Co(NH3) 6] 2+ has ∆ o = 10,100 cm –1 and B = 920 cm–1. How manyelectronic absorptions do you expect for the complex and at whatenergies?

The complex [Co(NH3)6]2+ is an octahedral complex. In octahedral complexes, the d orbitals split into two sets with different energy levels due to the crystal field splitting. The lower energy set is called t2g and the higher energy set is called eg.

The number of electronic absorptions in an octahedral complex is determined by the number of electrons that can be excited from the t2g set to the eg set. In the case of [Co(NH3)6]2+, the cobalt ion is in the +2 oxidation state. Cobalt is in the 3d series and has an atomic number of 27, so it has 7 3d electrons. When it loses 2 electrons to form the +2 ion, it is left with 5 3d electrons.

These 5 electrons will fill the t2g set, which can hold 6 electrons. Therefore, there is only one electron that can be excited to the eg set. So, we expect only one electronic absorption for this complex.

The energy of this absorption can be calculated using the formula for the energy of a photon, E = hv, where h is Planck's constant and v is the frequency of the light. However, in this problem, we are given the energy in terms of wavenumbers (∆o), so we can use the relationship between energy and wavenumbers, E = hc∆o, where c is the speed of light.

Substituting the given values, ∆o = 10,100 cm–1, h = 6.626 x 10^-34 J.s, and c = 3.00 x 10^10 cm/s, we get:

E = (6.626 x 10^-34 J.s)(3.00 x 10^10 cm/s)(10,100 cm–1) = 1.99 x 10^-19 J

To convert this energy to electron volts (eV), we use the conversion factor 1 eV = 1.602 x 10^-19 J, so:

E = (1.99 x 10^-19 J)(1 eV/1.602 x 10^-19 J) = 1.24 eV

So, we expect one electronic absorption at an energy of 1.24 eV.