Question

Write the shorthand valence shell electron configuration of each of the following elements, basing your answer on the element's location on the periodic table. a. nickel, \(Z=28\) b. niobium, \(Z=41\) c. hafnium, \(Z=72\) d. astatine, \(Z=85\)

Short answer

Ni: [Ar] 4s² 3d⁸ Nb: [Kr] 5s² 4d³ Hf: [Xe] 6s² 4f¹⁴ 5d² At: [Xe] 6s² 4f¹⁴ 5d¹⁰ 6p⁵

Step-by-step solution

Locate the elements on the periodic table and determine the preceding noble gases.

For each element, we need to first find its location on the periodic table and then find the noble gas in the previous row. a. Nickel (Ni, \(Z=28\)) is in period 4 of the transition metals. The noble gas that precedes it is Argon (Ar, \(Z=18\)). b. Niobium (Nb, \(Z=41\)) is in period 5 of the transition metals. The noble gas that precedes it is Krypton (Kr, \(Z=36\)). c. Hafnium (Hf, \(Z=72\)) is in period 6 of the transition metals. The noble gas that precedes it is Xenon (Xe, \(Z=54\)). d. Astatine (At, \(Z=85\)) is in period 6 of the halogens. The noble gas that precedes it is Xenon (Xe, \(Z=54\)).

Write the shortcut electron configurations.

Now, we write the shortcut electron configuration for each element, starting with the symbol of the preceding noble gas in brackets, followed by the electron configuration of the remaining electrons. a. Nickel (Ni, \(Z=28\)): The preceding noble gas is Argon (Ar, \(Z=18\)), so our starting point is [Ar]. Since nickel has 10 more electrons than argon, we add 4s², 3d⁸. \[ \text{Ni: [Ar] 4s^2 3d^8} \] b. Niobium (Nb, \(Z=41\)): The preceding noble gas is Krypton (Kr, \(Z=36\)), so our starting point is [Kr]. Since niobium has 5 more electrons than krypton, we add 5s², 4d³. \[ \text{Nb: [Kr] 5s^2 4d^3} \] c. Hafnium (Hf, \(Z=72\)): The preceding noble gas is Xenon (Xe, \(Z=54\)), so our starting point is [Xe]. Since hafnium has 18 more electrons than xenon, we add 6s², 4f¹⁴, 5d². \[ \text{Hf: [Xe] 6s^2 4f^{14} 5d^2} \] d. Astatine (At, \(Z=85\)): The preceding noble gas is Xenon (Xe, \(Z=54\)), so our starting point is [Xe]. Since astatine has 31 more electrons than xenon, we add 6s², 4f¹⁴, 5d¹⁰, 6p⁵. \[ \text{At: [Xe] 6s^2 4f^{14} 5d^{10} 6p^5} \]