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Which of the following is most likely structure of $CrCl_3,6H_2O $ if 1/3 of total chlorine of the compound is precipitated by adding $AgNO_3$ to its aqueous solution :

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Explanation

The compound $[CrCl_2(H_2O)_4]Cl.2H_2O$ is the most likely structure because when 1/3 of the total chlorine is precipitated, it suggests that one out of the three chlorine atoms is outside the coordination sphere and hence can react with $AgNO_3$ to form a precipitate of $AgCl$.

Which one of the following will be able to show cis-trans isomerism:

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Explanation

The complex $M(AA’)_2$ will be able to show cis-trans isomerism because it contains two different bidentate ligands (AA’), which can arrange themselves in different positions around the central metal atom, leading to the formation of cis and trans isomers.

$K_3CoF_6 $ is high spin complex.What is the hybrid state of Co atom in this complex:

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Explanation

The complex $K_3CoF_6$ is a high spin complex. Cobalt in this complex is in the +3 oxidation state, represented as $Co^{3+}$. In a high spin complex, the ligands (F- in this case) are weak field ligands, leading to minimal pairing of electrons. The electron configuration of $Co^{3+}$ is $[Ar] 3d^6$. The hybridization state of $Co^{3+}$ in such a high spin octahedral complex is $sp^3d^2$.

The type of isomerism shown by $[Co(en)_2(NCS)_2]Cl and [Co(en)_2(NCS)Cl]NCS$ is:

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Explanation

The complexes $[Co(en)_2(NCS)_2]Cl$ and $[Co(en)_2(NCS)Cl]NCS$ exhibit ionization isomerism. Ionization isomerism occurs when there is an exchange of anions between the coordination sphere and the counter ion. In this case, the anions NCS- and Cl- are exchanged between the inside and the outside of the coordination sphere.

The co-ordination number and oxidation number of X in $[X(SO_4)(NH_3)_4]Cl $ is :

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Explanation

For the complex $[X(SO_4)(NH_3)_4]Cl$, the coordination number is determined by the number of ligand donor atoms attached to the central metal ion. Here, $SO_4^{2-}$ is a bidentate ligand and $NH_3$ is monodentate. Thus, we have $1 imes 2 + 4 imes 1 = 6$. The oxidation number of X can be calculated as: Let the oxidation number of X be x. The overall charge of the complex ion $[X(SO_4)(NH_3)_4]^+$ is +1. Therefore, x + (-2) + 4(0) = +1, solving this gives x = +3. Hence, the coordination number is 6 and the oxidation number is 3.

The IUPAC name of the compound $[Cu (NH_3)_4](NO_3)_2 $ is :

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Hexafluoroferrate (III) ion is an outer orbital complex.The number of unpaired electrons present in it is :

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Explanation

Hexafluoroferrate(III) ion, $[FeF_6]^{3-}$, involves iron in the +3 oxidation state. Iron has an electronic configuration of [Ar] 3d^5 4s^1. In the +3 state, it loses three electrons, giving [Ar] 3d^5. Since fluoride is a weak field ligand, it doesn't pair up the electrons, resulting in 5 unpaired electrons.

Which of the following complex species involves $d^2sp^3$ hybridisation :

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How many ions are produced from $ [Co(NH_3)_6]Cl_2 $ in solution :

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Explanation

The compound $[Co(NH_3)_6]Cl_2$ dissociates in water to produce one complex ion $[Co(NH_3)_6]^{2+}$ and two chloride ions $Cl^-$. Thus, it produces a total of 3 ions in solution.

In the complex $ Fe(CO)_X $ , the value of x is :

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Explanation

In the complex $Fe(CO)_X$, the value of $X$ is 5, making the complex $Fe(CO)_5$. This is a well-known carbonyl complex of iron.

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