Answer
Verified
377.1k+ views
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.
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.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred
What is the stopping potential when the metal with class 12 physics JEE_Main
The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main
How do you arrange NH4 + BF3 H2O C2H2 in increasing class 11 chemistry CBSE
Is H mCT and q mCT the same thing If so which is more class 11 chemistry CBSE
Trending doubts
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE
Select the correct plural noun from the given singular class 10 english CBSE
What organs are located on the left side of your body class 11 biology CBSE
The sum of three consecutive multiples of 11 is 363 class 7 maths CBSE
What is the z value for a 90 95 and 99 percent confidence class 11 maths CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
How many squares are there in a chess board A 1296 class 11 maths CBSE