Question

Which one of the following is expected to be a paramagnetic complex
A.\[{{[Ni{{({{H}_{2}}O)}_{6}}]}^{2+}}\]
B.\[~[Ni{{(CO)}_{4}}]\]
C.\[~{{[Zn{{(N{{H}_{3}})}_{4}}]}^{2+}}\]
D.\[{{[Co{{(N{{H}_{3}})}_{6}}]}^{+3}}\]

Answer 1

Hint: Weak ligands favor the development of paramagnetic compounds because paramagnetic complexes are high spin complexes, meaning they have at least one unpaired electron. The one in the d orbital with three electrons will be paramagnetic.

Complete answer:According to the valence bond theory, orbital intermixing takes place. The ligand's filled orbital and the unoccupied orbitals of the central metal atom mix orbitals. Spectrochemical series are used to assess the potency of ligands. This series was established through experimentation. The strength of the ligand controls whether or not the electrons are paid. In contrast to weak ligands like iodide, bromide, fluoride, and chloride, strong field ligands like cyanide, carbonyl, ammonia, and EDTA are able to pair up the electrons.

For the first option which is Option A, we can see that the oxidation state of Ni is +2 in this case. If we try to write the electronic configuration of $N{{i}^{2+}}$then it will be $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}4{{s}^{0}}3{{d}^{8}}$and we can see there are 6 electrons in the d subshell which means there are 2 unpaired electrons and since water is a weak ligand it won’t force pairing up of electron.

In the second option that is Option B, the oxidation state of Ni is 0 so the electronic configuration of Ni will be$1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}4{{s}^{2}}3{{d}^{8}}$. Here, we will have 2 unpaired electrons but since carbonyl is a strong field ligand so it will force the pairing of the unpaired electrons and there will be no unpaired electrons.

In the third option that is option C zinc has an oxidation state of +2 and when we see the electronic configuration of$Z{{n}^{+2}}$, it can be written as \[1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}4{{s}^{0}}3{{d}^{10}}\]and we can see there are no unpaired electrons.

In the last option that is option D, cobalt has the oxidation state of +3 so the electronic configuration of $C{{o}^{+3}}$ will be \[1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}4{{s}^{0}}3{{d}^{6}}\]which means there are 4 unpaired electrons. In general $N{{H}_{3}}$is a weak ligand but with metals in a +3 oxidation state, it acts as a strong ligand. Therefore, it will force the pairing of electrons and there will be zero unpaired electrons.
Hence, the correct option is A.\[{{[Ni{{({{H}_{2}}O)}_{6}}]}^{2+}}\]

Note: The term "diamagnetic complexes" refers to complexes that do not include any unpaired electrons. Due to the absence of unpaired electrons, diamagnetic complexes lack the permanent magnet moment, are not strongly attracted by magnetic fields, and are instead weakly repelled by them. In the absence of the magnetic field, they have no impact on them because they don't contain any unpaired electrons.