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Question: In the complex \({{[Ni{{({{H}_{2}}O)}_{2}}{{(N{{H}_{3}})}_{4}}]}^{2+}}\) the magnetic moment if Ni i...

In the complex [Ni(H2O)2(NH3)4]2+{{[Ni{{({{H}_{2}}O)}_{2}}{{(N{{H}_{3}})}_{4}}]}^{2+}} the magnetic moment if Ni is:
A. Zero
B. 2.83 BM
C. 1.73 BM
D. 3.87 BM

Explanation

Solution

First find out the number of unpaired electrons that are there in the coordination compound, then use the formula Magnetic moment(μ)=n(n+2)B.MMagnetic\,moment(\mu )=\sqrt{n(n+2)}B.M.

Complete answer:
In order to answer the question, we need to learn about magnetic moments of coordination compounds. The magnetic moment of coordination compounds can be measured by the formula:
Magnetic moment(μ)=n(n+2)B.MMagnetic\,moment(\mu )=\sqrt{n(n+2)}B.M
where n is number of unpaired electrons in central metal atom/ion and B.M. (Bohr Magneton) is a unit of magnetic moment.
There are some complications in the study of magnetic behaviour of coordination compounds of the first transition series. For example, metal ions with upto three electrons in d orbitals like Ti3+(d1),V3+(d2)T{{i}^{3+}}({{d}^{1}}),{{V}^{3+}}({{d}^{2}})and Cr3+(d3)C{{r}^{3+}}({{d}^{3}}), two vacant orbitals are available for d2sp3{{d}^{2}}s{{p}^{3}} hybridisation. The magnetic behaviour of these free ions and their coordination entities are similar.
When more than three 3d electrons are present, the required pair of 3d orbitals for octahedral hybridization is not directly available. Thus, for d4,d5,d6{{d}^{4}},{{d}^{5}},{{d}^{6}}cases, a vacant pair of d-orbitals result only by pairing of 3d electrons which leaves two, one and zero unpaired electrons, respectively.
There are complications with species containing d and d configuration. For example, [Mn(CN)6]3−{{[Mn{{(CN)}_{6}}]}^{3-}} has magnetic moment of two unpaired electrons while [MnCl6]3−{{[MnC{{l}_{6}}]}^{3-}} has that of four unpaired electrons. [CoF6]3−{{[Co{{F}_{6}}]}^{3-}}is paramagnetic while [Co(C2O4)3]3−{{[Co{{({{C}_{2}}{{O}_{4}})}_{3}}]}^{3-}}is diamagnetic.
In the compound [Ni(H2O)2(NH3)4]2+{{[Ni{{({{H}_{2}}O)}_{2}}{{(N{{H}_{3}})}_{4}}]}^{2+}}the Ni is in +2 oxidation state and there consist 8 electrons in 3d subshell. In NH3N{{H}_{3}} the two unpaired electrons left gets paired up, as it is a strong field ligand and hence the number of unpaired electrons becomes 0. hence the magnetic moment n(n+2)\sqrt{n(n+2)} where n is the number of unpaired electrons comes out to be 0. Hence the correct option is A.

Note:
Magnetic moment depends on the total no of valence electrons present. A compound possessing more magnetic moment has the ability to behave like a magnet( attract or repel magnetic substances).