The energy difference between the 5d and 6s sublevels in gold accounts for its color. Assuming this energy difference is about 2.7 eV (electron volt; 1 eV1.602X10⁻¹⁹J), explain why gold has a warm yellow color.

which of the following atoms (or ions) has the smallest radius? a) K+  b)Ar   c) Cl-

An element has the following electron configuration, [Ar]3d104s²4p3. This element is a _-- O transition metal O metal noble gas nonmetal O metalloid

the electron affinity of the lutetium atomwas measured using photoelectron spectroscopy. In this experiment, a beam of lutetium negative ions, Lu was prepared and irradiated with a laser beam having a wavelength of 1064nm. If the emitted electron in this experiment has a kinetic energy of 0.825 eV (1 eV = 1.602 x 10^-22kJ ), What is the electronaffinity of lutetium?

5) Why does copper (Cu) easily lose and electron from its s subshell instead of its d shell when forming a Cu ion? What does this have to do with the stability of the atom? Explain the difference between electron affinity and first ionization energy. Why does ionization energy for an atom increase for each additional e

Given the following successive ionization energies, predict which element they correspond to. |₁ (kJ/mol) 800.0 Be N C Li 12 2430.0 13 3660.0 14 25020.0

What is the arrangement of these atoms in order of increasing atomic radius: C, Ca, Co, Cr, Cs, Cu ?

Calculate the change in energy for the following processes: Na(g) + Cl(g) → Na^+1(g) + Cl^-1(g) Ca(g) + 2Br(g) → Ca^+2(g) + 2Br-1(g)

Calculate the frequency and wavelength of electromagnetic radiation associated with the electron transition from the 8 th to the 11th quantum energy level in the Be3+ ion. Present both values in SI units using scientific notation. b) How much energy (in kJ) is carried by 1.75 mol of photons absorbed during this transition?

Write condensed electron configurations for the following atoms and ions, using appropriate noble-gas core abbreviations. Put superscripted electron count numbers in parentheses. Examples: Fe = [Ar]4s(2)3d(6), Sn2+ = [Kr]4d(10)5s(2), S2- = [Ar].  I- =