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Question: What is the formula of the coordination complexes when a \[{{N}}{{{i}}^{2 + }}\] ion is bound to two...

What is the formula of the coordination complexes when a Ni2+{{N}}{{{i}}^{2 + }} ion is bound to two water molecules and two bidentate oxalate ions?
A. [Ni(H20)2(C204)2]2−{{ }}{\left[ {{{Ni}}{{\left( {{{{H}}_{{2}}}{{0}}} \right)}_{{2}}}{{\left( {{{{C}}_{{2}}}{{{0}}_{{4}}}} \right)}_{{2}}}} \right]^{{{2 - }}}}
B. [Ni(H20)3(C204)2]2−{{ }}{\left[ {{{Ni}}{{\left( {{{{H}}_{{2}}}{{0}}} \right)}_3}{{\left( {{{{C}}_{{2}}}{{{0}}_{{4}}}} \right)}_{{2}}}} \right]^{{{2 - }}}}
C. [Ni(H20)2(C204)2]2+{{ }}{\left[ {{{Ni}}{{\left( {{{{H}}_{{2}}}{{0}}} \right)}_2}{{\left( {{{{C}}_{{2}}}{{{0}}_{{4}}}} \right)}_{{2}}}} \right]^{2 + }}
D. [Ni(H20)2(C204)4]2+{{ }}{\left[ {{{Ni}}{{\left( {{{{H}}_{{2}}}{{0}}} \right)}_2}{{\left( {{{{C}}_{{2}}}{{{0}}_{{4}}}} \right)}_4}} \right]^{2 + }}

Explanation

Solution

Ligands are ions or molecules that are directly attached to the metal atom. Here four ligands are present, two water and two oxalate ligands. Water has one donor atom and is a neutral ligand and oxalate has two donor atoms and is an anionic ligand. Formula can be found by adding the charge of the ligands and metal, this should be equal to the total charge of the coordination sphere.

Complete step by step answer:
Coordination compounds are compounds that consist of anions or neutral molecules that are bound to a central atom by means of coordinate covalent bonds. Coordination compounds are known as coordination complexes. The molecules or ions that are bound to the central metal atom are referred to as ligands. The coordination number of the central atom in the coordination compound is the total number of sigma bonds through which the ligands are bound to the central metal atom. Coordination compounds can be cationic or anionic depending upon the charge of the Coordination sphere.
Based on the number of donor atom, ligands are classified as monodentate, only one donor atom is present, bidentate,
two donor atoms are present and polydentate, more than two donor atoms are present
Here the given complex contains two ligands, water, and oxalate. Water is a neutral ligand whereas oxalate, C2O42−{{{C}}_{{2}}}{{O}}_{{4}}^{{{2 - }}} is
an anionic ligand with two donor atoms.
There are two water molecules and two oxalate ions as per the question.
Oxidation number of nickel is 2+{2^ + }, charge on oxalate is 2−{2^ - } and water being neutral is zero.
when two oxalates are bound the charge of the ligand becomes 4−{4^ - }. Adding the oxidation numbers,
{2^ + }$$$$ + $$$${4^ - } = 2− = \,{2^ - }
That is the complex must be anionic with charge 2−{2^ - }
The answer is option A [Ni(H20)2(C204)2]2−{{ }}{\left[ {{{Ni}}{{\left( {{{{H}}_{{2}}}{{0}}} \right)}_{{2}}}{{\left( {{{{C}}_{{2}}}{{{0}}_{{4}}}} \right)}_{{2}}}} \right]^{{{2 - }}}}

So, the correct answer is Option A.

Note: Denticity of a ligand is the number of ligating groups with which ligand is bound. The central atom together with the ligand enclosed in a square bracket is called the coordination sphere. The special arrangement with which ligands are bound to the central metal atom is coordination polyhedral.