Is Cd_2^+ diamagnetic or paramagnetic?

1 Answer
Mar 19, 2016

Interesting proposition... You may mean "Cd"^(2+) or an actual "Cd"_2^(+)... I'll do both.


CADMIUM(II) CATION

"Cd" has an atomic number of 48, so it has an electron configuration of

1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4s^2 4p^6 \mathbf(4d^10 5s^2).

Bolded are the valence electrons and their orbitals.

The valence atomic orbital energies are

  • 5s: color(green)("-8.99 eV, or -867.4 kJ/mol")
  • 4d: color(green)("-17.84 eV, or -1721.3 kJ/mol").

Therefore, any ionizations removing the first two electrons will remove from the 5s orbital without ambiguity. That means the electron configuration of "Cd"^(2+) is

1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4s^2 4p^6 \mathbf(4d^10).

There are no singly-occupied orbitals. Therefore, this cationic transition metal is diamagnetic.


CADMIUM DIATOMIC MOLECULE

In the off chance you mean a hypothetical, gas-phase diatomic cation... here is a molecular orbital diagram I constructed for the homonuclear diatomic molecule "Cd"_2, including its n = 4 and n = 5 orbitals (except 5p).

Overall, the condensed electron configuration of the neutral molecule would likely be:

color(blue)([KK_sigma][KK_pi] (sigma_(4d_(z^2)))^2 (pi_(4d_(xz)))^2 (pi_(4d_(yz)))^2 (delta_(4d_(x^2-y^2)))^2 (delta_(4d_(xy)))^2 (sigma_(5s))^2 (delta_(4d_(xy))^"*")^2 (delta_(4d_(x^2-y^2))^"*")^2 (pi_(4d_(xz))^"*")^2 (pi_(4d_(yz))^"*")^2 (sigma_(4d_(z^2))^"*")^2 (sigma_(5s)^"*")^2)

where KK_sigma stands in for the core sigma interactions and KK_pi stands in for the core pi interactions. Since these are not valence, they are not as relevant to describe the reactivity of "Cd".

All electrons are paired, making the neutral molecule "Cd"_2 diamagnetic. Hence, "Cd"_2^(+) , with one less electron from a fully-occupied orbital, is paramagnetic.

CHALLENGE: Why does Cd_2 only exist hypothetically? Also, if you were to singly ionize Cd_2, which orbital would you boot electrons out of first? Will that make the molecule more, or less stable? Why?

WHAT THE HECK IS A DELTA BOND?

With d orbitals, if you noticed the notation on the MO diagram, we introduced a \mathbf(delta) bond, which is when orbitals overlap via four lobes sidelong, rather than two lobes sidelong (pi) or one lobe head-on (sigma).

Inorganic Chemistry, Miessler et al., pg. 2

JUSTIFICATION OF ORBITAL ORDERING

In order of bond strength, \mathbf(delta < pi < sigma) bonds due to the extent of orbital overlap in each. (In general, the more lobes you have that overlap, the less the overlap between each lobe is.)

Therefore, we expected the delta bonding MOs to be less stabilized and therefore higher in energy than the pi bonding MOs, which are higher in energy than the sigma bonding MOs. The opposite order is expected for the antibonding MOs.

However, notice that since the 5s atomic orbital is highest in energy to begin with, there is an extra "8.85 eV" ("853.89 kJ/mol") positive difference in energy with respect to the 4d atomic orbitals, so I expect the sigma_(5s) bonding MO to still be higher in energy than either the pi or delta MOs, rather than lower.