Question

Ionisation energy and electron affinity of fluorine are respectively $17.42$ and $3.45\text{eV}.$ Calculate the electronegativity of fluorine atom on Pauling's scale.

• A. $2.726$
• B. $2.762$
• C. $3.726$
• D. $3.762$

Answer 1

The correct option is C $3.726$

Electronegativity is a measure of an atom's ability to attract the shared electrons to itself. On the periodic table, electronegativity generally increases as you move from left to right across a period and decreases as you move down a group. Many methods were proposed to calculate electronegativity. Mulliken and Pauling's scale are well known among them.
According to Mulliken's electronegativity formula:
$\left(X_M\right)=\frac{\text{Ionisation energy + Electron affinity}}{2}=\frac{17.42+3.45}{2}=10.435$
Therefore, electronegativity on Pauling's scale $\left(X_P\right)$ in eV $\frac{10.435}{2.8}=3.726$