Answer the following: [NiCl₄]²⁻ is paramagnetic while [Ni(CO)₄] is diamagnetic though both are…
Answer the following:
(a)[1.5]
$[\text{NiCl}_4]^{2-}$ is paramagnetic while $[\text{Ni}(\text{CO})_4]$ is diamagnetic though both are tetrahedral. Explain the statement by referring to the Valence Bond Theory.
(Atomic number of $\text{Ni} = 28$)
(b)[1.5]
Write the IUPAC name of complex compound $[\text{PtBr}_2(\text{en})_2]\text{Cl}_2$.
(c)[2]
Draw a diagram to show the splitting of d orbital in an octahedral crystal field for a complex having $d^6$ configuration in strong ligand field.
In $[\mathrm{NiCl_4}]^{2-}$, Ni is $+2$ ($3d^8$). $\mathrm{Cl^-}$ is a weak-field ligand, so the 3d electrons do not pair; two unpaired electrons remain and one $4s$ and three $4p$ orbitals hybridise to give $sp^3$ (tetrahedral): paramagnetic.
In $[\mathrm{Ni(CO)_4}]$, Ni is $0$ ($3d^8 4s^2$). CO is a strong-field ligand, so the $4s$ electrons shift to 3d, giving $3d^{10}$ with all electrons paired; $sp^3$ hybridisation (tetrahedral): diamagnetic.
Pt is $+4$: $x - 2 + 0 = +2$; ligands in alphabetical order (bromido, ethane-1,2-diamine).
Answer (c)
Octahedral crystal field splitting for $d^6$ in a strong ligand field (low spin): the five d orbitals split into a lower $t_{2g}$ set ($d_{xy}, d_{yz}, d_{zx}$) and a higher $e_g$ set ($d_{x^2-y^2}, d_{z^2}$) separated by $\Delta_o$; all six electrons pair up in $t_{2g}$ ($t_{2g}^6 e_g^0$), zero unpaired electrons (diamagnetic); CFSE $= -2.4\Delta_o$ (plus pairing energy).