An element in the 4th row of the periodic table has the following ionization energies: i1...

An element in the 4th

row of the periodic table has the following ionization energies:

i1 = 762kJ/mol,

i2 = 1021 kJ/mol,

i3 = 1203 kJ/mol,

i4 = 1402 kJ/mol,

i5 = 5002 kJ/mol,

i6 = 8321 kJ/mol,

i7 = 10243 kJ/mol,

i8 = 13210 kJ/mol

What is the identity of the element?

Solutions

Expert Solution

Ge---------> Ge^+ + e^ i1 = 762kJ/mol,

Ge^+ --------> Ge^2+ + e^- i2 = 1021 kJ/mol,

Ge^2+ -----------> Ge^3+ + e^- i3 = 1203 kJ/mol,

Ge^3+ ---------> Ge^4+ + e^- i4 = 1402 kJ/mol, Ge^+ 4 get stabel electronic configuration.

i5 = 5002 kJ/mol,

i6 = 8321 kJ/mol,

i7 = 10243 kJ/mol,

i8 = 13210 kJ/mol

I4 and I5 energy difference is more . so the element is Ge

queen_honey_blossom answered 1 year ago

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