When ${{O}_{2}}$ is absorbed on a metallic surface, electron transfer occurs from the metal to ${{O}_{2}}$. The true statement(s) regarding this adsorption is (are):
This question has multiple correct options
(a) ${{O}_{2}}$ is physisorbed
(b) heat is released
(c) occupancy of $\pi _{2p}^{*}$ of ${{O}_{2}}$is increased ${{O}_{2}}$
(d) bond length of ${{O}_{2}}$ is increased

Hint: Oxygen molecule when gets adsorbed ( i.e. the process of retention of molecules on the surface) on metallic surface , chemical reaction occurs between them and electron transfers takes place from the metallic surface to the oxygen in the $\pi $ molecular orbitals along with the release of energy. Now identify the correct statement.

Complete answer:
Oxygen is a non-metal present in the 16th group and belongs to the category of p-block elements and has the atomic number as 8 and mass number as 16.
Now, let’s discuss what is absorption. By the term absorption we mean, the phenomenon of attracting and retaining the molecules on the surface . It is of two types; physisorption and chemisorption. The molecules which get adsorbed on the surface is called adsorbate and the surface on which adsorption occurs is called adsorbent.
In physisorption , the particles of the adsorbate are held by weak Vander wall’s forces and can be easily reversed by either increasing temperature or decreasing pressure. On the other hand in chemisorption, the particles of the adsorbate are held by chemical forces i.e. chemical reaction takes place and this is irreversible.
When oxygen molecules are adsorbed on the metallic surface, there occurs a chemical reaction between them i.e. it involves chemisorption and the metallic surface gives electrons to the oxygen atom i.e. it gains electrons from the metallic surface and chemisorption is an exothermic process i.e. heat is released when oxygen molecules are adsorbed on the metallic surface.
The molecular orbital electronic configuration of oxygen molecule i.e.${{O}_{2}}$ ( atomic number=16) is;
     \[{{O}_{2}}=KK\text{ }\sigma 2{{s}^{2}}\text{ }{{\sigma }^{*}}2{{s}^{2}}\ \sigma 2p_{z}^{2}\text{ }\pi 2p_{x}^{2}\text{ }\pi 2p_{y}^{2}\text{ }{{\pi }^{*}}2p_{x}^{1}\text{ }{{\pi }^{*}}2p_{y}^{1}\]
Bond order= $\dfrac{\text{number of electrons in bonding molecular orbitals- number of electrons in antibonding molecular orbitals}}{2}$
Number of electrons in bonding molecular orbitals= 8
Number of electrons in antibonding molecular orbitals=4
Then,
Bond order= $\dfrac{8-4}{2}$ = 2
When its gains electron one electron, the molecular orbital electronic configuration of oxygen molecule i.e. $O_{2}^{=}$ ( atomic number=17) is;
     \[O_{2}^{-}=KK\text{ }\sigma 2{{s}^{2}}\text{ }{{\sigma }^{*}}2{{s}^{2}}\ \sigma 2p_{z}^{2}\text{ }\pi 2p_{x}^{2}\text{ }\pi 2p_{y}^{2}\text{ }{{\pi }^{*}}2p_{x}^{2}\text{ }{{\pi }^{*}}2p_{y}^{1}\]
Bond order= $\dfrac{\text{number of electrons in bonding molecular orbitals- number of electrons in antibonding molecular orbitals}}{2}$
Number of electrons in bonding molecular orbitals= 8
Number of electrons in antibonding molecular orbitals=5
Then,
Bond order= $\dfrac{8-5}{2}$ = 1
Since, the bond order of ${{O}_{2}}$is decreased in $O_{2}^{=}$,therefore , bond length of $O_{2}^{=}$is increased as comparative to ${{O}_{2}}$because bond order and bond length are inversely proportional to one another. If bond order is decreased, the bond length is increased and vice-versa.
So, we can now see that from the above given options ;

All are correct i.e.(b),(c) and (d) except the option(a).

Note:
Don’t get confused in the adsorption and absorption. Adsorption is a surface phenomenon in which the concentration of particles is more on the surface and its rate is rapid initially but decreases slowly. On the other hand, absorption is a bulk phenomenon i.e. occurs uniformly throughout the body of the surface and takes place at a uniform rate.