Samacheer Kalvi Class 12 Chemistry Solutions Chapter 5 Coordination Chemistry

Part - I Text Book Evaluation

I. Choose the Correct Answer

 

Question 1. The sum of primary valence and secondary valance of the metal M in the complex \( [M(en)_2 (Ox)] Cl \) is L
(a) 3
(d) 9
Answer: (d) 9
In simple words: The primary valence is the charge of the metal ion, and the secondary valence is its coordination number. When you add these two values for the complex, the total is 9. This sum helps determine the overall properties of the complex.

🎯 Exam Tip: Always remember that the primary valence equals the oxidation state of the metal, and the secondary valence equals its coordination number.

 

Question 2. An excess of silver nitrate is added to 100ml of \( 0.01M \) solution of pentaaquachloridochromium (III) chloride. The number of moles of \( AgCl \) precipitated would be
(a) 0.02
(b) 0.002
(c) 0.01
(d) 0.2
Answer: (b) 0.002
In simple words: The complex is \( [Cr(H_2O)_5Cl]Cl_2 \). Only the chloride ions outside the coordination sphere will react with silver nitrate to form \( AgCl \). Since there are two such chloride ions per molecule, 0.001 moles of the complex will produce 0.002 moles of \( AgCl \).

🎯 Exam Tip: For precipitation reactions involving coordination complexes, only the counter ions (those outside the square brackets) react, not the ligands or central metal within the complex ion.

 

Question 3. A complex has a molecular formula \( MSO_4Cl.6H_2O \). The aqueous solution of it gives white precipitate with Barium chloride solution and no precipitate is obtained when it is treated with silver nitrate solution. If the secondary valence of the metal is six, which one of the following correctly represents the complex?
(a) \( [M(H_2O)_4Cl] SO_4.2H_2O \)
(b) \( [M(H_2O)_6] SO_4 \)
(c) \( [M(H_2O)_5Cl]SO_4.H_2O \)
(d) \( [M (H_2O)_3Cl] SO_4.3H_2O \)
Answer: (c) \( [M(H_2O)_5Cl]SO_4.H_2O \)
In simple words: The complex gives a precipitate with barium chloride, meaning the sulfate ion (\( SO_4^{2-} \)) is outside the complex. It doesn't react with silver nitrate, so no chloride ions are outside the complex. With a secondary valence of six, five water molecules and one chloride ion are inside the complex, leaving one water molecule outside. This makes the correct formula \( [M(H_2O)_5Cl]SO_4.H_2O \).

🎯 Exam Tip: The reactivity of ions in a complex helps determine which ions are inside (ligands) and which are outside (counter ions) the coordination sphere.

 

Question 4. Oxidation state of Iron and the charge on the ligand NO in \( [Fe (H_2O)_5 NO] SO_4 \)
(a) +2 and 0 respectively
(b) +3 and 0 respectively
(c) +3 and -1 respectively
(d) +1 and +1 respectively
Answer: (d) +1 and +1 respectively
In simple words: In this complex, the sulfate ion has a -2 charge, and water is neutral. For the overall complex to be neutral (as it's a salt), the `[Fe(H2O)5NO]` part must have a +2 charge. The `NO` ligand in such complexes typically exists as `NO+`. If `Fe` is `+1` and `NO` is `+1`, the total charge of `[Fe(H2O)5NO]` is `+2`, which balances the `SO4` charge. This makes the oxidation state of iron +1 and the charge on the `NO` ligand +1.

🎯 Exam Tip: Remember that `NO` can act as `NO+` (nitrosonium), `NO` (neutral nitric oxide), or `NO-` (nitroxyl) depending on the complex, which significantly affects the metal's oxidation state.

 

Question 5. As per IUPAC guidelines, the name of the complex \( [Co(en)_2 (ONO)Cl] Cl \) is
(a) chlorobisethylenediaminenitritocobalt(III) chloride
(b) chi or id obis (e thane-1, 2-diamine) nitrito K-Ocobaltate(III) chloride
(c) chloridobis (ethane-1, 2-diammine) nitrito K -Ocobalt(II) chloride
(d) chloridobis (ethane-1, 2-diamine) nitrito K -Ocobalt(III) chloride
Answer: (d) chloridobis (ethane-1,2-diamine) nitrito K -Ocobalt(III) chloride
In simple words: When naming complexes, you list ligands alphabetically. 'en' (ethane-1,2-diamine) is a bidentate ligand, `ONO` is nitrito-O (or K-O), and `Cl` is chlorido. Cobalt is in the +3 oxidation state. The `Cl` outside the bracket is the counter ion. This follows the rules for naming coordination compounds.

🎯 Exam Tip: Pay close attention to alphabetical order of ligands, the use of prefixes like 'bis' or 'tris' for complex ligands, and specifying the coordination mode for ambidentate ligands (like `ONO` as nitrito-O).

 

Question 6. IUPAC name of the complex \( K_3[Al(C_2O_4)_3] \) is
(a) potassiumtrioxalatoaluminium(III)
(b) potassiumtrioxalatoaluminate(II)
(c) potassiumtrisoxalatoaluminate(III)
(d) potassiumtrioxalatoaluminate(III)
Answer: (d) potassiumtrioxalatoaluminate(III)
In simple words: This compound starts with potassium as the counter ion. Inside the complex, there are three oxalate ligands (\( C_2O_4 \)), and aluminum is the central metal. Since the complex ion is anionic, 'aluminum' becomes 'aluminate'. The oxidation state of aluminum is +3. The prefix 'tris' is used because `C2O4` is a complex ligand.

🎯 Exam Tip: Remember that if the complex ion is anionic, the metal name ends with '-ate', and if a ligand already has a numerical prefix (like 'di' in oxalate), use 'bis', 'tris', 'tetrakis' for multiples.

 

Question 7. A magnetic moment of 1.73BM will be shown by one among the following (NEET)
(a) \( TiCl_4 \)
(b) \( [CoCl_6]^{4-} \)
(c) \( [Cu(NH_3)_4]^{2+} \)
(d) \( [Ni(CN)_4]^{2-} \)
Answer: (c) \( [Cu(NH_3)_4]^{2+} \)
In simple words: A magnetic moment of 1.73 BM (Bohr magnetons) suggests the presence of one unpaired electron. In \( [Cu(NH_3)_4]^{2+} \), copper has an oxidation state of +2, which means it has a `d9` electron configuration. Due to crystal field splitting, this `d9` configuration will have one unpaired electron, matching the given magnetic moment. This is a common pattern to identify transition metal complexes.

🎯 Exam Tip: For quick calculations, a magnetic moment of approximately 1.73 BM corresponds to one unpaired electron (\( \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \)).

 

Question 8. Crystal field stabilization energy for high spin \( d^5 \) octahedral complex is
(a) \( -0.6\Deltaο \)
(b) 0
(c) \( 2(Ρ – \Deltaο) \)
(d) \( 2(Ρ + \Deltaο) \)
Answer: (b) 0
In simple words: In a high spin \( d^5 \) octahedral complex, all five `d` electrons are spread out, with one electron in each `t2g` and `eg` orbital. Since `t2g` orbitals are stabilized by \( -0.4\Deltaο \) and `eg` orbitals are destabilized by \( +0.6\Deltaο \), having 3 electrons in `t2g` and 2 in `eg` results in a total CFSE of \( (3 \times -0.4\Deltaο) + (2 \times +0.6\Deltaο) = -1.2\Deltaο + 1.2\Deltaο = 0 \). This means there's no overall stabilization from the crystal field.

🎯 Exam Tip: Remember that a high spin \( d^5 \) configuration fills each orbital once before pairing, leading to zero CFSE in an octahedral field because the stabilization and destabilization cancel each other out.

 

Question 9. In which of the following coordination entities the magnitude of \( \Delta o \) will be maximum?
(a) \( [Co(CN)_6]^{3-} \)
(b) \( [Co(C_2O_4)_3]^{3-} \)
(c) \( [Co(H_2O)_6]^{3+} \)
(d) \( [Co(NH_3)_6]^{3+} \)
Answer: (a) \( [Co(CN)_6]^{3-} \)
In simple words: The value of \( \Delta o \) (octahedral crystal field splitting energy) depends on the strength of the ligand. Strong field ligands cause greater splitting. According to the spectrochemical series, `CN-` (cyanide) is a very strong field ligand, much stronger than oxalate, water, or ammonia. So, the complex with `CN-` will have the largest \( \Delta o \) value.

🎯 Exam Tip: Memorize the spectrochemical series to quickly determine the relative strength of ligands and, thus, the magnitude of crystal field splitting energy (CFSE).

 

Question 10. Which one of the following will give a pair of enantiomorphs?
(a) \( [Cr(NH_3)_6][Co(CN)_6] \)
(b) \( [Co(en)_2Cl_2]Cl \)
(c) \( [Pt(NH_3)_4][FtCl_4] \)
(d) \( [CO(NH_3)_4Cl_2]NO_2 \)
Answer: (b) \( [Co(en)_2Cl_2]Cl \)
In simple words: Enantiomorphs are non-superimposable mirror images, meaning the complex is chiral. Octahedral complexes of the type `[M(AA)2B2]` or `[M(AA)3]` where `AA` is a bidentate ligand, can be chiral and thus form enantiomers. \( [Co(en)_2Cl_2]Cl \) fits this pattern in its cis form, while the trans form is achiral. This complex's cis-isomer is optically active.

🎯 Exam Tip: Chiral complexes (those forming enantiomers) are often found in octahedral structures with bidentate ligands, especially in their cis-isomeric forms.

 

Question 11. Which type of isomerism is exhibited by \( [Pt(NH_3)_2Cl_2] \)?
(a) Coordination isomerism
(b) Linkage isomerism
(c) Optical isomerism
(d) Geometrical isomerism
Answer: (d) Geometrical isomerism
In simple words: The complex \( [Pt(NH_3)_2Cl_2] \) is a square planar complex. It can exist in two forms: cis, where the two ammonia ligands are adjacent to each other, and trans, where they are opposite. These different arrangements in space give rise to geometrical isomerism.

🎯 Exam Tip: Square planar complexes of the type \( MA_2B_2 \) (like \( [Pt(NH_3)_2Cl_2] \)) are known to exhibit cis-trans isomerism, which is a type of geometrical isomerism.

 

Question 12. How many geometrical isomers are possible for \( [Pt(Py)(NH_3)(Br)(Cl)] \)?
(a) 3
(b) 4
(c) 0
(d) 15
Answer: (a) 3
In simple words: This is a square planar complex of the type `MABCD`, where all four ligands are different. For such complexes, there are always three possible geometrical isomers. One ligand (e.g., Py) is fixed, and then the other three are arranged in different positions relative to it, generating distinct isomers.

🎯 Exam Tip: For square planar complexes with four different monodentate ligands (MABCD), always remember that exactly three geometrical isomers are possible.

 

Question 13. Which one of the following pairs represents linkage isomers?
(a) \( [Cu(NH_3)_4] [PtCl_4] \) and \( [Pt (NH_3)_4] [CuCl_4] \)
(b) \( [Co(NH_3)_5 (NO_3)] SO_4 \) and \( [CO(NH_3)_5 (ONO)] \)
(c) \( [Co(NH_3)_4 (NCS)_2] Cl \) and \( [Co(NH_3)_4 (SCN)_2]Cl \)
(d) both (b) and (c)
Answer: (c) \( [Co(NH_3)_4 (NCS)_2] Cl \) and \( [Co(NH_3)_4 (SCN)_2]Cl \)
In simple words: Linkage isomers occur when an ambidentate ligand can attach to the central metal atom through different donor atoms. In option (c), the thiocyanate ligand (\( SCN^- \)) can bond through nitrogen (NCS) or sulfur (SCN), leading to linkage isomerism. This is a classic example where the bonding site changes.

🎯 Exam Tip: Look for ambidentate ligands (like `NCS-/SCN-`, `NO2-/ONO-`, `CN-/NC-`) to identify linkage isomerism. The difference lies only in which atom of the ligand bonds to the metal.

 

Question 14. Which kind of isomerism is possible for a complex \( [Co(NH_3)_4Br_2]Cl \)? (PTA -3)
(a) geometrical and ionization
(b) geometrical and optical
(c) optical and ionization
(d) geometrical only
Answer: (a) geometrical and ionization
In simple words: This complex can show geometrical isomerism because the two bromide ligands can be arranged in cis or trans positions around the central cobalt. It can also show ionization isomerism because the chloride ion outside the coordination sphere can swap places with a bromide ligand inside, forming \( [Co(NH_3)_4BrCl]Br \). This ability to swap ions or change spatial arrangement means it exhibits both types of isomerism.

🎯 Exam Tip: When identifying isomerism, first check for structural possibilities (ionization, linkage, hydrate, coordination), then for stereoisomers (geometrical, optical).

 

Question 15. Which one of the following complexes is not expected to exhibit isomerism?
(a) \( [Ni(NH_3)_4 (H_2O)_2] \) `[`
(b) \( [Pt(NH_3)_2Cl_2] \)
(c) \( [Co(NH_3)_5SO_4]Cl \)
(d) \( [FeCl_6]^{3-} \)
Answer: (d) \( [FeCl_6]^{3-} \)
In simple words: A complex will not show isomerism if its structure is simple and there are no different ways to arrange its parts. \( [FeCl_6]^{3-} \) is an octahedral complex with six identical chloride ligands. Since all ligands are the same and it's a simple structure, it cannot form geometrical or optical isomers. Other options like \( [Pt(NH_3)_2Cl_2] \) can show geometrical isomerism (cis/trans), and \( [Co(NH_3)_5SO_4]Cl \) can show ionization isomerism. This complex is very symmetrical, preventing different arrangements.

🎯 Exam Tip: Complexes with all identical monodentate ligands in highly symmetrical geometries (like octahedral `MA6` or tetrahedral `MA4`) typically do not exhibit any form of isomerism.

 

Question 16. A complex in which the oxidation number of the metal is zero is
(a) \( K_4 [Fe(CN)_6] \)
(b) \( [Fe(CN)_3 (NH_3)_3] \)
(c) \( [Fe(CO)_5] \)
(d) both (b) and (c)
Answer: (c) \( [Fe(CO)_5] \)
In simple words: Carbonyl ligands (`CO`) are neutral. In \( [Fe(CO)_5] \), since all five `CO` ligands are neutral, the oxidation state of the central iron metal must be zero for the entire complex to be neutral. This is a common feature of metal carbonyls, where metals often exist in low or zero oxidation states.

🎯 Exam Tip: Always remember that carbonyl (`CO`), ammonia (`NH3`), and water (`H2O`) are neutral ligands, meaning they do not contribute to the oxidation state calculation of the central metal.

 

Question 17. Formula of tris(ethane-1,2-diamine) iron (II) phosphate
(a) \( [Fe(CH_3–CH(NH_2)_2)_3](PO_4)_3 \)
(b) \( [Fe(H_2N – CH_2-CH_2-NH_2)_3] (PO_4) \)
(c) \( [Fe(H_2N-CH_2-CH_2-NH_2)_3](PO_4)_2 \)
(d) \( [Fe(H_2N-CH_2-CH_2-NH_2)_3]_3(PO_4)_2 \)
Answer: (d) \( [Fe(H_2N-CH_2-CH_2-NH_2)_3]_3(PO_4)_2 \)
In simple words: "Tris(ethane-1,2-diamine)" means there are three `en` (ethane-1,2-diamine) ligands. "Iron (II)" means `Fe` has a +2 oxidation state. "Phosphate" is \( PO_4^{3-} \). For the charges to balance, three iron(II) complexes, each with a `+2` charge (\( [Fe(en)_3]^{2+} \)), are needed to balance two phosphate ions, each with a `-3` charge. This creates a neutral compound.

🎯 Exam Tip: Correctly identifying the charge of the central metal and ligands (especially polyatomic ones like phosphate) is crucial for writing the correct formula for a coordination compound.

 

Question 18. Which of the following is paramagnetic in nature? (PTA -5)
(a) \( [Zn(NH_3)_4]^{2+} \)
(b) \( [CO (NH_3)_6]^{3+} \)
(c) \( [Ni(H_2O)_6]^{2+} \)
(d) \( [Ni (CN)_4]^{2-} \)
Answer: (c) \( [Ni(H_2O)_6]^{2+} \)
In simple words: A paramagnetic substance has unpaired electrons. Let's look at the options: (a) \( Zn^{2+} \) is `d10` (no unpaired electrons), (b) \( Co^{3+} \) with strong field \( NH_3 \) is `d6` low spin (no unpaired electrons), (d) \( Ni^{2+} \) with strong field \( CN^- \) is `d8` square planar (no unpaired electrons). However, (c) \( Ni^{2+} \) in \( [Ni(H_2O)_6]^{2+} \) is `d8` with weak field \( H_2O \). This results in a high spin octahedral complex with two unpaired electrons, making it paramagnetic.

🎯 Exam Tip: To determine paramagnetism, always find the oxidation state of the metal, its d-electron configuration, and consider the ligand's field strength (strong or weak) to predict electron pairing.

 

Question 19. Fac-mer isomerism is shown by
(a) \( [Co (en)_3]^{3+} \)
(b) \( [Co(NH_3)_4(Cl)_2]^{+} \)
(c) \( [Co (NH_3)_3 (Cl)_3] \)
(d) \( [Co (NH_3)_5 Cl]SO_4 \)
Answer: (c) \( [Co (NH_3)_3 (Cl)_3] \)
In simple words: Fac-mer isomerism (facial-meridional isomerism) is a type of geometrical isomerism found in octahedral complexes of the \( MA_3B_3 \) type. This complex, \( [Co(NH_3)_3 (Cl)_3] \), has three identical ammonia ligands and three identical chloride ligands, making it capable of forming both facial (all identical ligands on one face of the octahedron) and meridional (identical ligands forming a meridian around the center) isomers.

🎯 Exam Tip: Fac-mer isomerism is unique to octahedral complexes of the \( MA_3B_3 \) type, where 'fac' means ligands occupy positions on a triangular face and 'mer' means they lie on a meridian.

 

Question 20. Choose the correct statement.
(a) Square planar complexes are more stable than octahedral complexes
(b) The spin only magnetic moment of \( [Cu (Cl)_4]^{4-} \) is 1.732 BM and it has square planar structure.
(c) Crystal field splitting energy (\( \Delta o \)) of \( [FeF_6]^{4-} \) is higher than the (\( \Delta o \)) of \( [Fe (CN)_6]^{4-} \)
(d) Crystal field stabilization energy of \( [V(H_2O)_6]^{2+} \) is higher than the crystal field stabilization of \( [Ti(H_2O)_6]^{2+} \)
Answer: (d) Crystal field stabilization energy of \( [V(H_2O)_6]^{2+} \) is higher than the crystal field stabilization of \( [Ti(H_2O)_6]^{2+} \)
In simple words: Let's check each option. (a) Octahedral complexes are often more stable. (b) \( [CuCl_4]^{4-} \) is a `d9` complex, so it has one unpaired electron, leading to a magnetic moment of 1.732 BM, but its geometry is usually distorted tetrahedral, not square planar. (c) \( F^- \) is a weak field ligand while \( CN^- \) is a strong field ligand, so \( \Delta o \) for \( [Fe(CN)_6]^{4-} \) is much higher than for \( [FeF_6]^{4-} \). (d) \( V^{2+} \) is `d3` and \( Ti^{2+} \) is `d2`. For `d3` in an octahedral field, the CFSE is greater than for `d2`. This makes (d) the correct statement.

🎯 Exam Tip: Accurately evaluating statements requires a strong understanding of ligand field strength, d-electron configurations, magnetic moment calculations, and common coordination geometries.

