Question

consider these elements: n, mg, o, f, al.
a. write the ground - state electron configuration for each element.
b. arrange the elements in order of decreasing atomic radius.
c. arrange the elements in order of increasing ionization energy.
d. use the electron configurations in part (a) to explain the differences between your answers to parts (b) and (c).

Explanation:

Part (a)

Step 1: Determine atomic number

Find the atomic number (Z) of each element from the periodic table: N (Z=7), Mg (Z=12), O (Z=8), F (Z=9), Al (Z=13). The ground - state electron configuration is built by filling orbitals following the Aufbau principle (\(1s^2 < 2s^2 < 2p^6 < 3s^2 < 3p^1\) etc.), Pauli exclusion principle, and Hund's rule.

• For N (Z = 7):

The electron configuration is \(1s^22s^22p^3\) (we fill \(1s\) with 2 electrons, \(2s\) with 2 electrons, and then \(2p\) with 3 electrons since \(2 + 2+3=7\)).

• For Mg (Z = 12):

We fill orbitals as \(1s^22s^22p^63s^2\) ( \(1s^2+2s^2 + 2p^6+3s^2=2 + 2+6 + 2=12\)).

• For O (Z = 8):

The electron configuration is \(1s^22s^22p^4\) ( \(2+2 + 4=8\)).

• For F (Z = 9):

The electron configuration is \(1s^22s^22p^5\) ( \(2+2+5 = 9\)).

• For Al (Z = 13):

The electron configuration is \(1s^22s^22p^63s^23p^1\) ( \(2+2+6 + 2+1=13\)).

Part (b)

Step 1: Recall atomic radius trends

Atomic radius generally decreases across a period (left to right) and increases down a group (top to bottom). The elements N, O, F are in period 2 (n = 2), and Mg, Al are in period 3 (n = 3). In period 3, Mg is to the left of Al, so \(r(\text{Mg})>r(\text{Al})\). In period 2, N is to the left of O, which is to the left of F, so \(r(\text{N})>r(\text{O})>r(\text{F})\). Also, the radius of period 3 elements (n = 3) is larger than period 2 elements (n = 2) because the principal quantum number \(n\) is larger (electrons are in a shell further from the nucleus).
So the order of decreasing atomic radius is \(\text{Mg}>\text{Al}>\text{N}>\text{O}>\text{F}\)

Part (c)

Step 1: Recall ionization energy trends

Ionization energy (IE) generally increases across a period (left to right) and decreases down a group (top to bottom). Across a period, effective nuclear charge increases, so it is harder to remove an electron. Down a group, the outermost electron is in a higher - energy (larger) orbital, so it is easier to remove.

• Mg and Al are in period 3. Al has one electron in the \(3p\) orbital, and Mg has a filled \(3s\) orbital. The electron in the \(3p\) orbital of Al is easier to remove than the electrons in the filled \(3s\) orbital of Mg, so \(IE(\text{Al})<IE(\text{Mg})\).
• N, O, F are in period 2. N has a half - filled \(2p\) sub - shell (\(2p^3\)), which is relatively stable. O has \(2p^4\) (one more electron than the half - filled state), and F has \(2p^5\). The order across period 2 (from left to right) with some considerations for half - filled sub - shells: \(IE(\text{N})<IE(\text{O})\) (because the electron added to O to get from N to O is paired in a \(2p\) orbital, and there is electron - electron repulsion which makes it a bit easier to remove an electron from O compared to the stability of the half - filled sub - shell in N), and \(IE(\text{O})<IE(\text{F})\) (as we move from O to F, effective nuclear charge increases). Also, since Mg and Al are in period 3 (lower IE than period 2 elements in general, because their outermost electrons are in \(n = 3\) orbitals which are further from the nucleus than \(n=2\) orbitals of N, O, F), we have:

The order of increasing ionization energy is \(\text{Al}<\text{Mg}<\text{N}<\text{O}<\text{F}\)

Part (d)

Answer:

s:

Part (a)

• N: \(\boldsymbol{1s^22s^22p^3}\)
• Mg: \(\boldsymbol{1s^22s^22p^63s^2}\)
• O: \(\boldsymbol{1s^22s^22p^4}\)
• F: \(\boldsymbol{1s^22s^22p^5}\)
• Al: \(\boldsymbol{1s^22s^22p^63s^23p^1}\)

Part (b)

\(\boldsymbol{\text{Mg}>\text{Al}>\text{N}>\text{O}>\text{F}}\)

Part (c)

\(\boldsymbol{\text{Al}<\text{Mg}<\text{O}<\text{N}<\text{F}}\) (Note: The initial reasoning about N and O was corrected based on actual IE values. The key is that the half - filled sub - shell in N gives it a relatively high IE, but the paired electron in O's \(2p^4\) leads to lower IE than N)

Part (d)

• Atomic radius (part b): Determined by principal quantum number (\(n\)) of outermost electrons (larger \(n\) means larger radius for period 3 vs period 2) and effective nuclear charge (decreases across a period as \(Z\) increases, pulling electrons closer).
• Ionization energy (part c): Depends on electron configuration stability (e.g., filled \(3s^2\) in Mg, half - filled \(2p^3\) in N) and effective nuclear charge (increases across a period, increasing IE, with exceptions for electron - electron repulsion in O). Lower \(n\) (period 2) and stable configurations increase IE, while higher \(n\) (period 3) and less stable configurations (e.g., \(3p^1\) in Al) decrease IE. The difference arises because radius is about electron - cloud size (orbital size, \(n\), and \(Z_{eff}\)), while IE is about the energy to remove an electron (stability of configuration, \(n\), and \(Z_{eff}\)).

Final Answers (Summarized)

a.

• N: \(1s^22s^22p^3\)
• Mg: \(1s^22s^22p^63s^2\)
• O: \(1s^22s^22p^4\)
• F: \(1s^22s^22p^5\)
• Al: \(1s^22s^22p^63s^23p^1\)

b. \(\text{Mg}>\text{Al}>\text{N}>\text{O}>\text{F}\)

c. \(\text{Al}<\text{Mg}<\text{O}<\text{N}<\text{F}\)

d. Atomic radius depends on \(n\) (orbital size) and \(Z_{eff}\) (electron - cloud contraction across a period). Ionization energy depends on configuration stability (e.g., filled/\(half - filled\) sub - shells), \(n\) (orbital energy), and \(Z_{eff}\) (electron - removal energy). Period 3 elements (Mg, Al) have larger \(n\) (bigger radius) but lower \(IE\) (easier electron removal) than period 2 elements (N, O, F). Within periods, \(Z_{eff}\) drives radius decrease and (mostly) \(IE\) increase, with configuration - stability exceptions (e.g., N vs O).