Co-ordination Compounds | Question Bank for Class 12 Chemistry

Ans. The coordination number of central metal ion is 6.

Ans. Ligand is an atom or group of atoms which is either negatively charged or have lone pair of electrons to form co-ordinate bond with central metal ion.

Ans. Cobalt has $+3$ oxidation state. $[x+0-1-1+1-2=0] \Rightarrow x=3$

Ans. Ethylene diamine, oxalate ion.

Ans. $\left[P t\left(H_{2} O\right) p y(e n)\right]^{2+} .$ The charge on complex ion is $+2$ because ligands are neutral molecules.

Ans. Sodium tetrafluorodihydroxochromate (III)

Ans. Different.

Ans. $K_{4}\left[F e(C N)_{6}\right]$

Ans. (i) $\operatorname{cis}-\left[P t\left(N H_{3}\right)_{2} C l_{2}\right],$ trans-                      

Ans. $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{SO}_{4}\right]^{+} \mathrm{Br}^{-} \quad$ is ionisation isomer of $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Br}\right]^{2+} \mathrm{SO}_{4}^{2-}$

Ans. Co-ordination isomerism.

Ans. $d^{2} s p^{3}$

Ans. It is square planar.

Ans. - The hybridization is $d s p^{2}$ and shape is square planar.

Ans. $K_{3}\left[F e(C N)_{6}\right]$ is more stable than $K_{4}\left[F e(C N)_{6}\right]$

Ans. It is used as catalyst in polymerisation of alkanes as heterogeneous catalyst.

Ans. (i) $\mathrm{Mg}$ (ii) Fe (iii) Co (iv) Pt.

Ans. Hexadentate ligand is a ligand which has 6 donor atoms, e.g. EDTA                   EDTA forms complex with $C a^{2+}$ and $M g^{2+},$ therefore, it is used for measuring hardness of water.

Ans. Coordination number is defined as number of monodentate ligands surrounding a central metal atom or ion. The coordination number of $\mathrm{Co}$ in ( i ) is 6 and $\mathrm{Fe}$ in ( i ) is 4.

Ans. Explain geometrical isomerism with reference to square planar complexes giving one example. How is tetrahedral complexes with simple ligands do not exhibit geometrical isomerism?                               Tetrahedral complexes do not show geometrical isomerism because relative position of ligands are the same.

Ans. (a) Ligands are $N H_{3},$ the charge on it is zero. $H_{2} O$ has zero charge. $C l$ has negative charge $(-1)$ (b) It has octahedral shape.

Ans. The tetrahedral complexes do not show geometrical isomerism because the relative positions of the atoms with respect to each other will be same. The square planar complexes on the other hand show this kind of isomerism because if same kind of ligands occupy position adjacent to each other, it is called cis-form and if these are opposite to each other, it is called trans-form.

Ans. (i) $\quad \Delta_{0}$ is affected by nature of ligand. Greater the strength of ligand, greater the value of (ii) is also affected by oxidation state of central metal ion. Generally, the higher the ionic charge on the central metal ion, the greater will be the value of $\Delta_{0}$.

Ans. Cr has electronic configuration $C r(0)$ has electronic configuration $[A r] 4 s^{0} 3 d^{6} \because C O$ will cause pairing of electrons.                     $d^{2} s p^{3}$ hybridisation octahedral shape, diamagnetic.

Ans. $\mathrm{Co}(27) 1 \mathrm{s}^{2} 2 \mathrm{s}^{2} 2 \mathrm{p}^{6} 3 \mathrm{s}^{2} 3 \mathrm{p}^{6} 4 \mathrm{s}^{2} 3 \mathrm{d}^{7}$                            

Ans. $-\left[N i(C N)_{4}\right]^{2-}$ has square planar geometry. $\left[A g\left(N H_{3}\right)_{2} C l\right]$ remains colourless in aqueous solution because $A g^{+}$ has no unpaired electron, therefore it cannot undergo $d-d$ -transition.

Ans. (i) $\quad\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}$ is diamagnetic because $\mathrm{NH}_{3}$ is strong field ligand, causes pairing of electrons whereas $\left[\mathrm{CoF}_{6}\right]^{3-}$ is paramagnetic because $F^{-}$ is weak field ligand, does not cause pairing of electrons and there are unpaired electrons. (ii) $\left[F e\left(H_{2} O\right)_{6}\right]^{3+}$ has more unpaired electron than $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-},$ therefore, more paramagnetic.

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Ans. $\left[N i C N_{4}\right]^{2-}, C N^{-}$ cause pairing of electrons in $N i^{2+}$ ion.                