II. Answer the Following Questions

 

Question 1. Write the IUPAC names for the following complexes. (PTA -3)
1. \( Na_2 [Ni (EDTA)] \)
2. \( [Ag(CN)_2]^- \)
3. \( [Co(en)_3]_2 (SO_4)_3 \)
4. \( [Co (ONO) (NH_3)_5]^{2+} \)
Answer:
1. Sodium ethylenediaminetetraacetatonickelate(II)
2. Dicyanidoargentate(I) ion
3. Tris(ethane-1,2-diamine)cobalt(III) sulfate
4. Pentaamminenitrito-O-cobalt(III) ion
In simple words: Naming coordination compounds involves following specific rules: counter ions first, then ligands in alphabetical order (using bis/tris for complex ligands), followed by the metal and its oxidation state. If the complex is an anion, the metal name ends in -ate. For ambidentate ligands, the bonded atom is specified. Knowing these rules helps correctly name any complex.

🎯 Exam Tip: Always double-check the metal's oxidation state and whether the complex is cationic, anionic, or neutral, as this affects the naming of the metal.

 

Question 2. Write the formula for the following coordination compounds.
1. Potassiumhexacyanidoferrate(II)
2. Pentacarbonyl iron(0)
3. Pentaamminenitrito-N-cobalt(III) ion
4. Hexaamminecobalt(III) sulphate
5. Sodium tetrafluoridodihydroxidochromate(III)
Answer:
1. \( K_4[Fe(CN)_6] \)
2. \( [Fe(CO)_5] \)
3. \( [Co(NH_3)_5(NO_2)]^{2+} \)
4. \( [Co(NH_3)_6]_2(SO_4)_3 \)
5. \( Na_3[CrF_4(OH)_2] \)
In simple words: When writing formulas, list the central metal first, then ligands in alphabetical order. Use brackets for the coordination sphere and balance charges with counter ions. Remember that `CO` is neutral, and the oxidation state of the metal determines the complex charge. For example, for "pentaamminenitrito-N-cobalt(III) ion", the `Co` is `+3`, `NH3` is neutral, and `NO2` is `-1`, making the complex have a `+2` charge.

🎯 Exam Tip: Pay close attention to roman numerals indicating the metal's oxidation state and the specific donor atom for ambidentate ligands (e.g., nitrito-N vs. nitrito-O) when constructing the formula.

 

Question 3. Arrange the following in order of increasing molar conductivity
(i) \( Mg[Cr(NH_2)(Cl)_5] \)
(ii) \( [Cr(NH_3)_5Cl]_3[CoF_6]_2 \)
(iii) \( [Cr(NH_3)_3Cl_3] \)
Answer:
(i) \( Mg[Cr(NH_3)_5Cl_5] \rightarrow Mg^{2+} + [Cr(NH_3)_5Cl_5]^{2-} \) (2 ions)
(ii) \( [Cr(NH_3)_5Cl]_3[CoF_6]_2 \rightarrow 3[Cr(NH_3)_5Cl]^{2+} + 2[CoF_6]^{3-} \) (5 ions)
(iii) \( [Cr(NH_3)_3Cl_3] \) – Neutral complex – No ions
Molar conductivity increases as the number of ions increase.
\( \implies \) Molar conductivity is in the increasing order: \( [Cr(NH_3)_3Cl_3] < Mg[Cr(NH_3)_5Cl_5] < [Cr(NH_3)_5Cl]_3[CoF_6]_2 \)
In simple words: Molar conductivity depends on how many ions a complex forms when dissolved in water. A neutral complex like \( [Cr(NH_3)_3Cl_3] \) forms no ions, so it has very low conductivity. \( Mg[Cr(NH_3)_5Cl_5] \) forms two ions. \( [Cr(NH_3)_5Cl]_3[CoF_6]_2 \) forms five ions (3 cations and 2 anions). The more ions, the higher the conductivity, explaining the increasing order.

🎯 Exam Tip: Remember that molar conductivity is directly proportional to the total number of ions produced by the dissociation of a complex in solution.

 

Question 4. Give an example of a coordination compound used in medicine and two examples of biologically important coordination compounds.
Answer:
Medical uses of coordination compounds:
1. \( Ca-EDTA \) chelate is used in treating lead and radioactive poisoning. It helps remove harmful lead and radioactive metal ions from the body.
2. Cis-platin is an antitumor drug used in cancer treatment.
Biological importance of coordination compounds:
1. Red blood corpuscles (RBCs) contain a heme group, which is an \( Fe^{2+} \) porphyrin complex. This complex helps carry oxygen from the lungs to tissues and carbon dioxide from tissues to lungs.
2. Chlorophyll, the green pigment in plants, is a coordination complex with \( Mg^{2+} \) as the central metal ion. It is surrounded by a porphyrin ring and is vital for photosynthesis.
In simple words: Coordination compounds are very important in both medicine and living things. For example, specific metal complexes can treat poisoning or cancer. In our bodies and plants, they are key to processes like carrying oxygen (iron in blood) and making food (magnesium in chlorophyll). They are like tiny machines that make life work.

🎯 Exam Tip: When discussing biological roles, focus on specific examples like hemoglobin (iron), chlorophyll (magnesium), and vitamin B12 (cobalt), explaining their function concisely.

 

Question 5. Based on VB theory explain why \( [Cr(NH_3)_6]^{3+} \) is paramagnetic, while \( [Ni(CN)_4]^{2-} \) is diamagnetic.
Answer:

In simple words: The paramagnetism or diamagnetism of a complex depends on whether it has unpaired electrons. For \( [Cr(NH_3)_6]^{3+} \), chromium is in a `+3` state with `d3` electrons. Since `NH3` is a weak field ligand, these three electrons remain unpaired, making the complex paramagnetic. For \( [Ni(CN)_4]^{2-} \), nickel is in a `+2` state with `d8` electrons. `CN-` is a strong field ligand, forcing all `d8` electrons to pair up, resulting in no unpaired electrons and making the complex diamagnetic.

🎯 Exam Tip: To explain magnetic behavior using VB theory, clearly state the metal's oxidation state, the d-electron count, the ligand's field strength (to determine pairing), and the resulting hybridisation.

 

Question 6. Draw all possible geometrical isomers of the complex \( [Co(en)_2Cl_2]^{+} \) and identify the optically active isomer.
Answer:
In simple words: The complex \( [Co(en)_2Cl_2]^{+} \) can form two main geometrical isomers: cis and trans. In the trans isomer, the two chlorine atoms are on opposite sides, making the molecule symmetrical and therefore optically inactive. In the cis isomer, the two chlorine atoms are next to each other, creating a non-superimposable mirror image. This cis-isomer is chiral and thus optically active, meaning it can rotate plane-polarized light.

🎯 Exam Tip: For octahedral complexes with bidentate ligands, remember that cis-isomers are often optically active (chiral), while trans-isomers are typically optically inactive (achiral due to symmetry).

 

Question 7. \( [Ti (H_2O)_6]^{3+} \) is coloured, while \( [Sc (H_2O)_6]^{3+} \) is colourless – Explain. (PAT -4; MARCH 2020)
Answer:

• In \( [Ti (H_2O)_6]^{3+} \), the outer electronic configuration of titanium ion \( Ti^{3+} \) is \( 3d^1 \).
• This single electron is present in the lower energy \( t_{2g} \) orbitals in the octahedral aqua ligand field. It absorbs light and gets excited to a higher energy \( e_g \) level.
• This process is known as a d-d transition.
• The absorption maximum for this excitation is at \( 20000 \text{ cm}^{-1} \), corresponding to a crystal field stabilization energy (\( \Delta o \)) of \( 239.7 \text{ KJ mol}^{-1} \).
• The light transmitted is purple, so the complex appears purple in colour.
• But in \( [Sc(H_2O)_6]^{3+} \), the outer electronic configuration of scandium ion \( Sc^{3+} \) is \( 3d^0 \).
• Since there is no electron in the d orbital, a d-d transition is not possible. Thus, \( [Sc(H_2O)_6]^{3+} \) is colourless.

In simple words: A complex appears colored if its electrons can jump to higher energy levels by absorbing visible light (d-d transition). Titanium in \( [Ti(H_2O)_6]^{3+} \) has one d-electron, which can absorb light and move up, making the complex colored (purple). However, scandium in \( [Sc(H_2O)_6]^{3+} \) has no d-electrons, so no d-d transition can happen, and it remains colorless. The presence of d-electrons allows for color.

🎯 Exam Tip: Color in transition metal complexes is primarily due to d-d transitions. If a metal ion has a `d0` or `d10` configuration, it typically won't exhibit d-d transitions and will be colorless.

 

Question 8. Give an example for complex of the type \( [Ma_2b_2C_2] \) where a, b, c are monodentate ligands and give the possible isomers.
Answer:
Example for the complex of the type \( [Ma_2b_2c_2] \) is \( [Cr(NH_3)_2 (H_2O)_2Br_2]^{+} \).
There are totally 6 isomers possible for this type of complex: 5 geometrical isomers and 1 optically active pair of enantiomers.




In simple words: For a complex like \( [Cr(NH_3)_2 (H_2O)_2Br_2]^{+} \), which has three types of ligands, there are many ways to arrange them. It can have two of each ligand. We can draw 5 different geometrical arrangements. One of these arrangements, a specific cis-form, will be chiral, meaning it's non-superimposable on its mirror image, giving an optically active pair. The other forms are symmetrical and optically inactive.

🎯 Exam Tip: Systematically drawing isomers by fixing one ligand and then arranging others helps ensure all possibilities are covered without duplication. Pay extra attention to the symmetry elements to identify optical activity.

 

Question 9. Give one test to differentiate [Co(NH3)5Cl]SO4 and [Co(NH3)5 SO4]Cl
Answer: We can tell the difference between these two compounds by how they react with certain solutions. Let's look at how they behave.
1. The aqueous solution of [Co(NH3)5Cl]SO4 gives a white precipitate of BaSO4 when barium chloride (BaCl2) solution is added. This is because it releases sulfate ions (\( SO_4^{2-} \)) into the solution.
2. The aqueous solution of [Co(NH3)5 SO4]Cl gives a curdy white precipitate of silver chloride (AgCl) when silver nitrate (AgNO3) solution is added. This happens because it releases chloride ions (\( Cl^- \)) into the solution.
In simple words: To tell them apart, add barium chloride to the first one; it will form a white solid. Add silver nitrate to the second one; it will also form a white solid.

🎯 Exam Tip: Remember that sulfate ions react with barium to form barium sulfate, and chloride ions react with silver to form silver chloride. These are common precipitation tests for these ions.

 

Question 10. In an octahedral crystal field, draw the figure to show the splitting of d orbitals.
Answer:
In an octahedral crystal field, the five d orbitals split into two sets: two \( e_g \) orbitals ( \( d_{x^2-y^2} \) and \( d_{z^2} \) ) at higher energy, and three \( t_{2g} \) orbitals ( \( d_{xy} \), \( d_{yz} \) and \( d_{zx} \) ) at lower energy. This energy difference is called the crystal field splitting energy (\( \Delta_o \)). Ligands approach the metal ion along the axes, causing stronger repulsion with the \( e_g \) orbitals.
In simple words: When ligands come near a metal atom in an octahedral shape, the d-orbitals inside the metal split into two groups. Some go up in energy, and some go down. This happens because the ligands push away the electrons in the d-orbitals differently.

🎯 Exam Tip: Clearly label the axes, the d-orbitals in free ion, the barycenter (average energy), and the split \( t_{2g} \) and \( e_g \) levels, along with the \( \Delta_o \) value.

 

Question 11. What is linkage isomerism? Explain with an example.
Answer: Linkage isomerism happens when a coordination compound has the same overall formula but the ligand connects to the central metal atom through a different atom. This is possible with ambidentate ligands, which can bond in two or more ways.
For example, the ligand nitrite (\( NO_2^- \)) can bond to a metal either through the nitrogen atom (forming a nitro complex, -\( NO_2 \)) or through an oxygen atom (forming a nitrito complex, -ONO). These two forms are linkage isomers. A common example is \( [Co(NH_3)_5(NO_2)]^{2+} \) and \( [Co(NH_3)_5(ONO)]^{2+} \). The way the ligand attaches changes the properties of the compound, like its color or stability.
In simple words: Linkage isomerism is when a chemical compound has the same parts, but a specific part (called a ligand) can connect to the central metal atom in two different ways, using a different atom each time.

🎯 Exam Tip: Always identify the ambidentate ligand in the complex and clearly show the two different atoms through which it can bind to the central metal atom.

 

Question 12. Classify the following ligand based on the number of donor atoms,
a) NH3
b) en
c) ox\( ^{2-} \)
d) triaminotriethylamine
e) pyridine
Answer:

Ligands are classified based on how many atoms they use to bond with the central metal ion. Monodentate ligands use one atom, bidentate use two, and tetradentate use four. This determines how many "teeth" the ligand has to grip the metal.
In simple words: Ligands are grouped by how many connections they make to the metal atom. Some connect with one atom (monodentate), some with two (bidentate), and some with four (tetradentate).

🎯 Exam Tip: Remember common examples for each type of ligand, especially "en" (ethylenediamine) for bidentate and "oxalate" for bidentate, and ammonia/pyridine for monodentate.

 

Question 13. Give the difference between double salts and coordination compounds.
Answer: Double salts and coordination compounds are two types of addition compounds, but they behave very differently in solution. Here are their main differences:

Double salts are simpler as they break down completely into their individual ions when dissolved, acting just like a mixture of their components. In contrast, coordination compounds keep their complex structure (the coordination entity) even in solution, so they do not show reactions for all the ions inside that complex part.
In simple words: Double salts break into all their basic parts when put in water, while coordination compounds keep some of their parts together as a group.

🎯 Exam Tip: The key difference lies in their behavior in solution: double salts dissociate completely, while coordination compounds maintain their complex ion identity.

 

Question 14. Write the postulates of Werner's theory?
Answer: Werner's theory explains the structure and bonding in coordination compounds. It proposed that every metal atom in a complex has two types of valency: primary valency and secondary valency. These valencies help define how ligands attach to the central metal.

Werner's theory was groundbreaking because it explained the different ways ions bond in complexes and how the shape of these complexes is determined. It highlighted that metal ions have a fixed number of primary and secondary valencies, which direct the overall structure.
In simple words: Werner's theory says that metal atoms in complex compounds have two types of "stickiness": primary (like their usual charge) and secondary (which decides how many things can stick to them and in what shape).

🎯 Exam Tip: Focus on clearly defining primary and secondary valency, their nature (ionizable/non-ionizable, directional/non-directional), and what they represent (oxidation state/coordination number).

 

Question 15. Why tetrahedral complexes do not exhibit geometrical isomerism.
Answer: Tetrahedral complexes do not show geometrical isomerism (cis-trans isomerism) because all the ligands in a tetrahedral arrangement are equidistant from each other. In a tetrahedral shape, every position is next to every other position.
In a tetrahedral complex of the type \( [Ma_2b_2] \), there's no way to arrange the ligands such that two identical ligands are "next to" each other (cis) versus "opposite" each other (trans). All possible arrangements lead to the same relative positions of the ligands. Geometrical isomerism requires distinct arrangements where some ligands are adjacent (cis) and others are opposite (trans), which is not possible in a perfectly symmetrical tetrahedral shape.
In simple words: Imagine a tripod; no matter how you put two different colored balls on it, they will always look the same in relation to each other. Tetrahedral molecules are like this; all positions are "cis" to each other, so there's no "trans" option.

🎯 Exam Tip: Remember that tetrahedral geometry lacks the necessary fixed planes or distinct positions required to define cis and trans isomers, unlike square planar or octahedral complexes.

 

Question 16. Explain optical isomerism in coordination compounds with an example.
Answer: Optical isomerism happens when coordination compounds are mirror images of each other but cannot be placed on top of each other (they are non-superimposable). These mirror images are called enantiomers. They behave the same way chemically and physically, except for how they interact with plane-polarized light. One enantiomer rotates the light clockwise (dextro form, 'd'), and the other rotates it counter-clockwise (laevo form, 'l').
• Chiral coordination compounds are those that show optical isomerism.
• Their solutions can rotate plane-polarized light.
• If light rotates clockwise, it's the 'd' form; if counter-clockwise, it's the 'l' form.
• Octahedral complexes often show optical isomerism.
Example: The complex \( [CoCl_2(en)_2]^+ \) shows optical isomerism. It can exist as two cis isomers which are optically active and one trans isomer which is optically inactive.
The cis isomers are chiral (they don't have a plane of symmetry) and thus are optically active. The trans isomer has a plane of symmetry and is therefore achiral and optically inactive. This is an important distinction in coordination chemistry.
In simple words: Optical isomerism is like having a left hand and a right hand; they are mirror images but cannot fit perfectly on top of each other. These "handy" molecules can twist light in different directions.

🎯 Exam Tip: When drawing optical isomers, ensure the mirror image is clearly depicted and that the cis/trans labels are correctly applied. Always identify the presence or absence of a plane of symmetry to determine optical activity.

 

Question 17. Explain Solvate isomerism with an example. (PTA- 6; March 2020)
Answer: Solvate isomerism is a type of isomerism where the solvent molecules are part of the coordination sphere or are simply present as crystallization solvent. These isomers differ in how solvent molecules (like water or ammonia) are arranged within the complex. Specifically, they involve the exchange of free solvent molecules with a ligand in the coordination entity.
• If the solvent molecule is water, these isomers are called hydrate isomers.
• In these isomers, the total number of water molecules may remain the same, but their positions (inside or outside the coordination sphere) change.
• For example, \( CrCl_3.6H_2O \) can exist as three hydrate isomers:

These isomers have different physical properties like color and conductivity because the number of ions outside the coordination sphere changes. This type of isomerism highlights the dynamic nature of solvent-ligand interactions.
In simple words: Solvate isomerism is when a compound has the same parts, but water molecules (or other solvent molecules) are arranged differently—some inside the main complex, and some outside.

🎯 Exam Tip: Pay close attention to how many solvent molecules are inside and outside the coordination sphere, as this affects the number of ions produced in solution and thus conductivity.

 

Question 18. What is crystal field splitting energy?
Answer: Crystal field splitting energy (CFSE) is the energy difference that arises when the d-orbitals of a central metal ion in a complex split into different energy levels due to the presence of ligands. When ligands approach a metal ion, they cause the d-orbitals, which were initially all at the same energy (degenerate), to separate into higher and lower energy groups.
Specifically, the d-orbitals lying along the axes (\( d_{x^2-y^2} \) and \( d_{z^2} \)) experience strong repulsion from the ligands and thus increase in energy. In contrast, the d-orbitals with lobes directed between the axes (\( d_{xy} \), \( d_{yz} \), and \( d_{zx} \)) experience less repulsion and decrease in energy. This process of energy separation is called crystal field splitting. The amount of energy difference between these split sets of orbitals is the crystal field splitting energy.
In simple words: When ligands surround a metal atom, they push on the metal's d-electrons. This push is not even, so the d-orbitals split into different energy levels. The energy gap between these new levels is called crystal field splitting energy.

🎯 Exam Tip: Clearly define crystal field splitting as the degeneracy breaking of d-orbitals and crystal field splitting energy as the resulting energy gap, linking it to ligand repulsion.

 

Question 19. What is crystal field stabilization energy? (CFSE)? (PTA -1)
Answer: Crystal field stabilization energy (CFSE) is defined as the energy difference between the electronic configuration of a metal ion in the ligand field (ELF) and its electronic configuration in an isotropic (spherically symmetrical) field or barycentre (\( E_{iso} \)). In simpler terms, it's the net energy released when d-electrons fill the split d-orbitals in a coordination complex compared to their energy in a hypothetical spherical crystal field where all d-orbitals have the same energy.
The formula for CFSE in an octahedral field (\( \Delta_o \)) is:
\( CFSE (\Delta_o) = \{[nt_{2g}(-0.4) + n_{eg}(0.6)] \Delta_o + n_pP\} - \{n'_pP\} \)
Where:
\( nt_{2g} \) = Number of electrons in \( t_{2g} \) orbitals
\( n_{eg} \) = Number of electrons in \( e_g \) orbitals
\( n_p \) = Number of electron in the ligand field
\( n'_p \) = Number of electron pairs in the isotropic field (barycentre)
\( P \) = pairing energy
This stabilization energy contributes to the overall stability of a coordination complex. A higher negative CFSE value indicates greater stability.
In simple words: CFSE is the extra stability a metal atom gets when its electron shells split up in a complex. It's calculated by adding up how much energy the electrons lose or gain when they settle into these new energy levels.

🎯 Exam Tip: Understand that CFSE quantifies the stabilization achieved when d-electrons occupy the lower energy orbitals after splitting, and be able to apply the formula correctly based on electron configuration.

 

Question 20. A solution of [Ni(H2O)6]2+ is green, whereas a solution of [Ni(CN)4]2- is colorless - Explain.
Answer: The color of coordination compounds arises from d-d transitions, where electrons move between the split d-orbitals by absorbing light in the visible region.
• In \( [Ni(H_2O)_6]^{2+} \), the \( Ni^{2+} \) ion has an \( d^8 \) electron configuration. \( H_2O \) is a weak field ligand, so it does not cause electron pairing. This means there are two unpaired electrons in the d-orbitals.
• Because there are unpaired electrons, d-d electronic transitions are possible.
• When white light passes through the solution, \( [Ni(H_2O)_6]^{2+} \) absorbs red light (lower energy) and transmits the complementary color, which is green. Therefore, the solution appears green. The absorption of specific wavelengths causes the complex to be colored.
• In \( [Ni(CN)_4]^{2-} \), the \( Ni^{2+} \) ion also has an \( d^8 \) electron configuration. However, \( CN^- \) is a strong field ligand. It causes all the d-electrons to pair up.
• As a result, there are no unpaired electrons in \( [Ni(CN)_4]^{2-} \).
• With no unpaired electrons, d-d transitions are not possible. Thus, the complex does not absorb light in the visible region and appears colorless. This is why strong field ligands often lead to diamagnetic and colorless complexes when d-d transitions are prevented.
In simple words: The green complex has electrons that can jump between energy levels by absorbing red light, making it look green. The colorless complex has all its electrons paired up, so they can't jump, and no light is absorbed, making it clear.

🎯 Exam Tip: The key to explaining color in complexes is to consider the ligand strength (weak vs. strong field) and its effect on electron pairing and d-d transitions.

 

Question 21. Discuss briefly the nature of bonding in metal carbonyls. (PTA – 2)
Answer: In metal carbonyls, the bonding between the metal atom and the carbonyl (CO) ligand is unique and involves two main components, known as synergic bonding. This type of bonding strengthens both the metal-carbon bond and the carbon-oxygen bond in the ligand.
• The first component is a sigma ( \( \sigma \) ) bond. The carbon atom of the CO ligand donates an electron pair from its filled sigma orbital into a vacant d-orbital of the central metal atom. This forms a Metal \( \leftarrow \) CO sigma bond.
• This donation increases the electron density on the metal, making it electron-rich.
• To balance this increased electron density, the metal atom then donates electrons back to the CO ligand. This is the second component, called pi ( \( \pi \) ) back-bonding. A filled metal d-orbital interacts with the empty antibonding pi (\( \pi^* \)) orbital on the carbonyl ligand.
• Thus, electron density moves from the ligand to the metal through sigma bonding, and from the metal to the ligand through pi back-bonding.
• This synergistic effect mutually strengthens both the metal-carbon and carbon-oxygen bonds, leading to a strong M-CO bond in metal carbonyls. This unique electron delocalization stabilizes the complex.
In simple words: In metal carbonyls, the CO ligand first gives electrons to the metal, then the metal gives some back to the CO. This two-way electron sharing makes the bond very strong.

🎯 Exam Tip: Crucially, mention both the sigma donation from CO to metal and the pi back-donation from metal to CO, explaining how this synergic bonding stabilizes the complex.

 

Question 22. What is the coordination entity formed when an excess of liquid ammonia is added to an aqueous solution of copper sulphate?
Answer: When an excess of liquid ammonia is added to an aqueous solution of copper sulphate (CuSO4), a deep blue complex, tetraamminecopper(II) sulphate, is formed. The copper(II) ions react with ammonia molecules to form a stable coordination entity.
The reaction is:
\( CuSO_4 + 4NH_3 \rightarrow [Cu(NH_3)_4]SO_4 \)
The coordination entity formed is \( [Cu(NH_3)_4]^{2+} \). This complex ion is responsible for the characteristic deep blue color seen in this reaction, which is often used as a qualitative test for copper(II) ions.
In simple words: When a lot of ammonia is added to copper sulphate solution, it forms a deep blue compound. The special part of this blue compound is called the coordination entity, which is copper surrounded by four ammonia molecules.

🎯 Exam Tip: Remember this reaction as a classic example of complex formation and a qualitative test for copper(II) ions, always including the balanced equation and the correct formula of the coordination entity.

 

Question 23. On the basis of VB theory explain the nature of bonding in [Co(C2O4)3]3-
Answer: To explain the bonding in \( [Co(C_2O_4)_3]^{3-} \) using Valence Bond (VB) theory, we first determine the oxidation state of the central metal ion, its electron configuration, and then how ligands influence the hybridization.

In this complex, cobalt is in a +3 oxidation state, meaning it has a \( 3d^6 \) configuration. Since oxalate (\( C_2O_4^{2-} \)) is a weak field ligand, it does not force the pairing of electrons in the d-orbitals. As a result, the \( Co^{3+} \) uses its outer \( 4s \), \( 4p \), and \( 4d \) orbitals for hybridization, forming \( sp^3d^2 \) hybrid orbitals. This leads to an outer orbital octahedral complex with four unpaired electrons, making it paramagnetic.
In simple words: In this cobalt complex, the oxalate ligands are weak, so they don't force the cobalt's electrons to pair up. This means the cobalt uses its outer electron shells to bond, forming an octahedral shape and having unpaired electrons that make it magnetic.

🎯 Exam Tip: For VB theory explanations, always clearly state the metal's oxidation state, d-electron configuration, ligand strength (weak/strong field), resulting hybridization, geometry, and magnetic properties.

 

Question 24. What are the limitations of VB theory?
Answer: Valence Bond (VB) theory, while useful, has several limitations when explaining coordination compounds:
1. It does not explain the color of the complex. VB theory cannot account for why complexes absorb specific wavelengths of light, leading to observed colors.
2. It considers only the spin-only magnetic moments and does not consider orbital magnetic moments. This means it often gives incomplete information about the magnetic properties.
3. It does not provide a quantitative explanation as to why certain complexes are inner orbital complexes and others are outer orbital complexes for the same metal. For example, it struggles to explain why \( [Fe(CN)_6]^{4-} \) is diamagnetic (low spin) whereas \( [Fe(CN)_6]^{3-} \) is paramagnetic (high spin), even though both involve iron and cyanide ligands.
4. It does not explain why some complexes are kinetically inert (slow to react) and others are labile (fast to react).
5. It doesn't give a clear picture of the distortion in the geometry of certain complexes.
These limitations mean that while VB theory offers a simple picture of bonding, more advanced theories like Crystal Field Theory (CFT) and Ligand Field Theory (LFT) are needed for a complete understanding.
In simple words: VB theory can't explain why complexes have certain colors or why some are more stable than others. It also doesn't fully explain their magnetic behavior or how their exact shape is formed.

🎯 Exam Tip: When listing limitations, focus on what the theory *fails to explain*, such as color, quantitative magnetic properties, and the distinction between inner/outer orbital complexes without external data.

 

Question 25. Write the oxidation state, coordination number, nature of ligand, magnetic property and electronic configuration in octahedral crystal field for the complex K4[Mn(CN)6].
Answer: For the complex \( K_4[Mn(CN)_6] \), we can determine its key properties based on crystal field theory:

In this complex, manganese is in the \( +2 \) oxidation state and has a \( d^5 \) electron configuration. Since cyanide (\( CN^- \)) is a strong field ligand, it causes electron pairing. This results in a low-spin configuration where all five d-electrons occupy the lower energy \( t_{2g} \) orbitals, leaving one unpaired electron. This unpaired electron makes the complex paramagnetic.
In simple words: For the compound \( K_4[Mn(CN)_6] \), the manganese metal has a +2 charge. Six cyanide ligands attach to it. Since cyanide is a strong ligand, the electrons pair up, making the manganese \( t_{2g}^5e_g^0 \) and slightly magnetic (paramagnetic).

🎯 Exam Tip: For strong field ligands with \( d^5 \) metal ions, remember to draw the crystal field splitting diagram for octahedral geometry and show the low-spin configuration with one unpaired electron.

 

III. Evaluate yourself

 

Question 1. When a coordination compound CrCl3.4H2O is mixed with silver nitrate solution, one mole of silver chloride is precipitated per mole of the compound. There are no free solvent molecules in that compound. Assign the secondary valence to the metal and write the structural formula of the compound.
Answer: If one mole of silver chloride (AgCl) is precipitated per mole of the compound, it means that one chloride ion (\( Cl^- \)) is outside the coordination sphere and can ionize. The original compound is \( CrCl_3.4H_2O \). Since there are no free solvent molecules, all \( H_2O \) molecules must be inside the coordination sphere.
The central metal, Chromium (Cr), typically has a secondary valency (coordination number) of 6.
With four \( H_2O \) molecules inside, and one \( Cl^- \) ion outside, two \( Cl^- \) ions must be inside the coordination sphere to fulfill the coordination number of 6.
Therefore, the secondary valency of the metal (Cr) is 6.
The structural formula of the compound is \( [Cr(H_2O)_4Cl_2]Cl \).
In simple words: The compound \( CrCl_3.4H_2O \) forms one AgCl when reacted, meaning one chloride is outside the main complex. Since all water is inside, two more chlorides must be inside to make a total of six things around the chromium. So, the formula is \( [Cr(H_2O)_4Cl_2]Cl \), and the chromium has a secondary valency of 6.

🎯 Exam Tip: The amount of precipitate formed (e.g., AgCl) directly indicates the number of counter ions outside the coordination sphere. Use this to deduce the ligands inside the complex, keeping the coordination number in mind.

 

Question 2. In the complex, [Pt(NO2)(H2O)(NH3)2]Br, identify the following
i. Central metal atom/ion
ii. Ligand(s) and their types
iii. Coordination entity
iv. Oxidation number of the central metal ion
v. Coordination number
Answer: For the complex \( [Pt(NO_2)(H_2O)(NH_3)_2]Br \):
i. Central metal atom/ion: Platinum (Pt)
ii. Ligand(s) and their types:
     - \( NO_2^- \) (nitrito-N) – Anionic ligand, monodentate
     - \( H_2O \) (aqua) – Neutral ligand, monodentate
     - \( NH_3 \) (ammine) – Neutral ligand, monodentate
iii. Coordination entity: \( [Pt(NO_2)(H_2O)(NH_3)_2]^+ \)
iv. Oxidation number of the central metal ion:
     Let the oxidation state of Pt be x.
     \( x + (-1) + 0 + 2(0) = +1 \) (overall charge of coordination entity)
     \( x - 1 = +1 \)
     \( x = +2 \)
     So, Platinum is in the \( +2 \) oxidation state (Pt(II)).
v. Coordination number: The sum of monodentate ligands is \( 1 (NO_2) + 1 (H_2O) + 2 (NH_3) = 4 \). So, the coordination number is 4.
This compound is an example of a mixed-ligand complex where different types of ligands are attached to the central metal ion, leading to specific properties.
In simple words: In this complex, Platinum is the central metal. It has three types of ligands: nitrite, water, and ammonia. The whole complex part has a +1 charge. Platinum's charge is +2, and it has 4 ligands bonded, so its coordination number is 4.

🎯 Exam Tip: Always correctly identify the charge of each ligand when calculating the oxidation state of the central metal ion. Remember that water and ammonia are neutral ligands.

 

Question 3. Write the IUPAC name for the following compounds.
(i) K2[Fe(CN)3(Cl)2(NH3)]
(ii) [Cr(CN)2(H2O)4][Co(OX)2(en)]
(iii) [Cu(NH3)2Cl2]
(iv) [Cr(NH3)3(NC)2(H2O)]+
(v) [Fe(CN)6]4-
Answer: The IUPAC naming of coordination compounds follows specific rules, including naming ligands alphabetically and specifying oxidation states.
(i) \( K_2[Fe(CN)_3(Cl)_2(NH_3)] \)
    Potassium amminedichloridotricyanido-kCferrate(II)
(ii) \( [Cr(CN)_2(H_2O)_4][Co(OX)_2(en)] \)
    Tetraaquadi(cyanido-kC)chromium(III) bis(oxalato)ethylenediaminecobaltate(III)
(iii) \( [Cu(NH_3)_2Cl_2] \)
    Diamminedichloridocopper(II)
(iv) \( [Cr(NH_3)_3(NC)_2(H_2O)]^+ \)
    Aqua triammine di(isocyanato-kN)chromium(III) ion
(v) \( [Fe(CN)_6]^{4-} \)
    Hexacyanidoferrate(II) ion
IUPAC nomenclature provides a systematic way to name complex compounds, ensuring clarity and consistency in chemical communication. This helps identify the components and structure of coordination complexes.
In simple words: We give special names to these complex compounds. These names tell us what metal is in the middle, what other groups (ligands) are attached, and how many of each.

🎯 Exam Tip: Master the rules for alphabetical order of ligands, ending for anionic/cationic complexes, and correctly indicating oxidation states with Roman numerals. Pay attention to ambidentate ligands (kC vs kN).

 

Question 4. Give the structure for the following compounds.
(i) diamminesilver(I) dicyanidoargentate(I)
(ii) Pentaammine nitrito-KNcobalt (III) ion
(iii) hexafluorido cobaltate (III) ion
(iv) dichloridobis (ethylenediamine) Cobalt (III) sulphate
(v) Tetracarbonylnickel (0)
Answer: Writing the formula for coordination compounds requires understanding the metal's oxidation state, ligand charges, and the overall charge of the complex.
(i) Diamminesilver(I) dicyanidoargentate(I): \( [Ag(NH_3)_2][Ag(CN)_2] \)
(ii) Pentaammine nitrito-KNcobalt (III) ion: \( [Co(NH_3)_5(NO_2)]^{2+} \)
(iii) Hexafluorido cobaltate (III) ion: \( [CoF_6]^{3-} \)
(iv) Dichloridobis (ethylenediamine) Cobalt (III) sulphate: \( [Co(en)_2Cl_2]_2SO_4 \)
(v) Tetracarbonylnickel (0): \( [Ni(CO)_4] \)
These formulas correctly represent the components and charges of each coordination compound. The charges of the ligands and the central metal ion determine the overall charge of the complex, which balances the counter ions.
In simple words: These are the chemical recipes for the complex compounds. Each formula shows the central metal atom and all the groups (ligands) attached to it, including any ions that balance the charge.

🎯 Exam Tip: Pay attention to the oxidation state of the metal (indicated by Roman numerals) and the charges of the ligands to correctly balance the charges and write the overall formula. Remember the k-notation for ambidentate ligands.

 

Question 5. What should be the formula of isomer of the dissolved complex that gives yellow precipitate with AgNO3. What are the above isomers called?
Answer: Let's consider a complex with the empirical formula \( Co(NH_3)_4I_2Cl \).
• If the dissolved complex gives a yellow precipitate with \( AgNO_3 \), it means that a chloride ion (\( Cl^- \)) is present outside the coordination sphere. Therefore, the formula of the complex that would give a yellow precipitate with \( AgNO_3 \) is \( [Co(NH_3)_4I_2]Cl \).
• This complex has an iodide ion (\( I^- \)) inside the coordination sphere.
• The isomer mentioned above (where the \( Cl^- \) is outside the coordination sphere) and the original form \( [Co(NH_3)_4ICl]I \) (where \( I^- \) is outside and \( Cl^- \) is inside) are called ionisation isomers. They differ in which ion is outside the coordination sphere and can be precipitated.
Ionization isomers have the same molecular formula but produce different ions in solution. This difference in ionic species outside the coordination sphere can be detected by precipitation reactions.
In simple words: If a compound makes a yellow solid with silver nitrate, it means a chloride ion is freely available outside its main complex. These compounds that switch which ion is free are called ionisation isomers.

🎯 Exam Tip: Understand that ionisation isomers have the same composition but exchange ions between the coordination sphere and the counter-ion position, leading to different precipitation reactions.

 

Question 6. Three compounds A, B and C have empirical formula CrCl3.6H2O. They are kept in a container with a dehydrating agent and they lost water and attaining constant weight as shown below.
Answer: The empirical formula \( CrCl_3.6H_2O \) can exist as different hydrate isomers, which vary in the number of water molecules inside and outside the coordination sphere. The amount of water lost upon dehydration helps us identify these isomers.

The loss of water during dehydration tells us how many water molecules are outside the coordination sphere. For example, if a compound loses all its water (like compound C), it means all water molecules were inside the coordination sphere. If it loses some water, that means some were outside. This method allows us to deduce the specific structure of each hydrate isomer.
In simple words: We have three similar compounds with the same basic formula but different arrangements of water. By drying them and seeing how much water they lose, we can figure out exactly how many water molecules are attached inside the main complex and how many are just sitting outside.

🎯 Exam Tip: To solve this type of problem, calculate the molar mass of \( CrCl_3.6H_2O \) and the molar mass of water. Then, determine how many moles of water correspond to the given weight loss for each compound to find the number of water molecules outside the coordination sphere.

 

Question 7. Indicate the possible type of isomerism for the following complexes and draw their isomers,
i) [Co(en)3][Cr(CN)6]
ii) [Co(NH3)5(NO2)]2+
iii) [Pt(NH2)2(NO2)]Cl
Answer: The type of isomerism shown by these complexes depends on their structure and the nature of their ligands.
i) \( [Co(en)_3][Cr(CN)_6] \) and \( [Cr(en)_3][Co(CN)_6] \) are coordination isomers. Here, the ligands swap between the cationic and anionic complex parts.
ii) \( [Co(NH_3)_5(NO_2)]^{2+} \) and \( [Co(NH_3)_5(ONO)]^{2+} \) are linkage isomers. The nitrite ligand (\( NO_2^- \)) can bind through either nitrogen (nitro) or oxygen (nitrito).
iii) \( [Pt(NH_3)_2(NO_2)]Cl \) can exhibit linkage isomerism (nitrito vs nitro), and potentially geometrical isomerism if the complex is square planar. It also can form ionization isomers like \( [Pt(NH_3)_2Cl]NO_2 \).
Coordination isomers involve the exchange of ligands between a cationic and an anionic complex ion. Linkage isomers involve ambidentate ligands bonding through different atoms. Ionisation isomers involve the exchange of counter ions with ligands.
In simple words: These compounds can exist in different forms called isomers. The first one shows coordination isomerism (ligands swap between two parts). The second shows linkage isomerism (a ligand connects in two different ways). The third can show linkage or ionization isomerism, depending on how it's set up.

🎯 Exam Tip: For each complex, analyze the central metal's coordination number, the type of ligands (especially ambidentate ones), and the overall charge to determine all possible isomerism types.

Evaluate Yourself: 8

 

Question 1. Draw all possible stereoisomers of a complex \( \text{Ca[Co(NH3)Cl(OX)2]} \)
Answer: The complex \( \text{Ca[Co(NH3)Cl(Ox)2]} \) can be split into \( \text{Ca}^{2+} \) and \( \text{[Co(NH3)Cl(Ox)2]}^{2-} \). We can draw its stereoisomers, which include both geometrical isomers (cis and trans forms) and optical isomers (mirror images). The 'cis' forms are optically active and have mirror images, while the 'trans' form is optically inactive. This shows different ways the parts can be arranged in 3D space.
In simple words: This question asks to draw all the different 3D shapes for the given chemical compound. You need to show the 'cis' and 'trans' versions, and also their mirror images if they exist. The 'cis' form will have mirror images, but the 'trans' form won't.

