Explanation:
<h2>The number of energy levels (n) increases, so there is a greater distance between the nucleus and the outermost orbital.</h2>
Answer:
Cd(s) + AgNO₃(aq) → Cd(NO₃)₂ (aq) + Ag(s)
Oxidized: Cd
Reduced: Ag
Explanation:
Cd(s) + AgNO₃(aq) → Cd(NO₃)₂ (aq) + Ag(s)
Cd → Cd²⁺ + 2e⁻ Half reaction oxidation
1e⁻ + Ag⁺ → Ag Half reaction reduction
Ag changed oxidation number from +1 to 0
Cd changed oxidation number from 0 to +2
Let's ballance the electrons
( Cd → Cd²⁺ + 2e⁻ ) .1
( 1e⁻ + Ag⁺ → Ag ) .2
Cd + 2e⁻ + 2Ag⁺ → 2Ag + Cd²⁺ + 2e⁻
Finally the ballance equation is:
Cd(s) + 2AgNO₃(aq) → Cd(NO₃)₂ (aq) + 2Ag(s)
Answer:
Option-1 (O²⁻) is the correct answer.
Explanation:
All given anions contains same charge. So, we can ignore the effect of charge on these anions.
As we know all given compounds belongs to same group (Group 6) in periodic table. And from top to bottom along the group the elements are placed as,
Oxygen O
Sulfur S
Selenium Se
Tellurium Te
Hence, moving from top to bottom along the group the number of shells increases. And with increase in number of shell the atomic or ionic radii increases. As Oxygen is present at the top of the group, therefore, it has the smallest radius due to less number of shells.
