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Coordination Compounds question

2025 · 23 Jan · Shift 1 · Q6
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  5. /2025 · 23 Jan · Shift 1 · Q6

Coordination Compounds question

2025 · 23 Jan · Shift 1 · Q6

JEE MainChemistryCoordination CompoundsMCQ+4 / −1
CrCl3⋅xNH3\mathrm{CrCl}_3 \cdot \mathrm{xNH}_3CrCl3​⋅xNH3​ can exist as a complex. 0.1 molal aqueous solution of this complex shows a depression in freezing point of 0.558∘C0.558^{\circ} \mathrm{C}0.558∘C. Assuming 100%100 \%100% ionisation of this complex and coordination number of Cr is 6 , the complex will be (Given Kf=1.86 K kg mol−1\mathrm{K}_{\mathrm{f}}=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}Kf​=1.86 K kg mol−1 )
  1. A
    [Cr(NH3)5Cl]Cl2\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}\right] \mathrm{Cl}_2[Cr(NH3​)5​Cl]Cl2​
  2. B
    [Cr(NH3)4Cl2]Cl\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right] \mathrm{Cl}[Cr(NH3​)4​Cl2​]Cl
  3. C
    [Cr(NH3)6]Cl3\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_6\right] \mathrm{Cl}_3[Cr(NH3​)6​]Cl3​
  4. D
    [Cr(NH3)3Cl3]\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_3 \mathrm{Cl}_3\right][Cr(NH3​)3​Cl3​]
View written solutionFree

Correct answer: A

  1. Use freezing point depression formula

For an electrolyte,

ΔTf=iKfm\Delta T_f = i K_f mΔTf​=iKf​m

where:

  • ΔTf=0.558∘C\Delta T_f = 0.558^\circ \mathrm{C}ΔTf​=0.558∘C
  • Kf=1.86 K kg mol−1K_f = 1.86\, \mathrm{K\,kg\,mol^{-1}}Kf​=1.86Kkgmol−1
  • m=0.1m = 0.1m=0.1
  • iii = van’t Hoff factor

So,

i=ΔTfKfm=0.5581.86×0.1i = \frac{\Delta T_f}{K_f m} = \frac{0.558}{1.86 \times 0.1}i=Kf​mΔTf​​=1.86×0.10.558​ i=0.5580.186=3i = \frac{0.558}{0.186} = 3i=0.1860.558​=3

Thus, the complex produces 3 ions in aqueous solution.

  1. Check each option by ionisation

Since ionisation is assumed to be 100%100\%100%, the number of particles formed equals the number of ions obtained on dissociation.

Option A

[Cr(NH3)5Cl]Cl2\left[\mathrm{Cr}(\mathrm{NH}_3)_5\mathrm{Cl}\right]\mathrm{Cl}_2[Cr(NH3​)5​Cl]Cl2​

Ionises as:

[Cr(NH3)5Cl]Cl2→[Cr(NH3)5Cl]2++2Cl−\left[\mathrm{Cr}(\mathrm{NH}_3)_5\mathrm{Cl}\right]\mathrm{Cl}_2 \rightarrow \left[\mathrm{Cr}(\mathrm{NH}_3)_5\mathrm{Cl}\right]^{2+} + 2\mathrm{Cl}^-[Cr(NH3​)5​Cl]Cl2​→[Cr(NH3​)5​Cl]2++2Cl−

Total ions =3= 3=3

Also, coordination number of Cr =6= 6=6 because inside bracket there are 5 NH3\mathrm{NH}_3NH3​ and 1 Cl\mathrm{Cl}Cl.

So this fits.

Option B

[Cr(NH3)4Cl2]Cl\left[\mathrm{Cr}(\mathrm{NH}_3)_4\mathrm{Cl}_2\right]\mathrm{Cl}[Cr(NH3​)4​Cl2​]Cl

Ionises as:

[Cr(NH3)4Cl2]Cl→[Cr(NH3)4Cl2]++Cl−\left[\mathrm{Cr}(\mathrm{NH}_3)_4\mathrm{Cl}_2\right]\mathrm{Cl} \rightarrow \left[\mathrm{Cr}(\mathrm{NH}_3)_4\mathrm{Cl}_2\right]^+ + \mathrm{Cl}^-[Cr(NH3​)4​Cl2​]Cl→[Cr(NH3​)4​Cl2​]++Cl−

Total ions =2= 2=2

Does not fit.

Option C

[Cr(NH3)6]Cl3\left[\mathrm{Cr}(\mathrm{NH}_3)_6\right]\mathrm{Cl}_3[Cr(NH3​)6​]Cl3​

Ionises as:

[Cr(NH3)6]Cl3→[Cr(NH3)6]3++3Cl−\left[\mathrm{Cr}(\mathrm{NH}_3)_6\right]\mathrm{Cl}_3 \rightarrow \left[\mathrm{Cr}(\mathrm{NH}_3)_6\right]^{3+} + 3\mathrm{Cl}^-[Cr(NH3​)6​]Cl3​→[Cr(NH3​)6​]3++3Cl−

Total ions =4= 4=4

Does not fit.

Option D

[Cr(NH3)3Cl3]\left[\mathrm{Cr}(\mathrm{NH}_3)_3\mathrm{Cl}_3\right][Cr(NH3​)3​Cl3​]

This is a neutral complex, so it does not ionise into separate ions in solution.

Total ions =1= 1=1 particle only.

Does not fit.

  1. Select the correct option

Only Option A gives i=3i = 3i=3, matching the observed depression in freezing point.

Therefore, the complex is:

[Cr(NH3)5Cl]Cl2\boxed{\left[\mathrm{Cr}(\mathrm{NH}_3)_5\mathrm{Cl}\right]\mathrm{Cl}_2}[Cr(NH3​)5​Cl]Cl2​​
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