🎯 Exam Tip: When drawing stereoisomers, clearly distinguish between cis and trans forms. Remember that cis isomers of such complexes can often be optically active, while trans isomers are usually not.

 

Evaluate Yourself: 9

 

Question 1. The spin-only magnetic moment of the Tetrachloridomanganate(II)ion is 5.9 BM. On the basis of VBT, predict the type of hybridisation and geometry of the compound.
Answer: For \( \text{[MnCl4]}^{2-} \):
Central metal ion is \( \text{Mn}^{2+} \), which has a \( \text{3d}^5 \text{4S}^0 \) electronic configuration.
The ligand \( \text{Cl}^- \) is a weak field ligand, meaning it does not cause the \( \text{3d} \) electrons to pair up. This leads to 5 unpaired electrons. Since there are 5 unpaired electrons, the magnetic moment calculated as \( \mu_s = \sqrt{n(n+2)} = \sqrt{5(5+2)} = \sqrt{35} \approx 5.9 \) BM. The hybridization is \( \text{sp}^3 \), which results in a tetrahedral geometry.
In simple words: The central metal is Manganese with a charge of +2. It has 5 electrons in its d-shell. The chloride ligands are weak and don't push these electrons to pair up. Because of this, there are 5 single, unpaired electrons. This leads to an \( \text{sp}^3 \) hybridization and a tetrahedral shape for the complex.

🎯 Exam Tip: Remember that weak field ligands do not cause electron pairing, leading to more unpaired electrons and higher magnetic moments.

 

Question 2. Predict the number of unpaired electrons in \( \text{[CoCl}_4\text{]}^{2-} \) ion on the basis of VBT.
Answer: For \( \text{[CoCl}_4\text{]}^{2-} \):
Central metal ion is \( \text{Co}^{2+} \), which has a \( \text{3d}^7 \text{4S}^0 \) electronic configuration. The chloride ligand \( \text{Cl}^- \) is a weak field ligand, so it does not cause electron pairing in the \( \text{3d} \) orbitals. This complex will have 3 unpaired electrons.
In simple words: Cobalt in this complex has a +2 charge. Its d-shell has 7 electrons. Since chloride is a weak ligand, these electrons stay unpaired as much as possible, resulting in 3 unpaired electrons.

🎯 Exam Tip: When using VBT, always determine the oxidation state of the metal, its d-electron configuration, and the nature of the ligand (strong or weak field) to predict pairing and hybridization.

 

Question 3. A metal complex \( \text{Co(en)}_2\text{Cl}_2\text{Br} \) has been isolated in two forms A and B. (b) reacted with silver nitrate to give a white precipitate readily soluble in ammonium hydroxide. Whereas A gives a pale yellow precipitate. Write the formula of A and B. state the hybridization of Co in each and calculate their spin only magnetic moment.
Answer:
Since form B gives a white precipitate with silver nitrate (\( \text{AgNO}_3 \)), it means B contains a free chloride ion (\( \text{Cl}^- \)) outside the coordination sphere. Therefore, the formula for B is \( \text{[Co(en)}_2\text{ClBr]Cl} \).
Since form A gives a pale yellow precipitate with silver nitrate, it means A contains a free bromide ion (\( \text{Br}^- \)) outside the coordination sphere. Therefore, the formula for A is \( \text{[Co(en)}_2\text{Cl}_2\text{]Br} \).

For \( \text{[Co(en)}_2\text{Cl}_2\text{]Br} \):
Central metal ion is \( \text{Co}^{3+} \). Its electronic configuration is \( \text{3d}^6 \text{4S}^0 \).
The ligand 'en' (ethylenediamine) is a strong field ligand, which causes the \( \text{3d} \) electrons to pair up.
This leads to 0 unpaired electrons. The hybridization is \( \text{d}^2\text{sp}^3 \), which results in an octahedral geometry. The magnetic moment is \( \mu_s = \sqrt{0(0+2)} = 0 \). This complex is diamagnetic.
In simple words: Form B gives a white solid with silver nitrate, showing it has a free chloride ion. So B is \( \text{[Co(en)}_2\text{ClBr]Cl} \). Form A gives a yellow solid, showing it has a free bromide ion. So A is \( \text{[Co(en)}_2\text{Cl}_2\text{]Br} \). For complex A, Cobalt has a +3 charge and 6 d-electrons. Since 'en' is a strong ligand, all electrons pair up, making it diamagnetic with \( \text{d}^2\text{sp}^3 \) hybridization and an octahedral shape.

🎯 Exam Tip: Pay attention to the specific precipitates formed with \( \text{AgNO}_3 \), as this indicates which anion is outside the coordination sphere and helps determine the complex's formula.

 

Evaluate Yourself: 10

 

Question 1. The mean pairing energy and octahedral field splitting energy of \( \text{[Mn(CN)}_6\text{]}^{3-} \) are \( \text{28,800 cm}^{-1} \) and \( \text{38,500 cm}^{-1} \) respectively. Whether this complex is stable in low spin or high spin?
Answer: For \( \text{[Mn(CN)}_6\text{]}^{3-} \):
Central metal ion is \( \text{Mn}^{3+} \). Its electronic configuration is \( \text{3d}^4 \).
Mean pairing energy (\( \text{P} \)) = \( \text{28,800 cm}^{-1} \)
Octahedral field splitting energy (\( \Delta_0 \)) = \( \text{38,500 cm}^{-1} \)

High Spin \( \text{Mn}^{3+} \) complex:
\( \text{nt}_{2g} = 3 \), \( \text{n}_{eg} = 1 \), \( \text{n}_p = 0 \), \( \text{n}'_p = 0 \)
\( \text{CFSE}(\Delta_0) = [\text{nt}_{2g}(-0.4) + \text{n}_{eg}(0.6)]\Delta_0 + \text{n}_p\text{P} - \text{n}'_p\text{P} \)
\( \text{CFSE} = [3(-0.4) + 1(0.6)] \times 38,500 + 0 \times 28,800 - 0 \times 28,800 \)
\( \text{CFSE} = [-1.2 + 0.6] \times 38,500 = -0.6 \times 38,500 = \text{-23,100 cm}^{-1} \)

Low Spin \( \text{Mn}^{3+} \) complex:
\( \text{nt}_{2g} = 4 \), \( \text{n}_{eg} = 0 \), \( \text{n}_p = 1 \), \( \text{n}'_p = 0 \)
\( \text{CFSE}(\Delta_0) = [\text{nt}_{2g}(-0.4) + \text{n}_{eg}(0.6)]\Delta_0 + \text{n}_p\text{P} - \text{n}'_p\text{P} \)
\( \text{CFSE} = [4(-0.4) + 0(0.6)] \times 38,500 + 1 \times 28,800 - 0 \times 28,800 \)
\( \text{CFSE} = [-1.6 + 0] \times 38,500 + 28,800 = -1.6 \times 38,500 + 28,800 \)
\( \text{CFSE} = -61,600 + 28,800 = \text{-32,800 cm}^{-1} \)

Since the low spin \( \text{Mn}^{3+} \) complex has a more negative CFSE value ( \( \text{-32,800 cm}^{-1} \) vs \( \text{-23,100 cm}^{-1} \) ), it is more favored and stable.
In simple words: To find out if the complex is stable in high spin or low spin, we calculate its Crystal Field Stabilization Energy (CFSE) for both cases. For \( \text{[Mn(CN)}_6\text{]}^{3-} \), the low spin state has a more negative CFSE value than the high spin state. A more negative CFSE means the complex is more stable, so the low spin state is favored.

🎯 Exam Tip: When comparing stability, always look for the more negative CFSE value, as this indicates greater stabilization for the complex.

 

Question 2. Draw energy level diagram and indicate the number of electrons in each level for the complex \( \text{[Cu(H2O)}_6\text{]}^{2+} \). Whether the complex is paramagnetic or diamagnetic?
Answer: For \( \text{[Cu(H}_2\text{O)}_6\text{]}^{2+} \):
Central metal ion is \( \text{Cu}^{2+} \). Its electronic configuration is \( \text{3d}^9 \text{4S}^0 \).
The water ligand \( \text{H}_2\text{O} \) is a weak field ligand.


After crystal field splitting, the \( \text{t}_{2g} \) orbitals are filled with 6 electrons (3 pairs), and the \( \text{e}_{g} \) orbitals are filled with 3 electrons (1 pair and 1 unpaired electron). Since there is one unpaired electron present in the \( \text{e}_{g} \) orbital, the complex \( \text{[Cu(H}_2\text{O)}_6\text{]}^{2+} \) is paramagnetic. The coordination number is 6, and the geometry is octahedral.
In simple words: Copper has a +2 charge with 9 d-electrons. When water ligands attach, the d-orbitals split into two energy levels. In this complex, there is one electron left alone in the higher energy level. Because of this single, unpaired electron, the complex is paramagnetic, meaning it is attracted to magnets. The overall shape of the complex is octahedral.

🎯 Exam Tip: For d-electron counts like \( \text{d}^4 \) to \( \text{d}^7 \) in octahedral complexes, remember to consider both strong and weak field ligands, as they impact electron pairing and thus the magnetic property. For \( \text{d}^1 \) to \( \text{d}^3 \) and \( \text{d}^8 \) to \( \text{d}^{10} \), the magnetic properties are usually consistent regardless of ligand strength.

 

Question 3. For the \( \text{[CoF}_6\text{]}^{3-} \) ion the mean pairing energy is found to be \( \text{21000 cm}^{-1} \). The magnitude of \( \Delta_0 \) is \( \text{13000cm}^{-1} \). Calculate the crystal field stabilization energy for low spin and high spin states.
Answer: For \( \text{[CoF}_6\text{]}^{3-} \):
Central metal ion is \( \text{Co}^{3+} \). Its electronic configuration is \( \text{3d}^6 \).
Mean pairing energy (\( \text{P} \)) = \( \text{21,000 cm}^{-1} \)
Octahedral field splitting energy (\( \Delta_0 \)) = \( \text{13,000 cm}^{-1} \)

High Spin \( \text{Co}^{3+} \) complex:
\( \text{nt}_{2g} = 4 \), \( \text{n}_{eg} = 2 \), \( \text{n}_p = 1 \), \( \text{n}'_p = 1 \)
\( \text{CFSE} = [\text{nt}_{2g}(-0.4) + \text{n}_{eg}(0.6)]\Delta_0 + \text{n}_p\text{P} - \text{n}'_p\text{P} \)
\( \text{CFSE} = [4(-0.4) + 2(0.6)] \times 13,000 + 1 \times 21,000 - 1 \times 21,000 \)
\( \text{CFSE} = [-1.6 + 1.2] \times 13,000 + 0 \)
\( \text{CFSE} = -0.4 \times 13,000 = \text{-5,200 cm}^{-1} \)

Low Spin \( \text{Co}^{3+} \) complex:
\( \text{nt}_{2g} = 6 \), \( \text{n}_{eg} = 0 \), \( \text{n}_p = 3 \), \( \text{n}'_p = 1 \)
\( \text{CFSE} = [\text{nt}_{2g}(-0.4) + \text{n}_{eg}(0.6)]\Delta_0 + \text{n}_p\text{P} - \text{n}'_p\text{P} \)
\( \text{CFSE} = [6(-0.4) + 0(0.6)] \times 13,000 + 3 \times 21,000 - 1 \times 21,000 \)
\( \text{CFSE} = [-2.4 + 0] \times 13,000 + 63,000 - 21,000 \)
\( \text{CFSE} = -2.4 \times 13,000 + 42,000 = -31,200 + 42,000 = \text{+10,800 cm}^{-1} \)

Since the high spin complex has a negative CFSE value (\( \text{-5,200 cm}^{-1} \)) and the low spin complex has a positive CFSE value (\( \text{+10,800 cm}^{-1} \)), the high spin complex is more favored for \( \text{[CoF}_6\text{]}^{3-} \).
In simple words: We calculate the Crystal Field Stabilization Energy (CFSE) for both high spin and low spin states of \( \text{[CoF}_6\text{]}^{3-} \). The high spin complex has a negative CFSE, which means it is more stable. The low spin complex has a positive CFSE, meaning it is less stable. So, the high spin arrangement is preferred for this complex.

🎯 Exam Tip: Always compare the calculated CFSE values: a more negative value indicates greater stability for that spin state.

12th Chemistry Guide Coordination Chemistry Additional Questions

Part II – Additional Questions

I. Choose the best answer.

 

Question 1. A theory of co-ordination compounds was proposed by
(a) Rutherford
(b) J.J.Thomson
(c) Alfred Werner
(d) Neils Bohr
Answer: (c) Alfred Werner
In simple words: Alfred Werner was the scientist who came up with the basic ideas about how coordination compounds are put together.

🎯 Exam Tip: Remember Alfred Werner's name as the pioneer of coordination chemistry theory; it's a fundamental historical fact.

 

Question 2. The primary valence of a metal ion refers to
(a) atomic number
(b) oxidation state
(c) atomic mass
(d) Co-ordination number
Answer: (b) oxidation state
In simple words: The primary valence of a metal in a complex just means its oxidation state, which is like its electrical charge.

🎯 Exam Tip: Distinguish between primary and secondary valence: primary relates to oxidation state, while secondary relates to coordination number.

 

Question 3. The secondary valence of a metal ion refers to
(a) atomic number
(b) oxidation state
(c) atomic mass
(d) Co-ordination number
Answer: (d) Co-ordination number
In simple words: The secondary valence tells us how many ligands (or groups) are directly attached to the central metal atom.

🎯 Exam Tip: Understand that the secondary valence determines the geometry of the complex.

 

Question 4. Primary valence of a metal ion is always satisfied by..............
(a) Positive ions
(b) negative ions
(c) neutral molecules
(d) All of the options
Answer: (b) negative ions
In simple words: The main charge of a metal atom is usually balanced by negative ions.

🎯 Exam Tip: Primary valence is always satisfied by anions, while secondary valence can be satisfied by anions, neutral molecules, or even cations in some cases.

 

Question 5. Among the following complexes, which one shows Zero Crystal field stability: energy (CFSE) is (PTA- 6)
(a) \( \text{d}^{10} \) configuration
(b) \( \text{[Fe(H}_2\text{O)}_6\text{]}^{3+} \)
(c) \( \text{[Co(H}_2\text{O)}_6\text{]}^{2+} \)
(d) \( \text{[Co(H}_2\text{O)}_6\text{]}^{3+} \)
Answer: (b) \( \text{[Fe(H}_2\text{O)}_6\text{]}^{3+} \)
In simple words: For this complex, the Iron atom has a d5 electron configuration and water is a weak ligand, so the electrons do not pair up. This results in zero Crystal Field Stabilization Energy, meaning it doesn't gain extra stability from the splitting of d-orbitals.

🎯 Exam Tip: Remember that high-spin \( \text{d}^0 \), \( \text{d}^5 \), and \( \text{d}^{10} \) electron configurations in octahedral complexes always have zero CFSE.

 

Question 6. The sphere in which the metal ion and ligands are firmly attached is called
(a) outer sphere
(b) ionization sphere
(c) Coordination sphere
(d) None of the options
Answer: (c) Coordination sphere
In simple words: The part of a complex where the metal and its attached groups are tightly held together is called the coordination sphere.

🎯 Exam Tip: The coordination sphere (or inner sphere) contains the central metal and ligands directly bonded to it; it's a stable unit that does not ionize in solution.

 

Question 7. The sphere in which the metal ion and other ions are loosely bound is called
(a) Inner sphere
(b) Ionization sphere
(c) Coordination sphere
(d) None of the options
Answer: (b) Ionization sphere
In simple words: The outer part of a complex, where other ions are loosely attached and can easily break off in water, is known as the ionization sphere.

🎯 Exam Tip: The ionization sphere (or outer sphere) contains counter ions that balance the charge of the complex and can be easily removed or exchanged in solution.

 

Question 8. The groups which satisfy the secondary valence are called
(a) Central metal ion
(b) ligands
(c) Complexion
(d) Complex
Answer: (b) ligands
In simple words: The secondary valence of a metal atom is taken care of by the ligands, which are the molecules or ions directly attached to it.

🎯 Exam Tip: Ligands are the key components that fulfill the secondary valence and define the geometry around the central metal ion.

 

Question 9. In a complex a ligand acts as a
(a) Lewis acid
(b) Lewis base
(c) Bronsted acid
(d) Bronsted base
Answer: (a) Lewis acid
In simple words: In a complex, a ligand acts like a Lewis acid because it accepts electrons from the central metal atom to form bonds.

🎯 Exam Tip: Ligands in coordination chemistry typically act as Lewis bases (electron pair donors), but some can act as Lewis acids in specific bonding situations, such as \( \pi \)-back bonding.

 

Question 10. In a complex a ligand acts as a
(a) Lewis acid
(b) Lewis base
(c) Bronsted acid
(d) Bronsted base
Answer: (b) Lewis base
In simple words: Usually, in a complex, a ligand gives away a pair of electrons, so it acts as a Lewis base.

🎯 Exam Tip: Most ligands donate electron pairs to the central metal, classifying them as Lewis bases, while the metal acts as a Lewis acid.

 

Question 11. In a complex primary valence of a metal ion is
(a) ionizable
(b) non-ionisable
(c) directional
(d) Coordination number
Answer: (a) ionizable
In simple words: The primary valence of a metal ion can form bonds that easily break apart into ions when put in a solution.

🎯 Exam Tip: Primary valence is ionizable and non-directional, while secondary valence is non-ionizable and directional, determining the complex's geometry.

 

Question 12. In a complex secondary valence of a metal ion is
(a) ionizable
(b) non-ionisable
(c) non-directional
(d) Oxidation state
Answer: (b) non-ionisable
In simple words: The secondary valence of a metal ion creates strong bonds that do not easily break apart into ions.

🎯 Exam Tip: The bonds formed by secondary valence are strong, leading to the non-ionizable nature of the coordination sphere.

 

Question 13. In a complex primary valence of a metal ion is
(a) non-ionizable
(b) directional
(c) non-directional
(d) Co-ordination number
Answer: (c) non-directional
In simple words: The primary valence of a metal ion does not point in a specific direction in space; it just represents the overall charge.

🎯 Exam Tip: Recall that primary valence determines the oxidation state and is non-directional, whereas secondary valence determines geometry and is directional.

 

Question 14. According to spectrochemical series which of the following ligand produces strongest field and cause maximum splitting? (PTA -1)
(a) \( \text{F}^- \)
(b) CO
(c) \( \text{H}_2\text{O} \)
(d) \( \text{Cl}^- \)
Answer: (b) CO
In simple words: Among the options, carbon monoxide (CO) is known to create the strongest field around a metal atom, causing the largest energy difference between its d-orbitals.

🎯 Exam Tip: Familiarize yourself with the spectrochemical series, which ranks ligands based on their ability to cause crystal field splitting; strong field ligands like CO and \( \text{CN}^- \) cause large splitting.

 

Question 15. In the complex \( \text{K}_4\text{[Fe(CN)}_6\text{]} \) the central metal ion is
(a) \( \text{K}^+ \)
(b) \( \text{Fe}^{2+} \)
(c) \( \text{Fe}^{3+} \)
(d) \( \text{CN}^- \)
Answer: (b) \( \text{Fe}^{2+} \)
In simple words: In this compound, the metal ion at the center is Iron with a +2 charge.

🎯 Exam Tip: Always identify the central metal ion and calculate its oxidation state first when analyzing coordination complexes.

 

Question 16. In the complex \( \text{K}_4\text{[Fe(CN)}_6\text{]} \) the ligand is
(a) \( \text{K}^+ \)
(b) \( \text{Fe}^{2+} \)
(c) \( \text{Fe}^{3+} \)
(d) \( \text{CN}^- \)
Answer: (d) \( \text{CN}^- \)
In simple words: In this complex, the cyanide ion is the ligand, which is the part that attaches to the central metal.

🎯 Exam Tip: Ligands are the molecules or ions directly coordinated to the central metal within the square brackets of the complex formula.

 

Question 17. In the complex \( \text{K}_4\text{[Fe(CN)}_6\text{]} \) the primary valence of the central metal ion is
(a) +2
(b) +3
(c) +4
(d) +6
Answer: (a) +2
In simple words: For this complex, the primary valence of the iron metal, which is its charge, is +2.

🎯 Exam Tip: To find the primary valence (oxidation state), consider the charge of the counter ion and the total charge contribution from the ligands.

 

Question 18. In the complex \( \text{K}_4\text{[Fe(CN)}_6\text{]} \) the secondary valence of the central metal ion is
(a) 2
(b) 3
(c) 4
(d) 6
Answer: (d) 6
In simple words: The secondary valence for the iron in this complex is 6, meaning six cyanide ligands are directly attached to it.

🎯 Exam Tip: The secondary valence is simply the coordination number, which is the total number of ligand donor atoms attached to the central metal ion.

 

Question 19. The geometry of the complex \( \text{K}_4\text{[Fe(CN)}_6\text{]} \) is
(a) Square planar
(b) tetrahedral
(c) Octahedral
(d) trigonal bipyramidal
Answer: (c) Octahedral
In simple words: Because six ligands are attached to the central iron atom, the complex takes on an octahedral, or eight-sided, shape.

🎯 Exam Tip: A coordination number of 6 typically corresponds to an octahedral geometry.

 

Question 20. [Question text missing in source]
(a) Wilkinson Catalyst
(b) Zeiglar Natta catalyst
(c) Magnus's green salt
(d) Zeise's salt
Answer: (d) Zeise's salt
In simple words: Zeise's salt is a famous early example of an organometallic compound that contains a metal-alkene bond.

🎯 Exam Tip: Zeise's salt is a historically important organometallic complex involving platinum and ethene, known for its unique \( \pi \)-bonding nature.

 

Question 21. Which among the following is not a neutral ligand?
(a) aqua
(b) ammine
(c) Oxalato
(d) Pyridine
Answer: (c) Oxalato
In simple words: Among the choices, "oxalato" refers to the oxalate ion, which has a negative charge, making it a negatively charged ligand, not a neutral one.

🎯 Exam Tip: To identify neutral ligands, look for names that do not end in '-o' (for anionic) or '-ium' (for cationic), with common exceptions like aqua (\( \text{H}_2\text{O} \)) and ammine (\( \text{NH}_3 \)).

 

Question 22. Which is a monodentate ligand?
(a) Carbonato
(b) Oxalato
(c) Cyanido
(d) en
Answer: (c) Cyanido
In simple words: A monodentate ligand attaches to the metal at only one point, and the cyanido ligand (cyanide ion) does exactly that.

🎯 Exam Tip: Monodentate ligands have a single donor atom, while bidentate (like 'en' and oxalato) and polydentate ligands have multiple donor atoms.

 

Question 23. \( \text{[Co(NH}_3\text{)}_5\text{NO}_2\text{]}^{2+} \) ion exhibits
(a) Ionisation isomerism
(b) Linkage isomerism
(c) Coordination isomerism
(d) solvate isomerism
Answer: (b) Linkage isomerism
In simple words: This compound can show linkage isomerism because the nitrite group (\( \text{NO}_2 \)) can attach to the cobalt metal in two different ways, either through its nitrogen atom or its oxygen atom.

🎯 Exam Tip: Linkage isomerism occurs with ambidentate ligands (e.g., \( \text{NO}_2^- \), \( \text{SCN}^- \)) that can bind to the central metal through different donor atoms.

 

Question 24. \( \text{[Pt(NH}_3\text{)}_4\text{] [Pd(Cl)}_4\text{]} \) complex exhibits
(a) Ionisation isomerism
(b) Linkage isomerism
(c) Coordination isomerism
(d) solvate isomerism
Answer: (c) Coordination isomerism
In simple words: This complex can show coordination isomerism because the ligands can be swapped between the two metal ions (platinum and palladium) in the complex.

🎯 Exam Tip: Coordination isomerism happens when the cation and anion are both complex ions, and ligands can exchange between them.

 

Question 25. \( \text{[Cr(NH}_3\text{)}_4\text{ClBr]NO}_2 \) and \( \text{[Cr(NH}_3\text{)}_4\text{ClNO}_2\text{]Br} \) are
(a) Linkage isomers
(b) Ionisation isomers
(c) Coordination isomers
(d) solvate isomers
Answer: (b) Ionisation isomers
In simple words: These two compounds are ionization isomers because the nitrate and bromide ions can switch places, with one being inside the complex and the other outside as a free ion.

🎯 Exam Tip: Ionization isomers differ in which ion is present outside the coordination sphere and can be ionized in solution.

 

Question 26. Which among the following gives a curdy white precipitate with silver nitrate solution?
(a) \( \text{[Co(NH}_3\text{)}_4\text{ClBr]Br} \)
(b) \( \text{[Co(NH}_3\text{)}_4\text{Br}_2\text{]Cl} \)
(c) \( \text{[Cr(NH}_3\text{)}_4\text{ClBr]NO}_2 \)
(d) \( \text{[Cr(NH}_3\text{)}_4\text{ClNO}_2\text{]Br} \)
Answer: (b) \( \text{[Co(NH}_3\text{)}_4\text{Br}_2\text{]Cl} \)
In simple words: Only the compound with a free chloride ion outside the square brackets will react with silver nitrate to form a white solid (silver chloride). In this case, option (b) has a chloride ion that can be released.

🎯 Exam Tip: A curdy white precipitate with silver nitrate (\( \text{AgNO}_3 \)) indicates the presence of a free chloride ion (\( \text{Cl}^- \)) outside the coordination sphere, forming \( \text{AgCl} \).

 

Question 27. Which among the following gives three moles of silver chloride with silver nitrate solution?
(b) \( \text{[Cr(H}_2\text{O)}_5\text{Cl]Cl}_2\text{.2H}_2\text{O} \)
(c) \( \text{[Cr(H}_2\text{O)}_6\text{]Cl}_3 \)
(d) \( \text{[Cr(H}_2\text{O)}_3\text{Cl}_3\text{].3H}_2\text{O} \)
Answer: (c) \( \text{[Cr(H}_2\text{O)}_6\text{]Cl}_3 \)
In simple words: To get three moles of silver chloride, the original complex must have three chloride ions outside its coordination sphere that can react. Option (c) is the only one where all three chlorides are outside the brackets, making them available to form \( \text{AgCl} \).

🎯 Exam Tip: The number of moles of \( \text{AgCl} \) precipitated is equal to the number of chloride ions present outside the coordination sphere (i.e., not directly bonded to the metal).

 

Question 28. Which among the following gives a white precipitate with barium chloride solution?
(a) \( \text{[Co(NH}_3\text{)}_5\text{Cl]Cl}_2 \)
(b) \( \text{[Cu(NH}_3\text{)}_4\text{]SO}_4 \)
(c) \( \text{[Ag(NH}_3\text{)}_2\text{]Br} \)
(d) \( \text{[Co(NO}_2\text{)}_3\text{(NH}_3\text{)}_3\text{]} \)
Answer: (b) \( \text{[Cu(NH}_3\text{)}_4\text{]SO}_4 \)
In simple words: Barium chloride solution reacts to form a white solid with compounds that have a free sulfate ion (\( \text{SO}_4^{2-} \)). Only option (b) contains a sulfate ion outside the complex.

🎯 Exam Tip: A white precipitate with barium chloride (\( \text{BaCl}_2 \)) indicates the presence of a free sulfate ion (\( \text{SO}_4^{2-} \)) outside the coordination sphere, forming \( \text{BaSO}_4 \).

 

Question 29. According to VBT, in a Coordination complex the bonding between central metal ion and ligands is
(a) ionic
(b) covalent
(c) metallic
(d) vander waals force
Answer: (b) covalent
In simple words: Valence Bond Theory (VBT) explains that the connection between the metal and the ligands in a complex is a type of covalent bond, where electrons are shared.

🎯 Exam Tip: VBT describes the metal-ligand bond as coordinate covalent (dative) bonds, where ligands donate electron pairs to the metal.

 

Question 30. In which of the following complexes the central metal ion undergoes \( \text{sp}^3 \) hybridisation
(a) \( \text{[Ni(CN)}_4\text{]}^{2-} \)
(b) \( \text{[Pt(NH}_3\text{)}_4\text{]}^{2+} \)
(c) \( \text{[NiCl}_4\text{]}^{2-} \)
(d) \( \text{[Cu(NH}_3\text{)}_4\text{]}^{2+} \)
Answer: (c) \( \text{[NiCl}_4\text{]}^{2-} \)
In simple words: For \( \text{[NiCl}_4\text{]}^{2-} \), the nickel atom uses one s-orbital and three p-orbitals to form four new identical hybrid orbitals. This creates an \( \text{sp}^3 \) hybridization, leading to a tetrahedral shape.

🎯 Exam Tip: \( \text{sp}^3 \) hybridization results in tetrahedral geometry and is common in \( \text{d}^0 \), high spin \( \text{d}^5 \), high spin \( \text{d}^8 \) (if weak field ligands), and \( \text{d}^{10} \) complexes.

 

Question 31. If the (n-1) d orbitals are involved in hybridisation, the complex is called as
(a) Outer orbital complex
(b) Spin paired complex
(c) High spin complex
(d) Spin free complex
Answer: (b) Spin paired complex
In simple words: When the d-orbitals from the inner shell (n-1) of the metal are used for bonding, it often means the electrons have paired up, leading to a spin-paired complex.

🎯 Exam Tip: Involvement of (n-1)d orbitals indicates an inner orbital complex, which usually corresponds to low spin or spin-paired complexes due to strong field ligands.

 

Question 32. If the n d orbitals are involved in hybridisation, the complex is called as
(a) Inner orbital complex
(b) low spin complex
(c) high spin complex
(d) spin paired complex
Answer: (c) high spin complex
In simple words: When the d-orbitals from the outermost shell (n) of the metal are used for bonding, it means the electrons are spread out and not paired, leading to a high-spin complex.

🎯 Exam Tip: Involvement of nd orbitals indicates an outer orbital complex, typically associated with high spin or spin-free complexes, often due to weak field ligands.

 

Question 33. Which statement is incorrect?
(a) \( \text{[Ni(CO)}_4\text{]} \) – Tetrahedral, Paramagnetic
(b) \( \text{[Ni(CN)}_4\text{]}^{2-} \) – Square planar, diamagnetic
(c) \( \text{[Ni(CO)}_4\text{]} \) – Tetrahedral, diamagnetic
(d) \( \text{[Ni(Cl)}_4\text{]}^{2-} \) – Tetrahedral, Paramagnetic
Answer: (a) \( \text{[Ni(CO)}_4\text{]} \) – Tetrahedral, Paramagnetic
In simple words: The statement that \( \text{[Ni(CO)}_4\text{]} \) is paramagnetic is wrong. This complex is actually diamagnetic because all its electrons are paired up.

🎯 Exam Tip: Carbonyl ligands are strong field ligands, always causing pairing of electrons, making \( \text{[Ni(CO)}_4\text{]} \) diamagnetic, not paramagnetic.

 

Question 34. In \( \text{[Ni(CO)}_4\text{]} \), the hybridisation of central metal ion is
(a) \( \text{dsp}^2 \)
(b) \( \text{sp}^3 \)
(c) \( \text{d}^2\text{sp}^3 \)
(d) \( \text{dsp}^3 \)
Answer: (b) \( \text{sp}^3 \)
In simple words: For \( \text{[Ni(CO)}_4\text{]} \), the nickel atom uses one s-orbital and three p-orbitals to form new hybrid orbitals, leading to an \( \text{sp}^3 \) hybridization and a tetrahedral shape.

🎯 Exam Tip: Nickel carbonyl \( \text{[Ni(CO)}_4\text{]} \) is a classic example of an \( \text{sp}^3 \) hybridized tetrahedral complex, even though CO is a strong field ligand, because Ni is in 0 oxidation state (\( \text{3d}^{10} \)).

 

Question 35. Co-ordination number of Ni in \( \left[ \text{Ni}\left( {\text{C}}_{2}{\text{O}}_{4} \right)_{3} \right]^{4-} \) is (PTA - 4)
(a) 3
(b) 6
(c) 4
(d) 2
Answer: (b) 6
In simple words: The coordination number tells you how many atoms are directly attached to the central metal atom. Here, each oxalate ligand has two points of attachment, and there are three such ligands, making the total six.

🎯 Exam Tip: Remember that bidentate ligands like oxalate (\( {\text{C}}_{2}{\text{O}}_{4}^{2-} \)) contribute two coordination sites, not just one, to the central metal ion.

 

Question 36. According to CFT, the bonding between central metal ion and ligands is
(a) ionic
(b) covalent
(c) metallic
(d) Vanderwaal's force
Answer: (a) ionic
In simple words: Crystal Field Theory (CFT) sees the bond between a metal and its ligands as if they are attracting each other like magnets, with the metal being positive and the ligands being negative charges. This is like an ionic bond.

🎯 Exam Tip: CFT focuses on electrostatic interactions, picturing ligands as point charges that create an electric field around the central metal ion.

 

Question 37. Which is a strong field ligand?
(a) \( \text{I}^{-} \)
(b) \( \text{CN}^{-} \)
(c) \( \text{Cl}^{-} \)
(d) \( \text{S}^{2-} \)
Answer: (b) \( \text{CN}^{-} \)
In simple words: A strong field ligand makes the d-orbitals in the metal split apart a lot, creating a big energy gap. Cyanide (\( \text{CN}^{-} \)) is known to do this strongly.

🎯 Exam Tip: Strong field ligands cause a large crystal field splitting energy (Δ), leading to low-spin complexes, while weak field ligands cause a small splitting.

 

Question 38. Which is a weak field ligand?
(a) \( {\text{NO}}_{2}^{-} \)
(b) \( {\text{NH}}_{3} \)
(c) CO
(d) \( \text{SCN}^{-} \)
Answer: (d) \( \text{SCN}^{-} \)
In simple words: A weak field ligand causes less splitting of the d-orbitals in the metal, leading to a smaller energy gap. Thiocyanate (\( \text{SCN}^{-} \)), when bonded through sulfur, acts as a weak field ligand.

🎯 Exam Tip: Weak field ligands often result in high-spin complexes because electrons prefer to occupy separate orbitals before pairing up.

 

Question 39. The observed colour of a coordination compound can be explained using
(a) Valence bond theory
(b) Werner's theory
(c) Crystal field theory
(d) Molecular orbital theory
Answer: (c) Crystal field theory
In simple words: Crystal Field Theory (CFT) helps us understand why coordination compounds have different colors. It explains that the color comes from electrons jumping between different energy levels when light hits them.

🎯 Exam Tip: CFT describes d-d transitions, where electrons absorb specific wavelengths of light, and the observed color is the complementary color of the absorbed light.

 

Question 40. Which among the following is a mononuclear carbonyl?
(a) \( \left[ {\text{Co}}_{2}\left( \text{CO} \right)_{8} \right] \)
(b) \( \left[ {\text{Fe}}_{3}\left( \text{CO} \right)_{12} \right] \)
(c) \( \left[ \text{Fe}\left( \text{CO} \right)_{5} \right] \)
(d) \( \left[ {\text{Fe}}_{2}\left( \text{CO} \right)_{9} \right] \)
Answer: (c) \( \left[ \text{Fe}\left( \text{CO} \right)_{5} \right] \)
In simple words: A mononuclear carbonyl is a complex that has only one metal atom. Iron pentacarbonyl, \( \left[ \text{Fe}\left( \text{CO} \right)_{5} \right] \), fits this description as it has only one iron atom.

🎯 Exam Tip: Look for the subscript on the metal atom to quickly identify if it's mononuclear (one metal atom) or polynuclear (two or more metal atoms).

 

Question 41. If the instability constant value of \( \left[ \text{Cu}\left( {\text{NH}}_{3} \right)_{4} \right]^{2+} \) is \( 1.0 \times {10}^{-12} \), its stability constant value is:
(a) \( 1.0 \times {10}^{-12} \)
(b) \( 1.0 \times {10}^{12} \)
(c) 12
(d) -12
Answer: (b) \( 1.0 \times {10}^{12} \)
In simple words: The stability constant is simply the inverse of the instability constant. If you have the instability value, just take 1 divided by that number to get the stability constant. A low instability constant means a high stability constant.

🎯 Exam Tip: Remember that \( \text{Stability constant} = 1 / \text{Instability constant} \). A higher stability constant means the complex is more stable and less likely to break apart.

 

Question. From the above table which of the following complex is most stable?

(a) \( \left[ \text{Cu}\left( {\text{NH}}_{3} \right)_{4} \right]^{2+} \)
(b) \( \left[ \text{Fe}\left( \text{SCN} \right) \right]^{2+} \)
(c) \( \left[ \text{Co}\left( {\text{NH}}_{3} \right)_{6} \right]^{3+} \)
(d) \( \left[ \text{Ag}\left( \text{CN} \right)_{2} \right]^{-} \)
Answer: (c) \( \left[ \text{Co}\left( {\text{NH}}_{3} \right)_{6} \right]^{3+} \)
In simple words: The complex with the smallest instability constant is the most stable because it is least likely to break apart. Comparing the numbers, \( 6.2 \times {10}^{-36} \) is the smallest, meaning \( \left[ \text{Co}\left( {\text{NH}}_{3} \right)_{6} \right]^{3+} \) is the most stable.

🎯 Exam Tip: Remember that a small instability constant (or a large stability constant) indicates a highly stable complex.

 

Question 43. Phthalo blue – a bright blue pigment is a complex of
(a) \( {\text{Ni}}^{2+} \)
(b) \( {\text{Co}}^{3+} \)
(c) \( {\text{Cu}}^{2+} \)
(d) \( {\text{Ag}}^{+} \)
Answer: (c) \( {\text{Cu}}^{2+} \)
In simple words: Phthalo blue, a common and strong blue pigment, is a type of copper complex. It gets its vibrant color from the copper ion at its center.

🎯 Exam Tip: Some coordination complexes are well-known for their vivid colors, which makes them useful as pigments and dyes, like the copper phthalocyanine complex.

 

Question 44. For removing lead poisoning, the chelating ligand used is
(a) DMG
(b) EDTA
(c) en
(d) CO
Answer: (b) EDTA
In simple words: EDTA is a special type of chemical that can grab onto metal ions like lead very tightly. This helps to remove harmful lead from the body in cases of poisoning.

🎯 Exam Tip: EDTA is a polydentate ligand, meaning it can form multiple bonds with a metal ion, creating stable ring structures called chelates, which are effective in chelation therapy.

 

Question 45. The process in which coordination complexes are used in the extraction of silver and gold from their ores is
(a) Mond's process
(b) Alumino thermic process
(c) Mac-Arthur-Forrest cyanide process
(d) Bessimerisation
Answer: (c) Mac-Arthur-Forrest cyanide process
In simple words: This process uses cyanide to form a complex with gold or silver from their ore, making the metal soluble and easy to separate. It is a way to get precious metals out of the ground.

🎯 Exam Tip: The cyanide process is a hydrometallurgical technique that relies on the formation of stable soluble cyano complexes of gold and silver.

 

Question 46. Ni ions present in nickel chloride solution is estimated accurately using the ligand
(a) DMG
(b) EDTA
(c) en
(d) CO
Answer: (a) DMG
In simple words: DMG, or dimethylglyoxime, is a chemical that reacts specifically with nickel ions to form a colored complex that can be easily measured. This helps scientists figure out exactly how much nickel is in a solution.

🎯 Exam Tip: DMG is a well-known chelating agent used for the quantitative estimation of nickel, forming a bright red precipitate that is easy to filter and weigh.

