Which statement is incorrect about graphite?

Options

A. It is layered structure
B. Layers are heald by van der Waal force of attraction
C. It is used as a dry lubricant
D. Each carbon in hexagonal ring is `sp^3` hybridised

Answer

To determine which statement is incorrect about graphite, we will analyze the properties of graphite step by step.

### Step 1: Understand the Structure of Graphite
Graphite has a layered structure. Each layer consists of carbon atoms arranged in a hexagonal lattice. This layered arrangement allows the layers to slide over each other easily.

**Hint:** Remember that the layered structure is a key characteristic of graphite.

### Step 2: Conductivity of Graphite
Graphite is a good conductor of electricity. This is due to the presence of delocalized electrons that can move freely between the layers. Each carbon atom in graphite is bonded to three other carbon atoms, leaving one electron free to conduct electricity.

**Hint:** Focus on the role of delocalized electrons in conductivity.

### Step 3: Hybridization of Carbon in Graphite
In graphite, each carbon atom undergoes sp² hybridization. This means that each carbon atom forms three sigma bonds with neighboring carbon atoms and has one unhybridized p-orbital that contributes to the delocalized electron cloud.

**Hint:** Distinguish between sp² hybridization in graphite and sp³ hybridization in diamond.

### Step 4: Forces Holding Layers Together
The layers in graphite are held together by weak Van der Waals forces. This weak interaction allows the layers to slide over each other, making graphite useful as a lubricant.

**Hint:** Recall that the weak forces between layers contribute to the lubricating properties of graphite.

### Step 5: Analyze the Statements
Now, we will analyze each statement provided in the options to identify the incorrect one:

1. Graphite has a layered structure. (True)
2. Graphite is a poor conductor of electricity. (False - it is a good conductor)
3. Layers are held by Van der Waals forces. (True)
4. Each carbon in a hexagonal ring is sp³ hybridized. (False - they are sp² hybridized)

### Conclusion
The incorrect statement about graphite is that "Each carbon in a hexagonal ring is sp³ hybridized." The correct hybridization for carbon in graphite is sp².

**Final Answer:** The incorrect statement is that each carbon in a hexagonal ring is sp³ hybridized.

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