Given below are two statements: Statement I : The higher oxidation states are more stable down the group among transition elements unlike p-block elements. Statement II : Copper can not liberate hydrogen from weak acids. In the light of the above statements, choose the correct answer from the options given below :

We need to evaluate the truth of two statements about transition elements.

Statement I: "The higher oxidation states are more stable down the group among transition elements unlike p-block elements."

This statement is true. Here is why:

In transition elements, as we go down a group, the higher oxidation states become increasingly stable. This is because:

- The ionisation enthalpies of 4d and 5d elements are comparable (and sometimes lower than expected) due to poor shielding by intervening electron shells.

- The greater spatial extension of 4d and 5d orbitals compared to 3d orbitals allows better orbital overlap, which stabilises higher oxidation states through stronger bonding.

- For example, in Group 6: Cr commonly shows +3 and +6, but Mo and W preferentially show +6. In Group 8: Fe shows +2 and +3, while Os readily shows +8 (as in $$OsO_4$$).

This is opposite to p-block elements, where the lower oxidation states become more stable down the group due to the inert pair effect.

Statement II: "Copper cannot liberate hydrogen from weak acids."

This statement is true. Here is why:

The ability of a metal to liberate hydrogen from acids depends on its standard electrode potential. A metal can liberate $$H_2$$ only if its standard reduction potential is less than that of the $$H^+/H_2$$ couple ($$E° = 0.00$$ V).

Copper has a positive standard reduction potential: $$E°(Cu^{2+}/Cu) = +0.34$$ V.

Since $$E°_{Cu} > E°_{H^+/H_2}$$, copper is a less reactive (more noble) metal than hydrogen. Therefore, copper cannot reduce $$H^+$$ ions to $$H_2$$ gas, even from dilute strong acids, let alone from weak acids (which have even lower $$H^+$$ concentration). The reaction is thermodynamically unfavourable (positive $$\Delta G$$).

Since both statements are true, the correct answer is Option 1: Both Statement I and Statement II are true.