Problem 38 - Chemistry: The Central Science

Question

Write equations that show the process for (a) the first two ionization energies of lead and (b) the fourth ionization energy of zirconium.

Step-by-Step Solution

Answer

a) The first two ionization energies of lead (Pb) are represented by these equations: \(1^{st}\) Ionization energy: \(Pb(g) -> Pb^{+}(g) + e^{-}\) \(2^{nd}\) Ionization energy: \(Pb^{+}(g) -> Pb^{2+}(g) + e^{-}\) b) The fourth ionization energy of zirconium (Zr) is represented by this equation: \(4^{th}\) Ionization energy: \(Zr^{3+}(g) -> Zr^{4+}(g) + e^{-}\)

Step 1: a) First ionization energy of lead

To write the equation for the first ionization energy of lead, we need to remove one electron from the neutral lead atom: Pb(g) -> Pb^(+)(g) + e^(-) Here, X is the lead (Pb) atom, and n is 1 (first ionization energy).

Step 2: b) Second ionization energy of lead

To write the equation for the second ionization energy of lead, we need to remove another electron from the Pb^+ ion obtained in step a: Pb^(+)(g) -> Pb^(2+)(g) + e^(-) Here, X is the lead (Pb) atom and n is now 2 (second ionization energy).

Step 3: c) Fourth ionization energy of zirconium

To write the equation for the fourth ionization energy of zirconium, we need to remove four electrons from the neutral zirconium atom: Zr^(3+)(g) -> Zr^(4+)(g) + e^(-) Here, X is the zirconium (Zr) atom and n is 4 (fourth ionization energy).