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Coordination Compounds question

2022 · 25 Jul · Shift 1 · Q16
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Coordination Compounds question

2022 · 25 Jul · Shift 1 · Q16

JEE MainChemistryCoordination CompoundsNumerical+4 / −1
Consider the following metal complexes : [Co(NH3)6]3+[CoCl(NH3)5]2+[Co(CN)6]3−[Co(NH3)5(H2O)]3+\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}\left[\mathrm{CoCl}\left(\mathrm{NH}_{3}\right)_{5}\right]^{2+}\left[\mathrm{Co}(\mathrm{CN})_{6}\right]^{3-}\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5}\left(\mathrm{H}_{2} \mathrm{O}\right)\right]^{3+}[Co(NH3​)6​]3+[CoCl(NH3​)5​]2+[Co(CN)6​]3−[Co(NH3​)5​(H2​O)]3+ The spin-only magnetic moment value of the complex that absorbes light with shortest wavelength is ‾\underline{\hspace{2cm}}​ B. M. (Nearest integer)
Numerical answer
View written solutionFree

Correct answer: 0

  1. Find the oxidation state and d-electron count of Co in each complex

For cobalt, Z=27Z=27Z=27, so:

Co3+:3d6\text{Co}^{3+} : 3d^6Co3+:3d6

In all the given complexes, cobalt is in the +3+3+3 oxidation state.

Hence each complex is an octahedral d6d^6d6 system.

  1. Relate shortest wavelength to crystal field splitting

Absorption of light in coordination complexes generally corresponds to d→dd\to dd→d transition energy:

Δo=hcλ\Delta_o = \frac{hc}{\lambda}Δo​=λhc​

So, shortest wavelength means largest Δo\Delta_oΔo​.

Thus we must identify the complex with the strongest ligand field.

  1. Compare ligands using spectrochemical series

Relevant ligand strengths:

CN−>NH3>H2O>Cl−\mathrm{CN^-} > \mathrm{NH_3} > \mathrm{H_2O} > \mathrm{Cl^-}CN−>NH3​>H2​O>Cl−

Therefore, among the complexes,

  • [Co(CN)6]3−[\mathrm{Co}(\mathrm{CN})_6]^{3-}[Co(CN)6​]3− has the strongest field ligands.
  • So it has the largest Δo\Delta_oΔo​.
  • Hence it absorbs light of the shortest wavelength.
  1. Determine spin state of [Co(CN)6]3−[\mathrm{Co}(\mathrm{CN})_6]^{3-}[Co(CN)6​]3−

This is an octahedral d6d^6d6 complex with very strong-field ligand CN−\mathrm{CN^-}CN−, so it is low spin:

t2g6eg0t_{2g}^6 e_g^0t2g6​eg0​

Number of unpaired electrons, n=0n=0n=0.

  1. Calculate spin-only magnetic moment

Spin-only magnetic moment is:

μ=n(n+2) B.M.\mu = \sqrt{n(n+2)}\ \text{B.M.}μ=n(n+2)​ B.M.

With n=0n=0n=0:

μ=0(0+2)=0 B.M.\mu = \sqrt{0(0+2)}=0\ \text{B.M.}μ=0(0+2)​=0 B.M.

Nearest integer:

0\boxed{0}0​
  1. Comparison with stored answer

Stored correct answer = 000

Our derived answer = 000

So they agree.

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