Write the formulas for the following coordination compounds :
- Tetraamminediaquacobalt (lll) chloride
- Potassium tetracyanonickelate (ll)
- Tris(ethane-1,2-diamine) chromium (lll) chloride
- Amminebromidochloridonitrito-N- platinate (ll)
- Dichloridobis(ethane-1,2-diamine) platinum (IV) nitrate
- Iron (lll) hexacyanoferrate (ll)
Write the IUPAC names of the following coordination compounds :
- Hexaamminecobalt (III) chloride
- Pentaamminechloridocobalt (III) chloride (HO3)2
- Potassium hexacyanoferrate (III)
- Potassium trioxalatoferrate (III)
- Potassium tetrachloridopalladate (II)
- Diamminechlorido (methylamine) platinum (II) chloride
Indicate the types of isomerism exhibited by the following complexes and draw the structures for these isomers :
Give evidence that [CO(NH3)5CI] SO4 and [CO(NH3)5 SO4]Cl are ionisation isomers.
The ionisation isomers dissolve in water to yield different ions and thus react differently to various reagents :
Explain on the basis of valence bond theory that [Ni(CN)4]2- ion with square planar structure is diamagnetic and the [NiCI4]2- ion with tetrahedral geometry is paramagnetic.
[NiCl4]2- is paramagnetic while [Ni(CO)4] is diamagnetic though both are tetrahedral. Why?
In Ni(CO)4, Ni is in zero oxidation state whereas in NiCl42-, it is in +2 oxidation state. In the presence of CO ligand, the unpaired d-electrons of Ni pair up but Cl– being a weak ligand is unable to pair up the unpaired electrons.
[Fe(H2O)6]3+ is strongly paramagnetic whereas [Fe(CN)6]3- is weakly paramagnetic. Explain.
In presence of CN–, (a strong ligand), the 3d electrons pair up leaving only one unpaired electron. The hybridisation is d2 sp3 forming inner orbital complex. In the presence of H2O, (a weak ligand), 3d electrons do not pair up. The hybridisation is sp3 d2 forming an outer orbital complex containing five unpaired electrons hence, it is strongly paramagnetic.
Explain [CO(NH3)6]3+ is an inner orbital complex whereas [Ni(NH3)6]2+ is an outer orbital complex.
in the presence of NH3, the 3d electrons pair up leaving two d orbitals empty to be involved in d2 sp3 hybridisation forming inner orbital complex in case of [CO(NH3)6]3+.
In Ni(NH3)6]2+, Ni is in +2 oxidation state and has d3 configuration, the hybridisation involved is sp3 d2 forming outer orbital complex.
Predict the number of unpaired electrons in the square planer [Pt(CN)4]2- ion.
For square planer shape, the hybridisation is dsp2 . Hence, the unpaired electrons in 5d orbital pair up to make one f-orbital empty for dsp2 hybridisation. Thus there is no unpaired electron.
The hexaaquomanganese (ll) ion contains five unpaired electrons, while the hexacyano ion contains only one unpaired electron. Explain using Crystal Field Theory.
Mn(II) has 3d5 electronic configuration. Water is a weak field ligand and therefore ∆0 is small. Therefore, the hexaaqua complex will be high spin complex containing 5 unpaired electrons. On the other hand, CN– is a strong field ligand and therefore, ∆0 is large. Therefore, in its cyano complex, the electrons pair up and have only one unpaired electron.
Calculate the overall complex dissociation equilibrium constant for the Cu(NH3)42+ ion, given that β4 for this complex is 2.1 × 1013.