CBSE Class 12 Chemistry Chapter 9: Coordination Compounds NCERT Solutions

To write the formulas, we need to identify the central metal ion, its oxidation state, the ligands, and the counter ions. The coordination number helps determine the arrangement within the coordination sphere.

1. The central metal ion is Cobalt (Co) with an oxidation state of +3. The ligands are tetraammine (four NH_3 molecules, neutral) and diaqua (two H_2O molecules, neutral). The counter ion is chloride (Cl^-). The coordination number is 4 (ammine) + 2 (aqua) = 6. The complex ion is [Co(H_2O)_2(NH_3)_4]^{3+}. To balance the +3 charge, three chloride ions are needed. Formula: $[CO(H\_2O)\_2(NH\_3)\_4]CI\_3$

2. The central metal ion is Nickel (Ni) with an oxidation state of +2. The ligand is tetracyano (four CN^- ions). The counter ion is Potassium (K^+). The complex ion is [Ni(CN)_4]^{2-}. To balance the -2 charge, two potassium ions are needed. Formula: $K\_2[Ni(CN)\_4]$

3. The central metal ion is Chromium (Cr) with an oxidation state of +3. The ligand is tris(ethane-1,2-diamine), meaning three molecules of ethylenediamine (en, neutral bidentate ligand). The counter ion is chloride (Cl^-). The coordination number is 3 * 2 = 6. The complex ion is [Cr(en)_3]^{3+}. To balance the +3 charge, three chloride ions are needed. Formula: $[Cr(en)\_3]Cl\_3$

4. The central metal ion is Platinum (Pt) with an oxidation state of +2. The ligands are ammine (NH_3, neutral), bromo (Br^-, -1 charge), chloro (Cl^-, -1 charge), and nitrito-N (NO_2^-, -1 charge). The complex is an anion (platinate). The total charge from ligands is -1 + (-1) + (-1) = -3. For Pt(II), the complex ion charge is +2 - 3 = -1. Formula: $[Pt(NH\_3)BrCl(NO\_2)]^-$

5. The central metal ion is Platinum (Pt) with an oxidation state of +4. The ligands are dichlorido (two Cl^-, -1 each) and bis(ethane-1,2-diamine) (two en molecules, neutral). The counter ion is nitrate (NO_3^-, -1 charge). The total charge from ligands is 2*(-1) = -2. For Pt(IV), the complex ion charge is +4 - 2 = +2. The complex ion is [PtCl_2(en)_2]^{2+}. To balance the +2 charge, two nitrate ions are needed. Formula: $[PtCl\_2(en)\_2](NO\_3)\_2$

6. The central metal ion is Iron (Fe) with an oxidation state of +3. The ligand is hexacyano (six CN^- ions). The complex is an anion (ferrate). The complex ion is [Fe(CN)_6]^{4-}. The counter ion is Iron(III) (Fe^{3+}). The formula is written as Fe_x[Fe(CN)_6]_y. To balance the charges, we need 3 Fe^{3+} ions and 2 [Fe(CN)_6]^{4-} ions. Formula: $Fe\_3[Fe(CN)\_6]\_2$. (Note: The source provided Fe4[Fe(CN)6]3, which corresponds to Iron(III) hexacyanoferrate(III) or Iron(II) hexacyanoferrate(III) depending on interpretation. Assuming Iron(III) hexacyanoferrate(II) implies Fe(III) as counter and Fe(II) in complex, or vice versa. The common compound is Prussian blue, Fe_4[Fe(CN)_6]_3, which is Iron(III) hexacyanoferrate(II) if the central iron is +2 and the outer is +3, or vice versa. The provided solution Fe_4[Fe(CN)_6]_3 implies Fe(III) in the complex and Fe(II) outside, or vice versa. Let's follow the source's formula for Iron(III) hexacyanoferrate(II) which is typically written as Fe_3[Fe(CN)_6]_2 if Fe is +2 in complex and +3 outside, or Fe_4[Fe(CN)_6]_3 if Fe is +3 in complex and +2 outside. The source's answer is $Fe\_4[Fe(CN)\_6]\_3$, which implies Fe(III) in the complex and Fe(II) outside, or vice versa. Let's assume the question implies Fe(III) as the counter ion and Fe(II) as the central metal ion in the complex, leading to Fe_3[Fe(II)(CN)_6]_2. However, the provided answer is $Fe\_4[Fe(CN)\_6]\_3$. This formula corresponds to Iron(III) hexacyanoferrate(II) if the central iron is +2 and the outer iron is +3, or vice versa. Let's stick to the provided answer's formula. Formula: $Fe\_4[Fe(CN)\_6]\_3$)

To determine the IUPAC names, we identify the ligands, the central metal ion, its oxidation state, and the counter ions, following specific rules for naming.

1. Ligands: Six ammine (NH_3) groups. Central metal ion: Cobalt (Co). Counter ion: Chloride (Cl^-). The complex ion is [Co(NH_3)_6]^{3+}. Oxidation state of Co: x + 6(0) = +3 => x = +3. Name: Hexaamminecobalt(III) chloride.

2. Ligands: Five ammine (NH_3) groups and one chloro (Cl) group. