Two isomers X and Y with the formula Cr(H2O)5ClBr2 were taken for an experiment on depression in the freezing point. It was found that one mole of X gave depression corresponding to 2 moles of particles and one mole of Y gave depression due to 3 moles of particles. The structural formula of X and Y respectively, are:
1. [Cr(H2O)5Cl] Br2; [Cr(H2O)4Br2]Cl.H2O
2. [Cr(H2O)5Cl] Br2; [Cr(H2O)3ClBr2].2H2O
3. [Cr(H2O)5Br]BrCl; [Cr(H2O)4ClBr]Br.H2O
4. [Cr(H2O)4Br2]ClH2O; [Cr(H2O)5Cl]Br2
Identify the outer orbital complex among the following options.
(Atomic number Mn=25, Fe=26, Co=27, Ni=28)
1. [Fe(CN)6]4-
2. [Mn(CN)6]4-
3. [Co(NH3)6]3+
4. [Ni(NH3)6]2+
For a metal ion in an octahedral field, the correct electronic configuration is:
1. | \( t_{2 \mathrm{~g}}^4 e_g^0 \text { when } \Delta_O<P\) |
2. | \(e_{2 \mathrm{~g}}^2 t_g^2 \text { when } \Delta_O<\mathrm{P} \) |
3. | \(\mathrm{t}_{2 \mathrm{~g}}{ }^3 \mathrm{e}_{\mathrm{g}}{ }^1 \text { when }{\Delta}_{\mathrm{O}}<\mathrm{P}\) |
4. | \(\mathrm{t}_{2 \mathrm{~g}}{ }^3 \mathrm{e}_{\mathrm{g}}{ }^1 \text { when }{\Delta}_{\mathrm{O}}>\mathrm{P}\) |
Given below are two statements :
Statement I: | The identification of Ni2+ is carried out by dimethyl glyoxime in the presence of NH4OH. |
Statement II: | The dimethyl glyoxime is a bidentate neutral ligand. |
1. Statement I is false but Statement II is true.
2. Both Statement I and Statement II are false.
3. Statement I is true but Statement II is false.
4. Both Statement I and Statement II are true.
Ethylene diaminetetraacetate (EDTA) ion is:
1. | Bidentate ligand with two "N" donor atoms |
2. | Tridentate ligand with three "N" donor atoms |
3. | Hexadentate ligand with four "O" and two "N'' donor atoms |
4. | Unidentate ligand |
Optical isomerism is exhibited by among the following is/are:
1. | [Cr(Ox)3]3+ | 2. | [PtCl2(en)2]2+ |
3. | [Cr(NH3)2Cl2en]+ | 4. | All of the above |
The correct splitting diagram of d orbitals in an octahedral crystal field is:
1. | 2. | ||
3. | 4. | None of these. |
[Cr(NH3)6]3+ is paramagnetic ,while is diamagnetic because:
1. | Electrons in the 3d orbitals remain unpaired in |
2. | Electrons in the 3d orbitals remain unpaired in [Cr(NH3)6]3+ |
3. | Electrons in the 3p orbitals remain unpaired in [Cr(NH3)6]3+ |
4. | Electrons in the 3p orbitals remain unpaired in |