The electron affinity values for each case are determined as follows:
(A) Be + e$^-$ $\to$ Be$^-$: Electron affinity is negative (E.A. = --ive).
(B) N + e$^-$ $\to$ N$^-$: Electron affinity is negative (E.A. = --ive).
(C) O + e$^-$ $\to$ O$^-$: Electron affinity is positive for O $\to$ O$^-$ (E.A. = +ive).
(D) Na + e$^-$ $\to$ Na$^-$: Electron affinity is positive (E.A. = +ive).
(E) Al + e$^-$ $\to$ Al$^-$: Electron affinity is positive (E.A. = +ive).
Thus, the correct answer is: A, B, and C only.
20 mL of sodium iodide solution gave 4.74 g silver iodide when treated with excess of silver nitrate solution. The molarity of the sodium iodide solution is _____ M. (Nearest Integer value) (Given : Na = 23, I = 127, Ag = 108, N = 14, O = 16 g mol$^{-1}$)