Question

The pair having the same magnetic moment is:
\[\begin{array}{*{20}{l}}
  {\left[ {At.{\text{ }}No.:{\text{ }}Cr{\text{ }} = {\text{ }}24,{\text{ }}Mn{\text{ }} = {\text{ }}25,{\text{ }}Fe{\text{ }} = {\text{ }}26,{\text{ }}Co{\text{ }} = {\text{ }}27} \right]\;} \\
  {A:{\text{ }}{{\left[ {Cr{{\left( {{H_2}O} \right)}_6}} \right]}^{2 + }}\;and{\text{ }}{{\left[ {Fe{{\left( {{H_2}O} \right)}_6}} \right]}^{2 + }}} \\
  {B:{\text{ }}{{\left[ {Mn{{\left( {{H_2}O} \right)}_6}} \right]}^{2 + }}\;and{\text{ }}{{\left[ {Cr{{\left( {{H_2}O} \right)}_6}} \right]}^{2 + }}} \\
  {C:{\text{ }}{{\left[ {CoC{l_4}} \right]}^{2 - }}\;and{\text{ }}{{\left[ {Fe{{\left( {{H_2}O} \right)}_6}} \right]}^{2 + }}} \\
  {D:{\text{ }}{{\left[ {Cr{{\left( {{H_2}O} \right)}_6}} \right]}^{2 + }}\;and{\text{ }}{{\left[ {CoC{l_4}} \right]}^{2 - }}}
\end{array}\]

Answer 1

Hint: Magnetic properties develop from the spin as well as the orbital angular momentum of electrons that are present in a compound. Molecular compounds which consist of one or more unpaired electrons are considered to be paramagnetic. The magnitude of para-magnetism is generally expressed as an effective magnetic moment.

Step by step answer:Recall the Pauli Exclusion Principle which states that no two electrons in the same atom can possess a similar set of four quantum numbers and Hund’s Rule of Maximum Multiplicity states that electron pairing in the orbitals which belong to the same subshell (i.e. p, d or f) does not take occur until every orbital linking to that subshell has obtained one electron each.
The magnetic moment can be calculated with the following formula:
$\mu = \sqrt {n(n + 2)} $
Here, n = number of unpaired electrons
The details of the electronic configuration and number of unpaired electrons in the complexes given in the question have been mentioned in the following table:

Complex ion	Electronic configuration of metal ion	Number of unpaired electrons (n)
\[{\left[ {Cr{{\left( {{H_2}O} \right)}_6}} \right]^{2 + }}\]	\[C{r^{2 + }},{\text{ }}\left[ {Ar} \right]{\text{ }}3{d^4}\]	4
\[{\left[ {Fe{{\left( {{H_2}O} \right)}_6}} \right]^{2 + }}\]	\[F{e^{2 + }},{\text{ }}\left[ {Ar} \right]{\text{ }}3{d^6}\]	4
\[{\left[ {Mn{{\left( {{H_2}O} \right)}_6}} \right]^{2 + }}\]	\[M{n^{2 + }},{\text{ }}\left[ {Ar} \right]{\text{ }}3{d^5}\]	5
\[{\left[ {CoC{l_4}} \right]^{2 - }}\]	\[C{o^{2 + }},{\text{ }}\left[ {Ar} \right]{\text{ }}3{d^7}\]	3

As clearly visible from the table, Cr2+ and Fe2+ have the same number of unpaired electrons (n = 4). Thus, magnetic moment of \[{\left[ {Cr{{\left( {{H_2}O} \right)}_6}} \right]^{2 + }}\;and{\text{ }}{\left[ {Fe{{\left( {{H_2}O} \right)}_6}} \right]^{2 + }}\]will be same as n is same in this case which can be calculated as follows:
$\mu = \sqrt {4(4 + 2)} = 4.89$

Hence, the correct answer is Option A.

Note: Magnetic moments are often employed in integration with the electronic spectra in order to obtain information regarding the oxidation state as well as stereochemistry of the central metal ion with respect to coordination complexes.