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NCERT · Class 12 · Chemistry

Chapter 5: Coordination Compounds

Practice each question, then open the answer or worked solution.

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Exercise questions

32 questions
Q31Written answer

Amongst the following, the most stable complex is

(i) ${{\text{ }\!\![\!\!\text{ Fe(}{{\text{H}}_{\text{2}}}\text{O}{{\text{)}}_{\text{6}}}\text{ }\!\!]\!\!\text{ }}^{\text{3+}}}$ 

(ii) ${{\text{ }\!\![\!\!\text{ Fe(N}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{6}}}\text{ }\!\!]\!\!\text{ }}^{\text{3+}}}$ 

(iii) ${{\text{ }\!\![\!\!\text{ Fe(}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}{{\text{)}}_{\text{3}}}\text{ }\!\!]\!\!\text{ }}^{\text{3+}}}$ 

(iv) ${{\text{ }\!\![\!\!\text{ FeC}{{\text{l}}_{\text{6}}}\text{ }\!\!]\!\!\text{ }}^{\text{3-}}}$ 

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Worked solution

Ans: In the compounds, Fe is the central metal ion and its oxidation state +3. As ${{\text{C}}_{\text{2}}}\text{O}_{\text{4}}^{\text{2-}}$ is a didentate chelating ligand, it can form chelating ring and hence is the most stable complex.

Therefore, the answer is an option (iii).


Q32Written answer

What will be the correct order for the wavelengths of absorption in the visible region for the following: ${{\text{ }\!\![\!\!\text{ Ni(N}{{\text{O}}_{\text{2}}}{{\text{)}}_{\text{6}}}\text{ }\!\!]\!\!\text{ }}^{\text{4-}}}\text{,  }\!\![\!\!\text{ Ni(N}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{6}}}{{\text{ }\!\!]\!\!\text{ }}^{\text{2+}}}\text{,  }\!\![\!\!\text{ Ni(}{{\text{H}}_{\text{2}}}\text{O}{{\text{)}}_{\text{6}}}{{\text{ }\!\!]\!\!\text{ }}^{\text{2+}}}$ 

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Worked solution

Ans: In all three complexes, the core metal ion is identical. The absorption depends thus on the ligands in the visible area. In the spectrochemical series, the sequence in which the CFSE values of the ligands are increasing is:  ${{\text{H}}_{\text{2}}}\text{O}$ < $\text{N}{{\text{H}}_{\text{3}}}$ < $\text{N}{{\text{O}}_{\text{2}}}$.

Therefore, the amount of crystal-field splitting observed will be in the following order: 

$\text{ }{{\text{ }\!\!\Delta\!\!\text{ }}_{\text{0 (}{{\text{H}}_{2}}\text{O)}}}$ < $\text{ }{{\text{ }\!\!\Delta\!\!\text{ }}_{\text{0 (N}{{\text{H}}_{3}}\text{)}}}$  < $\text{ }{{\text{ }\!\!\Delta\!\!\text{ }}_{\text{0 (NO}_{2}^{-}\text{)}}}$ 

So, the wavelength of absorption in the visible region will be in the order: ${{\text{ }\!\![\!\!\text{ Ni(}{{\text{H}}_{\text{2}}}\text{O}{{\text{)}}_{\text{6}}}\text{ }\!\!]\!\!\text{ }}^{\text{2+}}}$ > ${{\text{ }\!\![\!\!\text{ Ni(N}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{6}}}\text{ }\!\!]\!\!\text{ }}^{\text{2+}}}$ > ${{\text{ }\!\![\!\!\text{ Ni(N}{{\text{O}}_{\text{2}}}{{\text{)}}_{\text{6}}}\text{ }\!\!]\!\!\text{ }}^{\text{2+}}}$.