Question

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

Short 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-by-step solution

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).

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).

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).