Chemistry - PERIODIC TABLE Question with Solution | TestHub

Electron affinity generally increases across a period and decreases down a group. However, there are exceptions.

Option A: $\mathrm{F}>\mathrm{Cl}$ is INCORRECT. Chlorine has a higher electron affinity than fluorine due to smaller electron-electron repulsion in the larger 3p subshell of Cl compared to the compact 2p subshell of F. So, $\mathrm{Cl}>\mathrm{F}$.

Option B: ${\mathrm{F}}^{+}>{\mathrm{Cl}}^{+}$ is CORRECT. This refers to the electron affinity of the positive ions. ${\mathrm{F}}^{+}$ has a stronger effective nuclear charge and smaller size than ${\mathrm{Cl}}^{+}$, making it more eager to accept an electron.

Option C: $\mathrm{O}>\mathrm{Se}$ is INCORRECT. Similar to F vs Cl, oxygen has a smaller electron affinity than sulfur and selenium due to electron-electron repulsion in its small 2p subshell. So, $\mathrm{Se}>\mathrm{O}$.

Option D: $\mathrm{N}>\mathrm{B}$ is INCORRECT. Boron has a higher electron affinity than nitrogen. Nitrogen has a stable half-filled 2p subshell ($2p^3$), making it less likely to accept an additional electron. So, $\mathrm{B}>\mathrm{N}$.

The final answer is $\boxed{B}$