This option is correct because it belongs to felspar mineral and having Ca and Al to form a complex CaAl₂Si₂O_8.

This option is correct because the felspars are most important rock minerals in which Ba⁺⁺, K⁺, Ca⁺⁺, are linked with three dimensional framework of the SiO₄ and AlO₄ in tetrahedron. Celsian and anorthite are best known examples of  felspar. (Celsian = BaAl₂Si₂O₈, and Anorthite = CaAl₂Si₂O₈)

This option is correct because in the case of borazine, B-N bond is polar due to electronegativity difference between B and N atoms while benzene is non-polar due to bonding between same atoms.

This option is correct because the colour is observed due LMCT spectra. In CrO₄^2-, electrons transfer by oxygen to the vacant orbital of Cr⁶⁺ ion. A little transition spectra is observed.

This option is correct because in SN¹ mechanism, the single reacting complex ions substitute the ligand due to metal ligand breaking to form five coordinated intermediate complex.

This option is correct because the compound [Ru (2, 2 bipyridyl) ₃] shows MLCT due to the metal electron transfer to the pi antibonding molecular orbital of ligand. It is produced orange in colour.

This option is correct because the electrons transfer from reductant to oxidant in outer sphere mechanism, the reductant and oxidant in the transition state remain intact

This option is correct because if the (PNCl₂)₃ reacts with lead fluoride, two tetramer derivatives P₄N₄Cl₄F₂ and P₄N₄Cl₂F₆ are formed.

This option is incorrect because this complex shows two isomers.

This option is correct because the volan and quilon polymer are synthesised on reduction of CrO₂Cl₂ by methanol or ethanol in presence of carboxylic acid. This yield a solution of synthetic polymer. (R = methacrylate group is called volan, when R = stearate and myristate are called quilon polymer)

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 when borazole is pyrolysed above 340⁰C, it produces analogues compounds in the form of diphenyl and naphthalene like B₆N₆H₁₀ and B₅N₅H_8.

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 mixture of Ce are treated with oxidising agent bromate in neutral medium to form a pracipitate Ce(OH)₄.

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 in redox reaction, an electron transfer occur through the intact coordination shell of the metal ion. This redox reaction may occur self exchange electron to homonuclear species.

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 complexes cis and trans [PtA₂X₂] react with thiourea to give substituted product [Pt(tu)₄]²⁺but Cl is replaced by NH₃ to form [Pt(NH₃)₂(tu)₂]²⁺_. This is used in the thiourea reaction to the assigned structure, which is known as Kurnakov test. 

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 in the ions of Cr₂O₇^2-, the Cr have six oxidation state, due to fact that all the electrons are removed from Cr atom. Therefore, the oxygen atoms are low lying MOs of Cr atom in this condition Cr₂O₇^2-_.

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.