Is Cd_2^+ diamagnetic or paramagnetic?
1 Answer
Interesting proposition... You may mean
CADMIUM(II) CATION
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
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
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 coresigma interactions andKK_pi stands in for the corepi 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
CHALLENGE: Why does
WHAT THE HECK IS A DELTA BOND?
With
JUSTIFICATION OF ORBITAL ORDERING
In order of bond strength,
Therefore, we expected the
However, notice that since the