting ions that form the ionic bond in sodium phosphide.

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QUESTION IMAGE

Question

Explanation:

Part A

a. Balance equation (predict the products and their state of matter)

Step 1: Predict products

Step 2: Balance the equation

b. Ionic equation

Step 1: Dissociate soluble compounds

Step 2: Write the ionic equation

c. Net ionic equation

Step 1: Identify spectator ions

Step 2: Remove spectator ions

Part B

Answer: