Oxygen Adsorption on Metallic Surface and Electron Transfer Effects

To determine the correct statements regarding the adsorption of $\text{O}_2$ gas on a metallic surface accompanied by electron transfer, let us analyze each option step-by-step:

1. Type of Adsorption (Statement A): When electron transfer occurs between the metallic surface and $\text{O}_2$, actual chemical bonds (with ionic or covalent character) are formed, yielding species like superoxide ($\text{O}_2^-$) or peroxide ($\text{O}_2^{2-}$). Processes involving electron transfer and chemical bond formation are classified as chemisorption (chemical adsorption), not physisorption. Therefore, Statement (A) is incorrect.

2. Thermodynamics of Adsorption (Statement B): Adsorption is an exothermic process ($\Delta H < 0$). During adsorption, the translational degrees of freedom of the gas molecules are restricted, leading to a decrease in entropy ($\Delta S < 0$). For the process to be spontaneous ($\Delta G < 0$), the enthalpy change $\Delta H$ must be negative ($\Delta H = \Delta G + T\Delta S < 0$). Furthermore, the formation of chemical bonds during chemisorption liberates energy. Thus, heat is released. Therefore, Statement (B) is correct.

3. Molecular Orbital Occupancy (Statement C): According to Molecular Orbital Theory (MOT), the valence electron configuration of neutral $\text{O}_2$ ($16$ electrons) is: \sigma_{1s}^2 \, \sigma^*_{1s}^2 \, \sigma_{2s}^2 \, \sigma^*_{2s}^2 \, \sigma_{2p_z}^2 \, (\pi_{2p_x}^2 = \pi_{2p_y}^2) \, (\pi^*_{2p_x}^1 = \pi^*_{2p_y}^1) The lowest available energy levels for incoming electrons transferred from the metal surface are the partially filled antibonding $\pi^*_{2p}$ orbitals. Hence, the transferred electrons occupy these $\pi^*_{2p}$ orbitals, increasing their electron occupancy. Therefore, Statement (C) is correct.

4. Bond Order and Bond Length (Statement D): The bond order ($\text{B.O.}$) of a diatomic molecule is calculated as: $\text{Bond Order} = \frac{N_b - N_a}{2}$ where $N_b$ is the number of bonding electrons and $N_a$ is the number of antibonding electrons.

   • For neutral $\text{O}_2$: $N_b = 10$, $N_a = 6 \implies \text{B.O.} = \frac{10 - 6}{2} = 2$.
   • When electrons are added to the antibonding $\pi^*_{2p}$ orbitals, $N_a$ increases, causing the bond order to decrease ($\text{B.O.} = 1.5$ for $\text{O}_2^-$ and $\text{B.O.} = 1$ for $\text{O}_2^{2-}$).

   Since bond length is inversely proportional to bond order: $\text{Bond Length} \propto \frac{1}{\text{Bond Order}}$ a decrease in bond order leads to an increase in the $\text{O-O}$ bond length. Therefore, Statement (D) is correct.

Conclusion: The correct statements are (B), (C), and (D).