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Question: Assign the charge on the following ions: \( a){[Hg{\left( {CN} \right)_4}]^x} \) \( b){[Co{\l...

Assign the charge on the following ions:
a)[Hg(CN)4]xa){[Hg{\left( {CN} \right)_4}]^x}
b)[Co(NH3)2Cl4]xb){[Co{\left( {N{H_3}} \right)_2}C{l_4}]^x}
c)[Fe(CN)6]xc){[Fe{\left( {CN} \right)_6}]^x}

Explanation

Solution

Hint : The charge on the coordination complex is always equal to the sum of the charges on the central metal and the ligands present around the central metal. The oxidation number of the atom is the charge on its ion.

Complete Step By Step Answer:
As we have seen in order to find the charge on the coordination complex we have to add the charges of central metal and all the ligands present.
a)[Hg(CN)4]xa){[Hg{\left( {CN} \right)_4}]^x}
In this complex, HgHg is present in +2+ 2 oxidation state and the cyanide (CN)\left( {CN} \right) ligands have a −1- 1 charge. Thus, in order to find the overall charge on the complex we’ll add them.
x=2+4(−1)x = 2 + 4\left( { - 1} \right)
x=−2x = - 2
Thus, the charge on the coordination complex is −2- 2
The complex becomes: [Hg(CN)4]2−{[Hg{\left( {CN} \right)_4}]^{2 - }}
b)[Co(NH3)2Cl4]xb){[Co{\left( {N{H_3}} \right)_2}C{l_4}]^x}
In this complex, the central metal ion is cobalt (Co)\left( {Co} \right) in +2+ 2 oxidation state , ammonia has no charge while chloride ions have −1- 1 . Thus we’ll add them:
x=2+(2×0)+4(−1)x = 2 + \left( {2 \times 0} \right) + 4\left( { - 1} \right)
x=−2x = - 2
Thus, the charge on the complex is −2- 2
The complex becomes: [Co(NH3)2Cl4]2−{[Co{\left( {N{H_3}} \right)_2}C{l_4}]^{2 - }}
c)[Fe(CN)6]xc){[Fe{\left( {CN} \right)_6}]^x}
In this complex, metal FeFe is present in +2+ 2 oxidation state, and the ligands (cyanide ions) are having a - 1 charge. Thus, we’ll add the charges of all the constituents ;
x=2+6(−1)x = 2 + 6\left( { - 1} \right)
x=−4x = - 4
Thus, the charge on the complex is −4- 4
The complex becomes: [Fe(CN)6]4−{[Fe{\left( {CN} \right)_6}]^{4 - }}
In this case, the central metal FeFe can also have +3+ 3 oxidation state. The overall charge on the complex will become −3- 3 . Thus, [Fe(CN)6]3−{[Fe{\left( {CN} \right)_6}]^{3 - }} complex can also be made.

Note :
In such types of questions, we just have to add all the charges. The important thing is that one should know the oxidation states of all the atoms in order to find the total charge on the coordination complex.