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Which one of the following will be able to show cis-trans isomerism:
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:
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:
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 :
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 :
Hexafluoroferrate (III) ion is an outer orbital complex.The number of unpaired electrons present in it is :
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 :
How many ions are produced from $ [Co(NH_3)_6]Cl_2 $ in solution :
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 :
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.
Which is the central ion in $ [Cu(H_2O)_4] ^{2+} $ ion :
In the ion $[Cu(H_2O)_4]^{2+}$, the central ion is $Cu^{2+}$. The water molecules act as ligands surrounding the central copper ion.
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