Question

Match List I with List II and select the correct answer using code given below the lists:

List I with Metal Ion		List II Magnetic Moment(BM)
A	$C{r^{3 + }}$	$1$	$\sqrt {35} $
B	$F{e^{2 + }}$	$2$	$\sqrt {30} $
C	$N{i^{2 + }}$	$3$	$\sqrt {24} $
D	$M{n^{2 + }}$	$4$	$\sqrt {15} $
		$5$	$\sqrt 8 $

a. A-1, B-3,C-5,D-4
b. A-2,B-3,C-5,D-1
c. A-4,B-3.C-5,D-1
d. A-4,B-5,C-3,D-1

Answer 1

Hint:The magnetic moment of a metal gives information about the oxidation number of the central metal ion in coordination compounds. It is represented by:
$\mu = \sqrt {n(n + 2)} $
Where n is the number of unpaired electrons of that metal. The unpaired electrons are calculated

Complete answer:
Simple crystal field theory states that during the electronic configuration the unpaired electrons in the d orbital does not degenerate themselves rather are split such that two orbitals, the \[{d_{xz - yz}}\] and ${d_{z - z}}$ are at higher energy than the \[{d_{xz}},{d_{xy}},{d_{yz}}\] orbitals.
The Magnetic moment of a metal is the magnetic strength and orientation of a magnet or other object that produces a magnetic field. The magnetic moment is due to the orbital motion caused by unpaired electrons in the outermost shell. Magnetic moment of a metal is
$\mu = \sqrt {n(n + 2)} $ in which ‘n ‘ is the number of unpaired electrons which is calculated by the number of electrons in the outermost shell during electronic configuration.

In the given question:
For $C{r^{3 + }}$the atomic number is 24 for which electronic configuration is $[Ar]3{d^5}4{s^2}$ and the number of unpaired electrons is 3. To calculate the BM
\[\mu = \sqrt {3(3 + 2)} \]
$\mu = 15BM$
Similarly, for $F{e^{2 + }}$ the atomic number is 26 $[Ar]3{d^6}4{s^1}$ and the number of unpaired electrons is 4
$\mu = \sqrt {4(4 + 2)} $
$\mu = 24BM$
For $N{i^{2 + }}$ the atomic number is 28 and the electronic configuration is $[Ar]3{d^8}$ the unpaired electron is 2
$\mu = \sqrt {2(2 + 2)} $
$\mu = 8BM$
For $M{n^{2 + }}$, the atomic number is 25 and the electronic configuration is $[Ar]3{d^5}$ and the unpaired electron left is 5.
 $\mu = \sqrt {5(5 + 2)} $
$\mu = 35BM$
Therefore the magnetic moment or the BM for the $C{r^{3 + }},F{e^{2 + }},N{i^{2 + }},M{n^{2 + }}$ is $15BM,24BM,8BM,35BM$.

Therefore the correct option is (c).

Note:
The magnetic moment is different in the compounds forming octahedral or tetrahedral structure. In the octahedral compounds the energy of the ‘d’ orbitals are no longer degenerate but are split into two in which their energy remains higher whereas in the tetrahedral compounds they split in such a way that the energy of two ‘d’ orbitals remains lower.