Question

An ion which carries a $ + 3 $ charge possesses a magnetic moment of $ 4.9{\text{ BM}} $ and its last electron possesses orbital angular momentum of $ \sqrt 6 h $ . Identify that element if the highest value of n in the electronic configuration of that ion is $ 3 $ .
A. $ Mn $
B. $ Fe $
C. $ Cu $
D. $ Zn $

Answer 1

Hint :In the chemical compounds, magnetic properties arise from the spin and orbital angular momentum of the electrons present in the compound. The extent or magnitude of paramagnetism can be expressed in terms of the number of unpaired electrons which is known as the magnetic moment of the element.

Complete Step By Step Answer:
As per question, the given information is as follows:
Magnetic moment of the ion $ = 4.9\;{\text{BM}} $
Orbital angular momentum $ = \sqrt 6 h $
As we know, the magnetic moment of an ion is expressed in terms of number of unpaired electrons as follows:
 $ \mu = \sqrt {n(n + 2)} $
Where, $ \mu $ represents magnetic moment and $ n $ represents the number of electrons of an atom.
Substituting value of magnetic moment in the expression, then the number of unpaired electrons of the ion will be as follows:
 $ \Rightarrow \sqrt {n(n + 2)} = 4.9 $
 $ \Rightarrow n(n + 2) = 24.01 $
 $ \Rightarrow {n^2} + 2n - 24.01 = 0 $
On solving the above quadratic equation, the value of n will be as follows:
 $ n = \dfrac{{ - 2 + \sqrt {4 + 4 \times 24.01} }}{2} $
 $ \Rightarrow n = 4 $
Hence, the number of unpaired electrons in the given ion $ = 4 $
Now, the expression for orbital angular momentum is given as follows:
 $ L = \sqrt {l(l + 1)} h $
Where, $ L $ represents the orbital angular momentum and $ l $ represents the azimuthal quantum number.
Substituting given value of orbital angular momentum:
 $ \Rightarrow \sqrt {l(l + 1)} h = \sqrt 6 h $
As h is common in both sides so it will be cancelled out and on squaring both sides of the expression:
 $ \Rightarrow l(l + 1) = 6 $
 $ \Rightarrow {l^2} + l - 6 = 0 $
 $ \Rightarrow (l - 2)(l + 3) = 0 $
As the value of $ l $ cannot be a negative value. Hence, the value of azimuthal quantum number $ = 2 $ that means the unpaired electrons are present in the d-subshell of the metal ion.
As it is given in the question that the highest value of n in electronic configuration of that ion is $ 3 $ which means the metal element belongs to the 3d series of the d-block and the unpaired electrons are present in the 3d orbital of the element.
Now comparing the electronic configuration of given metal in the options to the condition given in the question as follows:
Manganese:
 $ M{n^{3 + }} = [Ar]3{d^4}4{s^0} $
Value of n $ = 3 $
Value of $ l = 2 $
Number of unpaired electrons $ = 4 $
Iron:
 $ F{e^{3 + }} = [Ar]3{d^5}4{s^0} $
Value of n $ = 3 $
Value of $ l = 2 $
Number of unpaired electrons $ = 5 $
Copper:
 $ C{u^{3 + }} = [Ar]3{d^8}4{s^0} $
Value of n $ = 3 $
Value of $ l = 2 $
Number of unpaired electrons $ = 2 $
Zinc:
 $ Z{n^{3 + }} = [Ar]3{d^9}4{s^0} $
Value of n $ = 3 $
Value of $ l = 2 $
Number of unpaired electrons $ = 1 $
Hence, the only metal which satisfies the observed conditions according to question is manganese (Mn).
Thus, option (A) i.e., $ Mn $ is the right answer.

Note :
Do not get confused between the n used in the expression of magnetic moment and n used for electronic configuration as n in magnetic moment represents the number of unpaired electrons in the ion whereas n in electronic configuration represents the value of principal quantum number for the ion.