<h3>Answer:</h3>
Rubidium (Rb)
<h3>Explanation:</h3>
Ionization Energy is defined as, "the minimum energy required to knock out or remove the valence electron from valence shell of an atom".
<h3>Trends in Periodic table:</h3>
Along Periods:
Ionization Energy increases from left to right along the periods because moving from left to right in the same period the number of protons (atomic number) increases but the number of shells remain constant hence, resulting in strong nuclear interactions and electrons are more attracted to nucleus hence, requires more energy to knock them out.
Along Groups:
Ionization energy decreases from top to bottom along the groups because the number of shells increases and the distance between nucleus and valence electrons also increases along with increase in shielding effect provided by core electrons. Therefore, the valence electrons experience less nuclear attraction and are easily removed.
<h3>Conclusion:</h3>
Given elements belong to same group hence, Rubidium present at the bottom of remaining elements will have least ionization energy due to facts explained in trends of groups above.
Silver is an example of an element. True. Elements are pure substances that cannot be broken down into simpler substances. An atom is the smallest unit of an element that retains all properties of the element.
Answer: Option (B) is the correct answer.
Explanation:
Both oxygen and fluorine are period 2 elements and when we move across a period then there occurs a decrease in atomic size of the atoms. Hence, the atomic radius of a neutral fluorine atom is smaller than a neutral oxygen atom.
Moreover, atomic number of fluorine is 9 and it has higher nuclear charge due to which it will cause more attraction of electrons. As a result, size of a fluorine ion will be smaller.
On the other hand, size of oxygen atom is larger and has small nuclear charge due to which attraction of electrons by its nucleus will not be strong enough. Hence, the size of 
will be larger.
Thus, we can conclude that the statement 
has a larger nuclear charge than has, is correct for the fact that the ion is smaller than the ion.
Answer:
0.59 g
Explanation:
The reaction at the first cathode;
Ag^+(aq) + e -----> Ag
if Q = It = 1.5 * 10 * 60 = 900C
According to Faraday's second law of electolysis; if we pass the same quantity of electricity through different electrolytes, the mass of substances deposited on each cathode is proportional to the equivalent weights.
Equivalent weight E = Atomic mass/valency
For Ag = 108/1 = 108
For Ni = 59/2 = 29.5
Hence
Let m1 = mass of Ag deposited =2.16
Let m2 = mass of Ni deposited
Let E1 = equivalent weight of Ag
Let E2 = equivalent weight of Ni
m1/m2 = E1/E2
2.16/m2 = 108/29.5
m2 = 2.16 * 29.5/108
m2 = 0.59 g
