What Is the Correct Order of Ionization Potential for Different Element Sets?

[zorro]

Homework Statement

Arrange the following in the increasing order of ionisation potential.
1. N,F,O
2. C,O,N
3. Au,Cu,Ag

Homework Equations

Electronic Configurations

The Attempt at a Solution

1. The highest ionisation energy will be that of F. Next will be of O because of paired p electrons. Last of N because of 1 unpaired electron.
So N<O<F
But answer is O<N<F

2. On similar reasons, N<C<O
But again it is C<O<N

3. Cu<Ag<Au
Au has f electrons which cause poor shielding effect and hence greater nuclear pull.
Ag has 4d electrons and Cu has 3d electrons out of which 4d offer weaker screening effect.
Hence the above order.
But answer is Ag<Cu<Au

Please provide detailed explanation

 

[nate808]

for the correct order of ionisation potential for each set.

For the first set (N, F, O), the correct order of increasing ionization potential is N < O < F. This is because the ionization potential increases as you move across a period in the periodic table from left to right. This is due to the increase in nuclear charge and decrease in atomic radius. In this case, F has the highest nuclear charge and smallest atomic radius, making it the most difficult to remove an electron from. O has a slightly lower nuclear charge and slightly larger atomic radius than F, making it easier to remove an electron. N has the lowest nuclear charge and largest atomic radius, making it the easiest to remove an electron from.

For the second set (C, O, N), the correct order of increasing ionization potential is C < N < O. This is because the ionization potential also increases as you move up a group in the periodic table. This is due to the decrease in shielding effect from inner electrons as you move up a group. In this case, C has the highest shielding effect and therefore the lowest ionization potential. N has less shielding effect than C, making it slightly more difficult to remove an electron. O has the least shielding effect, making it the most difficult to remove an electron.

For the third set (Au, Cu, Ag), the correct order of increasing ionization potential is Cu < Ag < Au. This is because the ionization potential increases as you move from left to right in the d-block of the periodic table. This is due to the increase in nuclear charge and decrease in atomic radius. In this case, Cu has the smallest atomic radius and highest nuclear charge, making it the most difficult to remove an electron from. Ag has a slightly larger atomic radius and lower nuclear charge than Cu, making it easier to remove an electron. Au has the largest atomic radius and lowest nuclear charge, making it the easiest to remove an electron from. Additionally, as you mentioned, the f electrons in Au do contribute to a weaker shielding effect, but the effect is not significant enough to change the overall trend in ionization potential.

In summary, the correct order of increasing ionization potential for each set is: N < O < F, C < N < O, and Cu < Ag < Au. It is important to remember that ionization potential is affected by multiple factors, including nuclear charge, atomic radius, and shielding effect, and must be considered within