NEET Chemistry Questions: Co-ordination Compounds

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The complex $ [Co(NH_3)_5Br]SO_4 $ will give white ppt with :




No explanation available.
$ [Co(NH_3)_6]^{3+} $ ion is :




The ion $[Co(NH_3)_6]^{3+}$ is diamagnetic because cobalt in this complex is in the +3 oxidation state, which means it has an electronic configuration of $[Ar] 3d^6$. The ammonia ligands are strong field ligands, which cause pairing of all the electrons in the 3d orbitals, resulting in no unpaired electrons.
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 :




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:




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 :




No explanation available.
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 :




No explanation available.

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.
All ligands are :




Ligands are species that donate a pair of electrons to a central metal atom or ion to form a coordination complex. According to Lewis theory, substances that donate electron pairs are called Lewis bases. Therefore, all ligands are Lewis bases.
The hybridization in $ Ni(CO)_4 $ is :




In the complex $ Ni(CO)_4 $, the nickel metal has a 3d^8 4s^2 electron configuration. When it forms the complex, it uses its 4s and 4p orbitals to hybridize into four $ sp^3 $ orbitals to accommodate the four CO ligands. Therefore, the hybridization in $ Ni(CO)_4 $ is $ sp^3 $.