Question

When white ppt. of $\text{A}{{\text{g}}_{2}}\text{C}{{\text{O}}_{3}}$ is dissolved in $\text{N}{{\text{H}}_{3}}$ solution, then compound X is formed.
Find the value of the magnetic moment of X.
A. 0
B. 2
C. 1
D. 3

Answer 1

Hint: Write the equation of the reaction between $\text{A}{{\text{g}}_{\text{2}}}\text{C}{{\text{O}}_{3}}$ with aqueous $\text{N}{{\text{H}}_{3}}$. The compound formed will be X. The reaction is $\text{A}{{\text{g}}_{2}}\text{C}{{\text{O}}_{3}}+\text{N}{{\text{H}}_{3}}\to \text{X}$. Find the charge on the central metal atom and write its configuration and then, find the magnetic moment using the formula $\sqrt{\text{n}\left( \text{n}+\text{2} \right)}$ B.M.

Complete answer:
A complex is formed when silver carbonate reacts with ammonia solution is diammine silver $\left( \text{I} \right)$ and carbonate ions. The reaction involved is $\text{A}{{\text{g}}_{2}}\text{C}{{\text{O}}_{3}}\left( \text{s} \right)+\text{N}{{\text{H}}_{3}}\left( \text{aq}\text{.} \right)\to 2{{\left[ \text{Ag}{{\left( \text{N}{{\text{H}}_{3}} \right)}_{2}} \right]}^{+}}+\text{CO}_{3}^{2-}$.
The compound X formed is ${{\left[ \text{Ag}{{\left( \text{N}{{\text{H}}_{3}} \right)}_{2}} \right]}^{+}}$. This compound contains silver as its central metal atom. The charge on silver is +1 (as the overall charge on the compound is +1 and charge on ammonia ligands is zero because it is neutral ligand, thus charge on $\text{Ag}$ is 1 or $\text{A}{{\text{g}}^{+}}$). We are checking the central metal atom because the magnetic moment of the complex will be decided by checking the magnetic moment of the central metal atom.
The configuration of silver on its ground state is $\left[ \text{Kr} \right]4{{\text{d}}^{10}}5{{\text{s}}^{1}}$ with atomic number 47. The configuration of $\text{A}{{\text{g}}^{+}}$ will be $\left[ \text{Kr} \right]4{{\text{d}}^{10}}5{{\text{s}}^{0}}$. The electron will be removed from the 5s subshell which is the outermost shell.
The magnetic moment of $\text{A}{{\text{g}}^{+}}$ will be found by using the formula $\sqrt{\text{n}\left( \text{n}+\text{2} \right)}$ B.M. where n is the unpaired electrons. The configuration of ${{\text{d}}^{10}}$ subshell is $\begin{matrix}
   \uparrow \downarrow & \uparrow \downarrow & \uparrow \downarrow & \uparrow \downarrow & \uparrow \downarrow \\
\end{matrix}$. All the electrons of this configuration are paired. So, the number of unpaired electrons will be zero. The magnetic moment will be $\sqrt{0\left( 0+2 \right)}=\sqrt{0}=0$ is zero.

The magnetic moment of compound X will be 0, which is option ‘a’.

Note:
The electrons will be removed from the outermost shell only, otherwise the configuration will be wrong, hence, magnetic moment. The formula of magnetic moment is $\sqrt{\text{n}\left( \text{n}+\text{2} \right)}$ where n are unpaired electrons not the number of electrons.