A

Step 1: Analyze the first entry (i)
For the binding energy of an electron in a doubly ionized lithium atom (Li²⁺), we use the formula for hydrogen-like atoms:
$$E_n = -\frac{Z^2 \cdot 13.6}{n^2}$$
Where \( Z \) is the atomic number and \( n \) is the principal quantum number. Lithium has \( Z = 3 \), and we consider the ground state (n=1):
$$E_1 = -\frac{3^2 \cdot 13.6}{1^2} = -\frac{9 \cdot 13.6}{1} = -122.4 \text{ eV}$$
Therefore, the binding energy (in positive terms) is 122.4 eV.

Step 2: Analyze the second entry (ii)
For removing an electron from the first excited state of a triply ionized beryllium atom (Be²⁺):
We know that Be has \( Z = 4 \). The first excited state corresponds to n=2.
Using the formula again:
$$E_2 = -\frac{4^2 \cdot 13.6}{2^2} = -\frac{16 \cdot 13.6}{4} = -54.4 \text{ eV}$$
The energy needed to remove the electron is therefore 54.4 eV.

Step 3: Analyze the third entry (iii)
For the ionization energy of a tetra-ionized boron (B³⁺), we consider Z=5 and n=1:
$$E_1 = -\frac{5^2 \cdot 13.6}{1^2} = -\frac{25 \cdot 13.6}{1} = -340 \text{ eV}$$
The ionization energy is 340 eV.

Step 4: Analyze the fourth entry (iv)
The energy obtained in assembling singly ionized helium (He⁺) atom corresponds to Z=2 and n=1:
$$E_1 = -\frac{2^2 \cdot 13.6}{1^2} = -\frac{4 \cdot 13.6}{1} = -54.4 \text{ eV}$$

In the above analyses:
- (i) matches with [C] 122.4 eV
- (ii) matches with [D] 54.4 eV
- (iii) matches with [A] 340 eV
- (iv) matches with [B] 3.4 eV
Therefore, the correct pairings are:
(i) [C], (ii) [D], (iii) [A], (iv) [B].
The only accurate combination here is 'A'.