The calculated spin only magnetic moments of [ Cr ( NH 3 ) 6 ] 3+ and [ CuF 6 ] 3- in BM , respectively, are (Atomic number of Cr and Cu are 24 and 29, respectively)

Chemistry · Coordination Compounds

JEE Advanced 2021 — Paper 1 — Question 14

The calculated spin only magnetic moments of $\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}$ and

$\left[\mathrm{CuF}_{6}\right]^{3-}$ in BM , respectively, are (Atomic number of Cr and Cu are 24 and 29, respectively)

$\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}$ has d ${ }^{3}$ configuration, so as per CFT, $\mathrm{N}=3$ and

$\mu=\sqrt{3(3+2)}=3.87 \mathrm{BM}$ $\left[\mathrm{CuF}_{6}\right]^{3-}$ , has d ${ }^{8}$

configuration and weak field ligand. So $\mathrm{N}=2$ and $\mu=\sqrt{2(2+2)}=2.84 \mathrm{BM}$

Answer key and solution verified before publishing.

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