Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

The geometry of \[{\left[ {Ag{{(N{H_3})}_2}} \right]^ + }\]and \[{\left[ {Cu{{\left( {N{H_3}} \right)}_4}} \right]^{2 + }}\] are respectively:

seo-qna
Last updated date: 24th Apr 2024
Total views: 264.3k
Views today: 3.64k
Answer
VerifiedVerified
264.3k+ views
Hint: We will first look at the hybridization of each molecule. We will first look at the hybridization of \[{\left[ {Cu{{\left( {N{H_3}} \right)}_4}} \right]^{2 + }}\]and then later the hybridization of \[{\left[ {Ag{{(N{H_3})}_2}} \right]^ + }\]. Using the hybridisations we can answer the geometry of both.

Complete answer:
Let's look at the electronic configuration of \[{\left[ {Cu{{\left( {N{H_3}} \right)}_4}} \right]^{2 + }}\]
\[_{29}Cu = 1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}3{d^{10}}4{s^1}\]
\[C{u^{2 + }} = \left[ {Ar} \right]3{d^9}4{s^0}4{p^0}\]
This will give us \[ds{p^2}\]hybridization as it has 1 empty d-orbital, 1 empty s-orbital and 2 p-orbitals.
Hence, geometry \[{\left[ {Cu{{\left( {N{H_3}} \right)}_4}} \right]^{2 + }}\]ion is square-planar and it is paramagnetic ion, as it has one unpaired electron.
Now about \[{\left[ {Ag{{(N{H_3})}_2}} \right]^ + }\]:
\[Ag = 4{d^9}4{s^2}\]
Hence, \[A{g^ + } = 4{d^{10}}\]
\[4{d^{10}}\]means no empty d-orbital, hence 2 \[N{H_3}\]molecules will occupy 1s-orbital and 1 p-orbital.
Hence its hybridization is \[sp\], which gives linear geometry.

ADDITIONAL INFORMATION :
The compound \[{\left[ {Cu{{\left( {N{H_3}} \right)}_4}} \right]^{2 + }}\]will be named as :
Name the ligand (ammine-\[N{H_3}\]) first, then add the prefix (tetra-4) to indicate the number of ligands. 2. Using the suffix ion, write the centre atom (copper-\[Cu\]) followed by its oxidation state (ii) in roman letters (since it is not a neutral entity).
Tetraamminecopper(ii)ion
We can also say that , because \[N{H_3}\] is a strong field ligand and is monodentate, it will only offer a single lone pair to \[Cu\]. However, because there are four \[N{H_3}\], they will each give four lone pairs to \[Cu\].
As a result, this compound's hybridization is dsp2.
It also has a square planar form.
And the compound[\[{\left[ {Ag{{(N{H_3})}_2}} \right]^ + }\]will be named as: Diamminesilver(I) hydroxide.

Note:
The compound \[{\left[ {Ag{{(N{H_3})}_2}} \right]^ + }\] It's also known as Tollen’s reagent.Tollen's reagent is a chemical reagent used to detect aldehydes, aromatic aldehydes, and alpha-hydroxy ketone functional groups.
Recently Updated Pages