Is Cd_2^+ diamagnetic or paramagnetic?

Mar 19, 2016

Interesting proposition... You may mean ${\text{Cd}}^{2 +}$ or an actual ${\text{Cd}}_{2}^{+}$... I'll do both.

$\text{Cd}$ has an atomic number of $48$, so it has an electron configuration of

$1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{6} 3 {d}^{10} 4 {s}^{2} 4 {p}^{6} \setminus m a t h b f \left(4 {d}^{10} 5 {s}^{2}\right)$.

Bolded are the valence electrons and their orbitals.

The valence atomic orbital energies are

• $5 s : \textcolor{g r e e n}{\text{-8.99 eV, or -867.4 kJ/mol}}$
• $4 d : \textcolor{g r e e n}{\text{-17.84 eV, or -1721.3 kJ/mol}}$.

Therefore, any ionizations removing the first two electrons will remove from the $5 s$ orbital without ambiguity. That means the electron configuration of ${\text{Cd}}^{2 +}$ is

$1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{6} 3 {d}^{10} 4 {s}^{2} 4 {p}^{6} \setminus m a t h b f \left(4 {d}^{10}\right)$.

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

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 ${\text{Cd}}_{2}$, including its $n = 4$ and $n = 5$ orbitals (except $5 p$).

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

$\textcolor{b l u e}{\left[K {K}_{\sigma}\right] \left[K {K}_{\pi}\right] {\left({\sigma}_{4 {d}_{{z}^{2}}}\right)}^{2} {\left({\pi}_{4 {d}_{x z}}\right)}^{2} {\left({\pi}_{4 {d}_{y z}}\right)}^{2} {\left({\delta}_{4 {d}_{{x}^{2} - {y}^{2}}}\right)}^{2} {\left({\delta}_{4 {d}_{x y}}\right)}^{2} {\left({\sigma}_{5 s}\right)}^{2} {\left({\delta}_{4 {d}_{x y}}^{\text{*")^2 (delta_(4d_(x^2-y^2))^"*")^2 (pi_(4d_(xz))^"*")^2 (pi_(4d_(yz))^"*")^2 (sigma_(4d_(z^2))^"*")^2 (sigma_(5s)^"*}}\right)}^{2}}$

where $K {K}_{\sigma}$ stands in for the core $\sigma$ interactions and $K {K}_{\pi}$ stands in for the core $\pi$ interactions. Since these are not valence, they are not as relevant to describe the reactivity of $\text{Cd}$.

All electrons are paired, making the neutral molecule ${\text{Cd}}_{2}$ diamagnetic. Hence, ${\text{Cd}}_{2}^{+}$ , with one less electron from a fully-occupied orbital, is paramagnetic.

CHALLENGE: Why does $C {d}_{2}$ only exist hypothetically? Also, if you were to singly ionize $C {d}_{2}$, 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 $\setminus m a t h b f \left(\delta\right)$ bond, which is when orbitals overlap via four lobes sidelong, rather than two lobes sidelong ($\pi$) or one lobe head-on ($\sigma$).

JUSTIFICATION OF ORBITAL ORDERING

In order of bond strength, $\setminus m a t h b f \left(\delta < \pi < \sigma\right)$ 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 $5 s$ atomic orbital is highest in energy to begin with, there is an extra $\text{8.85 eV}$ ($\text{853.89 kJ/mol}$) positive difference in energy with respect to the $4 d$ atomic orbitals, so I expect the ${\sigma}_{5 s}$ bonding MO to still be higher in energy than either the $\pi$ or $\delta$ MOs, rather than lower.