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:

To determine the larger species in each pair, we analyze the electron configuration and the effect of charge on atomic/ionic radius:

Part a: $\boldsymbol{\ce{Li}}$ or $\boldsymbol{\ce{Li^+}}$

• $\ce{Li}$ has an electron configuration of $1s^2 2s^1$.
• $\ce{Li^+}$ is formed when $\ce{Li}$ loses its $2s$ electron, so its electron configuration is $1s^2$.
• 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). This pulls the electrons closer to the nucleus, reducing the radius.
• Thus, $\ce{Li}$ (neutral atom) is larger than $\ce{Li^+}$ (cation).

Part b: $\boldsymbol{\ce{I^-}}$ or $\boldsymbol{\ce{Cs^+}}$

• First, find the number of electrons:
• $\ce{I^-}$: Iodine (atomic number 53) gains 1 electron, so it has $53 + 1 = 54$ electrons.
• $\ce{Cs^+}$: Cesium (atomic number 55) loses 1 electron, so it has $55 - 1 = 54$ electrons.
• Both species have the same number of electrons (isoelectronic), so we compare their nuclear charges.
• Nuclear charge of $\ce{I^-}$: +53 (from the nucleus of iodine).
• Nuclear charge of $\ce{Cs^+}$: +55 (from the nucleus of cesium).
• A higher nuclear charge pulls the electrons more strongly towards the nucleus, resulting in a smaller radius.
• Since $\ce{I^-}$ has a smaller nuclear charge, its electrons are less strongly pulled, so $\ce{I^-}$ is larger than $\ce{Cs^+}$.

Part c: $\boldsymbol{\ce{Cr}}$ or $\boldsymbol{\ce{Cr^{3+}}}$

• $\ce{Cr}$ (neutral atom) has an electron configuration (simplified) that includes electrons in the 4s and 3d orbitals.
• $\ce{Cr^{3+}}$ is formed when $\ce{Cr}$ loses 3 electrons (from the 4s and 3d orbitals).
• Losing electrons to form a cation reduces the electron - electron repulsion and increases the effective nuclear charge experienced by the remaining electrons. This causes the electrons to be pulled closer to the nucleus, decreasing the radius.
• Therefore, the neutral atom $\ce{Cr}$ is larger than the cation $\ce{Cr^{3+}}$.

Part d: $\boldsymbol{\ce{O}}$ or $\boldsymbol{\ce{O^{2-}}}$

• $\ce{O}$ has an electron configuration of $1s^2 2s^2 2p^4$.
• $\ce{O^{2-}}$ is formed when $\ce{O}$ gains 2 electrons, so its electron configuration is $1s^2 2s^2 2p^6$.
• When an atom gains electrons to form an anion, the added electrons increase electron - electron repulsion. Also, the effective nuclear charge per electron decreases (since there are more electrons to shield the nucleus). This causes the electrons to spread out more, increasing the radius.
• Thus, the anion $\ce{O^{2-}}$ is larger than the neutral atom $\ce{O}$.

Final Answers:

a. $\boldsymbol{\ce{Li}}$
b. $\boldsymbol{\ce{I^-}}$
c. $\boldsymbol{\ce{Cr}}$
d. $\boldsymbol{\ce{O^{2-}}}$

Answer:

To determine the larger species in each pair, we analyze the electron configuration and the effect of charge on atomic/ionic radius:

Part a: $\boldsymbol{\ce{Li}}$ or $\boldsymbol{\ce{Li^+}}$

• $\ce{Li}$ has an electron configuration of $1s^2 2s^1$.
• $\ce{Li^+}$ is formed when $\ce{Li}$ loses its $2s$ electron, so its electron configuration is $1s^2$.
• 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). This pulls the electrons closer to the nucleus, reducing the radius.
• Thus, $\ce{Li}$ (neutral atom) is larger than $\ce{Li^+}$ (cation).

Part b: $\boldsymbol{\ce{I^-}}$ or $\boldsymbol{\ce{Cs^+}}$

• First, find the number of electrons:
• $\ce{I^-}$: Iodine (atomic number 53) gains 1 electron, so it has $53 + 1 = 54$ electrons.
• $\ce{Cs^+}$: Cesium (atomic number 55) loses 1 electron, so it has $55 - 1 = 54$ electrons.
• Both species have the same number of electrons (isoelectronic), so we compare their nuclear charges.
• Nuclear charge of $\ce{I^-}$: +53 (from the nucleus of iodine).
• Nuclear charge of $\ce{Cs^+}$: +55 (from the nucleus of cesium).
• A higher nuclear charge pulls the electrons more strongly towards the nucleus, resulting in a smaller radius.
• Since $\ce{I^-}$ has a smaller nuclear charge, its electrons are less strongly pulled, so $\ce{I^-}$ is larger than $\ce{Cs^+}$.

Part c: $\boldsymbol{\ce{Cr}}$ or $\boldsymbol{\ce{Cr^{3+}}}$

• $\ce{Cr}$ (neutral atom) has an electron configuration (simplified) that includes electrons in the 4s and 3d orbitals.
• $\ce{Cr^{3+}}$ is formed when $\ce{Cr}$ loses 3 electrons (from the 4s and 3d orbitals).
• Losing electrons to form a cation reduces the electron - electron repulsion and increases the effective nuclear charge experienced by the remaining electrons. This causes the electrons to be pulled closer to the nucleus, decreasing the radius.
• Therefore, the neutral atom $\ce{Cr}$ is larger than the cation $\ce{Cr^{3+}}$.

Part d: $\boldsymbol{\ce{O}}$ or $\boldsymbol{\ce{O^{2-}}}$

• $\ce{O}$ has an electron configuration of $1s^2 2s^2 2p^4$.
• $\ce{O^{2-}}$ is formed when $\ce{O}$ gains 2 electrons, so its electron configuration is $1s^2 2s^2 2p^6$.
• When an atom gains electrons to form an anion, the added electrons increase electron - electron repulsion. Also, the effective nuclear charge per electron decreases (since there are more electrons to shield the nucleus). This causes the electrons to spread out more, increasing the radius.
• Thus, the anion $\ce{O^{2-}}$ is larger than the neutral atom $\ce{O}$.

Final Answers:

a. $\boldsymbol{\ce{Li}}$
b. $\boldsymbol{\ce{I^-}}$
c. $\boldsymbol{\ce{Cr}}$
d. $\boldsymbol{\ce{O^{2-}}}$