Answer :
The equation showing process for ionization energies -
a) For lead,
First Ionization energy : Pb (g) + IE₁ → Pb+ (g) + e-
Second Ionization energy: Pb+ (g) + IE₂ → Pb₂+ (g) + e-
b) Fourth ionization energy of zirconium -
Zr₃+(g) → Zr₄+(g)+e−
To remove an electron from a gaseous atom then the energy that is required is known as Ionization Energy.
Amount of energy required to remove the most loose bounded electrons form an isolated gaseous atom to form a gaseous ion, it is called the ionization enthalpy or ionization energy. It is usually represented in the units of kJ mol-1
The first two ionization energies of lead -
Lead is a heavy metal whose atomic number is 82. It has electronic configuration : [Xe] 4f14 5d10 6s2 6p2
Thee general formula to write the equation .
First : Pb (g) + IE1 → Pb+ (g) + e-
Second: Pb+ (g) + IE2 → Pb2+ (g) + e-
The first and second ionization energies of Lead, Pb is equal to =
First = 715.6 kJ mol⁻¹
Second = 1450.5 kJ mol⁻¹
The fourth ionization energy of zirconium is:
Zr₃+(g) → Zr₄+(g)+e−
The fourth ionization energy is 3313 kJ mol⁻¹
To learn more about ionization energy,
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