Question

M(g) = M +(g) + e , delta H = 100 ev; M(g) = M2 +(g) + 2e , delta H = 250 ev; then what is the ionisation potential of M(g) ?​

Answer 1

Answer:

M → M⁺ + e⁻ ( this is the first IP of M, as ionisation potential is the energy required to form a positive ion by release of an electron )

The value is already provided to us as 100 eV.

The next piece of information is unnecessary, unless they want the IP of M⁺.

If that is the question, then please note that the first and second IPs of M have been provided.

M → M⁺ + e⁻ ---- (1)

M → M⁺² + 2e⁻ -----(2)

Inverting the first equation, we have-

M⁺ + e⁻ → M ------(3)

If we consider the first and third equation, M cancels out (being on opposite sides of the equation) and so does one electron.

Our final equation can then be written as-

M⁺ → M⁺² + e⁻

This is the equation for the first IP of M⁺.

We use the changes in the equation for changes in the heats given. Since we have inverted the first equation, total heat becomes = - H₁ + H₂= 150 ev, which is the first IP of M⁺.