 

Question 47. Catalyst used for hydrogenation of alkenes is
(a) Ziegler-Natta Catalyst
(b) Zeise's Salt
(c) Wilkinson's catalyst
(d) Magnus's green Salt
Answer: (c) Wilkinson's catalyst
In simple words: Wilkinson's catalyst is a special chemical mixture that helps to add hydrogen to double bonds in organic molecules, turning alkenes into alkanes. It makes this reaction happen faster and more easily.

🎯 Exam Tip: Wilkinson's catalyst (\( \left[ \text{RhCl}\left( {\text{PPh}}_{3} \right)_{3} \right] \)) is a homogeneous catalyst famous for its selective hydrogenation of alkenes and alkynes.

 

Question 49. The complex used as an antitumor drug in cancer treatment is
(a) Ca-EDTA Chelate
(b) Cisplatin
(c) Trans-Platin
(d) Cyano cobalamine
Answer: (b) Cisplatin
In simple words: Cisplatin is an important medicine used to fight cancer. It works by interfering with the DNA in cancer cells, stopping them from growing and spreading.

🎯 Exam Tip: Cisplatin's anti-cancer activity is due to its ability to bind to DNA, forming cross-links that prevent DNA replication and lead to programmed cell death in cancer cells.

 

Question 50. RBC is composed of
(a) \( {\text{Mg}}^{2+} \)
(b) \( {\text{Fe}}^{2+} \)
(c) \( {\text{Fe}}^{3+} \)
(d) Co
Answer: (b) \( {\text{Fe}}^{2+} \)
In simple words: Red Blood Cells (RBCs) contain a special part called hemoglobin, which has iron in it. This iron is in the \( {\text{Fe}}^{2+} \) form and is crucial for carrying oxygen around the body.

🎯 Exam Tip: The iron in hemoglobin must be in the \( {\text{Fe}}^{2+} \) (ferrous) state to bind oxygen; if it oxidizes to \( {\text{Fe}}^{3+} \) (ferric), it forms methemoglobin which cannot transport oxygen effectively.

 

Question 51. In RBC and chlorophyll, the ligand is
(a) EDTA
(b) DMG
(c) Porphyrin
(d) en
Answer: (c) Porphyrin
In simple words: Both red blood cells and chlorophyll, which gives plants their green color, have a similar ring-like molecule called porphyrin. This porphyrin ligand holds the central metal ion (iron in RBCs, magnesium in chlorophyll) in place.

🎯 Exam Tip: Porphyrin is a large, cyclic tetradentate ligand that is vital in biological systems for holding metal ions, enabling functions like oxygen transport and photosynthesis.

 

II. Match the following

 

Question 1.

Answer:
(i) \( {\text{K}}_{2}\left[ \text{Ni}\left( \text{CN} \right)_{4} \right] \) - (c) anionic complex
(ii) \( \left[ \text{Co}\left( {\text{NH}}_{3} \right)_{5}{\text{NO}}_{2} \right]{\text{Cl}}_{2} \) - (d) Cationic complex
(iii) \( \left[ \text{Co}\left( {\text{NH}}_{3} \right)_{3}{\text{Cl}}_{3} \right] \) - (a) Neutral complex
(iv) \( \left[ \text{Ni}\left( \text{CO} \right)_{4} \right] \) - (b) Metal carbonyl
In simple words: We classify complexes based on their overall charge and the types of ligands they have. A metal carbonyl complex is one where the ligand is carbon monoxide.

🎯 Exam Tip: To classify complexes, check the overall charge on the coordination sphere: positive for cationic, negative for anionic, and zero for neutral. Metal carbonyls specifically involve CO ligands.

 

Question 2.

Answer:
(i) Coordination isomers - (c) Interchange of one or more ligands
(ii) Linkage isomers - (d) Bonding through different donor atoms
(iii) Ionisation isomers - (b) Different ions in solution
(iv) Solvate isomers - (a) Exchange of free solvent molecules with ligand
In simple words: Isomers are molecules with the same formula but different arrangements. Each type of isomerism is caused by a different kind of change, like how ligands attach or what ions are outside the complex.

🎯 Exam Tip: Understand the key characteristic of each isomerism type: linkage involves ambidentate ligands, ionization involves exchange of ions between coordination sphere and counter ions, solvate involves solvent molecules, and coordination involves exchange of ligands between complex cation and anion.

 

Question 3.

Answer:
(i) \( \left[ \text{Ni}\left( \text{CO} \right)_{4} \right] \) - (c) Mononuclear carbonyl
(ii) \( \left[ {\text{Fe}}_{3}\left( \text{CO} \right)_{12} \right] \) - (a) Polynuclear carbonyl
(iii) \( {\text{Fe}}_{2}\left( \text{CO} \right)_{9} \) - (b) Bridged carbonyl
In simple words: Carbonyl complexes are grouped by how many metal atoms they have and how the CO ligands are attached. Mononuclear means one metal, polynuclear means many metals, and bridged means CO ligands connect two metal atoms.

🎯 Exam Tip: Identify mononuclear carbonyls by the presence of a single metal atom, and polynuclear carbonyls by multiple metal atoms, often with bridging CO ligands between them.

 

Question 4.

Answer:
(i) Chromium hexacarbonyl - (b) Octahedral
(ii) Iron pentacarbonyl - (c) Trigonal bipyramidal
(iii) Nickel tetra carbonyl - (a) Tetrahedral
In simple words: The shape of a metal carbonyl complex depends on how many ligands are attached to the central metal atom. Six ligands usually make an octahedral shape, five make a trigonal bipyramidal shape, and four make a tetrahedral shape.

🎯 Exam Tip: Recall VSEPR theory and hybridization rules: 4 ligands typically lead to tetrahedral or square planar, 5 to trigonal bipyramidal, and 6 to octahedral geometry.

 

Question 5.

Answer:
(i) sp - (e) Lineal
(ii) \( \text{sp}^{2} \) - (d) Trigonal planar
(iii) \( \text{sp}^{3} \) - (f) Tetrahedral
(iv) \( \text{dsp}^{2} \) - (a) Square planar
(v) \( \text{dsp}^{3} \) - (c) Trigonal bipyramidal
(vi) \( {\text{d}}^{2}\text{sp}^{3} \) - (b) Octahedral
In simple words: The type of hybridization around a central atom tells us about the shape of the molecule. Each hybridization combination leads to a specific geometric arrangement of the atoms.

🎯 Exam Tip: Memorize the common hybridization types and their corresponding geometries, as this is fundamental to understanding molecular structures in coordination chemistry.

 

III. Pick the odd man out

 

Question 1. Pick the odd man out w.r.t complexion
(a) \( \left[ \text{Co}\left( {\text{NH}}_{3} \right)_{6} \right]{\text{Cl}}_{3} \)
(b) \( \left[ \text{Fe}\left( {\text{H}}_{2}\text{O} \right)_{6} \right]{\text{Cl}}_{2} \)
(c) \( \left[ \text{Cu}\left( {\text{NH}}_{3} \right)_{4} \right]{\text{SO}}_{4} \)
(d) \( {\text{K}}_{4}\left[ \text{Fe}\left( \text{CN} \right)_{6} \right] \)
Answer: (d) \( {\text{K}}_{4}\left[ \text{Fe}\left( \text{CN} \right)_{6} \right] \)
In simple words: Most of the listed complexes are cationic, meaning they have a positive charge. However, \( {\text{K}}_{4}\left[ \text{Fe}\left( \text{CN} \right)_{6} \right] \) is an anionic complex because the \( \left[ \text{Fe}\left( \text{CN} \right)_{6} \right] \) part has a negative charge, making it different from the others.

🎯 Exam Tip: To determine the charge of a complex ion, consider the charges of the metal ion and all the ligands, then subtract any counter-ions outside the square brackets.

 

Question 2. Pick the odd man out w.r.t geometry
(a) \( \left[ \text{Co}\left( {\text{NH}}_{3} \right)_{6} \right]{\text{Cl}}_{3} \)
(b) \( \left[ \text{Fe}\left( {\text{H}}_{2}\text{O} \right)_{6} \right]{\text{Cl}}_{2} \)
(c) \( \left[ \text{Cu}\left( {\text{NH}}_{3} \right)_{4} \right]{\text{SO}}_{4} \)
(d) \( {\text{K}}_{4}\left[ \text{Fe}\left( \text{CN} \right)_{6} \right] \)
Answer: (c) \( \left[ \text{Cu}\left( {\text{NH}}_{3} \right)_{4} \right]{\text{SO}}_{4} \)
In simple words: The complexes \( \left[ \text{Co}\left( {\text{NH}}_{3} \right)_{6} \right]{\text{Cl}}_{3} \), \( \left[ \text{Fe}\left( {\text{H}}_{2}\text{O} \right)_{6} \right]{\text{Cl}}_{2} \), and \( {\text{K}}_{4}\left[ \text{Fe}\left( \text{CN} \right)_{6} \right] \) all have six ligands and are octahedral. But \( \left[ \text{Cu}\left( {\text{NH}}_{3} \right)_{4} \right]{\text{SO}}_{4} \) has only four ligands and is square planar, making it the different one.

🎯 Exam Tip: The geometry of a coordination complex is primarily determined by its coordination number (number of ligands attached to the central metal ion) and the electronic configuration of the metal.

 

Question 3. Pick the odd man out w.r.t primary valence of the metal ion
(b) \( {\text{K}}_{3}\left[ \text{Fe}\left( \text{CN} \right)_{6} \right] \)
(c) \( \left[ \text{Fe}\left( \text{CO} \right)_{5} \right] \)
(d) \( \left[ \text{Co}\left( {\text{NH}}_{3} \right)_{3}{\text{Cl}}_{3} \right] \)
Answer: (c) \( \left[ \text{Fe}\left( \text{CO} \right)_{5} \right] \) – Primary valence is zero, in other complexes the primary valence is +3
In simple words: The primary valence is the oxidation state of the metal. In most of these complexes, the metal has an oxidation state of +3. However, in \( \left[ \text{Fe}\left( \text{CO} \right)_{5} \right] \), the iron metal has an oxidation state of zero, making it different.

🎯 Exam Tip: Always calculate the oxidation state of the central metal ion to determine its primary valence. Remember that neutral ligands like CO do not contribute to the charge.

 

Question 4. Pick the odd man out w.r.t the nature of the ligand
(a) aqua
(b) carbonyl
(c) nitrosyl
(d) nitrato
Answer: (d) nitrato – negative ligand others are neutral ligand
In simple words: Aqua, carbonyl, and nitrosyl are all neutral ligands, meaning they don't carry a charge. But nitrato is a negatively charged ligand. This difference in charge makes nitrato the odd one out.

🎯 Exam Tip: Carefully identify whether a ligand is neutral or charged, as this affects the overall charge of the complex and the oxidation state of the metal. Common neutral ligands include \( {\text{H}}_{2}\text{O} \), \( {\text{NH}}_{3} \), CO, and NO.

 

IV. Pick the wrong statement

 

Question 1.
(i) The secondary valence of a metal ion is satisfied by ligands.
(ii) The secondary valence determines the geometry of the complex.
(iii) The secondary valence is oxidation state of the metal ion.
(iv) If the secondary valence is six, the geometry is tetrahedral.
(a) (i) & (ii)
(b) (i) & (iii)
(c) (ii) & (iii)
(d) (iii) & (iv)
Answer: (d) (iii) & (iv)
In simple words: The secondary valence is the coordination number, not the oxidation state. Also, if there are six ligands, the shape is usually octahedral, not tetrahedral. So, statements (iii) and (iv) are incorrect.

🎯 Exam Tip: Remember Werner's theory: primary valence is the oxidation state, satisfied by anions, and non-directional. Secondary valence is the coordination number, satisfied by ligands, and determines the geometry of the complex.

 

Question 2.
(i) Central metal ion and ligands are bonded through ionic bond.
(ii) Central metal ion accepts electron pairs donated by ligands.
(iii) Ligands act as lewis acids
(iv) Charge on the coordination sphere is the net change inside the coordination sphere.
(a) (i) & (ii)
(b) (i) & (iii)
(c) (ii) & (iii)
(d) (iii) & (iv)
Answer: (b) (i) & (iii)
In simple words: In coordination complexes, the metal and ligands share electrons in a special way, like a coordinate covalent bond, not a simple ionic bond. Also, ligands are electron donors, making them Lewis bases, not Lewis acids. So, statements (i) and (iii) are wrong.

🎯 Exam Tip: Ligands are Lewis bases (electron pair donors), and metal ions are Lewis acids (electron pair acceptors) in coordination complexes, forming coordinate covalent bonds.

 

Question 3.
(i) Trans \( \left[ \text{Co}\text{Cl}_{2}{\left( \text{en} \right)}_{2} \right]^{+} \) is optically active.
(ii) In octahedral complexes if nd orbitals are involved in hybridisation, they are called as low spin complexes.
(iii) In a complex if all the electrons are paired, it is diamagnetic
(iv) Ligands which cause the pairing of electrons are called strong field ligands.
(a) (i) &(ii)
(b) (i) & (iii)
(c) (ii) & (iii)
(d) (iii) & (iv)
Answer: (a) (i) & (ii)
In simple words: A trans isomer is usually symmetrical and not optically active. Also, if the "nd" orbitals are used (meaning outer d-orbitals), the complex is called high spin, not low spin. So, statements (i) and (ii) are incorrect.

🎯 Exam Tip: Optical activity requires a complex to be chiral (non-superimposable on its mirror image). Trans isomers are generally symmetrical and achiral. Low spin complexes involve inner d-orbitals, while high spin complexes involve outer d-orbitals.

 

Question 4.
(i) The ligands present on the right side of the spectrochemical series are called strong field ligands.
(ii) The ligands present on the left side of the spectrochemical series are called weak field ligands.
(iii) CFSE (\( \Delta \text{E} = {\text{E}}_{\text{iso}} - {\text{E}}_{\text{LF}} \)
(iv) \( {\mu }_{\text{s}} = \sqrt{\text{n}\left( \text{n} - 2 \right)} \)
(a) (i) & (ii)
(b) (i) & (iii)
(c) (ii) & (iii)
(d) (iii) & (iv)
Answer: (c) (ii) & (iii)
In simple words: The formula for Crystal Field Stabilization Energy (CFSE) is \( \Delta {\text{E}}_{0} = {\text{E}}_{\text{LF}} - {\text{E}}_{\text{iso}} \), not the other way around. Also, the spin-only magnetic moment formula is \( {\mu }_{\text{s}} = \sqrt{\text{n}\left( \text{n} + 2 \right)} \), not \( \sqrt{\text{n}\left( \text{n} - 2 \right)} \). Therefore, statements (iii) and (iv) are incorrect.

🎯 Exam Tip: Ensure you remember the correct formulas for CFSE and spin-only magnetic moment. The spectrochemical series arranges ligands based on their ability to cause crystal field splitting, from weak field (left) to strong field (right).

 

V. Pick the Correct statement

 

Question 1.
(i) The primary valence of a metal ion is positive in most cases and zero in certain cases.
(ii) The primary valences are directional
(iii) The primary valence is satisfied by ligands in certain cases.
(iv) The primary valence is co-ordination number.
(a) (i) & (ii)
(b) (i) & (iii)
(c) (ii) & (iii)
(d) (iii) & (iv)
Answer: (b) (i) & (iii)
In simple words: The primary valence of a metal is its oxidation state, which can be positive or sometimes zero, and is satisfied by negatively charged ions. It is not directional and is not the coordination number. So, statements (i) and (iii) are correct.

🎯 Exam Tip: Primary valence (oxidation state) is typically positive and non-directional, while secondary valence (coordination number) is directional and determines geometry.

 

Question 2.
(i) Coordination entity consists of central metal ion and ligands.
(ii) In \( {\text{K}}_{4}\left[ \text{Fe}\left( \text{CN} \right)_{6} \right] \), the coordination entity is \( \left[ \text{Fe}\left( \text{CN} \right)_{6} \right]^{4-} \)
(iii) The groups present inside the coordination entity can be ionised
(iv) The coordination entity is not responsible for the geometry of the complex.
(a) (i) & (ii)
(b) (i) & (iii)
(c) (ii) & (iii)
(d) (iii) & (iv)
Answer: (a) (i) & (ii)
In simple words: The coordination entity is the central metal ion and its attached ligands. For \( {\text{K}}_{4}\left[ \text{Fe}\left( \text{CN} \right)_{6} \right] \), the \( \left[ \text{Fe}\left( \text{CN} \right)_{6} \right]^{4-} \) part is the coordination entity. Things inside this entity usually do not ionize, and the coordination entity *is* responsible for the shape. So, (i) and (ii) are correct.

🎯 Exam Tip: The coordination entity (inner sphere) is the non-ionizable part of a coordination compound and its shape dictates the overall geometry of the complex.

 

Question 3.
(i) In thiocyanate ligand if sulphur forms a coordination bond with metal, the ligand is named thiocyanato-KN
(ii) The ligand \( {\text{NH}}_{3} \) is named amine.
(iii) Ethylenediamine is a neutral ligand.
(iv) The name of cationic ligand ends with – ium
(a) (i) &(ii)
(b) (i) & (iii)
(c) (ii) & (iii)
(d) (iii) & (iv)
Answer: (c) (ii) & (iii)
In simple words: If sulfur bonds in thiocyanate, it's called thiocyanato-kS. The ligand \( {\text{NH}}_{3} \) is correctly called ammine, and ethylenediamine is a neutral ligand. Cationic ligands typically end in "-ium". So, (ii) and (iii) are correct.

🎯 Exam Tip: Be precise with ligand nomenclature, especially for ambidentate ligands (like thiocyanate) where the bonding atom is specified (kS or kN). Remember common neutral ligands like ammine and ethylenediamine.

 

Question 4.
(i) Linkage isomer is not possible with nitrite ligand.
(ii) Coordination isomerism is found in coordination compounds in which both cation and anion are complexions.
(iii) Ionisation isomer arises when an ionisable counter ion itself can act as a ligand.
(iv) In solvate isomer if the solvent molecule is water it is called a hydrate isomer.
(a) (i) &(ii)
(b) (i) & (iii)
(c) (iii) & (iv)
(d) (ii) & (iv)
Answer: (d) (ii) & (iv)
In simple words: Linkage isomerism *is* possible with nitrite, as it can bond through oxygen or nitrogen. Ionization isomerism happens when the counter-ion exchanges with a ligand, not when it acts as a ligand itself. But coordination isomerism needs both the positive and negative parts to be complex ions. And a solvate isomer involving water is called a hydrate isomer. So, (ii) and (iv) are correct.

🎯 Exam Tip: Nitrite (\( {\text{NO}}_{2}^{-} \)) is an ambidentate ligand, meaning it can bond in two ways (through N or O), hence it can show linkage isomerism. Solvate isomerism is a specific type of ionization isomerism where solvent molecules are involved.

 

VI. Assertion and Reason

 

Question 1. Assertion (A): [Co(NH3)4 Br2] Cl and [CO(NH3)4 CI Br] Br are ionisation isomers. Reason (R) : In solution, they give free Cl¯ and Br¯ ions respectively
Answer: i) Both A and R are correct, R explains A.
In simple words: This question asks about isomerism. The assertion correctly identifies two compounds as ionization isomers because they release different ions in solution. The reason explains why, stating they produce free chloride and bromide ions.

🎯 Exam Tip: For assertion-reason questions, first determine if each statement is true, then check if the reason correctly explains the assertion.

 

Question 2. Assertion (A): Geometrical isomerism exists in homoleptic complexes. Reason (R): Geometrical isomerism is due to different possible three-dimensional spatial arrangements of the ligands around the central metal atom.
Answer: iii) A is wrong, R is correct. Correct Assertion (A): Geometrical isomerism exists in heteroleptic complexes.
In simple words: Geometrical isomerism happens because ligands can be arranged in different ways in 3D space. It occurs in complexes with different types of ligands (heteroleptic), not just one type (homoleptic).

