Chemistry Q136 – AP-EAMCET 2025

Home Chemistry Chemical BondingQuestion #211

In which of following set, both the substances have same hybridisation ?

Clue 1 of 3: Hybridization Concept

Hybridization is the concept of mixing atomic orbitals to form new hybrid orbitals suitable for the pairing of electrons to form chemical bonds in valence bond theory. The type of hybridization ( $s p$ , $s p^{2}$ , $s p^{3}$ ) depends on the number of sigma bonds and lone pairs around the central atom.

🧠 Full Solution Path

Step 1: Understand Hybridization and its Relation to Structure✦ Active

Hybridization is the mixing of atomic orbitals to form new hybrid orbitals. The type of hybridization ( $s p$ , $s p^{2}$ , $s p^{3}$ ) dictates the geometry around an atom. Specifically:

* $s p$ hybridization: 2 electron domains (linear geometry).

* $s p^{2}$ hybridization: 3 electron domains (trigonal planar geometry).

* $s p^{3}$ hybridization: 4 electron domains (tetrahedral geometry).

💡 Teacher's Secret Hint

Remember that multiple bonds (double or triple bonds) count as only one electron domain for the purpose of determining hybridization.

Step 2: Determine Hybridization for Each Substance○ Expand

Let's determine the hybridization of the carbon atoms in each substance:

* **Diamond:** Each carbon atom is bonded to four other carbon atoms in a tetrahedral arrangement. This means each carbon atom has 4 sigma bonds and no lone pairs, leading to $s p^{3}$ hybridization.

* **Buckminster fullerene (C60):** Each carbon atom is bonded to three other carbon atoms, forming a network of pentagons and hexagons. There are 3 sigma bonds and no lone pairs around each carbon, leading to $s p^{2}$ hybridization.

* **Graphite:** Each carbon atom is bonded to three other carbon atoms in a planar hexagonal layer. There are 3 sigma bonds and no lone pairs around each carbon, leading to $s p^{2}$ hybridization.

* **Carbon dioxide ( $C O_{2}$ ):** The central carbon atom forms two double bonds with two oxygen atoms ( $O = C = O$ ). Each double bond consists of one sigma bond and one pi bond. Therefore, the carbon atom has 2 sigma bonds and no lone pairs, leading to $s p$ hybridization.

💡 Teacher's Secret Hint

For allotropes of carbon, visualize the local bonding environment of a carbon atom. For molecules, draw the Lewis structure to count sigma bonds and lone pairs.

Step 3: Evaluate the Given Sets○ Expand

Now, let's check each option to see which set has both substances with the same hybridization:

* **Option 1: Diamond ( $s p^{3}$ ), Buckminster fullerene ( $s p^{2}$ )** - Different hybridization.

* **Option 2: Graphite ( $s p^{2}$ ), Buckminster fullerene ( $s p^{2}$ )** - Same hybridization.

* **Option 3: Carbon dioxide ( $s p$ ), Graphite ( $s p^{2}$ )** - Different hybridization.

* **Option 4: Diamond ( $s p^{3}$ ), Carbon dioxide ( $s p$ )** - Different hybridization.

💡 Teacher's Secret Hint

Carefully compare the hybridization types for both substances in each set.

Step 4: Identify the Correct Set○ Expand

Based on the evaluation, the set containing Graphite and Buckminster fullerene both have $s p^{2}$ hybridization.

💡 Teacher's Secret Hint

This question tests your knowledge of the structures and bonding in common carbon allotropes and simple inorganic molecules.

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Chemistry Question 136 – AP-EAMCET 2025

Clue 1 of 3: Hybridization Concept

🧠 Full Solution Path