Question

1. the statue of libertys green color comes from a patina, $\ce{cu2so4(oh)6}$. this patina is formed through many different reactions, but the first step is the formation of copper oxide, $\ce{cu2o}$. the patina cannot form unless $\ce{cu2o}$ is formed first. $\ce{cu2o}$ forms when pure copper ($\ce{cu}$) reacts with oxygen ($\ce{o}$) in the air. while the statue of libertys exterior is made of pure copper and bronze, another statues material is made of copper ($\ce{cu}$) mixed with tin ($\ce{sn}$). tin also forms a compound with oxygen in the air, $\ce{sno}$ (or $\ce{sno2}$).

a. which element is likely to have a larger first ionization energy, copper or tin?

Explanation:

Step1: Recall Ionization Energy Trend

Ionization energy generally increases across a period (left to right) in the periodic table because the effective nuclear charge increases, holding electrons more tightly. Copper (Cu) and tin (Sn) are in the same period (period 5). Cu is at atomic number 29, Sn at 50? Wait, no: Cu is [Ar] 3d¹⁰4s¹, Sn is [Kr] 4d¹⁰5s²5p². Wait, period 4 and 5? Wait, Cu is in period 4 (atomic number 29), Sn is in period 5 (atomic number 50). Wait, no, period number is the highest principal quantum number. Cu: n=4 (4s), Sn: n=5 (5s,5p). Wait, actually, when moving across a period (same n), ionization energy increases. But Cu is in period 4, Sn in period 5. Wait, no, let's check their positions. Copper is in group 11, period 4. Tin is in group 14, period 5. Wait, first ionization energy: as we move from left to right across a period, IE increases. But also, as we move down a group, IE decreases (because electrons are further from nucleus). Wait, Cu and Sn: Cu is in period 4, Sn in period 5 (down a group from... Wait, Cu is group 11, Sn is group 14. Wait, maybe better to check their electron configurations. Cu: [Ar] 3d¹⁰ 4s¹. Sn: [Kr] 4d¹⁰ 5s² 5p². The first ionization energy is the energy to remove the outermost electron. For Cu, the outermost electron is in 4s¹. For Sn, outermost electrons are in 5s²5p². Wait, the principal quantum number for Cu's outermost electron is n=4, for Sn it's n=5. Electrons in higher n (larger orbit) are easier to remove, so IE is lower. So Cu is in n=4, Sn in n=5. So Cu has higher first ionization energy than Sn.

Step2: Compare Period and Group Trends

• Across a period (same n), IE increases (left to right) because effective nuclear charge increases.
• Down a group (increasing n), IE decreases because electrons are further from nucleus, less tightly held.

Cu is in period 4, Sn in period 5 (down a group relative to... Wait, Cu and Sn are not in the same group. Cu is group 11, Sn group 14. But the principal quantum number of the outermost electron: Cu's outermost is 4s (n=4), Sn's is 5s/5p (n=5). Since n=5 is larger than n=4, the outermost electron of Sn is further from the nucleus than that of Cu. Therefore, it's easier to remove Sn's outermost electron, so Sn has lower first ionization energy, meaning Cu has larger first ionization energy.

Answer:

Copper (Cu) is likely to have a larger first ionization energy than tin (Sn).