🎯 Exam Tip: Remember that homoleptic complexes have only one type of ligand, while heteroleptic complexes have multiple types, which allows for various spatial arrangements and isomerism.

 

Question 3. Assertion (A) : According to VBT the ligand \( \leftarrow \) metal bond in a coordination complex, is covalent in nature. Reason (R) : Mutually shared electrons are provided equally by the central metal atom and the ligand
Answer: ii) A is correct, R is wrong Correct Reason: Mutually shared electrons are provided by ligands to the central metal atom.
In simple words: The bond between a ligand and a metal in a complex is covalent, meaning electrons are shared. However, the sharing is not equal; the ligands donate both electrons to the metal atom.

🎯 Exam Tip: In coordination complexes, the bond is a dative covalent bond (or coordinate bond) where the ligand acts as a Lewis base (electron pair donor) and the metal acts as a Lewis acid (electron pair acceptor).

 

VII. Two Mark Questions

 

Question 1. Write any two medicinal uses of coordination compounds? (PTA - 6)
Answer: Coordination compounds have several important medical applications.
1. Ca-EDTA chelate is used to treat lead poisoning and radioactive metal poisoning. It works by removing harmful lead and other radioactive metal ions from the body.
2. Cisplatin is an effective anti-tumor drug used in cancer treatment. This compound targets and damages cancer cells, helping to stop their growth.
In simple words: Coordination compounds help treat poisoning by removing bad metals from the body, and some, like cisplatin, are used as medicines to fight cancer.

🎯 Exam Tip: When listing uses, always specify the compound and its exact role or mechanism of action to earn full marks.

 

Question 2. What is a central metal ion?
Answer: A central metal ion is the metal atom or ion in a complex that accepts electron pairs from ligands. It acts as a Lewis acid in the complex. This ion is key to the structure and properties of the coordination compound.
In simple words: The central metal ion is the main metal part in a complex that takes electron pairs from other surrounding molecules. It's like the center of attention in the compound.

🎯 Exam Tip: Always remember that the central metal ion is the electron-pair acceptor (Lewis acid) in a coordination complex.

 

Question 3. What are ligands?
Answer: Ligands are ions or molecules that donate electron pairs to the central metal ion in a coordination complex. They can be negatively charged ions, neutral molecules, or even positively charged ions. Ligands are essential for forming the bonds that create a complex.
In simple words: Ligands are molecules or ions that give electrons to the central metal atom. They act like electron donors to form a bond.

🎯 Exam Tip: Ligands are also known as Lewis bases, as they donate electron pairs to form coordinate covalent bonds with the central metal ion.

 

Question 4. [Fe(CN)6]4- and [Fe(H2O)6]2+ are of different colours in dilute solutions, why? (PTA - 2)
Answer: Both complexes, [Fe(CN)6]4- and [Fe(H2O)6]2+, show different colors in dilute solutions due to the nature of their ligands.

• CN¯ is a strong field ligand, while H2O is a weak field ligand.
• Strong field ligands cause a larger crystal field splitting energy (\( \Delta_o \)) compared to weak field ligands.
• This difference in \( \Delta_o \) means that the two complexes absorb different wavelengths of light, leading to the emission of different complementary colors. For instance, strong field ligands often lead to low spin complexes, affecting their color.

In simple words: These two compounds have different colors because the atoms attached to the iron (ligands) are different. Some ligands pull electrons strongly, and others weakly. This changes how much light energy the compounds absorb, making them look like different colors.

🎯 Exam Tip: The color of coordination complexes is directly related to the crystal field splitting energy, which is influenced by the strength of the ligands involved.

 

Question 5. Define coordination number.
Answer: The coordination number of a metal is the total number of ligand donor atoms directly bonded to the central metal atom or ion in a coordination complex. It tells us how many points of attachment the metal has to the surrounding ligands. This number determines the complex's geometry.
In simple words: The coordination number is simply how many other atoms are directly stuck to the central metal atom in a complex.

🎯 Exam Tip: When determining the coordination number, count only the donor atoms, not the entire ligand molecule, especially for polydentate ligands.

 

Question 6. What is the coordination sphere?
Answer: The coordination sphere is the central metal atom or ion along with all the ligands directly attached to it, wrapped within a square bracket in chemical formulas. This entire unit acts as a single entity and is also called the inner sphere. Everything outside this sphere is considered counter ions.
In simple words: The coordination sphere is the central metal and all the atoms directly connected to it, acting as one unit.

🎯 Exam Tip: Remember that anything inside the square brackets of a complex formula is part of the coordination sphere and does not ionize in solution, while counter ions outside the brackets do.

 

Question 7. What are inert and labile complexes? (PTA – 4)
Answer: Complexes can be categorized based on how easily their ligands can be replaced.

• **Labile complexes:** These are complexes where ligands can be quickly and easily replaced by other ligands through a substitution reaction.
• **Inert complexes:** These are complexes where ligand substitution reactions occur very slowly, or sometimes not at all. Ligands in these complexes are not easily replaced.

A good example of an inert complex is the [Co(NH3)6]3+ ion, which is very stable.
In simple words: Labile complexes swap their attached parts easily, like changing clothes quickly. Inert complexes hold onto their attached parts tightly, making them hard to change.

🎯 Exam Tip: Do not confuse thermodynamic stability (how likely a complex is to form or dissociate) with kinetic lability/inertness (how fast its ligands exchange).

 

Question 8. What is d-d transition?
Answer: A d-d transition is when an electron in a central metal ion absorbs energy (often in the form of light) and moves from a lower energy d-orbital to a higher energy d-orbital within the same subshell. This process is responsible for the characteristic colors observed in many transition metal complexes.
In simple words: A d-d transition is when an electron in a metal atom jumps from one d-level to a higher d-level by soaking up light. This makes the chemical compound colorful.

🎯 Exam Tip: d-d transitions are forbidden by selection rules but become allowed through vibronic coupling, making them common in colored transition metal complexes.

 

Question 9. What are metallic carbonyls?
Answer: Metallic carbonyls are a type of coordination compound where carbon monoxide (CO) molecules act as ligands and are bonded to a central transition metal atom. These compounds feature a special type of bonding between the metal and the carbon monoxide. They are important in catalysis and synthesis.
In simple words: Metallic carbonyls are special chemical compounds where a metal is attached to carbon monoxide molecules.

🎯 Exam Tip: Remember that in metallic carbonyls, carbon monoxide bonds through its carbon atom, forming a metal-carbon bond, and also exhibits back-bonding.

 

Question 10. What is stability constant? (PTA – 5)
Answer: The stability constant, also known as the formation constant (\( \beta \)), is a measure of the equilibrium between a metal ion and ligands in solution that leads to the formation of a coordination complex. It is the reciprocal of the dissociation equilibrium constant or instability constant. A higher stability constant means the complex is more stable and forms more readily.
In simple words: The stability constant tells us how strong a bond is between a metal and the parts attached to it. A bigger number means the bond is stronger.

🎯 Exam Tip: The stability constant helps predict how much of a complex will form at equilibrium; a larger value indicates a more stable complex.

 

Question 11. Calculate the magnetic moment and magnetic property of [CoF6]³- (MARCH 2020)
Answer:

This complex is high spin due to the weak field ligand \( \text{F}^- \), which leads to four unpaired electrons. The magnetic moment calculation confirms its paramagnetic nature.
In simple words: For this complex, we find that it has 4 electrons that are not paired up. Because of these single electrons, it is called "paramagnetic," meaning it gets pulled towards a magnet. We can also calculate a number called the magnetic moment to show how strong this magnetic pull is.

🎯 Exam Tip: When calculating magnetic moment, always determine the number of unpaired electrons first, which depends on the ligand field strength (strong or weak field).

 

VIII. Three Mark questions

 

Question 1. Explain geometrical isomerism in square planar complexes.
Answer: Geometrical isomerism happens in coordination complexes that have different arrangements of ligands in space. In square planar complexes, this isomerism is particularly noticeable.
When dealing with square planar complexes of the type \( \text{[MA}_2\text{B}_2]\text{n}^\pm \) or \( \text{[MA}_2\text{BC}]\text{n}^\pm \), if similar groups are located on the same side of the central metal atom, the isomer is called a *cis-isomer*. If these similar groups are on opposite sides, it is called a *trans-isomer*. This difference in arrangement leads to distinct chemical and physical properties. For example, \( \text{Pt(NH}_3\text{)}_2\text{Cl}_2 \) (Cisplatin) shows geometrical isomerism. The cis-isomer has two \( \text{NH}_3 \) ligands on the same side, while the trans-isomer has them on opposite sides.
In simple words: Geometrical isomerism in flat square complexes means the same chemical parts can be arranged in two ways: either next to each other (cis) or across from each other (trans). This changes how the complex behaves.

🎯 Exam Tip: Remember that square planar complexes with formula \( \text{MA}_2\text{B}_2 \) and \( \text{MA}_2\text{BC} \) commonly exhibit cis-trans isomerism.

 

Question 2. Explain geometrical isomerism in octahedral complexes.
Answer: Geometrical isomerism also occurs in octahedral complexes, where ligands are arranged around a central metal atom in an octahedral shape. For complexes of the type \( \text{[MA}_2\text{B}_4]\text{n}^\pm \) or \( \text{[M(XX)}_2\text{B}_2]\text{n}^\pm \) (where A and B are monodentate ligands, and XX is a bidentate ligand with two similar donor atoms), cis- and trans-isomers are observed.

• A **cis-isomer** forms when two identical ligands are next to each other (at 90 degrees) in the octahedral structure.
• A **trans-isomer** forms when two identical ligands are opposite each other (at 180 degrees).

This spatial arrangement profoundly influences the complex's properties. For instance, \( \text{[Co(NH}_3\text{)}_4\text{Cl}_2\text{]}^+ \) shows both cis and trans forms. The specific positions are important; if two similar groups are present in positions like (1,2), (1,3), etc., it's cis, but if they are in (1,6), (2,4), (3,5), it's trans.
In simple words: In 3D octahedral shapes, we can also have cis and trans isomers. Cis means similar parts are close together, and trans means they are far apart. These different arrangements make the compounds slightly different.

🎯 Exam Tip: In octahedral complexes, cis-trans isomerism requires at least two pairs of identical ligands or specific bidentate ligands to be present.

 

Question 3. Write about facial and meridional isomers.
Answer: Facial (fac) and meridional (mer) isomerism are specific types of geometrical isomerism observed in octahedral complexes of the type \( \text{[MA}_3\text{B}_3]\text{n}^\pm \).

• A **facial (fac) isomer** occurs when the three identical ligands (A) are positioned at the corners of one triangular face of the octahedron, and the other three identical ligands (B) are at the corners of the opposing triangular face. Imagine a triangular face of the octahedron; if all three 'A' ligands occupy this face, it's a fac isomer.
• A **meridional (mer) isomer** occurs when the three identical ligands (A) lie along a meridian (an imaginary semicircle passing through the central metal atom and two opposite apices) of the octahedron. This means the ligands are arranged in a straight line around the "equator" of the complex.

These distinct spatial arrangements lead to different properties for the isomers. For example, \( \text{[Co(NH}_3\text{)}_3\text{Cl}_3\text{]} \) can exist as both fac and mer isomers. This is called meridional because each set of ligands can be regarded as lying on a meridian of an octahedron.
In simple words: In some 3D complexes, if three identical parts are on one "face" (a triangle) of the shape, it's a "fac" isomer. If these three parts are in a line around the middle "belt" of the shape, it's a "mer" isomer.

🎯 Exam Tip: Fac-mer isomerism is unique to \( \text{MA}_3\text{B}_3 \) type octahedral complexes and describes how three identical ligands are arranged relative to each other.

 

Question 4. In a tetrahedral crystal field, draw the figure to show splitting of d-orbitals (PTA – 6)
Answer:In a tetrahedral crystal field, the d-orbitals split into two sets: the lower energy set of three orbitals (\( \text{t}_2 \)) and the higher energy set of two orbitals (\( \text{e} \)). This is the opposite of octahedral splitting. The energy gap between these two sets is called the crystal field splitting energy for tetrahedral complexes (\( \Delta_t \)). This splitting occurs because the ligands approach the metal ion from specific directions, causing some d-orbitals to experience more repulsion than others. The \( \text{d}_{xy}\text{, d}_{yz}\text{, d}_{zx} \) orbitals are oriented closer to the ligand direction in a tetrahedral field, causing them to be higher in energy than the \( \text{d}_{x^2-y^2}\text{, d}_{z^2} \) orbitals.
In simple words: When a metal atom is in a tetrahedral shape surrounded by other atoms, its five d-orbitals split into two groups. Three d-orbitals go a bit higher in energy, and two d-orbitals go a bit lower. The energy difference between these groups is important for the complex's properties.

🎯 Exam Tip: Remember that in tetrahedral splitting, the \( \text{t}_2 \) set is at a higher energy level than the \( \text{e} \) set, which is the reverse of octahedral splitting, and \( \Delta_t \) is generally smaller than \( \Delta_o \).

 

Question 5. Write the use of complex formation in photography.
Answer: In photography, complex formation plays a crucial role during the fixing process. After film development, the unexposed silver bromide (AgBr) on the film needs to be removed. Sodium thiosulphate solution, often called "hypo," is used for this purpose.
The AgBr reacts with sodium thiosulphate to form a soluble complex called sodium dithiosulphatoargentate(I). This soluble complex can then be easily washed away with water, fixing the negative image on the film.
\( \text{AgBr} + 2\text{Na}_2\text{S}_2\text{O}_3 \rightarrow \text{Na}_3\text{[Ag}(\text{S}_2\text{O}_3)_2] + 2\text{NaBr} \)
This step ensures that the unexposed areas of the film do not darken over time, preserving the image.
In simple words: In photography, a special chemical liquid helps remove unused parts from the film by turning them into a water-soluble complex. This makes sure the picture stays clear and doesn't fade or get dark later.

🎯 Exam Tip: Understand the role of "hypo" (sodium thiosulphate) in fixing photographic film by forming a soluble silver complex, thereby stabilizing the image.

 

Question 6. Write the use of metal complexes in biological systems.
Answer: Metal complexes are vital in many biological systems, playing diverse and essential roles:

• **Red Blood Corpuscles (RBC):** Hemoglobin, found in RBCs, contains a heme group, which is an \( \text{Fe}^{2+} \) porphyrin complex. This complex is crucial for carrying oxygen from the lungs to body tissues and transporting carbon dioxide back to the lungs.
• **Chlorophyll:** This green pigment in plants and algae is a coordination complex containing \( \text{Mg}^{2+} \) as the central metal ion, surrounded by a porphyrin ring. Chlorophyll is essential for photosynthesis, where plants convert carbon dioxide and water into carbohydrates and oxygen.
• **Vitamin B12:** Cyanocobalamin (Vitamin B12) is the only vitamin that contains a metal ion, \( \text{Co}^+ \), surrounded by a porphyrin-like ligand. It plays a critical role in various metabolic processes.
• **Enzymes:** Many enzymes, which regulate biological processes, are metal complexes. For example, carboxypeptidase, an enzyme important in digestion, contains a zinc ion coordinated to the protein, helping to break down proteins.

In simple words: Metal complexes are super important for life. They help blood carry oxygen, make plants green for food, are part of essential vitamins, and help enzymes do their work in our bodies.

🎯 Exam Tip: When discussing biological roles, always link the specific metal complex (e.g., hemoglobin, chlorophyll) to its metal ion and its primary biological function.

 

Question 7. How is Cisplatin used as an antitumour drug in cancer treatment?
Answer: Cisplatin (cis-[Pt(NH3)2Cl2]) is a platinum-based anti-cancer drug that works effectively against various tumors. Its mechanism of action involves several key steps:

• It undergoes hydrolysis inside cancer cells, where it loses its chloride ligands and becomes an active species.
• This active platinum complex then reacts with DNA, forming cross-links within and between DNA strands.
• These cross-links hinder the DNA replication and transcription processes that are essential for cell growth.
• By preventing DNA replication and transcription, cisplatin ultimately inhibits cell growth and leads to the death of cancer cells.
• It also forms cross-links with cellular proteins, further disrupting cell functions and inhibiting mitosis (cell division).

In simple words: Cisplatin is a cancer medicine made of platinum. It works by getting into cancer cells and messing up their DNA. This stops the cancer cells from growing and multiplying, causing them to die.

🎯 Exam Tip: Highlight that cisplatin's anti-cancer action stems from its ability to bind to DNA, forming cross-links that disrupt cell replication and transcription.

 

IX. Five Mark Questions

 

Question 1. How are coordination complexes classified?
Answer: Coordination complexes can be classified in two main ways: based on the net charge on the complex and based on the type of ligands present.
**I. Classification based on the net charge on the complex:**
A coordination compound's complexion (the central metal atom/ion and its ligands) can carry a net charge.

• **Cationic complexes:** These have a net positive charge. Example: \( \text{[Ag(NH}_3\text{)}_2\text{]}^+ \)
• **Anionic complexes:** These have a net negative charge. Example: \( \text{[Ag(CN)}_2\text{]}^- \)
• **Neutral complexes:** These have no net charge. Example: \( \text{[Ni(CO)}_4\text{]} \)

**II. Classification based on the kind of ligands:**
The ligands attached to the central metal atom/ion also define a classification.

• **Homoleptic complexes:** In these, the central metal atom/ion is coordinated to only one kind of ligand. Example: \( \text{[Ag(NH}_3\text{)}_2\text{]}^+ \)
• **Heteroleptic complexes:** In these, the central metal atom/ion is coordinated to more than one kind of ligand. Example: \( \text{[Co(NH}_3\text{)}_5\text{Cl]}^{2+} \)

The classification helps in understanding the complex's properties and its behavior in reactions.
In simple words: We sort coordination complexes by two things: first, if the main part of the complex has a positive, negative, or no electrical charge. Second, if the metal in the middle is attached to only one type of molecule or many different types.

🎯 Exam Tip: When classifying, ensure you correctly identify the charge on the coordination sphere (not just the overall compound) and distinguish between compounds with one ligand type vs. multiple ligand types.

 

Question 2. Write the steps involved in the IUPAC nomenclature of coordination compounds.
Answer: Naming coordination compounds using IUPAC (International Union of Pure and Applied Chemistry) guidelines ensures clear and consistent communication. Here are the steps involved:
**General Rules:**

• The simple ions (counter ions) are named first, just like in other ionic compounds.
• When naming the complexion (the central metal and ligands), the ligands are named first, followed by the central metal atom/ion. If there's more than one type of ligand, they are named in alphabetical order.

**(a) Naming the ligands:**

• **Anionic ligands:** Their names usually end with 'o' (e.g., chloride becomes chlorido, cyanide becomes cyanido).
• **Cationic ligands:** Their names end with 'ium' (e.g., hydrazinium).
• **Neutral ligands:** Most are called by their molecular names (e.g., ethylenediamine, pyridine). However, there are a few exceptions, such as \( \text{H}_2\text{O} \) (aqua), \( \text{NH}_3 \) (ammine), \( \text{CO} \) (carbonyl), and \( \text{NO} \) (nitrosyl).
• **Ambidentate ligands:** A 'k-term' is used to show which atom of an ambidentate ligand is bonded to the metal (e.g., nitrito-k N or nitrito-k O).
• **Multiple ligands:** If there's more than one of the same ligand, prefixes like di-, tri-, tetra- are used. If the ligand's name already includes a Greek prefix (e.g., ethylenediamine), then alternative prefixes like bis-, tris-, tetrakis- are used instead.
• These numerical prefixes (di, tri, tetra, bis, tris, tetrakis) are not considered when alphabetizing ligands.

**(b) Naming the central metal atom/ion:**

• **Cationic or neutral complexes:** The metal's name is used as is (e.g., cobalt, platinum).
• **Anionic complexes:** A suffix 'ate' is added to the metal's name (e.g., cobaltate, ferrate for iron, cuprate for copper). Sometimes the Latin name of the metal is used before adding 'ate'.

