Sum of first three ionization energies of Al is 53.0 eV atom -1 and the sum of first two ionization energies of Na is 52.2 eV atom -1. Out of Al (III) and Na (II)

Question

Sum of first three ionization energies of Al is 53.0 eV atom -1 and the sum of first two ionization energies of Na is 52.2 eV atom -1. Out of Al (III) and Na (II) (A) Al (III) is more stable than Na (II) (B) Na (II) is more stable than Al (III) (C) Both are equally unstable (D) Both are equally stable

Tardigrade Admin · Accepted Answer

As ionization energy is not the only criteria for the stability of an oxidation state.

Q. Sum of first three ionization energies of $A l$ is $53.0 e V$ atom $^{- 1}$ and the sum of first two ionization energies of $N a$ is $52.2 e V$ atom $^{- 1}$ . Out of $A l$ (III) and $N a$ (II)

$A l$ (III) is more stable than $N a$ (II)

$N a$ (II) is more stable than $A l$ (III)

Both are equally unstable

Both are equally stable

As ionization energy is not the only criteria for the stability of an oxidation state.

Q. Sum of first three ionization energies of Al is 53.0eV atom −1 and the sum of first two ionization energies of Na is 52.2eV atom −1. Out of Al (III) and Na (II)

Al (III) is more stable than Na (II)

Na (II) is more stable than Al (III)

Both are equally unstable

Both are equally stable

As ionization energy is not the only criteria for the stability of an oxidation state.

Q. Sum of first three ionization energies of $A l$ is $53.0 e V$ atom $^{- 1}$ and the sum of first two ionization energies of $N a$ is $52.2 e V$ atom $^{- 1}$ . Out of $A l$ (III) and $N a$ (II)

$A l$ (III) is more stable than $N a$ (II)

$N a$ (II) is more stable than $A l$ (III)