Ans. Ionization isomerism and Geometrical isomerism is exhibited by $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\right] \mathrm{Br}$ The ionisation isomers are $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\right] \mathrm{Br}$ and $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl} \mathrm{Br}\right] \mathrm{Cl}$ The geometrical isomers are                   The central atom has $d^{2} s p^{3}$ hybridization.

Ans. Stability of co-ordination compound in aqueous solution is determined with the help of stability constant. Higher the value of stability constant, greater will be stability. Smaller the cation, higher the charge on the cation, more stable will be the complex.

Ans. Chelating complex is more stable than unchelated complex because there is strong force of attraction between cation and polydentate ligand as compared to monodentate ligand.

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Ans. $\begin{array}{llll}{-(1)} & {\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\right] \mathrm{Cl}} & {\text { Its }} & {\text { IUPAC }} & {\text { name }} & {\text { is }}\end{array}$ tetraamminedichlorocobalt (III) chloride. (ii) Ethylene diamine (en) is bidentate ligand. $\left[\mathrm{Co}(e n)_{3}\right]^{3+} .$ Its IUPAC name is tris (ethylene diamine) cobalt (MI) ion.

Ans. The IUPAC name is Dichlorobis (ethylene diamine) cobalt (III) chloride.                  

Ans. (i) Bidentate ligand has two donor atoms which can form two sigma bonds whereas monodentate ligand can form only one  bond (ii) Diammine dichloro platinum (IV) chloride.                      

Ans. $-(1) \quad$ Tetrahedral complexes of $N i$ (II) are paramagnetic due to presence of unpaired electrons whereas square planar complexes of $N i^{2+}$ are diamagnetic due to absence of unpaired electrons. (ii) Iransition metals have vacant $d$ -orbitals as well as pair of electrons in $d$ -orbitals, therefore, they can form $\pi$ bonded complexes.

Ans. (i) $s p^{3}$ has tetrahedral shape, e.g., $\left[Z n\left(N H_{3}\right)_{4}\right]^{2+}$ is hybridised. (ii) has square planar, e.g., is hybridised. (iii) has octahedral shape, e.g., is hybridised.

Ans. (i) $\quad$ (a) small size of cation and higher charge. (b) presence of vacant $d$ -orbitals. (ii) $\quad K\left[A g(C N)_{2}\right]$ is used for silver plating and $K\left[A u(C N)_{2}\right]$ is used for gold plating. (iii) Chlorobis (ethylene diamine) nitrito cobalt (III) ion.

Ans. Formation of complex is an exothermic process because there is force of attraction between central metal ion and ligands. Stability of complex decreases with the increases in temperature because formation of complex is exothermic process. On heating, coordination bond between central metal ion and ligand will break.

Ans. The metal-carbon bond in metal carbonyls possess both $s$ and $p$ character. The $M-C \sigma$ bond is formed by the donation of lone pair of electrons on the carbonyl carbon into a vacant orbital of the metal. The $M-C \pi$ -bond is formed by the donation of a pair of electrons from a filled $d$ -orbital of metal into the vacant antibonding $\pi^{*}$ orbital of carbon monoxide. The metal to ligand bonding creates a synergic effect which strengthens the bond between $C O$ and the metal.

Ans. $-M n$ in $+2$ state has the configuration $3 d^{5} .$ In presence of $H_{2} O$ as ligand, the distribution of these five electrons is $t_{2 g}^{3} e_{g}^{2},$ i.e., all the electrons remain unpaired. In presence of $\mathrm{CN}^{-}$ as ligand, the distribution is $t_{2 g}^{5} e_{g}^{0},$ i.e. , two $t_{2 g}$ orbitals contain paired electrons while the third orbital contains one unpaired electron.

Ans. The main postulates of Werner's theory of coordination compounds are as follows: (i) Metal possess two types of valences called (a) primary valence which are ionisable. (b) secondary valence which are non-ionisable. (ii) Primary valency is satisfied by the negative ions and it is that which a metal exhibits in the formation of its simple salts. For example, in the coordination compound, $\mathrm{CoCl}_{3} \cdot 6 \mathrm{NH}_{3}$ valency between $\mathrm{Co}$ and $\mathrm{Cl}$ is primary valency and valency between $\mathrm{Co}$ and $\mathrm{NH}_{3}$ is secondary. In terms of modern theories based on electronic configuration, the primary valency is now referred to as "oxidation state' of metal. (iii) Secondary valencies are satisfied by neutral ligand or negative ligand and are those which metal exercises in the formation of its complex ions. Every cation has a fixed number of secondary valencies and are directed in space about central metal ion in certain fixed directions. The six valencies are regarded as directed to corners of a regular octahedron whereas four may be arranged either in square planar manner or tetrahedral manner. This explains the shape of coordination compound (or complex ion or compound). For example, in coordination compound $\left[\mathrm{CoCl}_{3} \cdot 6 \mathrm{NH}_{3}\right],$ six ammonia molecules linked to $\mathrm{Co}$ by secondary valencies are directed to six corners of a regular octahedron and thus account for structure of as follows:

