On heating, lead (II) nitrate gives a brown gas (A). The gas (A) on cooling changes to a colourless

Q: Why does $NO_2$ dimerize on cooling?

Nitrogen dioxide ($NO_2$) is an odd-electron molecule, meaning it has an unpaired electron. This makes it a free radical and quite reactive. On cooling, the kinetic energy decreases, allowing two $NO_2$ molecules to pair their odd electrons and form a stable N-N bond, resulting in the diamagnetic dimer $N_2O_4$.

Q: Is $N_2O_3$ always a solid?

No, $N_2O_3$ is only stable as a deep blue solid or liquid at very low temperatures (below $-30^\circ\text{C}$). At room temperature, it readily dissociates back into $NO$ and $NO_2$.

Q: Why is the oxidation state of oxygen taken as -2?

In almost all of its compounds, oxygen is highly electronegative and pulls two electrons towards itself to complete its octet, resulting in an oxidation state of -2. The exceptions are peroxides (like $H_2O_2$), superoxides (like $KO_2$), and compounds with fluorine (like $OF_2$).

Q: Can lead nitrate decompose into something else?

Heavy metal nitrates (like those of Pb, Cu, Zn) characteristically decompose upon heating to yield the corresponding metal oxide, nitrogen dioxide ($NO_2$), and oxygen gas ($O_2$). Alkali metal nitrates (like $NaNO_3$), however, decompose to give the metal nitrite and oxygen.

Q: What is the structure of $N_2O_3$?

Dinitrogen trioxide has an asymmetric structure with an N-N bond. It can be thought of as an anhydride of nitrous acid ($HNO_2$). Its connectivity is $ON-NO_2$, where one nitrogen is bonded to one oxygen, and the other nitrogen is bonded to two oxygens.

Question

On heating, lead (II) nitrate gives a brown gas (A). The gas (A) on cooling changes to a colourless solid/liquid (B). (B) on heating with NO changes to a blue solid (C). The oxidation number of nitrogen in solid (C) is

Options:

(A) + 5

(B) + 4

(C) + 2

(D) + 3

Sigma Percentile · Accepted Answer

### The Colorful Chemistry of Nitrogen Oxides

This problem takes us on a fascinating journey through the chemistry of p-block elements, specifically focusing on the oxides of nitrogen. It's a classic sequence of reactions that tests your memory of qualitative analysis and the physical properties of these compounds.

### Step 1: The Thermal Decomposition of Lead Nitrate

We begin with the thermal decomposition of lead (II) nitrate, $Pb(NO_3)_2$. When heavy metal nitrates are heated, they typically decompose to yield the metal oxide, nitrogen dioxide gas, and oxygen gas.

The balanced chemical equation for this process is:
$$2Pb(NO_3)_2 \xrightarrow{\Delta} 2PbO + 4NO_2 \uparrow + O_2 \uparrow$$

The key observation here is the evolution of a **brown gas**. In the realm of inorganic chemistry, a brown gas evolved upon heating a nitrate is almost always nitrogen dioxide ($NO_2$). Therefore, we can confidently identify our compound (A) as $NO_2$.

### Step 2: The Dimerization of Nitrogen Dioxide

Nitrogen dioxide is an interesting molecule because it possesses an odd number of valence electrons (it's a free radical). This makes it highly reactive. When the temperature is lowered (cooling), these odd-electron molecules pair up to form a more stable dimer.

$$2NO_2 \xrightarrow{	ext{Cooling}} N_2O_4$$

The resulting compound, dinitrogen tetroxide ($N_2O_4$), is a **colourless solid or liquid**, depending on the exact temperature. This perfectly matches the description of compound (B).

### Step 3: Formation of the Blue Solid

The final step in our reaction sequence involves heating the colourless $N_2O_4$ with nitric oxide ($NO$). These two oxides of nitrogen react to form a mixed anhydride.

$$N_2O_4 + 2NO \xrightarrow{\Delta} 2N_2O_3$$

Dinitrogen trioxide ($N_2O_3$) is known to exist as a **deep blue solid** at very low temperatures. Thus, we have successfully identified compound (C) as $N_2O_3$.

### Step 4: Calculating the Oxidation State

Now that we know the identity of the blue solid (C) is $N_2O_3$, the final task is to determine the oxidation state of nitrogen within it.

Let the oxidation state of nitrogen be $x$. We know that the standard oxidation state of oxygen in most of its compounds (excluding peroxides and superoxides) is $-2$. Since the molecule is electrically neutral, the sum of the oxidation states of all atoms must equal zero.

$$2x + 3(-2) = 0$$

Solving for $x$:
$$2x - 6 = 0$$
$$2x = +6$$
$$x = +3$$

**Conclusion:**
The oxidation state of nitrogen in the blue solid $N_2O_3$ is **+3**. This elegant problem beautifully connects observational chemistry with fundamental redox concepts.

Animated Solution for Chemistry - s and p-Block Elements: On heating, lead (II) nitrate gives a brown gas (A). The gas (A) on cooling changes to a colourless solid/liquid (B). (B) on heating with NO changes to a blue solid (C). The oxidation number of nitrogen in solid (C) is

Select Answer:

Visualized Solution

$R e a c t i o n S e q u e n ce$

$T h er ma l D eco m p os i t i o n o f P b (N O_{3})_{2}$

$D im er i z a t i o n o f N O_{2}$

$R e a c t i o n w i t h N O$

$O x i d a t i o n S t a t e S e t u p$

$C a l c u l a t in g x$

$F ina l O x i d a t i o n S t a t e$

The Sigma Insight: Group 15 Elements

The Colorful Chemistry of Nitrogen Oxides

Step 1

Step 2

Step 3

Step 4

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Comprehension Passage

$W$ and $X$ are, respectively

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Animated Solution for Chemistry - s and p-Block Elements: On heating, lead (II) nitrate gives a brown gas (A). The gas (A) on cooling changes to a colourless solid/liquid (B). (B) on heating with NO changes to a blue solid (C). The oxidation number of nitrogen in solid (C) is

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Reaction Sequence

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Thermal Decomposition of Pb(NO3​)2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Dimerization of NO2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Reaction with NO

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Oxidation State Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating x

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Oxidation State