Question

The d electron configurations of $C{r^{2 + }}{{ }},{{ }}M{n^{2 + }},{{ }}F{e^{2 + }}$ and $N{i^{ + 2}}$ are $3{d^4},{{ }}3{d^5},{{ }}3{d^6}$ and $3{d^8}$ respectively which one of the following aqua complexes will exhibit the minimum paramagnetic behaviour
A. ${\left[ {Fe{{\left( {{H_2}O} \right)}_6}} \right]^{ + 2}}$
B. ${\left[ {Ni{{\left( {{H_2}O} \right)}_6}} \right]^{ + 2}}$
C. ${\left[ {Cr{{\left( {{H_2}O} \right)}_6}} \right]^{ + 2}}$
D. ${\left[ {Mn{{\left( {{H_2}O} \right)}_6}} \right]^{ + 2}}$

Answer 1

Hint: The paramagnetic character is measured depending on the number of unpaired electrons present. Therefore, the more the number of unpaired electrons, the more is the paramagnetic character. The d orbital has $5$ orbitals therefore, all the compounds mentioned in the question contain unpaired electrons.

Formula used: Paramagnetic character, $P = \sqrt {n\left( {n + 2} \right)} BM$
Where $n$ is the number of unpaired electrons.

Complete step by step answer:
Paramagnetism can be defined as the character of a substance that is weakly attracted to a magnetic field. This is seen in substances that contain unpaired electrons. Therefore, it is required to find the minimum paramagnetic character in the metals given above.
In chromium, we see a dipositive ion. The electronic configuration is $1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^1}3{d^5}$ . therefore, when two electrons are removed, one electron is removed from the $4s$ orbital and one from $3d$ orbital. Therefore, the d orbital will have $4$ unpaired electrons. This is demonstrated below. We have taken only the final orbital into consideration.

$ \uparrow $

$ \uparrow $

$ \uparrow $

$ \uparrow $

$3{d^4}$
Now putting this information in the formula above we get, $n = 4$
 $P = \sqrt {4\left( {4 + 2} \right)} $
$ \Rightarrow \sqrt {4\left( {4 + 2} \right)} $
Adding the numbers in the bracket,
$ \Rightarrow \sqrt {4\left( 6 \right)} $
$ \Rightarrow \sqrt {24} $
therefore, the degree of paramagnetism is $\sqrt {24} BM$
If we take, manganese metal, we know that the number of electrons in the outermost orbital is $5{e^ - }$ that is total number of electrons in the d orbital is $5{e^ - }$ . the electronic configuration of the metal is $1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^2}3{d^5}$ Therefore, since the metal loses two electrons when it forms a dipositive ion, the electrons are removed from the s orbital. Therefore, the new electronic configuration is, $1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^0}3{d^5}$ .

$ \uparrow $

$ \uparrow $

$ \uparrow $

$ \uparrow $

$ \uparrow $

$n = 5$ $3{d^5}$
$P = \sqrt {5\left( {5 + 2} \right)} $
Adding the numbers in the bracket,
$ \Rightarrow \sqrt {5\left( 7 \right)} $
$ \Rightarrow \sqrt {35} $
therefore, the degree of paramagnetism is $\sqrt {35} BM$

Therefore, there are $5$ unpaired electrons.
In dipositive ferrous ion, the electronic configuration changes from $1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^2}3{d^6}$ to $1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^0}3{d^6}$ and therefore, there will be $4$ electrons.

$ \uparrow $

$ \uparrow $

$ \uparrow $

$ \uparrow $

$n = 4$ $3{d^4}$
$P = \sqrt {4\left( {4 + 2} \right)} $
$ \Rightarrow \sqrt {4\left( {4 + 2} \right)} $
Adding the numbers in the bracket,
$ \Rightarrow \sqrt {4\left( 6 \right)} $
$ \Rightarrow \sqrt {24} $
therefore, the degree of paramagnetism is $\sqrt {24} BM$ also.
For Nickel ion, there are also $2{e^ - }$ which are removed. Therefore, electronic configuration is changed from $1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^2}3{d^8}$ to $1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^0}3{d^8}$ and there will be only $3$ unpaired electrons. Since Nickel ion has the least number of unpaired electrons it will also have the minimum paramagnetic character.

$ \uparrow $$ \downarrow $

$ \uparrow $$ \downarrow $

$ \uparrow $$ \downarrow $

$ \uparrow $

$ \uparrow $

$n = 2$ $3{d^8}$
$P = \sqrt {2\left( {2 + 2} \right)} $
Adding the numbers in the bracket,
$ \Rightarrow \sqrt {2\left( 4 \right)} $
$ \Rightarrow \sqrt 8 $
therefore, the degree of paramagnetism is $\sqrt 8 BM$
Therefore, the answer to the question will be option b that is ${\left[ {Ni{{\left( {{H_2}O} \right)}_6}} \right]^{ + 2}}$ since it has the lowest value for degree of paramagnetism.

Note: Degree of paramagnetism is based on the number of electrons that are unpaired.
The more unpaired electrons that are present in a metal ,more is the paramagnetic character.
For d block elements the first electrons are usually taken first from the $4s$ orbitals and then from the $3d$ orbitals.