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Ans. (i) Hexammine cobalt (III) chloride (ii) Tetrammine chloronitro cobalt (III) chloride (iii) Hexammine nickel (II) chloride (iv) Diamminechloro (methylamine) platinum (II) chloride (vi) Tris $(1,2$ -ethane diamine) cobalt (III) ion (vii) Hexaqua titanium (III) ion

Ans. In $\left[C r\left(N H_{3}\right)_{6}\right]^{3+},$ the chromium ion is in $+3$ oxidation state and has electronic configuration of $3 d^{\beta},$ as shown below :

Ans. Two principal types of isomerism are known among coordination compounds: (i) Stereo isomerism: (a) Geometrical isomerism, (b) Optical isomerism (ii) Structural isomerism: (a) Linkage isomerism, (b) Coordination isomerism, (c) Ionisation isomerism, (d) Solvate isomerism (i) Stereo isomerism: (a) Geometrical isomerism: This type of isomerism arises in heteroleptic complexes due to different possible geometric arrangements of the ligands. For example, in a square planar complex of formula $\left[M X_{2} L_{2}\right]$ the two ligands $X$ may be arranged adjacent to each other in a cis isomer, or opposite to each other in a trans isomer. (b) Optical isomerism: Optical isomers are mirror images that cannot be super imposed on one another. There are called as enantiomers. The two forms are called dextro (d) and laevo $(l)$ depending upon direction they rotate the plane of polarised light in polarimeter. Optical isomers $(d \text { and } l)$ of $\left[C o(e n)_{3}\right]^{3+}$ (ii) Structural isomerism: (a) Linkage isomerism: Linkage isomerism arises in a coordination compound containing ambidentate ligand. A simple example is provided by complexes containing the thiocynate ligand, $N C S^{-}$ which may find through the nitrogen to give $M \leftarrow N C S$ or through sulphur to give $M \leftarrow S C N$ (b) Coordination isomerism: This type of isomerism arises from the interchange of ligands between cationic and anionic entities of different metal ions present in a complex. Example : $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]\left[\mathrm{Cr}(\mathrm{CN})_{6}\right]$ and $\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}\right]\left[\mathrm{Co}(\mathrm{CN})_{6}\right]$ (c) Ionisation isomerism: This form of isomerism arises when the counter ion in a complex salt is itself a potential ligand and can displace a ligand which can then become counter ion. Example : $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{SO}_{4}\right] \mathrm{Br}$ and $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Br}\right] \mathrm{SO}_{4}$ (d) Solvate isomerism: This form of isomerism is known as "hydrate isomerism'. Solvate isomers differ by whether or not a solvent molecule is directly bonded to the metal ion or merely present as free solvent molecules in the crystal lattice. Example: $\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Cl}_{3} \&\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{Cl}\right] \mathrm{Cl}_{2} \cdot \mathrm{H}_{2} \mathrm{O}$

Ans. In $d^{1}$ coordination entity, $d^{1}$ electron occupies $t_{2 g}$ orbitals. In $d^{2}$ and $d^{3}$ also, electrons occupy orbitals singly in accordance with the Hund's rule. For $d^{4}$ ions two possibility exist. If is less than $P($ energy) required for electron pairing in a single orbital, we have weak field, high spin situation and the fourth electron enters one of the orbitals giving. Fifth electron also enters orbitals, i.e., . When, we have strong field, low spin situation, and pairing will occur in the level with orbitals remain unoccupied in to ions. Greater the value greater the chances of pairing.

Ans. Ligands have different field strengths and as a result the crystal field splitting, $\Delta_{0}$ or $\Delta_{t}$, depends upon the field produced by the ligand and charge on metal ions. Some ligands are able to produce strong fields in which case the splitting will be large whereas others produce weak fields and consequently result in small splitting of $d$ -orbitals. The arrangement of the ligands in order of increasing field strength is known as spectrochemical series. $I^{-} $