Question

Explain formation of

(1) $O_2$ (2) $N_2$ using VBT

Answer 1

The formation of $O_2$ involves one sigma ($\sigma$) bond and one pi ($\pi$) bond, forming a double bond.

The formation of $N_2$ involves one sigma ($\sigma$) bond and two pi ($\pi$) bonds, forming a triple bond.

Answer 2

The formation of $O_2$ and $N_2$ molecules using Valence Bond Theory (VBT) can be explained by the overlap of atomic orbitals.

1. Formation of $O_2$ Molecule

• Electronic Configuration of Oxygen: The ground state electronic configuration of an oxygen atom (atomic number 8) is $1s^2 2s^2 2p^4$. In terms of valence orbitals, it can be written as: $O: [He] 2s^2 2p_x^2 2p_y^1 2p_z^1$ Each oxygen atom has two half-filled $2p$ orbitals ($2p_y$ and $2p_z$) that can participate in bonding.

• Bond Formation:

  1. Sigma ($\sigma$) Bond: One half-filled $2p$ orbital from each oxygen atom (e.g., $2p_y$) undergoes head-on (axial) overlap along the internuclear axis to form a sigma ($\sigma$) bond.
  2. Pi ($\pi$) Bond: The other half-filled $2p$ orbital from each oxygen atom (e.g., $2p_z$) undergoes sideways (lateral) overlap above and below the internuclear axis to form a pi ($\pi$) bond. The axes of these $p$ orbitals are parallel to each other and perpendicular to the internuclear axis.
  3. Lone Pairs: The $2s$ orbital and the $2p_x$ orbital on each oxygen atom contain lone pairs of electrons and do not participate in bonding.

• Result: The $O_2$ molecule is formed by one $\sigma$ bond and one $\pi$ bond between the two oxygen atoms, resulting in a double bond. $O=O$

2. Formation of $N_2$ Molecule

• Electronic Configuration of Nitrogen: The ground state electronic configuration of a nitrogen atom (atomic number 7) is $1s^2 2s^2 2p^3$. In terms of valence orbitals, it can be written as: $N: [He] 2s^2 2p_x^1 2p_y^1 2p_z^1$ Each nitrogen atom has three half-filled $2p$ orbitals ($2p_x$, $2p_y$, and $2p_z$) that can participate in bonding.

• Bond Formation:

  1. Sigma ($\sigma$) Bond: One half-filled $2p$ orbital from each nitrogen atom (e.g., $2p_x$) undergoes head-on (axial) overlap along the internuclear axis to form a sigma ($\sigma$) bond.
  2. Pi ($\pi$) Bonds: The remaining two half-filled $2p$ orbitals from each nitrogen atom ($2p_y$ and $2p_z$) undergo sideways (lateral) overlap to form two pi ($\pi$) bonds. These two $\pi$ bonds are mutually perpendicular to each other and to the $\sigma$ bond.
  3. Lone Pairs: The $2s$ orbital on each nitrogen atom contains a lone pair of electrons and does not participate in bonding.

• Result: The $N_2$ molecule is formed by one $\sigma$ bond and two $\pi$ bonds between the two nitrogen atoms, resulting in a triple bond. $N \equiv N$

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Explanation of the solution:

For $O_2$: Each oxygen atom has two half-filled $2p$ orbitals. One $2p$ orbital from each oxygen forms a $\sigma$ bond via axial overlap. The other $2p$ orbital from each oxygen forms a $\pi$ bond via lateral overlap. This results in an $O=O$ double bond (one $\sigma$, one $\pi$). Lone pairs reside in $2s$ and the remaining $2p$ orbital.

For $N_2$: Each nitrogen atom has three half-filled $2p$ orbitals. One $2p$ orbital from each nitrogen forms a $\sigma$ bond via axial overlap. The remaining two $2p$ orbitals from each nitrogen form two $\pi$ bonds via lateral overlap. This results in an $N \equiv N$ triple bond (one $\sigma$, two $\pi$). A lone pair resides in the $2s$ orbital.