Question

Which of the following ions have the highest magnetic moment?
(A) $ M{n^{3 + }} $
(B) $ Z{n^{2 + }} $
(C) $ S{c^{3 + }} $
(D) $ T{i^{3 + }} $

Answer 1

Hint : The elements from group $ 13 $ to group $ 18 $ are known as p-block elements or main-group elements. Group $ 18 $ elements are known as noble gases as they are very stable and inert in nature. D-block elements are known to form coordination complexes. Magnetic moment of an ion depends upon the number of unpaired electrons present.

Complete Step By Step Answer:
Magnetic moment of an ion is defined in terms of torque that the ion experiences in the presence of a magnetic field. The spin magnetic moment is a magnetic moment caused due to the spin of particles. The SI unit of Magnetic moment is Bohr magneton (B.M.) which is equivalent to $ 9.27 \times {10^{ - 24}}A{m^2} $ .
Magnetic moment of an ion depends upon the number of unpaired electrons present. Magnet moment is directly proportional to the number of unpaired electrons in an ion.
 Formula for calculating the magnetic moment of an ion is given as:-
 $ MagneticMoment,\mu = \sqrt {n \times (n + 2)} $
Where, n is the number of unpaired electrons.
Magnesium, Mn has electronic configuration: $ 1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^5}4{s^2} $
Electronic configuration of $ M{n^{3 + }} = 1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^4} $
Here, $ M{n^{3 + }} $ have 4 unpaired electrons.
Magnetic moment for $ M{n^{3 + }} = \sqrt {4 \times (4 + 2)} $
 $ = \sqrt {24} $
 $ = 4.89B.M. $
Zinc, Zn has electronic configuration: $ 1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^{10}}4{s^2} $
Electronic configuration of $ Z{n^{2 + }} = 1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^9} $
Hence, there is 1 unpaired electron.
Magnetic for $ Z{n^{2 + }} = \sqrt {1 \times (1 + 2)} $
 $ = \sqrt 3 $
 $ = 1.73B.M. $
Scandium, Sc has electronic configuration: $ 1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^1}4{s^2} $
Electronic configuration of $ S{c^{3 + }} = 1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^0} $
Hence, there are 0 unpaired electrons.
Magnetic for $ S{c^{3 + }} = \sqrt {0 \times (0 + 2)} $
 $ = 0B.M. $
Titanium, Ti has electronic configuration: $ 1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^2}4{s^2} $
Electronic configuration of $ T{i^{3 + }} = 1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^1} $
Hence, there is 1 unpaired electron.
Magnetic for $ T{i^{3 + }} = \sqrt {1 \times (1 + 2)} $
 $ = \sqrt 3 $
 $ = 1.73B.M. $
Hence, from the above calculations of dipole moments, we can say that, $ M{n^{3 + }} $ has the highest dipole moment among the given ions.

Note :
Spin only magnetic moment is always directly proportional to the number of unpaired electrons. While writing the electronic configurations of ions, the electrons must be removed from the outermost valence orbital first which is the fourth shell $ 4s $ and then from the inner orbitals $ 3d $ . Hence, this should be taken care of by writing electronic configuration for any given ion.