Match List - I with List - II.

Choose the correct answer from the options given below:

(A)-(III), (B)-(IV), (C)-(II), (D)-(I)

The correct answer is Option 4. (A)-(III), (B)-(IV), (C)-(II), (D)-(I).

To explain how each of the matches was made between List I and List II, let us break down each property and its reasoning, based on the periodic trends and specific characteristics of the elements involved.

(A) Element with highest second ionisation enthalpy :

Ionisation enthalpy refers to the energy required to remove an electron from an atom or ion. Second ionisation enthalpy refers to the energy required to remove an electron from a singly charged ion (after the first electron has already been removed).  Copper (Cu) has the highest second ionisation enthalpy among the given elements. Copper's electron configuration is \([Ar] 3d^{10} 4s^1\). When the first electron (from the 4s orbital) is removed, Cu becomes Cu⁺ with the stable \(3d^{10}\) configuration. Removing the second electron would require disrupting this stable, fully filled \(3d^{10}\) configuration, which requires a lot of energy, making its second ionisation enthalpy very high.

Thus, (A) matches with (III) Cu.

(B) Element with highest third ionisation enthalpy:

Third ionisation enthalpy refers to the energy required to remove an electron from a doubly charged ion (after two electrons have already been removed). Zinc (Zn) has the highest third ionisation enthalpy. The electron configuration of neutral Zn is \([Ar] 3d^{10} 4s^2\). After losing two electrons (Zn → Zn²⁺), it achieves the stable \(3d^{10}\) configuration. Removing a third electron would disrupt this very stable configuration, and hence, the third ionisation enthalpy is extremely high for zinc. Thus, (B) matches with (IV) Zn.

(C) M in \(M(CO)_6\):

The symbol M represents a transition metal that forms a stable carbonyl complex, \([M(CO)_6]\). Carbonyl complexes are formed by transition metals due to their ability to accept electron density from the carbonyl ligands via σ-donation and π-back donation. Chromium (Cr) forms the well-known complex \([Cr(CO)_6]\) (hexacarbonylchromium), which is a stable and common organometallic compound. Chromium in \([Cr(CO)_6]\) is in the 0 oxidation state and is well-suited to engage in π-backbonding with CO ligands, making it a stable complex. The complex \([Cr(CO)_6]\) is widely studied and is a classic example of a metal carbonyl complex. Thus, (C) matches with (II) Cr.

(D) Element with highest heat of atomisation

Heat of atomisation refers to the energy required to break all the bonds in one mole of a substance to convert it into individual atoms in the gaseous phase. A higher heat of atomisation indicates stronger metallic bonding or interactions in the solid state. Cobalt (Co) has the highest heat of atomisation among the given elements. This is because cobalt, as a transition metal, exhibits strong metallic bonding due to the overlap of its \(3d\) orbitals. Cobalt forms a tightly packed metallic lattice where the \(3d\) electrons contribute to strong bonding. Breaking these bonds to form gaseous atoms requires a large amount of energy, hence a high heat of atomisation.

Thus, (D) matches with (I) Co.

Thus, the correct answer is option 4. (A)-(III), (B)-(IV), (C)-(II), (D)-(I)