Question

A magnetic moment of 1.73BM will be shown by which 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. $TiC{l_4}$
D. ${\left[ {CoC{l_6}} \right]^{4 - }}$

Answer 1

Hint: Magnetic moment is $\mu = \sqrt {n\left( {n + 2} \right)} $, where n is the no. of unpaired electrons of the compound. Equate the magnetic moment to this formula and find the value of n. And then find the no. of unpaired electrons present in each of the given options. If their no. of unpaired electrons matches with n then that compound will have 1.73BM magnetic moment.

Complete step by step answer:
$\mu = \sqrt {n\left( {n + 2} \right)} \\
\mu = 1.73 \\
\Rightarrow 1.73 = \sqrt {n\left( {n + 2} \right)} \\
\Rightarrow n\left( {n + 2} \right) = {1.73^2} \\
\Rightarrow {n^2} + 2n = 2.9929 \approx 3 \\
\Rightarrow {n^2} + 2n - 3 = 0 \\
\Rightarrow {n^2} + 3n - n - 3 = 0 \\
\Rightarrow n\left( {n + 3} \right) - 1\left( {n + 3} \right) = 0 \\
\Rightarrow \left( {n + 3} \right)\left( {n - 1} \right) = 0 \\
\therefore n = - 3, n = 1 \\ $
The no. of unpaired electrons cannot be negative. Therefore, the value of n is 1.
(i) ${\left[ {Cu{{\left( {N{H_3}} \right)}_4}} \right]^{2 + }}$
The atomic number of Copper (Cu) is 29. But the charge present on Cu here is +2 as ammonia does not have any charge (neutral). Copper loses 2 electrons.
So the total number of electrons present in $C{u^{2 + }}$ is 29 - 2 = 27.
The electronic configuration of $C{u^{2 + }}$ can be written as $1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^9} = \left[ {Argon} \right]3{d^9}$
The no. of unpaired electrons in $C{u^{2 + }}$ is 1.
(ii) ${\left[ {Ni{{\left( {CN} \right)}_4}} \right]^{2 - }}$
The atomic number of Nickel (Ni) is 28. But the charge present on Ni here is +2 as the charge present on cyanide is -1. Nickel loses 2 electrons.
So the total number of electrons present in $N{i^{2 + }}$ is 28 - 2 = 26.
The electronic configuration of $N{i^{2 + }}$ can be written as $1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^8} = \left[ {Argon} \right]3{d^8}$
Therefore, no. of unpaired electrons in $N{i^{2 + }}$ is 2.
(iii) $TiC{l_4}$
The atomic number of Titanium (Ti) is 22. But the charge present on Ti here is +4 as the charge present on chloride ion is -1. Titanium loses 4 electrons.
So the total number of electrons present in $T{i^{4 + }}$ is 22 - 4 = 18.
The electronic configuration of $T{i^{4 + }}$ can be written as $1{s^2}2{s^2}2{p^6}3{s^2}3{p^6} = \left[ {Argon} \right]$
The electronic configuration of $T{i^{4 + }}$ is the same as Argon. Argon is a stable element and $T{i^{4 + }}$ is also stable. So the no. of unpaired electrons is 0.
(iv) ${\left[ {CoC{l_6}} \right]^{4 - }}$
The atomic number of Cobalt (Co) is 27. But the charge present on Co here is +2 as the charge present on Chloride is -1. Cobalt loses 2 electrons.
So the total number of electrons present in $C{o^{2 + }}$ is 27 - 2 = 25.
The electronic configuration of $C{o^{2 + }}$ can be written as $1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^7} = \left[ {Argon} \right]3{d^7}$
Therefore, no. of unpaired electrons in $C{o^{2 + }}$ is 3.
Therefore, ${\left[ {Cu{{\left( {N{H_3}} \right)}_4}} \right]^{2 + }}$ shows a magnetic moment of 1.73BM.

So the correct answer is Option A.

Note: A magnetic dipole is a physical thing whereas a magnetic moment is a number which is used to quantify the strength of the thing’s dipole nature. Do not confuse between magnetic moment and magnetic dipole.