This systematic approach ensures that each coordination compound has a unique and unambiguous name.
In simple words: To name these compounds, you first name the simple parts around the main complex. Then, inside the complex, you name all the attached molecules (ligands) in alphabetical order. Ligands often have special endings like 'o' or 'ium', or they have common names. Finally, you name the central metal, sometimes with an 'ate' ending if the whole complex is negatively charged.

🎯 Exam Tip: Pay close attention to alphabetical order of ligands, correct prefixes for multiple ligands, and the 'ate' suffix for anionic complexes, as these are common areas for errors.

 

Question 3. Explain structural isomerism exhibited by coordination compounds.
Answer: Structural isomers, also known as constitutional isomers, are coordination compounds that have the same chemical formula but differ in the way their atoms are connected. These differences in bonding lead to distinct properties. There are four main types of structural isomerism in coordination compounds:
**i) Linkage Isomerism:**
This type arises when an ambidentate ligand (one that can bond to the metal through two different donor atoms) attaches to the central metal atom/ion via a different atom. For example, the nitrite ion (NO2-) can bond through nitrogen (nitrito-k N) or oxygen (nitrito-k O).
Example: \( \text{[Co(NH}_3\text{)}_5\text{NO}_2\text{]}^{2+} \) (nitrito-k N, N-attached) and \( \text{[Co(NH}_3\text{)}_5\text{ONO]}^{2+} \) (nitrito-k O, O-attached).
**ii) Coordination Isomerism:**
This occurs in compounds where both the cation and anion are complex ions. The isomers differ by the exchange of ligands between the cationic and anionic coordination entities.
Example: \( \text{[Co(NH}_3\text{)}_6\text{][Cr(CN)}_6\text{]} \) and \( \text{[Cr(NH}_3\text{)}_6\text{][Co(CN)}_6\text{]} \). Here, the metal ions have swapped ligands.
**iii) Ionisation Isomerism:**
This type results from the exchange of an ionisable counter ion with a ligand inside the coordination sphere. The isomers produce different ions in solution.
Example: \( \text{[Co(NH}_3\text{)}_5\text{Cl]SO}_4 \) and \( \text{[Co(NH}_3\text{)}_5\text{SO}_4\text{]Cl} \). The first releases \( \text{SO}_4^{2-} \) ions, while the second releases \( \text{Cl}^- \) ions.
**iv) Solvate Isomerism (or Hydrate Isomerism):**
This arises from the exchange of solvent molecules (like water, ammonia, or alcohol) between the coordination sphere and the crystal lattice outside the sphere. If the solvent molecule involved is water, it is specifically called hydrate isomerism.
Example: \( \text{[Cr(H}_2\text{O)}_6\text{]Cl}_3 \), \( \text{[Cr(H}_2\text{O)}_5\text{Cl]Cl}_2\text{.H}_2\text{O} \), and \( \text{[Cr(H}_2\text{O)}_4\text{Cl}_2\text{]Cl.2H}_2\text{O} \). These compounds have different numbers of water molecules inside and outside the coordination sphere.
In simple words: Structural isomers are compounds that have the same recipe (formula) but are put together differently. There are types like: if a connecting part attaches in two ways, if parts of the complex swap places, if a molecule inside the complex swaps with one outside, or if water molecules move in or out of the main complex.

🎯 Exam Tip: To differentiate between structural isomers, analyze the chemical formula for ambidentate ligands (linkage), multiple complex ions (coordination), exchangeable ions (ionization), or solvent molecules (solvate).

 

Question 4. Explain the main assumption of VBT of coordination compounds (PTA - 1)
Answer: The Valence Bond Theory (VBT) provides several key assumptions to explain bonding in coordination compounds:

• The bond formed between the central metal atom/ion and the ligands is primarily a dative covalent bond. This means the ligands donate an electron pair to the metal atom.
• Each ligand has at least one filled orbital that contains a lone pair of electrons available for donation.
• To accept these electron pairs from the ligands, the central metal atom/ion must have vacant orbitals. The number of vacant orbitals must be equal to its coordination number.
• These vacant orbitals of the central metal atom undergo hybridization. Hybridization helps to arrange the orbitals in a specific geometry, which is determined by the coordination number.
• The hybridized orbitals of the central metal ion then linearly overlap with the filled orbitals of the ligands. This overlap forms coordinate covalent sigma bonds.
• These hybridized orbitals are directional, giving the complex a definite geometry (e.g., octahedral, tetrahedral, square planar).
• In octahedral complexes, if the inner \( (\text{n}-1)\text{d} \) orbitals are involved in hybridization, the complexes are called inner orbital complexes (or low spin/spin paired complexes). If the outer \( \text{nd} \) orbitals are involved, they are called outer orbital complexes (or high spin/spin-free complexes).
• If the metal ion contains unpaired electrons, the complex is paramagnetic. If all electrons are paired, it is diamagnetic.
• The strength of the bond is greater with increased overlapping between the ligand orbitals and the hybridized metal orbital.

In simple words: VBT says that ligands give electrons to the metal, and the metal uses empty orbitals to accept them. These orbitals mix together (hybridize) to form new orbitals that point in certain directions, giving the complex a specific shape. If there are single electrons left over, the complex is magnetic.

🎯 Exam Tip: Focus on the key VBT points: ligand-metal dative bond, availability of empty metal orbitals, hybridization, definite geometry, and the relationship between unpaired electrons and magnetic properties.

 

Question 5. Write the salient features of Crystal Field Theory.
Answer: Crystal Field Theory (CFT) explains the bonding in coordination compounds by considering the interaction between the central metal ion and the ligands as purely electrostatic. Here are its main features:

• **Electrostatic Interaction:** CFT assumes the bond between the central metal atom and the ligands is purely ionic. This means it's an electrostatic attraction between the positively charged metal ion and the negatively charged ligands (or the negative end of polar neutral ligands).
• **Point Charges/Dipoles:** Ligands are treated as point charges if they are anions, or as electric dipoles if they are neutral molecules.
• **d-Orbital Degeneracy:** In an isolated gaseous metal ion, all five d-orbitals are degenerate (have the same energy). When ligands approach, the electrons in these d-orbitals experience repulsion.
• **Crystal Field Splitting:** As ligands approach the central metal ion, their negative fields cause the d-orbitals to split into different energy levels. This removal of degeneracy is called crystal field splitting. The pattern and magnitude of splitting depend on the geometry of the complex and the nature of the ligands.
• **No Overlap:** CFT does not consider any orbital overlap between the metal and the ligands.
• **Weak Field vs. Strong Field Ligands:** Ligands are classified based on their ability to cause crystal field splitting. Strong field ligands cause a large splitting (high \( \Delta \)), while weak field ligands cause a small splitting (low \( \Delta \)). This is organized in the spectrochemical series.
• **Color and Magnetic Properties:** The color of coordination compounds is explained by d-d electronic transitions occurring between the split d-orbitals when light is absorbed. The magnetic properties (paramagnetic or diamagnetic) are explained by the number of unpaired electrons in the split d-orbitals.

**Hypothetical Steps of Complex Formation (as per CFT):**
**Step-1:** In an isolated gaseous state, all five 'd' orbitals are degenerate. When ligands initially form a spherical field of negative charge around the metal, the energies of all five d-orbitals increase due to repulsion.
**Step-2:** As the ligands approach the metal atom in actual bond directions (e.g., along axes for octahedral complexes), the repulsion becomes specific. Orbitals pointing towards the ligands experience more repulsion and rise in energy (e.g., \( \text{e}_g \) in octahedral), while orbitals pointing between the ligands experience less repulsion and decrease in energy (e.g., \( \text{t}_{2g} \) in octahedral). This is crystal field splitting.
**Step-3:** Initially, the complex formation may not seem favorable due to repulsion. However, the overall attraction between the negatively charged ligands and the positively charged metal ion results in a net decrease in energy, which drives the complex formation.
In a tetrahedral complex, the \( \text{d}_{xy}\text{, d}_{yz}\text{, d}_{zx} \) (\( \text{t}_{2g} \)) orbitals are nearer to the ligands and thus have higher energy than the \( \text{d}_{x^2-y^2}\text{, d}_{z^2} \) (\( \text{e}_g \)) orbitals. This is the reverse of octahedral splitting.
In simple words: CFT explains how metal atoms and surrounding molecules (ligands) bond by thinking of them as tiny magnets. When the ligands get close, they push on the metal's d-electrons, splitting them into different energy levels. This splitting makes compounds colorful and magnetic. Stronger pushing from ligands creates bigger energy gaps.

🎯 Exam Tip: Emphasize that CFT treats ligands as point charges/dipoles, focuses on electrostatic interactions, and uses d-orbital splitting to explain color, magnetism, and stability without considering orbital overlap.

Question 6. Explain the classification of metallic carbonyls.
Answer: Metallic carbonyls are complex compounds formed between a metal atom and carbon monoxide (CO) ligands. They are typically classified in two main ways: based on the number of metal atoms and based on their structure.

I. Classification based on the number of metal atoms present:
a) Mononuclear carbonyls: These compounds have only one metal atom in their structure. For example, tetracarbonylnickel(0) \( [Ni(CO)_4] \) is a mononuclear carbonyl. These types of carbonyls are often simple in their molecular structure.
b) Polynuclear carbonyls: These compounds contain two or more metal atoms. They can be either homonuclear (all metal atoms are the same, like \( [Fe_3(CO)_{12}] \)) or heteronuclear (different metal atoms, like \( [MnCo(CO)_9] \)). These structures involve metal-metal bonding.

II. Classification based on the structure:
a) Non-bridged metal carbonyls: These carbonyls do not have carbon monoxide ligands that connect two or more metal atoms. All CO ligands are bonded to a single metal atom. An example is tetracarbonylnickel(0) \( [Ni(CO)_4] \), where all CO groups are terminal, meaning they are attached to only one metal atom. Many simple metal carbonyls fall into this category, showing straightforward terminal bonding.

These compounds also involve metal-metal bonds where two metal atoms are directly connected, like in decacarbonyldimanganese \( [Mn_2(CO)_{10}] \), where the metal atoms are joined and each has terminal CO ligands. This direct metal-metal bond is crucial for their stability.

b) Bridged carbonyls: These carbonyls have one or more CO ligands that connect two or more metal atoms. These bridging ligands are in addition to any terminal CO ligands. An example is diiron nonacarbonyl \( [Fe_2(CO)_9] \), where some CO groups bridge between the two iron atoms, creating a more complex and robust structure. The bridging creates extra stability by sharing electron density between metal centers.
In simple words: Metallic carbonyls are compounds made of metal atoms and carbon monoxide. They can be grouped by how many metal atoms they have (one or many) or by how the carbon monoxide is attached (just to one metal, or bridging between several metals). The structure shows how these parts are connected.

🎯 Exam Tip: When classifying, always provide an example for each type and clearly state the defining characteristic of that class (e.g., number of metal atoms or presence of bridging ligands).

 

Question 7. Write the IUPAC names for the following complexes. (MARCH 2020)
1. \( [Co(NH_3)_5Cl]^{2+} \)
2. \( K_3[Fe(C_2O_4)_3] \)
3. \( [Co(NH_3)_6][Cr(CN)_6] \)
4. \( [Cr(H_2O)_4Cl_2]Cl \cdot 2H_2O \)
5. \( [Pt(Py)_4][PtCl_4] \)
6. \( [Zn(NCS)_4]^{2-} \)
7. \( [Ag(NH_3)_2]^{2+} \)
Answer:
1. The IUPAC name for \( [Co(NH_3)_5Cl]^{2+} \) is Pentaamminechloridocobalt (III)ion. This means there are five ammonia ligands and one chloride ligand attached to a cobalt metal ion, which has an oxidation state of +3.
2. The IUPAC name for \( K_3[Fe(C_2O_4)_3] \) is Potassium trioxalato ferrate (III). Here, three potassium ions are associated with a complex ion where iron in its +3 oxidation state is bonded to three oxalate ligands.
3. The IUPAC name for \( [Co(NH_3)_6][Cr(CN)_6] \) is Hexaamminecobalt (III) hexacyanido-kC-chromate (III). This complex contains a positively charged cobalt complex and a negatively charged chromium complex, both with metal ions in the +3 oxidation state.
4. The IUPAC name for \( [Cr(H_2O)_4Cl_2]Cl \cdot 2H_2O \) is Tetraaquadichloridochromium (III) chloride dihydrate. This compound has a chromium ion in the +3 oxidation state, bonded to four water molecules and two chloride ions within the coordination sphere, with an additional chloride ion and two water molecules outside.
5. The IUPAC name for \( [Pt(Py)_4][PtCl_4] \) is Tetrapyridineplatinum (II) tetrachloridoplatinate (II). This is a coordination isomer pair, with one platinum in a positive complex with pyridine and another in a negative complex with chloride, both in the +2 oxidation state.
6. The IUPAC name for \( [Zn(NCS)_4]^{2-} \) is Tetrathiocyanato-kN-zincate (II) ion. Here, a zinc metal ion in the +2 oxidation state is bonded to four thiocyanate ligands through the nitrogen atom.
7. The IUPAC name for \( [Ag(NH_3)_2]^{2+} \) is Diamminesilver (I) ion. In this complex, a silver ion in the +1 oxidation state is coordinated to two ammonia ligands. This is a common linear complex.
In simple words: Each complex compound has a special name that tells us what metal is in the middle, what other atoms or groups are attached to it (called ligands), and how many of each ligand there are. The name also tells us the charge of the central metal.

🎯 Exam Tip: Remember to identify the central metal, its oxidation state, and the type and number of ligands correctly. Pay attention to whether the complex is cationic, anionic, or neutral, as this affects the metal's name (e.g., '-ate' suffix for anionic complexes).

 

Question 8. Write the formula for the following coordination compounds.
1. Tris (ethylenediamine) Chromium (III) Chloride
2. Potassium tetracyanido K-C nickelate (II)
3. Ammine bromido chlorido nitrito – kN platinate (II) ion
4. Dichlorido bis (ethane -1, 2 – diamine) Platinum (IV) nitrate
5. Hexa aqua manganese (II) Phosphate
Answer:
1. The formula for Tris (ethylenediamine) Chromium (III) Chloride is \( [Cr(en)_3]Cl_3 \). Here, chromium has three ethylenediamine ligands, and the overall complex is balanced by three chloride ions outside the coordination sphere.
2. The formula for Potassium tetracyanido-kC-nickelate (II) is \( K_2[Ni(CN)_4] \). In this compound, two potassium ions balance the negative charge of the complex, where nickel in the +2 oxidation state is bonded to four cyanide ligands through carbon.
3. The formula for Ammine bromido chlorido nitrito-kN platinate (II) ion is \( [Pt(NH_3)BrCl(NO_2)]^- \). This complex ion contains platinum(II) with one ammonia, one bromide, one chloride, and one nitrito ligand bonded through nitrogen.
4. The formula for Dichlorido bis (ethane-1,2-diamine) Platinum (IV) nitrate is \( [Pt(en)_2Cl_2](NO_3)_2 \). This complex has platinum in the +4 oxidation state, bonded to two ethylenediamine ligands and two chloride ligands, with two nitrate ions balancing the charge outside.
5. The formula for Hexaaquamanganese (II) Phosphate is \( [Mn(H_2O)_6]_3(PO_4)_2 \). In this compound, three manganese(II) ions, each coordinated to six water molecules, are associated with two phosphate ions to form a neutral compound.
In simple words: When given the name of a complex compound, we write its chemical formula by identifying the central metal, its charge, and all the ligands attached to it. Then, we add any balancing ions to make the whole compound neutral.

🎯 Exam Tip: Always make sure the overall charge of the complex matches the sum of the oxidation state of the metal and the charges of all ligands. Remember to use parentheses for polydentate ligands and for neutral molecules when they are part of the coordination sphere.

 

Question 9. What will be the correct order for the wavelengths of absorption in the visible region and explain for the following (PTA – 3) \( [Ni(NO_2)_6]^{4-} \); \( [Ni(NH_3)_6]^{2+} \); \( [Ni(H_2O)_6]^{2+} \)
Answer: The central metal ion in all three complexes is nickel (II), which is \( Ni^{2+} \). Therefore, the absorption of light in the visible region depends on the strength of the ligands. The order in which the Crystal Field Splitting Energy (CFSE) values of the ligands increase, based on the spectrochemical series, is as follows:

\( H_2O < NH_3 < NO_2^- \)

This means the amount of crystal-field splitting (Δ0) observed will be in the following order:

\( \Delta_0(H_2O) < \Delta_0(NH_3) < \Delta_0(NO_2^-) \)

A larger crystal field splitting energy means that higher energy light (shorter wavelength) is absorbed. Since the energy of absorbed light is inversely proportional to its wavelength, the wavelengths of absorption in the visible region will be in the inverse order of the CFSE:

\( [Ni(NO_2)_6]^{4-} < [Ni(NH_3)_6]^{2+} < [Ni(H_2O)_6]^{2+} \)

Therefore, the complex with \( NO_2^- \) (strongest ligand) will absorb light of the shortest wavelength, and the complex with \( H_2O \) (weakest ligand) will absorb light of the longest wavelength. This explains why different ligands lead to different colors in complexes.
In simple words: The type of ligand attached to a metal changes how much energy the complex absorbs. Stronger ligands cause more energy to be absorbed, which means the light they absorb has a shorter wavelength. So, weak ligands absorb longer wavelengths (like red), and strong ligands absorb shorter wavelengths (like blue).

🎯 Exam Tip: Remember the spectrochemical series for common ligands: strong field ligands like \( CN^- \) and \( CO \) cause large splitting, while weak field ligands like halides and water cause smaller splitting. A larger Δ0 (splitting energy) corresponds to shorter absorbed wavelength and higher energy light.

 

Question 10. Answer all the questions for the complex \( [Fe(en)_2Cl_2]Cl_2 \) (PTA -6)
1. Oxidation number of Fe
2. Hybridisation and shape
3. Magnetic behaviour
4. Number of geometric isomers
5. Whether there may be optical isomer also?
6. IUPAC name
Answer:
1. The oxidation number of Iron (Fe) in \( [Fe(en)_2Cl_2]Cl_2 \) is +3. This is calculated by considering the ethylenediamine (en) as neutral, the two inner sphere chlorides as -1 each, and the two outer sphere chlorides as -1 each.
2. The hybridisation of the central metal ion is \( d^2sp^3 \), which corresponds to an octahedral shape. Iron forms bonds using its inner d-orbitals in this complex.
3. The complex exhibits paramagnetic behaviour due to the presence of unpaired electrons in the d-orbitals of the \( Fe^{3+} \) ion. The strength of the ligands determines the number of unpaired electrons.
4. There are two geometric isomers possible for this complex: a cis isomer and a trans isomer. These isomers differ in the spatial arrangement of the ligands around the central metal ion.
5. Yes, the cis isomer of \( [Fe(en)_2Cl_2]^{2+} \) can also show optical isomerism. The cis form is chiral, meaning it is non-superimposable on its mirror image, which leads to optical activity.
6. The IUPAC name for this compound is Dichloridobis(ethane-1,2-diamine)iron(III) chloride. This name clearly describes the ligands, the metal, its oxidation state, and the counter ion.
In simple words: For this iron complex, the iron atom has a +3 charge and uses \( d^2sp^3 \) bonding, giving it an octahedral shape. It is magnetic because it has unpaired electrons. You can find two different shapes (cis and trans), and the cis shape can also twist light. The full chemical name tells us all these details.

🎯 Exam Tip: When dealing with isomers, remember that geometric isomerism refers to different spatial arrangements, while optical isomerism relates to non-superimposable mirror images (chirality). For octahedral complexes like \( [Ma_2b_2c_2]^n \) or \( [M(AA)_2B_2]^n \), cis isomers often exhibit optical isomerism.