Question

Which of the following homoleptic complex has a square planer geometry:
1.${{\text{[Ag(}}{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}}{{\text{)}}_{\text{2}}}{\text{]}}^{{\text{ - 3}}}}$
2. ${{\text{[Co(EDTA)]}}^{{\text{ - 2}}}}$
3. ${{\text{[Pt(trien)]}}^{{\text{ + 2}}}}$
4. ${[{\text{Ni(}}{{\text{H}}_2}{\text{O}}{{\text{)}}_4}{\text{]}}^{ + 2}}$

Answer 1

Hint: Geometry or structure of a complex mostly depends on the number of bonded electrons and the number of lone pairs present in the complex. This is done by following the valence bond theory. A square planar geometry has 4 bonded pairs and zero lone pairs. Homoleptic complexes have identical ligands bonded to the central metal atom/ion.

Complete Step by step answer: As mentioned in question all complexes have identical ligands or are homoleptic. Now we need to know the electronic configuration of each central metal atom to find out ${\text{ds}}{{\text{p}}^2}$ hybridization for any metal ion, as according to Valence bond theory the hybridization of a square planer complex is ${\text{ds}}{{\text{p}}^2}$ Let’s now identify the hybridization of each central metal.

i. Ag in ${{\text{[Ag(}}{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}}{{\text{)}}_{\text{2}}}{\text{]}}^{{\text{ - 3}}}}$ has the oxidation state +1 and the electronic configuration;
$^{47}{\text{A}}{{\text{g}}^ + } \to [{\text{Kr}}]{\text{4}}{{\text{d}}^{10}}5{{\text{s}}^0}5{{\text{p}}^0}$, therefore the hybridization is sp, and the geometry is linear.

ii. Co in ${{\text{[Co(EDTA)]}}^{{\text{ - 2}}}}$ has the oxidation state +2 and the electronic configuration;
$^{27}{\text{C}}{{\text{o}}^{ + 2}} \to {\text{[Ar]3}}{{\text{d}}^7}{\text{4}}{{\text{s}}^0}4{{\text{p}}^0}4{{\text{d}}^0}$, therefore the hybridization is ${\text{s}}{{\text{p}}^3}{{\text{d}}^2}$, and the geometry is octahedral.

iii. Pt in ${{\text{[Pt(trien)]}}^{{\text{ + 2}}}}$ has the oxidation number +2, and the electronic configuration;
$^{78}{\text{P}}{{\text{t}}^{ + 2}} \to {\text{[xe]4}}{{\text{f}}^{14}}{\text{5}}{{\text{d}}^8}6{{\text{s}}^0}6{{\text{p}}^0}$, therefore the hybridization is ${\text{s}}{{\text{p}}^3}$ , and the geometry is tetrahedral
iv. Ni in ${[{\text{Ni(}}{{\text{H}}_2}{\text{O}}{{\text{)}}_4}{\text{]}}^{ + 2}}$ has the oxidation number +2 and the electronic configuration;
$^{28}{\text{N}}{{\text{i}}^{ + 2}} \to {\text{[Ar]3}}{{\text{d}}^8}{\text{4}}{{\text{s}}^0}{\text{4}}{{\text{p}}^0}$, therefore the hybridization is ${\text{ds}}{{\text{p}}^2}$ , and the geometry is square planar.

Hence, the correct answer is option 4 i.e., ${[{\text{Ni(}}{{\text{H}}_2}{\text{O}}{{\text{)}}_4}{\text{]}}^{ + 2}}$

Note: Coordination of the Pt and Ni is equal in ${[{\text{Ni(}}{{\text{H}}_2}{\text{O}}{{\text{)}}_4}{\text{]}}^{ + 2}}$ and ${{\text{[Pt(trien)]}}^{{\text{ + 2}}}}$ that is 4 but there is two geometry that corresponds to coordination number 4, one in tetrahedral and the other is square planar, hence knowing the hybridization becomes important before concluding the geometry based on coordination number.