Characterisation of inorganic compounds: by IR, Raman, NMR, EPR, Mössbauer, UV-vis, NQR, MS, electron spectroscopy and microscopic techniques.
|No.of questions appeared||7||5||4||5||7||4||3||5||4||5||2|
1.As a ligand Cl– is:
(a) Only a σ – donor
(b) Only a π – donor
(c) Both a σ – donor and a π – donor
(d) A σ – donor and a σ – acceptor
Ligand Cl– is both σ – donor→and π -donor→
According to LFT, each ligand is always σ – donor some ligands like : NH3which has only one lone pair of electrons is only σ – donor. Some ligands like F–, Cl–, Si–, OH– etc which hence two or more lone pair of electrons are σ – as well as π – donors Some ligands like CO, CN– , SH4bpy, phenetc how vacant π* – orbitals so these are σ – donor but π – acceptors.
2.The correct d-electron configuration showing spin-orbit coupling is
Spin-orbit coupling of d-electrons takes place in which t2gorbitals are unsymmetricaly filled.
3.For the reaction, trans-[IrCl (CO) (PPh3)2] + Cl2→ trans-[IrCl3(CO) (PPh3)2], the correct observation
(a) VCO(product) > VCO(reactant)
(b) VCO(product) < VCO(reactant)
(c) VCO(product) = VCO(reactant)
(d) VCO(product) = VCO(free CO)
Since in product metal Ir is present in the higher oxidation state.So, there is less back bonding to π*orbital of CO.
Hence, V(CO) product >V(CO) reactant
4.Among the following the strongest oxidizing agent is:
In [WO]2-,[CrO]2-, [MoO]2-and [ReO]1-all the metals have same oxidation state but Cr6+is of smallest size. Therefore it has strongest tendency to attract electrons. Therefore strongest oxidizing agent.
5.The least basic among the following is:
(a) Al (OH)3
(b) La (OH)3
(c) Ce (OH)3
(d) Lu (OH)3
Al (OH)3→Amphoteric whereas hydroxides of lanthanoids are all base.
6.The carbonyl resonance in13C NMR spectrum of [(h5– C5 H5)Rh(CO)]3 (103Rh, nuclear spin, I=1/2, 100%) shows a triplet at –65º C owing to the presence of
(a) Terminal CO
(b) m2– CO
Doublet of carbonyl carbon indicates the each CO is attached with chemically equivalent two Rh- atoms–that means μ2-CO complex.
Actual structure is: