Question

The half-life period of a radioactive element is $\mathrm{1.5\times10^{10}}$ years. Calculate the time in which the activity of the element is reduced to 75% of its original value. [Given : log 2 = 0·30, log 3 = 0·48, log 4 = 0·60]

Hint:

First order: t = (2.303/k) log(100/75), with k = 0.693/t½ and log(4/3)=0.12.

Answer 1

Concept: Whether a complex ends up low spin (electrons forced to pair) or high spin (electrons stay unpaired) is a tug of war between two energies: the crystal-field splitting $\Delta$ (how far apart the d-orbitals get pushed) and the pairing energy P (the energy cost of squeezing two electrons into the same orbital). Low spin needs $\Delta$ to be bigger than P.

Step 1: Why tetrahedral splitting is small
In a tetrahedral arrangement, the ligands do not point straight at the d-orbitals, so they split the d-orbitals only a little. In fact the tetrahedral splitting is only about four-ninths of the octahedral one: \[ \Delta_t = \tfrac{4}{9}\,\Delta_o \] So $\Delta_t$ is naturally a small number.

Step 2: Compare with the pairing energy
For a complex to be low spin, the splitting $\Delta_t$ would have to be larger than the pairing energy P, so that electrons prefer to pair up rather than jump to the higher orbitals. But $\Delta_t$ is so small that it is almost always less than P. Because of this, the electrons would rather stay unpaired and spread out in the higher orbitals (high spin) than pair up. So tetrahedral complexes come out high spin, and low-spin tetrahedral complexes are hardly ever formed.

Answer: Because the tetrahedral splitting $\Delta_t$ (about $\tfrac{4}{9}\Delta_o$) is too small, it is almost always less than the pairing energy, so electrons stay unpaired (high spin) and low-spin tetrahedral complexes are rarely formed.