15. potassiums (ks) first ionization energy is 419 kj/mol. its second ionization energy is 3052 kj/mol, and its third ionization energy is 4420 kj/mol. explain why the second and third ionization energies are closer in value than the first and second ionization energies but are not identical to one another.
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16. why does fluorine pull electrons more strongly than carbon, even though both are in the same row?
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17. why does atomic size decrease across a row of the periodic table?
your answer
18. what is ionization energy, and how does a larger electrostatic force between the nucleus and an electron affect the amount of energy needed to remove that electron?

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### Question 15
# Brief Explanations:
Potassium has an electron configuration of $[Ar] 4s^1$. The first ionization energy removes the $4s^1$ electron, which is in a valence shell far from the nucleus with shielding from inner electrons. After removing the first electron, $K^+$ has a configuration of $[Ar]$, so the second and third ionizations remove electrons from the $Ar$-like (noble gas) core. The second and third electrons are in the same principal energy level (n=3 for Ar core), so their shielding and distance from the nucleus are more similar, leading to closer ionization energies. They aren't identical because each subsequent electron removal from the same shell experiences a slightly increasing effective nuclear charge (due to fewer electrons remaining, so less electron - electron repulsion), making the third ionization energy a bit higher than the second.
# Answer:
The first ionization of K removes the 4s¹ electron (valence, far from nucleus, shielded). After first ionization, K⁺ has [Ar] configuration. Second and third ionizations remove electrons from the Ar - like core (same principal energy level, similar shielding/distance), so their energies are closer. They differ as each subsequent electron removal from the same shell has a slightly higher effective nuclear charge (less electron - electron repulsion), increasing the third ionization energy slightly.

### Question 16
# Brief Explanations:
Fluorine and carbon are in the same period (row) of the periodic table. Electronegativity (ability to pull electrons) increases across a period. This is because as we move from left to right across a period, the effective nuclear charge ($Z_{eff}$) increases. The number of protons in the nucleus increases, while the shielding from inner electrons remains relatively constant (since electrons are added to the same principal energy level). Fluorine has more protons (atomic number 9) than carbon (atomic number 6). The higher effective nuclear charge of fluorine means it has a stronger attraction for electrons (both bonding and non - bonding) than carbon.
# Answer:
Fluorine and carbon are in the same period. Across a period, effective nuclear charge ($Z_{eff}$) increases (more protons, same shielding). Fluorine (Z = 9) has a higher $Z_{eff}$ than carbon (Z = 6), so it exerts a stronger pull on electrons.

### Question 17
# Brief Explanations:
As we move across a row (period) of the periodic table from left to right, the number of protons in the nucleus (atomic number) increases. Electrons are added to the same principal energy level (n remains constant). The effective nuclear charge ($Z_{eff}$) experienced by the outermost electrons increases because the increase in proton number is not fully shielded by the addition of electrons to the same shell. The outermost electrons are pulled more strongly towards the nucleus by the increasing positive charge. Since the electrons are pulled closer to the nucleus, the atomic radius (size) decreases.
# Answer:
Across a period, atomic number (protons) increases, electrons add to the same principal energy level. Effective nuclear charge ($Z_{eff}$) increases (same - shell electrons don't fully shield the increasing proton charge). Outermost electrons are pulled closer to the nucleus, decreasing atomic size.

### Question 18
# Brief Explanations:
Ionization energy is the energy required to remove an electron from a gaseous atom or ion. The electrostatic force between the nucleus (positive charge) and an electron (negative charge) is given by Coulomb's law ($F=\frac{kq_1q_2}{r^2}$, where $k$ is a constant, $q_1$ is nuclear charge, $q_2$ is electron charge, and $r$ is distance). A larger electrostatic force (due to a higher effective nuclear charge or smaller distance) means the electron is more strongly attracted to the nucleus. To remove an electron with a stronger attraction to the nucleus, more energy is required. So, a larger electrostatic force between the nucleus and an electron increases the ionization energy (energy needed to remove the electron).
# Answer:
Ionization energy is the energy to remove an electron from a gaseous atom/ion. A larger electrostatic force (between nucleus and electron) means the electron is more strongly attracted to the nucleus. Thus, more energy is needed to overcome this attraction and remove the electron (ionization energy increases).

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