write the complete ground state electronic configuration for Cd3+ . state wheather it is paramagnetic or diamagnetic. if there are unpaired electron how many indicates.
Answers
Answer:
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#).