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Coordination Compounds question

2014 · Shift 0 · Q18

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JEE Main
Chemistry
Coordination Compounds
2014 · Shift 0 · Q18

Coordination Compounds question

2014 · Shift 0 · Q18

JEE MainChemistryCoordination CompoundsMCQ+4 / −1

The octahedral complex of a metal ion

M^{3+}

with four monodentate ligands

L_1

,

L_2

,

L_3

and

L_4

absorb wavelengths in the region of red, green, yellow and blue, respectively. The increasing order of ligand strength of the four ligands is :

A
$L_3$ < $L_2$ < $L_4$ < $L_1$
B
$L_1$ < $L_2$ < $L_4$ < $L_3$
C
$L_4$ < $L_3$ < $L_2$ < $L_1$
D
$L_1$ < $L_3$ < $L_2$ < $L_4$

View written solutionFree

Correct answer: D

For an octahedral complex, the crystal field splitting energy $\Delta_o$ is related to the absorbed light by

\Delta_o = h\nu = \frac{hc}{\lambda}

So, smaller wavelength means higher energy absorbed, hence larger $\Delta_o$ .

Stronger ligands produce larger crystal field splitting. Therefore:

larger $\Delta_o$ $\Rightarrow$ stronger ligand
smaller $\Delta_o$ $\Rightarrow$ weaker ligand

Given absorptions:

$L_1$ absorbs red
$L_2$ absorbs green
$L_3$ absorbs yellow
$L_4$ absorbs blue

Now use the approximate wavelength order in visible light:

\lambda(\text{red}) > \lambda(\text{yellow}) > \lambda(\text{green}) > \lambda(\text{blue})

Hence energy order is opposite:

E(\text{red}) < E(\text{yellow}) < E(\text{green}) < E(\text{blue})

Therefore crystal field splitting order is:

\Delta_o(L_1) < \Delta_o(L_3) < \Delta_o(L_2) < \Delta_o(L_4)

Since ligand strength increases with $\Delta_o$ , the increasing order of ligand strength is

L_1 < L_3 < L_2 < L_4

Checking options:

A: incorrect
B: incorrect
C: incorrect
D: correct

Therefore, the correct answer is Option D.

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