[Ni(NH3)6]Cl2 Paramagnetic But [Co(NH3)6]Cl3 Is Diamagnetic

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Tuesday, May 19, 2020

Home / [Ni(NH3)6]Cl2 paramagnetic but [Co(NH3)6]Cl3 is diamagnetic / INORGANIC / [Ni(NH3)6]Cl2 paramagnetic but [Co(NH3)6]Cl3 is diamagnetic [Ni(NH3)6]Cl2 paramagnetic but [Co(NH3)6]Cl3 is diamagnetic by BY P GHOSH on May 19, 2020 in [Ni(NH3)6]Cl2 paramagnetic but [Co(NH3)6]Cl3 is diamagnetic, INORGANIC

Why is [Ni(NH3)6]Cl2 paramagnetic but [Co(NH3)6]Cl3 is diamagnetic ?

In case of [Ni(NH3)6]Cl2 complex, the oxidation state of nickel atom is +2 .The atomic number of nickel atom, 28 and that of Ni(II)ion is 26 . Valence shell electronic configuration of Ni(II) ion is [ Ar ] d8 . Coordination number of central metal Ni(II)ion : 6 . Arrangement of [Ni(NH3)6]Cl2 complex is Octahedral. Now, the paramagnetic character of above said octahedral Ni(II) complexdoes not explained on the basis of crystal field theory. Because, the ligand NH3 is a strong field ligand. So, under the influence of the octahedral crystal field, the five degenerate d-orbitals of Ni(II) ion are splitted into two sets ofenergetically different orbitals. These two sets of orbitals are energetically lower t2g orbital and energetically higher eg orbital. Now, the ligand NH3 is a strong field ligand and hence it is a low spin complex. Therefore, the electronic arrangement should be t2g6 eg2 . Here, 10Dqo > P (pairingenergy) and hence all the electrons are paired. Consequently, octahedral Ni(II) complex with strong field should be diamagnetic. But,actually the [Ni(NH3)6]Cl2 complex is paramagnetic in nature. Therefore, the paramagnetic character of [Ni(NH3)6]Cl2 complex can be explained on the basis valence bond theory. According to valence bond theory, the electronic arrangement of Ni(II)ion is as follows, Under the strong field effect, the two unpaired electrons of 3d-orbital has to be shifted to higher 4d-orbitals in order to form low spin inner orbital complex. This transfer of electrons from lower 3d to higher 4d-orbital is not energetically feasible. Thus only outer orbital high spin complex is formed in Ni(II) six coordinated complex is formed through sp3d2 hybridization. Under this condition, nickel ion in [Ni(NH3)6]Cl2 complex contain two unpaired electrons. Hence [Ni(NH3)6]Cl2complex is paramagnetic. On the other hand,in case of [Co(NH3)6]Cl3complex, the oxidation state of cobalt is +3 .The atomic number of cobalt : 27 and that of Co(III) ion : 24 The ligand NH3, which is a strongfield ligand. Co-ordination number of central metal Co(III) ion : 6 Valence shell electronic configuration of Co(III) ion : [ Ar ] d6 . Since, ligand NH3 is a strong field ligand, hence the complex is low spin one. Therefore, under the influence of the octahedral crystal field, the possible electronic arrangement of Co(III) ion is t2g6, eg0. The octahedral crystal field splitting of [Co(NH3)6]Cl3complex is as follows, From the above crystal field splitting diagram of Co(III) ion, it is evidentlyshown that, the Co(III) ion have no unpaired electrons in its outer 3d-orbital. That is, all electrons are paired, hence [Co(NH3)6]Cl3complex is diamagnetic in nature.

• Why is [Ni(NH3)6]Cl2 complex paramagnetic ?
• Why is [Co(NH3)6]Cl3 complex diamagnetic ?
• Why is [Ni(NH3)6]Cl2paramagnetic but [Co(NH3)6]Cl3 is a diamagnetic?
• Calculate the magnetic moment of [Ni(NH3)6]Cl2 complex .
• Calculate the number of electron of [Ni(NH3)6]Cl2 complex .
• Calculate the number of electrons of [Co(NH3)6]Cl3 complex .

Outer orbital complex definition with examples Tags # [Ni(NH3)6]Cl2 paramagnetic but [Co(NH3)6]Cl3 is diamagnetic # INORGANIC

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