Question

Beryllium is placed above magnesium in the second group. Beryllium dust, therefore when added to $NH_{4}OH$solution will.

• A. Have no effect
• B. Precipitate $\Delta º_{NH_{4}OH} = \Delta º_{Ba(OH)_{2}} + \Delta º_{NH_{4}Cl} - \Delta º_{BaCl_{2}}$ metal
• C. Precipitate $\Delta º_{NH_{4}OH} = \Delta º_{BaCl_{2}} + \Delta º_{NH_{4}Cl} - \Delta º_{Ba(OH)_{2}}$
• D. Lead to dissolution of $\Delta º_{NH_{4}OH} = \frac{\Delta º_{Ba(OH)_{2}} + 2\Delta º_{NH_{4}Cl} - \Delta º_{BaCl_{2}}}{2}$metal

Answer 1

Answer: (A)

Have no effect

Answer 2

No doubt $Be$ is placed above $Mg$ in the second group of periodic table but it is below $Mg$ in electrochemical series.