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Answer the following.
Answer the following: is paramagnetic while is diamagnetic though both are tetrahedral. Explain the…
Answer the following:
(a)[1.6666666666666667]
$[\text{NiCl}_4]^{2-}$ 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)[1.6666666666666667]
Write the IUPAC name of complex compound $[\text{PtBr}_2(\text{en})_2]\text{Cl}_2$.
(c)[1.6666666666666667]
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.
Answer
Answer (a)
AIIn $[\mathrm{NiCl_4}]^{2-}$, 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.
Answer (b)
AI| IUPAC name | dibromidobis(ethane-1,2-diamine)platinum(IV) chloride |
Pt is $+4$: $x - 2 + 0 = +2$; ligands in alphabetical order (bromido, ethane-1,2-diamine).
From ISC 2026 Improvement Chemistry Paper 1, question 21(ii).
Check your working with the Coordination complex analyser: hybridisation, shape, magnetic moment, CFSE, isomers and the IUPAC name of a complex.