JEE Advanced 2021 Paper 1 Q04 Chemistry Coordination Chemistry Crystal Field Theory Medium

JEE Advanced 2021 Paper 1 · Q04 · Crystal Field Theory

The calculated spin only magnetic moments of [Cr(NH$_3$)$_6$]$^{3+}$ and [CuF$_6$]$^{3-}$ in BM, respectively, are:

(Atomic numbers of Cr and Cu are 24 and 29, respectively.)

  1. A. 3.87 and 2.84
  2. B. 4.90 and 1.73
  3. C. 3.87 and 1.73
  4. D. 4.90 and 2.84
Reveal answer + step-by-step solution

Correct answer:A

Solution

Cr is atomic number 24, so Cr$^{3+}$ has electron configuration [Ar] 3d$^3$. In [Cr(NH$_3$)$_6$]$^{3+}$ the d$^3$ ion has three unpaired electrons (t$_{2g}^3$) irrespective of field strength. $\mu_{\text{spin}} = \sqrt{n(n+2)} = \sqrt{3(5)} = \sqrt{15} \approx 3.87$ BM.

Cu is atomic number 29, so Cu$^{3+}$ has configuration [Ar] 3d$^8$. F$^-$ is a weak-field ligand, so in [CuF$_6$]$^{3-}$ the d$^8$ ion adopts the high-spin arrangement t$_{2g}^6 e_g^2$ with two unpaired electrons. $\mu_{\text{spin}} = \sqrt{2(4)} = \sqrt{8} \approx 2.84$ BM.

The pair (3.87, 2.84) corresponds to option (A).

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