Question

the number of theoretical ionisation potential values of oxide ion (O^-2) is 'X', then value of x/2 is..​

Answer 1

Answer:

X+→ X+2 + e- Δ H = I.P of X+ . X+2 + e-→ X+ Δ H = - E.A of X+2 Thus, I.P of X+ = E.A of X+2

HOPE IT HELPS YOU

PLZ MARK AS BRAINLIEST

Answer 2

Concept: The potential difference equivalent to the energy in electron volts required to ionize a gas molecule or remove an electron entirely from an atom is known as the ionization potential.

Ionization energy, which is measured in electron volts, is the energy equivalent to ionization potential.

Given: Theoretical ionisation potential value of oxide ion is X.

To find: Theoretical ionisation potential values of oxide ion

Solution:

The electronic configuration of oxygen $O_{2}$ is $1s^{2} 2s^{2} 2p^{4}$, therefore the electronic configuration of $O^{-2}$ will be $1s^{2} 2s^{2} 2p^{6}$ (because 2 electrons are being added in oxide ion).  There are total 10 electrons in $O^{-2}$ ion (6+2+2 = 10). So, theoretical ionization potential value that can exist = 10 i.e., X= 10.

Therefore, $\frac{X}{2}$ = $\frac{10}{2}$ = 5.

Hence, the value of $\frac{X}{2}$ is 5.

#SPJ2