Question

which is the larger species in each pair?
a. li or li+
b. i^- or cs^+
c. cr or cr^{3+}
d. o or o^{2-}

Explanation:

Part a: Li or \( \text{Li}^+ \)

Step1: Analyze electron configuration

Li has an electron configuration of \( 1s^2 2s^1 \), while \( \text{Li}^+ \) has lost its outermost (2s) electron, so its electron configuration is \( 1s^2 \).

Step2: Compare radii based on electrons

When an atom loses an electron to form a cation, the remaining electrons experience a greater effective nuclear charge (since there are fewer electrons to shield the nucleus's positive charge). This pulls the electrons closer to the nucleus, decreasing the ionic radius compared to the atomic radius. So, Li (the atom) is larger than \( \text{Li}^+ \) (the cation).

Part b: \( \text{I}^- \) or \( \text{Cs}^+ \)

Step1: Determine electron configurations

Both \( \text{I}^- \) and \( \text{Cs}^+ \) are isoelectronic (they have the same number of electrons, 54). The electron configuration for both is similar to that of Xe (\( 1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^2 4p^6 4d^{10} 5s^2 5p^6 \)).

Step2: Compare nuclear charges

The nuclear charge of \( \text{I}^- \) (atomic number of I is 53) is \( +53 \), and for \( \text{Cs}^+ \) (atomic number of Cs is 55), it is \( +55 \). A higher nuclear charge pulls the electrons more tightly. Since \( \text{I}^- \) has a smaller nuclear charge, its electrons are less tightly held, so \( \text{I}^- \) has a larger radius than \( \text{Cs}^+ \).

Part c: Cr or \( \text{Cr}^{3+} \)

Step1: Analyze electron loss

Cr (atomic number 24) has an electron configuration (simplified) that includes electrons in the 4s and 3d orbitals. \( \text{Cr}^{3+} \) is formed when Cr loses 3 electrons (typically 1 from 4s and 2 from 3d).

Step2: Compare radii due to electron loss

Losing electrons to form a cation reduces the number of electron - electron repulsions and increases the effective nuclear charge experienced by the remaining electrons. This causes the electrons to be pulled closer to the nucleus, so the atomic radius of Cr is larger than the ionic radius of \( \text{Cr}^{3+} \).

Part d: O or \( \text{O}^{2-} \)

Answer:

Step1: Analyze electron gain

O (atomic number 8) has an electron configuration of \( 1s^2 2s^2 2p^4 \). \( \text{O}^{2-} \) is formed when O gains 2 electrons, resulting in an electron configuration of \( 1s^2 2s^2 2p^6 \) (isoelectronic with Ne).

Step2: Compare radii due to electron gain

Gaining electrons increases electron - electron repulsions and also, since the nuclear charge remains the same (\( +8 \)) while the number of electrons increases, the effective nuclear charge per electron decreases. This causes the electrons to spread out more, so the ionic radius of \( \text{O}^{2-} \) is larger than the atomic radius of O.

Final Answers:

a. \( \boldsymbol{\text{Li}} \)
b. \( \boldsymbol{\text{I}^-} \)
c. \( \boldsymbol{\text{Cr}} \)
d. \( \boldsymbol{\text{O}^{2-}} \)