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Question: A magnetic moment of \[1.73{\text{ }}BM\] will be shown by one among the following: A) \[{\left[ {Cu...

A magnetic moment of 1.73 BM1.73{\text{ }}BM will be shown by one among the following: A) [Cu(NH3)4]2+{\left[ {Cu{{\left( {N{H_3}} \right)}_4}} \right]^{2 + }} B) [Ni(CN)4]2−{\left[ {Ni{{\left( {CN} \right)}_4}} \right]^{2 - }} C) [TiCl4]\left[ {TiC{l_4}} \right] D) [CoCl6]4−{\left[ {CoC{l_6}} \right]^{4 - }}

Explanation

Solution

Magnetic moment is μ= n(n+2)\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= μ=n( n+2)\mu = \sqrt {n\left( {{\text{ }}n + 2} \right)} and its units are Bohr Magneton. (B.M.)
Now first we calculate the value of (n):
μ=n( n+2)\mu = \sqrt {n\left( {{\text{ }}n + 2} \right)}
μ=1.73\mu = 1.73 (Given)
1.73 =n( n+2)1.73{\text{ }} = \sqrt {n\left( {{\text{ }}n + 2} \right)}
n(n+2)=(1.73)2n\left( {n + 2} \right) = {\left( {1.73} \right)^2}
n2 +2n = 2.9929≈ 3{n^2}{\text{ }} + 2n{\text{ }} = {\text{ }}2.9929 \approx {\text{ }}3
n2+ 2n− 3= 0{n^2} + {\text{ }}2n-{\text{ }}3 = {\text{ }}0
by solving this equation we get value of nn
∴n=−3 , n= 1\therefore n = - 3{\text{ }},{\text{ }}n = {\text{ }}1
the number of electrons cannot we negative so n=1n = 1
-Now we find number of unpaired electrons in each option:

  1. [Cu(NH3)4]2+.{\left[ {Cu\left( {NH3} \right)4} \right]^{2 + }}.the atomic number of copper is2929 . But charge on Cu=+2Cu = + 2
    so total number of electrons in cu2+=29−2=27c{u^{2 + }} = 29 - 2 = 27
    the electronic configuration of Cu2+  ion  = [Ar]183d94s0  C{u^2}^ + \;ion\; = {\text{ }}{\left[ {Ar} \right]^{18}}3{d^9}4{s^0}\;
    the number of unpaired electrons in cu2+=1c{u^{2 + }} = 1
  2. [Ni(CN)4]2−{\left[ {Ni{{\left( {CN} \right)}_4}} \right]^{2 - }} the atomic number of Nickel is  28\;28 . But charge on Ni=+2Ni = + 2
    so total number of electrons in Ni2+=28−2=26N{i^{2 + }} = 28 - 2 = 26
    the electronic configuration of Ni2+=[Ar]183d8N{i^{2 + }} = {\left[ {Ar} \right]^{18}}3{d^8}
    so the number of unpaired electrons in Ni2+= 2N{i^{2 + }} = {\text{ }}2
  3. [TiCl4]\left[ {TiC{l_4}} \right] the atomic number of Titanium is 2222 . But charge on Ti=+4Ti = + 4
    so total number of electrons in Ti4+=22−4= 18T{i^{4 + }} = 22 - 4 = {\text{ }}18
    the electronic configuration of Ni2+=[Ar]18N{i^{2 + }} = {\left[ {Ar} \right]^{18}} same as argon.
    so the number of unpaired electrons in Ti4+= 0T{i^{4 + }} = {\text{ }}0
  4. [CoCl6]4−{\left[ {CoC{l_6}} \right]^{4 - }} the atomic number of Cobalt= 2727 . But charge on Co=+2Co = + 2
    so total number of electrons in Co2+= 27 − 2= 25C{o^{2 + }} = {\text{ }}27{\text{ }} - {\text{ }}2 = {\text{ }}25
    the electronic configuration of Co2+=[Ar]183d7C{o^{2 + }} = {\left[ {Ar} \right]^{{\text{18}}}}3{d^7}
    the number of unpaired electrons in Co2+= 3C{o^{2 + }} = {\text{ }}3
    so the [Cu(NH3)4]2+{\left[ {Cu\left( {NH3} \right)4} \right]^{2 + }}No. of unpaired electron is=1 = 1, then it shows a magnetic moment of 1.73 BM1.73{\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.