The spin-only magnetic moment value of M^{n+} | vedclass

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(A) $1$. Identify the number of diamagnetic complexes $(n)$: $K_2[NiCl_4]$: $Ni^{2+}$ is $d^8$, $sp^3$ hybridization, paramagnetic. $[Zn(H_2O)_6]Cl_2$

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(A) $1$. Identify the number of diamagnetic complexes $(n)$: $K_2[NiCl_4]$: $Ni^{2+}$ is $d^8$, $sp^3$ hybridization, paramagnetic. $[Zn(H_2O)_6]Cl_2$: $Zn^{2+}$ is $d^{10}$, $sp^3d^2$ hybridization, diamagnetic. $K_3[Mn(CN)_6]$: $Mn^{3+}$ is $d^4$, $d^2sp^3$ hybridization, paramagnetic. $[Cu(PPh_3)_3I]$: $Cu^+$ is $d^{10}$, $sp^3$ hybridization, diamagnetic. Thus, $n = 2$. $2$. Identify the metal with the least enthalpy of atomisation: Among $Ni$, $Zn$, $Mn$, and $Cu$, $Zn$ has the least enthalpy of atomisation due to the absence of unpaired $d$-electrons for metallic bonding. $3$. Calculate the spin-only magnetic moment of $Zn^{2+}$: $Zn^{2+}$ has the electronic configuration $[Ar] 3d^{10}$. Since there are no unpaired electrons, the number of unpaired electrons $(x)$ is $0$. $\mu = \sqrt{x(x+2)} = \sqrt{0(0+2)} = 0 \ BM$.

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