The electronegativity of cesium is 0.7 and that of flourine is 4.0 The bond formed between the two is:

Options

A. covalent
B. electrovalent
C. co-ordinate
D. metallic.

Answer

To determine the type of bond formed between cesium (Cs) and fluorine (F), we can follow these steps:

### Step 1: Identify the Electronegativity Values
- The electronegativity of cesium (Cs) is 0.7.
- The electronegativity of fluorine (F) is 4.0.

### Step 2: Calculate the Electronegativity Difference
- To find the difference in electronegativity, we subtract the electronegativity of cesium from that of fluorine:
\[ \text{Electronegativity Difference} = \text{Electronegativity of F} - \text{Electronegativity of Cs} = 4.0 - 0.7 = 3.3 \]

### Step 3: Analyze the Electronegativity Difference
- A large electronegativity difference (generally greater than 1.7) indicates that the bond formed is likely ionic. In this case, the difference is 3.3, which is quite significant.

### Step 4: Determine the Nature of the Bond
- Since cesium is a metal and has a low electronegativity, it tends to lose electrons and form a positive ion (Cs⁺).
- Fluorine, being a non-metal with a high electronegativity, tends to gain electrons and form a negative ion (F⁻).
- The transfer of electrons from cesium to fluorine leads to the formation of an ionic bond.

### Conclusion
- The bond formed between cesium and fluorine is an **ionic bond** (also known as an electrovalent bond).

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