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Assuming \([Ni(Cl)_4]^{2-}\) as complex \(A\) and \([Ni(CO)_4]\) as complex \(B\) choose the correct statement. |
State of hybridization of \(Ni\) in complex \(A\) is \(dsp^2\) and in complex \(B\) is \(sp^3\) Complex \(A\) is paramagnetic and complex \(B\) is diamagnetic Oxidation state of \(Ni\) in both complex \(+2\) \(CO\) is a weaker ligand than \(Cl^-\) |
Complex \(A\) is paramagnetic and complex \(B\) is diamagnetic |
The correct answer is option 2. Complex \(A\) is paramagnetic and complex \(B\) is diamagnetic. \([Ni(Cl)_4]^{2-}\): complex \(A\) Again in \([Ni(Cl)_4]^{2-}\), there is \(Ni^{2+}\) ion, However, in presence of weak field \(Cl^-\) ligands, \(NO\) pairing of \(d\)-electrons occurs. Therefore, \(Ni^{2+}\) undergoes \(sp^3\) hybridization to make bonds with \(Cl^-\) ligands in tetrahedral geometry. As there are unpaired electrons in the \(d\)-orbitals, \([Ni(Cl)_4]^{2-}\) is paramagnetic and is referred to as a high spin outer orbital complex.
\([Ni(CO)_4]\): complex \(B\) The valence shell electronic configuration of ground state Ni atom is \(3d^8 4s^2\). All of these \(10\) electrons are pushed into \(3d\) orbitals and get paired up when strong field \(CO\) ligands approach \(Ni\) atom. The empty \(4s\) and three \(4p\) orbitals undergo \(sp^3\) hybridization and form bonds with \(CO\) ligands to give \(Ni(CO)_4\). Thus \(Ni(CO)_4\) is diamagnetic.
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