Boron is unable to form `BF_(6)^(3-)` because of

Options

A. High electronegativily of boron
B. high electronegativity of fluorine
C. Lack of d-orbitals in boron
D. less difference in electronegativity between B and F

Answer

To solve the question of why boron is unable to form \( BF_6^{3-} \), we can analyze the properties of boron and its compounds step by step.

### Step 1: Understanding Boron's Electron Configuration
Boron (B) has an atomic number of 5, which means it has 5 electrons. Its electron configuration is \( 1s^2 2s^2 2p^1 \). This means boron has three valence electrons.

**Hint:** Remember that boron has a limited number of valence electrons, which affects its ability to form complex ions.

### Step 2: The Nature of Boron Compounds
Boron typically forms covalent compounds and can form stable complexes with other elements. However, it tends to form compounds with an incomplete octet due to its electron deficiency.

**Hint:** Consider how boron behaves in its compounds and its tendency to form covalent bonds.

### Step 3: Analyzing the \( BF_6^{3-} \) Ion
The \( BF_6^{3-} \) ion would require boron to expand its octet to accommodate six fluorine atoms. This would mean that boron would need to have more than eight electrons in its valence shell, which is not feasible for boron due to its small size and the absence of d-orbitals.

**Hint:** Think about the octet rule and how it applies to elements like boron, which cannot expand their valence shell.

### Step 4: Charge Consideration
The formation of \( BF_6^{3-} \) would also imply a significant negative charge on the boron atom. However, boron is a group 13 element and typically has a +3 oxidation state in its compounds. The stability of such a highly charged species is questionable.

**Hint:** Reflect on the stability of charged species and how boron's oxidation states influence its ability to form certain compounds.

### Conclusion
Boron is unable to form \( BF_6^{3-} \) primarily due to its inability to expand its octet beyond four electrons and the instability associated with having a high negative charge in this context.

### Summary of Steps:
1. **Electron Configuration**: Boron has 5 electrons, limiting its valence shell capacity.
2. **Nature of Compounds**: Boron forms covalent compounds and typically has an incomplete octet.
3. **Octet Rule**: \( BF_6^{3-} \) would require boron to exceed the octet rule, which is not possible.
4. **Charge Stability**: The high negative charge on boron in \( BF_6^{3-} \) makes it unstable.