Question

Consider the electronic structure of the element bismuth:
(a) The first ionization energy of bismuth is Ei1 = +703 kJ/mol. What is the longest possible wavelength of light that could ionize an atom of bismuth?
Answer:   nm (round to the ones place)
(b) Write the electron configurations of neutral Bi and the Bi+ cation. Use noble gas shorthand notation and "^" to denote exponents.
Answer:   
(c) What are the n and l quantum numbers of the electron removed when Bi is ionized to Bi+?
Answer:   
(d) Would you expect element 115 to have an ionization energy "greater than", "equal to", or "less than" that of bismuth?
Answer:  

Answer 1

Answer – a) Given, energy, E = 703 kJ/mol

λ = ?

We need to do conversion of kJ/mol to J

So, E = 703 kJ/mol*1000 J*kJ *1 mole / 6.023*10²³

          = 1.17*10^-18 J

We know formula,

E = h*C/ λ

So, λ = h*C / E

          = 6.626*10^-34 J.s * 3.0*10⁸ m.s^-1 / 1.17*10^-18 J

          = 1.70*10^-7 m

So, 1 m = 1*10⁹ nm

So, 1.70*10^-7 m = ?

= 170 nm

b) The electron configurations of neutral Bi and the Bi⁺ cation –

Bi has atomic number 83

Electron configurations of neutral Bi = [Xe] 45f^14 5d^10 6s^2 6p^3

Electron configurations of neutral Bi⁺ = [Xe] 45f^14 5d^10 6s^2 6p^2

c) The removed electron from the Bi is from the 6p orbital and in the pz orbital

so, n = 6, l = 1 , ml = -1 , ms = +1/2

d) The ionization energy of element 115 is less than the Bi, since we know as we goes from the top to bottom in the periodic table then there is ionization energy gets decreased.