[14%] Consider the properties of the group 1 elements, whose valence shell electron configuration is — Atomic Structure and Periodicity Chemistry Question
Problem Context
[14%] Consider the properties of the group 1 elements, whose valence shell electron configuration is ns1, in the table below.
Rationalize the observed trend in first ionization energies with increasing n.
Model Answer
Electron energies are proportional to –Zeff^2/n^2. For the valence electrons, Zeff only increases slightly going down a column of the periodic table, so electrons with larger values of n have higher energies and thus are easier to ionize.
Suppose a hydrogen atom were excited to its 2s1 state. If that excited state atom were to transfer its electron to Cs+ to form a ground-state Cs atom, how much energy would that reaction absorb or release?
Model Answer
The 2s1 state of H has an energy = –1312 kJ mol^-1 + 984 kJ mol^-1 = –328 kJ mol^-1. This is higher in energy than the valence electron in Cs at –376 kJ mol^-1, so the electron transfer to Cs+ will release 48 kJ mol^-1 of energy.
All but one of the atoms listed in the table have an excited state that is significantly lower in energy than the (n+1)s1 state described in the table. Explain this observation, noting which atom is the exception and why.
Model Answer
In all cases except H, the ns electron can be promoted to an np orbital, which is significantly lower in energy than the (n+1)s orbital. There is no 1p orbital, so this is not possible for H.
All but one of the atoms listed in the table have an excited state that is modestly higher in energy (38 – 55 kJ mol-1) than the (n+1)s1 state described in the table. Explain this observation, noting which atom is the exception and why.
Model Answer
In all cases except H, the (n+1)p orbital is modestly higher in energy than the (n+1)s orbital, giving rise to an excited state of modestly higher energy. For H, there is a 2p orbital, but in one-electron atoms the energy depends only on n, not l, so its energy is the same as that of the 2s state.
The compounds MCl(s) show a smooth decrease in their molar densities, except that HCl(s) is less dense than expected from the trend and CsCl(s) is more dense than expected. Explain this periodic trend, and give reasons for the two exceptions to the trend.
Model Answer
For Li – Rb, the MCl(s) compounds form ionic lattices with the rock salt structure. As the cations increase in size going down the column, the lattices expand and hence the molar density decreases. HCl(s) is a molecular crystal, and the weaker intermolecular forces give relatively long intermolecular distances and hence a less dense packing. CsCl(s) adopts a different lattice than the rock salt lattice (it is commonly called the CsCl lattice), which allows it to pack more densely than the rock salt lattice with a cation the size of cesium would.
137Cs (136.9070895 amu) undergoes radioactive decay to give a stable product whose atomic mass is 136.9058274 amu. What type of radioactive decay is this, and what is the identity of the decay product?
Model Answer
β– decay forms 137Ba.
Calculate the energy, in kJ mol-1, released by the radioactive decay of 137Cs.
Model Answer
E = ∆mc^2. ∆m = –1.26 × 10^-3 amu × (1.661 × 10^-27 kg amu^-1) = –2.09 × 10^-30 kg
E = (2.09 × 10^-30 kg)(2.998 × 10^8 m s^-1)^2 = 1.88 × 10^-13 J = 1.88 × 10^-16 kJ
(1.88 × 10^-16 kJ)(6.022 × 10^23 mol^-1) = 1.13 × 10^8 kJ mol^-1