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Question: An ion \({{M}^{+2}}\) forms the complexes as \({{[M{{({{H}_{2}}O)}_{6}}]}^{2+}}\), \({{[M{{(en)}_{3}...

An ion M+2{{M}^{+2}} forms the complexes as [M(H2O)6]2+{{[M{{({{H}_{2}}O)}_{6}}]}^{2+}}, [M(en)3]2+{{[M{{(en)}_{3}}]}^{2+}} and [MBr6]4−{{[MB{{r}_{6}}]}^{4-}}.
Match the complex with the appropriate colour respectively.
A. Green, blue and red
B. Blue, red and green
C. Green, red and blue
D. Red, blue and green

Explanation

Solution

To find out the appropriate answer of this question, figure out which is the stronger ligand among the complexes. Remember the wavelength increases from left to right in VIBGYOR series present in the visible spectrum.

Complete answer:
First of all, let us know about crystal splitting energy.
The d-orbitals of the central metal ion in an octahedral complex generally divide into two sets of different energies in presence of a ligand. And this splitting of the energies is called the crystal field splitting energy. We should know that, more is the splitting energy of a complex, it will require less wavelength to occupy the lower set of orbitals and will reflect a colour having higher wavelength value. In short, the splitting energy is directly proportional to energy and is inversely proportional to the wavelength.
So, here the complexes given are [M(H2O)6]2+{{[M{{({{H}_{2}}O)}_{6}}]}^{2+}}, [M(en)3]2+{{[M{{(en)}_{3}}]}^{2+}} and [MBr6]4−{{[MB{{r}_{6}}]}^{4-}} . All the complexes are having ligands attached to the metal ion M+2{{M}^{+2}}.
Among the given complexes, the strength of the ligands is bromine is less than the water which is further less than the enen ligand. Remember, more the ligand is a strong field ligand, more will be the value of the splitting energy and lower will be its wavelength. And the wavelength increases down the VIBGYOR series. So, here as the emitted colours are red, green and blue; the absorbed colours will be green, red and blue respectively.
Thus, the complexes [M(H2O)6]2+{{[M{{({{H}_{2}}O)}_{6}}]}^{2+}}, [M(en)3]2+{{[M{{(en)}_{3}}]}^{2+}} and [MBr6]4−{{[MB{{r}_{6}}]}^{4-}} are blue, red and green in colour respectively.

Hence, the correct option is B.

Note:
It is important to note that the reason for splitting in these complexes is due to the electrostatic interactions that occur between the electrons of the ligand and the lobes of the d-orbitals. Crystal splitting energy is helpful in spectrometry, etc.