Which of the following statements are correct ?
`(I)` In `ICl_(2), ClF_(3)` and `TeCl_(4)`, the number of lone pair(s) of electrons on central atoms are `3,2` and 1 respectively.
`(II)` Amongst `CO,CO_(2),CO_(3)^(2-),CH_(3)OH` the correct order from the weakest to the strongest carbon`-`oxygen bond `-=CH_(3)OH lt CO_(3)^(2-) lt CO_(2) lt CO`.
`(III)` The hybridisation of boron in `BF_(3)` is the same which nitrogen has in `ClNO` molecule.

Explanation

To determine which of the statements are correct, let's analyze each statement step by step.

### Statement (I):
**In ICl₂⁻, ClF₃, and TeCl₄, the number of lone pair(s) of electrons on central atoms are 3, 2, and 1 respectively.**

1. **ICl₂⁻ (Iodine Dichloride Ion)**:
- Iodine (I) has 7 valence electrons.
- In ICl₂⁻, Iodine forms 2 bonds with Chlorine, using 2 of its valence electrons.
- Remaining electrons = 7 - 2 = 5 electrons.
- These 5 electrons correspond to 3 lone pairs (since 2 electrons form 1 bond pair).
- **Lone pairs = 3**.

2. **ClF₃ (Chlorine Trifluoride)**:
- Chlorine (Cl) has 7 valence electrons.
- In ClF₃, Chlorine forms 3 bonds with Fluorine, using 3 of its valence electrons.
- Remaining electrons = 7 - 3 = 4 electrons.
- These 4 electrons correspond to 2 lone pairs.
- **Lone pairs = 2**.

3. **TeCl₄ (Tellurium Tetrachloride)**:
- Tellurium (Te) has 6 valence electrons.
- In TeCl₄, Tellurium forms 4 bonds with Chlorine, using 4 of its valence electrons.
- Remaining electrons = 6 - 4 = 2 electrons.
- These 2 electrons correspond to 1 lone pair.
- **Lone pairs = 1**.

Thus, the statement is correct: ICl₂⁻ has 3 lone pairs, ClF₃ has 2 lone pairs, and TeCl₄ has 1 lone pair.

### Statement (II):
**Amongst CO, CO₂, CO₃²⁻, CH₃OH, the correct order from the weakest to the strongest carbon-oxygen bond is CH₃OH < CO₃²⁻ < CO₂ < CO.**

1. **CO (Carbon Monoxide)**:
- Has a triple bond (C≡O).
- Strongest bond.

2. **CO₂ (Carbon Dioxide)**:
- Has a double bond (C=O).
- Weaker than CO but stronger than the others.

3. **CO₃²⁻ (Carbonate Ion)**:
- Has resonance structures with a bond order of 4/3 (approximately 1.33).
- Weaker than CO₂.

4. **CH₃OH (Methanol)**:
- Has a single bond (C-O).
- Weakest bond.

Thus, the order from weakest to strongest bond is indeed: CH₃OH < CO₃²⁻ < CO₂ < CO. This statement is correct.

### Statement (III):
**The hybridization of boron in BF₃ is the same as nitrogen in ClNO.**

1. **BF₃ (Boron Trifluoride)**:
- Boron has 3 valence electrons and forms 3 bonds with Fluorine.
- Hybridization = sp² (3 regions of electron density).

2. **ClNO (Chlorine Nitroxide)**:
- Nitrogen is the central atom.
- It has 3 regions of electron density (1 double bond with O and 1 single bond with Cl).
- Hybridization = sp².

Since both boron in BF₃ and nitrogen in ClNO have sp² hybridization, this statement is also correct.

### Conclusion:
All three statements are correct.

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