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.
For the reaction, \[ H_2(g) + I_2(g) \rightleftharpoons 2HI(g) \] Attainment of equilibrium is predicted correctly by:
For the reaction, \[ H_2(g) + I_2(g) \rightleftharpoons 2HI(g) \]
Attainment of equilibrium is predicted correctly by:
If \[ \frac{dy}{dx} + 2y \sec^2 x = 2 \sec^2 x + 3 \tan x \cdot \sec^2 x \] and
and \( f(0) = \frac{5}{4} \), then the value of \[ 12 \left( y \left( \frac{\pi}{4} \right) - \frac{1}{e^2} \right) \] equals to: