Among the species given below, the spin-only | vedclass

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(A) To find the spin-only magnetic moment, we calculate the number of unpaired electrons $(n)$ in each complex.$1$. $[Mn(CN)_6]^{3-}$: $Mn$ is in $+3$

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$[Mn(CN)_6]^{3-}$

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(A) To find the spin-only magnetic moment, we calculate the number of unpaired electrons $(n)$ in each complex.$1$. $[Mn(CN)_6]^{3-}$: $Mn$ is in $+3$ oxidation state $(3d^4)$. $CN^-$ is a strong field ligand, causing pairing. Configuration: $t_{2g}^4 e_g^0$. Unpaired electrons $n = 2$. Magnetic moment $\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83 \ BM$.$2$. $[Fe(CN)_6]^{3-}$: $Fe$ is in $+3$ oxidation state $(3d^5)$. $CN^-$ is a strong field ligand. Configuration: $t_{2g}^5 e_g^0$. Unpaired electrons $n = 1$. Magnetic moment $\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ BM$.$3$. $[Co(NH_3)_6]^{3+}$: $Co$ is in $+3$ oxidation state $(3d^6)$. $NH_3$ is a strong field ligand. Configuration: $t_{2g}^6 e_g^0$. Unpaired electrons $n = 0$. Magnetic moment $\mu = 0 \ BM$.$4$. $[Ti(H_2O)_6]^{3+}$: $Ti$ is in $+3$ oxidation state $(3d^1)$. $H_2O$ is a weak field ligand. Configuration: $t_{2g}^1 e_g^0$. Unpaired electrons $n = 1$. Magnetic moment $\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ BM$.Comparing the values, $[Mn(CN)_6]^{3-}$ has the highest number of unpaired electrons $(n=2)$, hence the highest magnetic moment.

Among the species given below, the spin-only magnetic moment is highest for
(Given: Atomic number of $Ti = 22, Mn = 25, Fe = 26$ and $Co = 27$)

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Among the species given below, the spin-only magnetic moment is highest for(Given: Atomic number of $Ti = 22, Mn = 25, Fe = 26$ and $Co = 27$)