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Consider Hexaqua manganese (II) :- In this compound, the oxidation state of Mn is +2 and its electronic configuration is d5. H2O is a weak ligand and crystal field is octahedral so the arrangement of electrons will be   t23g eg2. So the total number of unpaired electrons is 5. Now consider hexacyanoion:-  In this compound, the oxidation state of Mn is +2. It is surrounded by the strong ligands...
The oxidation state of Pt in the given compound is +2. Also, it is given that the compound has square planar geometry i.e., it has dsp2 hybridisation (d8). CN-  is a strong ligand so the pairing of electron will occur. So there are be no unpaired electrons in the given compound.
Firstly consider     :-        The oxidation state of cobalt is +3. So the electronic configuration of it will be d6. Since (NH3) is a strong ligand so the pairing of electron will be there.                                                      So, it has d2sp3 hybridisation and an inner orbital complex.   Now consider,     The oxidation state of nickel is +3. So its electronic configuration...
In both the compounds Fe has +3 oxidation state i.e., d5 configuration.                                                 In the case of strong ligand (CN-), the pairing of electron will occur. So number of electrons left unpaired will be 1. In case of weak ligand (H2O), pairing of electron will not there. Thus number of electrons unpaired will be 5.   We know that paramagnetic strength is...
The difference in the magnetic behaviour is due to the nature of ligands present. In case of    the oxidation state of nickel is +2 and also Cl- is a weak ligand. Thus its configuration becomes:-                                                          So it is paramagnetic and tetrahedral in nature. In the case of    ,  the oxidation state of nickel is 0. So its configuration is 3d8 4s2. We...
In both the compounds, the oxidation state of Nickel is +2. So it has d8 configuration.  Now, on the basis of ligand pairing of electrons will occur. Since CN- is a strong ligand so pairing will occur but in case of  Cl- pairing will not be there as it is a weak ligand. So, the configuration of both the compounds looks like :-                                                                    ...
The ionisation isomers dissolve in water to yield different ions and thus react differently to various reagents: [Co(NH3)5Br]SO4 + Ba2+ → BaSO4 (s) [Co(NH3)5SO4]Br + Ba2+ → No reaction [Co(NH3)5Br]SO4 + Ag+ → No reaction [Co(NH3)5SO4]Br + Ag+ → AgBr (s)
The ionisation isomers dissolve in water to yield different ions and thus react differently to various reagents: [Co(NH3)5Br]SO4 + Ba2+ → BaSO4 (s) [Co(NH3)5SO4]Br + Ba2+ → No reaction [Co(NH3)5Br]SO4 + Ag+ → No reaction [Co(NH3)5SO4]Br + Ag+ → AgBr (s)
Geometrical (cis-, trans-) isomers can exists for
There are 10 possible isomers.There are geometrical, ionisation and linkage isomers possible. A pair of optical isomer it also shows ionization isomerism    AND  It can also show linkage isomerism   AND
Two optical isomers can exist, Dextro and Laevo.
Both geometrical (cis-, trans-) and optical isomers exists for the given compound.
The IUPAC name of the following coordination compound   is : Diamminechlorido(methylamine)platinum(II) chloride.
The IUPAC names of the coordination compound    is  Potassium tetrachloridopalladate(II).
The IUPAC name of the following coordination compound     is   Potassium trioxalatoferrate(III)
The IUPAC name of the coordination compound   is  Potassium hexacyanoferrate(III).
The IUPAC name of the coordination compound  is Pentaamminechloridocobalt(III) chloride
The IUPAC name of the coordination compound     is : Hexaamminecobalt(III) chloride
The formula for the coordination compound  Iron(III) hexacyanidoferrate(II):
The chemical formula for the coordination compounds: Dichloridobis(ethane–1,2–diamine)platinum(IV) nitrate
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