Explain why Na tends to form ionic compounds while B forms covalent compounds ?

Explanation

To explain why sodium (Na) tends to form ionic compounds while boron (B) forms covalent compounds, we can analyze the electronic configurations and properties of both elements step by step.

### Step 1: Understand the Electronic Configuration
- Sodium (Na) has an atomic number of 11, which means it has 11 electrons. Its electronic configuration is 1s² 2s² 2p⁶ 3s¹. This can also be represented as [Ne] 3s¹, indicating that it has one electron in its outermost shell (the third shell).
- Boron (B), on the other hand, has an atomic number of 5, giving it 5 electrons. Its electronic configuration is 1s² 2s² 2p¹, or simply [He] 2s² 2p¹, showing that it has three electrons in its outermost shell (the second shell).

### Step 2: Analyze the Tendency to Lose or Gain Electrons
- Sodium has one electron in its outermost shell. To achieve a stable noble gas configuration (similar to neon), it can easily lose this one electron. The energy required to remove this electron is low due to its larger atomic size and low ionization enthalpy.
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