Question

Total number of species from the following with central atom utilising $2p^2$ hybrid orbitals for bonding is _______.
$\text{NH}_3, \, \text{SO}_2, \, \text{SiO}_2, \, \text{BeCl}_2, \, \text{C}_2\text{H}_2, \, \text{C}_2\text{H}_4, \, \text{BCl}_3, \, \text{HCHO}, \, \text{C}_6\text{H}_6, \, \text{BF}_3, \, \text{C}_2\text{H}_4\text{Cl}_2$

Answer 1

Correct Answer: 6

The problem asks to determine the total number of species from the given list where the central atom is \(sp^2\) hybridized.

Concept Used:

The hybridization of a central atom in a molecule can be determined using the Steric Number (SN) method. The steric number is the sum of the number of sigma (\(\sigma\)) bonds formed by the central atom and the number of lone pairs of electrons on it.

\[ \text{Steric Number (SN)} = (\text{Number of sigma bonds}) + (\text{Number of lone pairs}) \]

The hybridization is then determined based on the steric number:

• If SN = 2, the hybridization is \(sp\).
• If SN = 3, the hybridization is \(sp^2\).
• If SN = 4, the hybridization is \(sp^3\).

For molecules with multiple "central" atoms (like hydrocarbons), the hybridization of each of those atoms is considered.

Step-by-Step Solution:

Step 1: Analyze the hybridization of each species in the list.

\(\text{NH}_3\): The central atom is Nitrogen (N). It forms 3 \(\sigma\) bonds with H atoms and has 1 lone pair. \[ \text{SN} = 3 (\sigma \text{ bonds}) + 1 (\text{lone pair}) = 4 \implies sp^3 \text{ hybridization.} \]

\(\text{SO}_2\): The central atom is Sulfur (S). It forms 2 \(\sigma\) bonds with O atoms and has 1 lone pair. \[ \text{SN} = 2 (\sigma \text{ bonds}) + 1 (\text{lone pair}) = 3 \implies sp^2 \text{ hybridization.} \]

\(\text{SiO}_2\): Silicon dioxide has a giant covalent network structure where each Silicon (Si) atom is the central atom bonded to 4 Oxygen atoms tetrahedrally. \[ \text{SN} = 4 (\sigma \text{ bonds}) + 0 (\text{lone pairs}) = 4 \implies sp^3 \text{ hybridization.} \]

\(\text{BeCl}_2\): The central atom is Beryllium (Be). It forms 2 \(\sigma\) bonds with Cl atoms and has 0 lone pairs. \[ \text{SN} = 2 (\sigma \text{ bonds}) + 0 (\text{lone pairs}) = 2 \implies sp \text{ hybridization.} \]

\(\text{C}_2\text{H}_2\): In acetylene (H-C≡C-H), each Carbon (C) atom forms 2 \(\sigma\) bonds (one with H, one with C) and has 0 lone pairs. \[ \text{SN} = 2 (\sigma \text{ bonds}) + 0 (\text{lone pairs}) = 2 \implies sp \text{ hybridization for both C atoms.} \]

\(\text{C}_2\text{H}_4\): In ethene (\(H_2C=CH_2\)), each Carbon (C) atom forms 3 \(\sigma\) bonds (two with H, one with C) and has 0 lone pairs. \[ \text{SN} = 3 (\sigma \text{ bonds}) + 0 (\text{lone pairs}) = 3 \implies sp^2 \text{ hybridization for both C atoms.} \]

\(\text{BCl}_3\): The central atom is Boron (B). It forms 3 \(\sigma\) bonds with Cl atoms and has 0 lone pairs. \[ \text{SN} = 3 (\sigma \text{ bonds}) + 0 (\text{lone pairs}) = 3 \implies sp^2 \text{ hybridization.} \]

\(\text{HCHO}\): In formaldehyde, the central atom is Carbon (C). It forms 3 \(\sigma\) bonds (two with H, one with O) and has 0 lone pairs. \[ \text{SN} = 3 (\sigma \text{ bonds}) + 0 (\text{lone pairs}) = 3 \implies sp^2 \text{ hybridization.} \]

\(\text{C}_6\text{H}_6\): In benzene, each of the 6 Carbon (C) atoms in the ring forms 3 \(\sigma\) bonds (one with H, two with adjacent C atoms) and has 0 lone pairs. \[ \text{SN} = 3 (\sigma \text{ bonds}) + 0 (\text{lone pairs}) = 3 \implies sp^2 \text{ hybridization for all 6 C atoms.} \]

\(\text{BF}_3\): The central atom is Boron (B). It forms 3 \(\sigma\) bonds with F atoms and has 0 lone pairs. \[ \text{SN} = 3 (\sigma \text{ bonds}) + 0 (\text{lone pairs}) = 3 \implies sp^2 \text{ hybridization.} \]

\(\text{C}_2\text{H}_4\text{Cl}_2\): In 1,2-dichloroethane (\(ClH_2C-CH_2Cl\)), each Carbon (C) atom forms 4 \(\sigma\) bonds and has 0 lone pairs. \[ \text{SN} = 4 (\sigma \text{ bonds}) + 0 (\text{lone pairs}) = 4 \implies sp^3 \text{ hybridization for both C atoms.} \]

Final Computation & Result:

Step 2: Count the number of species with \(sp^2\) hybridized central atoms.

The species from the list with \(sp^2\) hybridization are:

1. \(\text{SO}_2\)
2. \(\text{C}_2\text{H}_4\)
3. \(\text{BCl}_3\)
4. \(\text{HCHO}\)
5. \(\text{C}_6\text{H}_6\)
6. \(\text{BF}_3\)

The total number of species with a central atom utilising \(sp^2\) hybrid orbitals is 6.

Answer 2

Central atom utilising \( sp^2 \) hybrid orbitals:

\[ \text{SO}_2, \, \text{C}_2\text{H}_4, \, \text{BCl}_3, \, \text{HCHO}, \, \text{C}_6\text{H}_6, \, \text{BF}_3 \]