The number of correct options is
(a) `1^(Theta) gt Br^(Theta) gt CI^(Theta) gt F^(Theta)` (polarisability)
(b) `Li^(o+) gt Na^(o+) gt K^(o+) gt Rb^(o+)` (polarisation power)
(c ) `H_(2)Ogt H_(2)S gtH_(2)Se gtH_(2)` Te (order of b.pt)
(d) `H_(2)^(Theta) lt H_(2)^(o+)` (order of stability) .

Explanation

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

### Step 1: Analyze Option (a)
**Statement:** `1^(Theta) gt Br^(Theta) gt CI^(Theta) gt F^(Theta)` (polarisability)

**Explanation:**
- Polarizability refers to the ability of an anion to distort its electron cloud in the presence of a cation.
- It is influenced by the size and charge of the anion. Larger anions with lower effective nuclear charge can be more easily distorted.
- The size of the halogens increases from fluorine to iodine (F < Cl < Br < I).
- Therefore, the order of polarizability is:
\[ I^- > Br^- > Cl^- > F^- \]
- This means option (a) is **correct**.

### Step 2: Analyze Option (b)
**Statement:** `Li^(o+) gt Na^(o+) gt K^(o+) gt Rb^(o+)` (polarisation power)

**Explanation:**
- Polarization power is the ability of a cation to distort the electron cloud of an anion.
- It is directly proportional to the charge of the cation and inversely proportional to its size.
- The size of the alkali metals increases from Li to Rb (Li < Na < K < Rb).
- Therefore, the order of polarization power is:
\[ Li^+ > Na^+ > K^+ > Rb^+ \]
- This means option (b) is **correct**.

### Step 3: Analyze Option (c)
**Statement:** `H_(2)O gt H_(2)S gt H_(2)Se gt H_(2)Te` (order of b.pt)

**Explanation:**
- The boiling points of hydrides of group 16 elements (O, S, Se, Te) are influenced by hydrogen bonding.
- Water (H2O) has the highest boiling point due to strong hydrogen bonding.
- H2S has a lower boiling point than H2O but higher than H2Se and H2Te due to weaker Van der Waals forces.
- The correct order is:
\[ H_2O > H_2S > H_2Se > H_2Te \]
- This means option (c) is **correct**.

### Step 4: Analyze Option (d)
**Statement:** `H_(2)^(Theta) lt H_(2)^(o+)` (order of stability)

**Explanation:**
- H2^- (hydride ion) has two electrons in the bonding molecular orbital and one in the antibonding molecular orbital, making it less stable than H2+ (which has one electron in the bonding orbital).
- H2+ is more stable because it has no electrons in the antibonding orbital.
- Therefore, the statement is correct:
\[ H_2^- < H_2^+ \]
- This means option (d) is **correct**.

### Conclusion
After analyzing all four options, we find that options (a), (b), (c), and (d) are all correct. Therefore, the number of correct options is **4**.