Question

Consider the following reactions :
$O_{\left(g\right)}+e^{-}\to O_{\left(g\right)}^{-},\Delta H_1$
$F_{\left(g\right)}+e^{-}\to F_{\left(g\right)}^{-},\Delta H_2$
$C{l}_{\left(g\right)}+e^{-}\to C{l}_{\left(g\right)}^{-},\Delta H_3$
$O_{\left(g\right)}^{-}+e^{-}\to O_{\left(g\right)}^{2-},\Delta H_4$
The correct statement(s) is/are

• A. $\Delta H_3$ and $\Delta H_2$ both are negative
• B. $\Delta H_1$ and $\Delta H_4$ both are positive
• C. $\Delta H_1$, $\Delta H_2$ and $\Delta H_3$ are negative
• D. $\Delta H_4$ is negative
  

Answer 1

Answer: (A), (C)

$\Delta H_1$, $\Delta H_2$ and $\Delta H_3$ are negative

a) $\Delta H_3$ and $\Delta H_2$ both are negative because both $F$ and $Cl$ is getting a stable electronic configuration by accepting an electron. So energy will release.
b) $\Delta H_1$ will have a negative value since due to the effective nuclear charge of $O$ atom, electron will receive an attractive force and electron will release energy to occupy a lower energy orbital. But $\Delta H_4$ is positive because here we are adding an electron to an ion which is already negatively charged, so to overcome the repulsion we need to provide energy.
c) For the already mentioned reasons we can say that $\Delta H_1$, $\Delta H_2$ and $\Delta H_3$ are negative.
d) $\Delta H_4$ is positive as it is a endothermic reaction as mentioned above.