Question

A magnetic moment of \[1.73{\text{ }}BM\] will be shown by one among the following: A) \[{\left[ {Cu{{\left( {N{H_3}} \right)}_4}} \right]^{2 + }}\] B) \[{\left[ {Ni{{\left( {CN} \right)}_4}} \right]^{2 - }}\] C) $\left[ {TiC{l_4}} \right]$ D) \[{\left[ {CoC{l_6}} \right]^{4 - }}\]

Answer 1

Hint:Magnetic moment is \[\mu = {\text{ }}\sqrt n\left( {n + 2} \right)\] where n = no of unpaired electrons of the compound. Equate the magnetic moment to this formula, find the n, if there are a number of unpaired electrons matched with the n then that compound will have 1.73 magnetic moment. First we find the number of unpaired electrons in each option.

Complete step by step answer:
Any molecule has a well-defined magnitude of magnetic moment. Its magnetic moments due to its unpaired electron spins and the effect of the orbital magnetic moment is negligible due to a non-spherical environment. magnetic moment= \[\mu = \sqrt {n\left( {{\text{ }}n + 2} \right)} \] and its units are Bohr Magneton. (B.M.)
 Now first we calculate the value of (n):
\[\mu = \sqrt {n\left( {{\text{ }}n + 2} \right)} \]
\[\mu = 1.73\] (Given)
\[1.73{\text{ }} = \sqrt {n\left( {{\text{ }}n + 2} \right)} \]
\[n\left( {n + 2} \right) = {\left( {1.73} \right)^2}\]
\[{n^2}{\text{ }} + 2n{\text{ }} = {\text{ }}2.9929 \approx {\text{ }}3\]
\[{n^2} + {\text{ }}2n-{\text{ }}3 = {\text{ }}0\]
by solving this equation we get value of \[n\]
 \[\therefore n = - 3{\text{ }},{\text{ }}n = {\text{ }}1\]
the number of electrons cannot we negative so \[n = 1\]
-Now we find number of unpaired electrons in each option:
1) \[{\left[ {Cu\left( {NH3} \right)4} \right]^{2 + }}.\]the atomic number of copper is\[29\] . But charge on \[Cu = + 2\]
so total number of electrons in \[c{u^{2 + }} = 29 - 2 = 27\]
the electronic configuration of \[C{u^2}^ + \;ion\; = {\text{ }}{\left[ {Ar} \right]^{18}}3{d^9}4{s^0}\;\]
the number of unpaired electrons in \[c{u^{2 + }} = 1\]
2) \[{\left[ {Ni{{\left( {CN} \right)}_4}} \right]^{2 - }}\] the atomic number of Nickel is\[\;28\] . But charge on \[Ni = + 2\]
so total number of electrons in \[N{i^{2 + }} = 28 - 2 = 26\]
the electronic configuration of \[N{i^{2 + }} = {\left[ {Ar} \right]^{18}}3{d^8}\]
so the number of unpaired electrons in \[N{i^{2 + }} = {\text{ }}2\]
3) $\left[ {TiC{l_4}} \right]$ the atomic number of Titanium is \[22\] . But charge on \[Ti = + 4\]
so total number of electrons in \[T{i^{4 + }} = 22 - 4 = {\text{ }}18\]
the electronic configuration of \[N{i^{2 + }} = {\left[ {Ar} \right]^{18}}\] same as argon.
 so the number of unpaired electrons in \[T{i^{4 + }} = {\text{ }}0\]
4) \[{\left[ {CoC{l_6}} \right]^{4 - }}\] the atomic number of Cobalt= \[27\] . But charge on \[Co = + 2\]
so total number of electrons in \[C{o^{2 + }} = {\text{ }}27{\text{ }} - {\text{ }}2 = {\text{ }}25\]
the electronic configuration of \[C{o^{2 + }} = {\left[ {Ar} \right]^{{\text{18}}}}3{d^7}\]
the number of unpaired electrons in \[C{o^{2 + }} = {\text{ }}3\]
so the \[{\left[ {Cu\left( {NH3} \right)4} \right]^{2 + }}\]No. of unpaired electron is\[ = 1\], then it shows a magnetic moment of \[1.73{\text{ }}BM\]
So,option (A) is correct.

Note:A magnetic dipole is a physical thing whereas magnetic moment is a number which is used to quantify the strength of the dipole nature. So don’t confuse magnetic moment and magnetic dipole.