The correct statement is/are:

Options

A. Zirconium (Zr) and hafnium (Hf) have almost same size
B. Correct order of ionisation energy of coinage metals is CugtAg`ltAu`
C. Carbon atom in `C Cl_(4)` is more electronegative than carbon atom in `CF_(4)`
D. `Pb^(2+)` is more stable than `Pb^(4+)`

Explanation

To determine the correct statements from the given options, let's analyze each statement step by step.

### Step 1: Analyze the first statement
**Statement:** Zirconium and hafnium have almost the same size.

- **Explanation:** Zirconium (Zr) has an atomic number of 40, while hafnium (Hf) has an atomic number of 72. Despite being in different periods, their sizes are similar due to the lanthanide contraction. The filling of the 4f orbitals in the lanthanide series causes increased nuclear charge, which pulls the outer electrons closer to the nucleus, reducing the atomic size. Therefore, this statement is **true**.

### Step 2: Analyze the second statement
**Statement:** The correct order of ionization energies of coinage metals is copper, silver, then gold.

- **Explanation:** The order of ionization energies for coinage metals is influenced by their electronic configurations. Copper (Cu) has a filled 3d subshell, silver (Ag) has a filled 4d subshell, and gold (Au) has filled 4f and 5d subshells. The effective nuclear charge increases due to the filling of inner orbitals, making it harder to remove electrons. Thus, the order of ionization energy is indeed Cu > Ag > Au, making this statement **true**.

### Step 3: Analyze the third statement
**Statement:** Carbon atoms in CCl4 are more electronegative than carbon in CF4.

- **Explanation:** In CF4, the carbon atom is bonded to four highly electronegative fluorine atoms, which pulls electron density towards them, making the carbon atom less electronegative. In CCl4, chlorine is less electronegative than fluorine, so the carbon retains more electron density. Therefore, carbon in CCl4 is actually more electronegative than carbon in CF4, making this statement **false**.

### Step 4: Analyze the fourth statement
**Statement:** Pb+2 is more stable than Pb+4.

- **Explanation:** Lead (Pb) exhibits an inert pair effect, where the s-electrons do not participate in bonding as effectively as the p-electrons when the 4f orbitals are filled. This results in Pb+2 being more stable than Pb+4 because the Pb+4 state requires participation from the s-electrons, which are less available. Thus, this statement is **true**.

### Conclusion
The correct statements are:
- Statement 1: True
- Statement 2: True
- Statement 3: False
- Statement 4: True

### Final Answer
The correct statements are: 1, 2, and 4.

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