Question

The hybridisation of Ag in the linear complex \[{\left[ {{\rm{Ag(N}}{{\rm{H}}_{\rm{3}}}{)_2}} \right]^ + }\] is:
A) \[ds{p^2}\]
B) \[sp\]
C) \[s{p^2}\]
D) \[s{p^3}\]

Answer 1

Hint: To find out the hybridization of the Ag atom in the given complex, first we have to calculate the oxidation state of the silver (Ag) atom. Then, we have to find its hybridization using the valence bond theory.

Complete Step by Step Answer:
Let’s first calculate the hybridization of silver atom of \[{\left[ {{\rm{Ag(N}}{{\rm{H}}_{\rm{3}}}{)_2}} \right]^ + }\]. We take x as the oxidation state of silver. As we know, the ammonia molecule has zero oxidation state. So, we can write,
\[x + 0 = 1\]
\[x = 1\]
So, the silver atom of the given complex has an oxidation state of +1.

We know that, the electronic configuration of a silver atom is as follows:
\[{\left[ {{\rm{Ar}}} \right]^{18}}4{s^2}3{d^{10}}4{p^6}5{s^1}4{d^{10}}\]

Now, the configuration of the \[{\rm{A}}{{\rm{g}}^{\rm{ + }}}\] ion is,
\[{\left[ {{\rm{Ar}}} \right]^{18}}4{s^2}3{d^{10}}4{p^6}4{d^{10}}5{s^0}\]
So, the orbital for the silver ion can be written as follows:

Image: Formation of the complex \[{\left[ {{\rm{Ag(N}}{{\rm{H}}_{\rm{3}}}{)_2}} \right]^ + }\]
As the 4d orbitals are fully paired, the molecule of ammonia occupies the sp orbitals. Therefore, the hybridization of the given compound is \[sp\] .
Hence, option B is right.

Note: According to the valence bond theory, under the influence of a ligand, the metal ion or atom can utilise (-1)d, ns, np or ns, np and nd orbitals for the process of hybridization to give a set of orbitals of specific geometry such tetrahedral, octahedral, square planar etc. The overlapping of these hybrid orbitals with the ligand orbitals donates pairs of electrons for the formation of bonds. The VBT can predict the geometry of a complex considering the magnetic behaviour of the complex.