Question

The calculated spin only magnetic moments of $\left[Cr\right(N{H}_3{)}_6{]}^{3+}$ and $[Cu{F}_6{]}^{3-}$ in $BM$, respectively, are
(Atomic number of $Cr$ and $Cu$ are $24$ and $29$, respectively)

• A. $3.87$ and $2.84$
• B. $4.90$ and $1.73$
• C. $3.87$ and $1.73$
• D. $4.90$ and $2.84$

Answer 1

The correct option is A $3.87$ and $2.84$

In $\left[Cr\right(N{H}_3{)}_6{]}^{3+}$ $Cr$ is in $+3$ oxidation state and thus has $d^3$ configuration, so number of unpaired electron in $Cr$ is $3$.
So as per $CFT$,
spin only magnetic momemnt $\mu =\sqrt{3(3+2)}=3.87BM$
In $[Cu{F}_6{]}^{3-}$, $Cu$ in in $+3$ oxidation state and has $d^8$ electronic configuration. Since $F^{-}$ is a weakfiled ligand, thenumber of unpaired electron in $Cu$ is $2$
So spin only magnetic moment,$\mu =\sqrt{2(2+2)}\approx 2.84BM$