## Filters

Clear All
In both compounds, the oxidation state of Fe is +2. Also, in  we have weak field ligand whereas in    we have strong field ligand. So there is a difference in CFSE value in both the compounds. As a result, the colour shown by both compounds is different.
In the case of   , we have CN- as a strong field ligand. So, the pairing of electrons will occur. The electronic configuration of Ni+2  is   d6. Since all the electrons will be paired thus d-d electronic transition is not possible on this case. Whereas in case of    we have a weak field ligand (H2O). The pairing of electrons will not occur. Thus electrons from a lower state of energy can...
In   the oxidation state of the compound is +3. Its electronic configuration is d3.  Also, NH3 is a weak field ligand so the pairing of electrons will not occur.                                     So this compound is paramagnetic in nature. In  the oxidation state of the Ni is +2.  Its electronic configuration is d8.   Also, CN- is a strong field ligand so the pairing of electrons will...
It is known that the degenerated d-orbitals split into two levels -  eg and t2g . The splitting of the degenerate levels due to the presence of ligands is called the crystal-field splitting and the energy difference between the two levels (eg and t2g) is called the crystal-field splitting energy (CFSE).  The CFSE is denoted by Δo. After splitting of orbitals, the filling of the electrons...
The arrangement of ligands in the increasing order of their crystal-field splitting energy (CFSE) values is known as spectrochemical series.                            The ligands on the right side of the series strong field ligands are present whereas on the left-hand side weak field ligands are present. The strong field ligands are capable of splitting d orbitals to a higher extent as...
The splitting of d orbital is shown below:-                              In this splitting dx2y2 and  dz2 experience a rise in energy and make the eg level, while dxy, dyzand dzx experience a fall in energy and generate the t2g level.
(iv) The oxidation state of Co in the given comound is +3.  The electronic configuration of the compound is d6.        Also, F- is a weak field ligand so no pairing of electrons will occur.                                    Hence the geometry of compound is octahedral and it is paramagnetic in nature.
(iii) In the given compound, the oxidation state of Co is +3. Its electronic configuration thus becomes d6. Also, oxalate is a weak field ligand therefore pairing of electrons will not occur.                                        Hence the complex is octahedral and paramgnetic in nature.
(ii) In the given complex the oxidation state of Fe is +3.  Its electronic configuration is d5.   Also, the F- ions are weak field ligands, therefore, the pairing of electrons will not occur.                               Thus its geometry is octahedral and it is paramagnetic in nature.
In the given compound oxidation state of Fe is +2. The electronic configuration of this compound is 3d6. Also CN- is a strong field ligand so it will cause the pairing of electrons.                                             Therefore the complex is diamagnetic and its geometry is octahedral.
When KCN is passed through an aqueous solution of copper sulphate, CN- being a strong ligand, will replace water and form   . It is known that in stable coordination compounds, the individual identity of each constituent is lost i.e.,  Cu+2 is not available freely.  Thus no precipitate of copper sulphide is obtained in the given conditions.
We know that strong ligands can replace weak ligands from its solution. (i) In this case F- ions can replace H2O from aqueous copper sulphate solution.                                            (ii) In this compound, Cl- being stronger ligand will replace H2O and give a bright green solution.
The geometrical isomers of the compound are given below:-                                    We know that the given compound has tetrahedral geometry, so it can be optically active only when it has unsymmetric chelating agents. Hence the given compound doesn't have any optically active isomer.
The possible isomers of the given compound are as follows :-
The isomers of the given compound are :-
The configurational isomers are:-
The optical isomers of the given compound are:-
The optical isomers of the given compound are given below  :-
The optical isomers of the given compound are:-
The facial (fac) and meridional (mer) isomers are possible for the given compound.
Exams
Articles
Questions