Question

If the magnetic moment of ${{\left[ Ag{{\left( CN \right)}_{2}} \right]}^{-1}}$ is zero, then the number of unpaired electrons will be:
A) 1
B) 2
C) 3
D) Zero

Answer 1

Hint: Magnetic moment is a way to express the paramagnetism of a coordination compound in terms of Bohr magneton. As it is the property of paramagnetism, the central metal ion must have some unpaired electrons in its valence shell to get the value. If the value of the magnetic moment is zero, it depicts the diamagnetic nature of the complex.

Formula Used:$\mu =\sqrt{n\left( n+2 \right)}\ \ \text{BM}$, where \[\mu \] is the magnetic moment, n is the number of unpaired electrons and BM (Bohr magneton) is the unit of the magnetic moment.

Complete answer:Magnetic moment is a term used in coordination chemistry to get information about the complex related to oxidation number and stereochemistry of the central metal ion. It is also used to determine the paramagnetism of a coordination complex and can be calculated if the number of unpaired electrons is known.

As per the data given in the question, the Magnetic moment of the given complex ${{\left[ Ag{{\left( CN \right)}_{2}} \right]}^{-1}}$ is zero, so according to the given formula of magnetic moment, the number of unpaired electrons can be calculated as follows:
$\mu =\sqrt{n\left( n+2 \right)}\ \ \text{BM}$
$\Rightarrow 0=\sqrt{n\left( n+2 \right)}$
Squaring both sides of the equation:
$\Rightarrow n\left( n+2 \right)=0$
As per the above equation, the possible values of n are as follows:
$\Rightarrow n=0\ ;\ n=-2$
As the number of unpaired electrons cannot be negative. Therefore, the number of unpaired electrons for the given complex is zero.

Option ‘D’ is correctoption (D) is correct.

Note: We can calculate the number of unpaired electrons using an alternative way i.e., by calculating the oxidation state of the silver ion and then, applying the valence bond theory to get the actual number of unpaired electrons as shown below:
Electronic configuration of $Ag=\ \left[ Kr \right]4{{d}^{10}}5{{s}^{1}}$
The oxidation state of silver ion: $+1$
Electronic configuration of $A{{g}^{+}}=\ \left[ Kr \right]4{{d}^{10}}5{{s}^{0}}$
So, all the electrons in the valence d orbitals are paired. Thus, the complex consists of zero unpaired electrons.