Given: The heat of sublimation of `K(s)` is `89 kJ mol^(-1)`.
`K(g) rarrK^(o+)(g)+e^(-), DeltaH^(Theta) = 419 kJ`
`F_(2)(g) rarr 2F(g),DeltaH^(Theta) = 155 kJ`
The lattice enegry of `KF(s)` is `-813kJ mol^(-1)`, the heat of formation of `KF(s)` is `-563 kJ mol^(-1)`. the `E_(A)` of `F(g)` is

Options

A. `-413`
B. `-336`
C. `-1149`
D. `+413`

Explanation

To calculate the electron affinity \( E_A \) of \( F(g) \), we can use the given thermodynamic data and apply Hess's law. Here's the step-by-step solution:

### Step 1: Write the formation reaction for \( KF(s) \)
The heat of formation \( \Delta H_f \) of \( KF(s) \) can be expressed as:
\[ \Delta H_f = \Delta H_{sublimation} + \Delta H_{ionization} + \Delta H_{electron\ affinity} + \Delta H_{lattice\ energy} \]

...