Correct increasing order for the wavelengths of absorption in the visible region for the complexes of ${Co}^{3 +}$ is

Difficult

A

${[Co {(H_{2} O)}_{6}]}^{3 +}, {[Co (en)_{3}]}^{3 +}, {[Co {({NH}_{3})}_{6}]}^{3 +}$
B

${[Co {(H_{2} O)}_{6}]}^{3 +}, {[Co {({NH}_{3})}_{6}]}^{3 +}, {[Co (en)_{3}]}^{3 +}$
C

${[Co {({NH}_{3})}_{6}]}^{3 +}, {[Co (en)_{3}]}^{3 +}, {[Co {(H_{2} O)}_{6}]}^{3 +}$
D

${[Co (en)_{3}]}^{3 +}, {[Co {({NH}_{3})}_{6}]}^{3 +}, {[Co {(H_{2} O)}_{6}]}^{3 +}$

Solution

Increasing order of crystal field splitting energy is: $H_{2} O < {NH}_{3} < en$
Thus, increasing order of energy for the given complexes is:
${[Co {(H_{2} O)}_{6}]}^{3 +} < {[Co {({NH}_{3})}_{6}]}^{3 +} < {[Co (en)_{3}]}^{3 +}$
$As, E = \frac{hc}{λ}$
Thus, increasing order of wavelength of absorption is:
${[Co (en)_{3}]}^{3 +} < {[Co {({NH}_{3})}_{6}]}^{3 +} < {[Co {(H_{2} O)}_{6}]}^{3 +}$

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