A tabela a seguir mostra algumas diferenças entre o diamante e a grafite, que são formas alotrópicas do carbono. }(c) Forma alotrópica & Dureza & Condução de calor & Condução de Diamante & Duro & Condutor & Isolante Grafite & Mole & Isolante & Conductor Essas propriedades são justificadas pelas e ruturas dos alótopos. Nessas estruturas, tomos de carbono do diamante e da grafi im, respectivamente, hibridização sp e mathrm(sp)^2 . d) mathrm(sp)^2 e mathrm(sp)^3 . mathrm(sp)^3 e mathrm(sp)^2 . e) mathrm(sp)^2 e sp. sp e mathrm(sp)^3 .

## Step 1: Understanding Allotropic Forms
### The question involves two allotropic forms of carbon: diamond and graphite. Each form has distinct properties due to differences in atomic structure and bonding.

## Step 2: Analyzing the Table Data
### From the table, we see that diamond is hard, conducts heat but is an electrical insulator. Graphite, on the other hand, is soft, does not conduct heat well, but is a good conductor of electricity.

## Step 3: Relating Properties to Hybridization
### The hardness and insulating properties of diamond are due to its three-dimensional tetrahedral structure where each carbon atom is $\mathrm{sp}^{3}$ hybridized, forming strong covalent bonds with four other carbon atoms.
### Graphite's softness and electrical conductivity arise from its planar structure where each carbon atom is $\mathrm{sp}^{2}$ hybridized, forming layers that can slide over each other. The delocalized electrons between these layers allow for electrical conduction.

## Step 4: Matching Hybridization with Options
### Based on the above analysis, diamond corresponds to $\mathrm{sp}^{3}$ hybridization, and graphite corresponds to $\mathrm{sp}^{2}$ hybridization.