How many pairs are, in which first species has lower ionisation enegy than second species:
(i) N and O
(ii) Br and K
(iii) Be and B
(iv) I and `I^(-)`
(V) Li and `Li^(+)`
(vi) and O and S
(vii) Ba and Sr.

Explanation

To determine how many pairs of species have the first species with lower ionization energy than the second species, we will analyze each pair one by one.

### Step-by-Step Solution:

1. **Pair (i): N and O**
- **Analysis**: Nitrogen (N) has a configuration of \(1s^2 2s^2 2p^3\) and Oxygen (O) has \(1s^2 2s^2 2p^4\). Nitrogen has a half-filled p subshell, which is more stable, leading to a higher ionization energy than oxygen.
- **Conclusion**: N has a higher ionization energy than O. **(Not a valid pair)**

2. **Pair (ii): Br and K**
- **Analysis**: Bromine (Br) has a configuration of \( [Ar] 4s^2 3d^{10} 4p^5\) and Potassium (K) has \( [Ar] 4s^1\). It is easier to remove an electron from K than from Br, making the ionization energy of K lower than that of Br.
- **Conclusion**: K has a lower ionization energy than Br. **(Valid pair)**

3. **Pair (iii): Be and B**
- **Analysis**: Beryllium (Be) has a configuration of \(1s^2 2s^2\) and Boron (B) has \(1s^2 2s^2 2p^1\). Removing an electron from the 2p subshell of B is easier than from the filled 2s subshell of Be.
- **Conclusion**: Be has a higher ionization energy than B. **(Not a valid pair)**

4. **Pair (iv): I and I⁻**
- **Analysis**: Iodine (I) has a configuration of \( [Kr] 5s^2 4d^{10} 5p^5\) and I⁻ has \( [Kr] 5s^2 4d^{10} 5p^6\). I⁻ is a negatively charged ion and has a fully filled subshell, making it harder to remove an electron compared to neutral I.
- **Conclusion**: I has a lower ionization energy than I⁻. **(Valid pair)**

5. **Pair (v): Li and Li⁺**
- **Analysis**: Lithium (Li) has a configuration of \(1s^2 2s^1\) and Li⁺ has \(1s^2\). Removing an electron from Li is easier than from the fully filled Li⁺.
- **Conclusion**: Li has a lower ionization energy than Li⁺. **(Valid pair)**

6. **Pair (vi): O and S**
- **Analysis**: Oxygen (O) has a configuration of \(1s^2 2s^2 2p^4\) and Sulfur (S) has \(1s^2 2s^2 2p^6 3s^2 3p^4\). As we move down the group, the ionization energy decreases due to increased distance from the nucleus.
- **Conclusion**: O has a higher ionization energy than S. **(Not a valid pair)**

7. **Pair (vii): Ba and Sr**
- **Analysis**: Barium (Ba) has a configuration of \( [Xe] 6s^2\) and Strontium (Sr) has \( [Kr] 5s^2\). As we move down the group, the ionization energy decreases.
- **Conclusion**: Ba has a higher ionization energy than Sr. **(Not a valid pair)**

### Final Count of Valid Pairs:
- The valid pairs where the first species has lower ionization energy than the second are:
- (ii) Br and K
- (iv) I and I⁻
- (v) Li and Li⁺

Thus, there are **3 valid pairs**.