Coordination Chemistry

This option is incorrect because this complex shows two isomers.

This option is correct because the central metal atom of the complexes belong to the 3d⁵ configuration, which is correlated to ground state term of S⁶. These terms are split in to 11 excited states from which 4 is quartets excited and remaining 7 is doublet excited.

This option is correct because the K₂NiF₆ shows the high spin character but actually cannot exhibit zero magnetic moment. Hence, these complexes behave as low spin in nature, the electrons are distributed in ⁶t_2g and ⁰e_g due to distribution of electron. No unpaired electron occur, i.e. magnetic moment is zero in low spin.

This option is correct, because the complex  X is labile, they contain the unhybridised metal orbital through which the ligand is attached more quickly in this orbital. But Y and Z are inert due to lack of availability unhybridised metal orbital the ligand cannot be attached in this orbital. (Lability shows more fast reaction and inert shows slow  reaction).

This option is correct, because the [Mn(H₂O)₆]³⁺ compound is high spin compound. It is related to the 3d⁴ configuration and splits into ³t_2g symmetrical and ¹e_g unsymmetrical, which exhibit the strong Jahn – Teller distortion.

It is the correct answer.

This option is correct because the cationic charge of the complex increases but the lability of the complexe decreases.

This option is correct  because the Cis [Pt(NH₃)₂Cl₂] complex ions is formed when [Pt(Cl)₄]^2-react with NH₃ to form [Pt(NH₃)(Cl)₃]^- and again it reacts with NH₃ to form Cis [Pt(Cl)₂(NH₃)₂] complex, due to the stronger effect Cl^- than NH₃.

This option is correct because [Co(H₂O)₆]²⁺ complexes ion in Co²⁺ are related to 3d⁷ electronic configuration and have F⁴ ground state term. It splits in to ⁴T_1g(F),⁴T_2g(F), ⁴A_2g(F) and ⁴T_1g(P), respectively, i.e. show 3 bands but other complexes ion spin allowed transition in ⁵T_2g and ⁵E_g one absorption band.

[Fe(CN)₆]^4-: d²sp³ hybridisation and inner orbital complex [Ni(NH₃)₆]²⁺: sp³d² hybridisation and outer orbital complex Hence, [Fe(CN)₆]^4- inner orbital complex and [Ni(NH₃)₆]²⁺ outer orbital complex.

Ground state term for Mn⁴⁺ is ⁶S_5/2.  Ground state term for Fe²⁺ is ⁵D₄. Ground state term for Co³⁺ is ⁵D₄.

Both R and S have dsp³ hybridisation and trigonal bipyramidal geometry.

Fe³⁺ and Fe²⁺ have d⁵ and d⁶ electronic configurations and both are in high spin ligand system. The term symbol is calculated by the formula- ^2s+1L_J For d⁵, n = 5s = n/2 = 5/2 2s+1 = 6 L = 0 = S J = (L+s) to (L-s) = 5/2 Possible term symbol for Fe³⁺ = ⁶S_5/2 Similarly, possible term symbols for Fe²⁺ = ⁵D₄, ⁵D₃, ⁵D₂, ⁵D₁, ⁵D₀
The ground state term for Fe³⁺ is ⁶S_5/2. The ground state term for Fe²⁺ is ⁵D₄.

V²⁺ has d³ electronic configuration and it is empowered in a weak field ligand.The term symbol is calculated by the formula- ^2s+1L_J For d³, n = 3s = n/2 = 3/2 2s+1 = 4 L = 3 = F J = (L+s) to (L-s) = 2, 1, 0 Possible term symbols = ⁴F_9/2, ⁴F_7/2, ⁴F_5/2, ⁴F_3/2 The ground state term for V²⁺ is ⁴F_^3/2.

In a weak field ligand system the electron filling of splitting of d-orbital is corresponds to Hund’s rule, followed by pairing occur. In strong field ligand system the electron filling of splitting of d-orbital is corresponds to pairing at initial stage.For Fe³⁺ electronic configuration is d⁵. For d⁵ in weak field ligand system, CFSE = 3 (-4) + 2 (6) = -12 + 12 = 0Dq (All electrons is unpaired, 3 in t2g and 2 in eg) For d⁵ in strong field ligand system, CFSE = 5 (-4) + 0 = -20Dq + 2P (two paired electrons and one electron is unpaired in t_2g and 0 electron in _eg orbital) Hence, the CFSE for octahedral complexes of Fe³⁺ in a weak field and strong field ligand systems are 0 and -20 Dq+2P respectively.