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Question: The type of magnetism exhibited by \({{\left[ Mn{{\left( {{H}_{2}}O \right)}_{6}} \right]}^{2+}}\) i...

The type of magnetism exhibited by [Mn(H2O)6]2+{{\left[ Mn{{\left( {{H}_{2}}O \right)}_{6}} \right]}^{2+}} ion is:
A. paramagnetism
B. diamagnetism
C. both a and b
D. none of these

Explanation

Solution

To find the type of magnetism, we need the number of unpaired electrons of the central atom of the compound. So, always ensure that we are counting the unpaired electrons of the given atom or ion only while finding it and not all the valence electrons.

Complete step by step solution:
-There can enter 2 electrons in opposite orientation in 1 orbital. Individual electron spins are added to get a total spin and individual angular momenta are added to give total angular momentum.
-Based on magnetic behaviour, atoms are classified as paramagnetic and diamagnetic. Diamagnetic atoms or ions have all paired electrons in their shell while paramagnetic atoms or ions have 1 or more unpaired electrons.
-The given compound is a coordination compound. The magnetic properties of such compounds depend on the central atom of the compounds. The central atom here in this compound is Mn. The ligand present here is water which is neutral and so the charge of the ion is the charge of the central atom itself.
-The atomic number of Mn is 25. Now we see that 2 electrons have been removed in Mn atom to form a cation with charge +2. We know that the electrons are removed from the outermost shell. So we need to see the electronic configuration of Mn atom first which is
1s22s22p63s23p64s23d51{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}4{{s}^{2}}3{{d}^{5}}
The outermost shell is 4s. So, when 2 electrons are removed, the configuration will become 1s22s22p63s23p63d51{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}3{{d}^{5}} .
-We know that the d subshell has 5 orbitals and electrons are filled first as unpaired and then are paired. So 5 electrons will occupy 5 different orbitals of the subshell. It can be shown as

So we can see that there are 5 unpaired electrons.
-Since there are unpaired electrons present in the central atom, we can say that the given ion is paramagnetic. Not only it is paramagnetic but it has the maximum value of magnetic moment which is found by the formula
μ=n(n+2)\mu =\sqrt{n\left( n+2 \right)}

Therefore the correct option is A.

Note: The charge of the ion is not always the charge of the central atom. It is only in the case of the neutral ligands. For other compounds, we find the charge on the central atom by the concept that the sum of all the charges of the atoms on the ion is equal to the charge present on that compound.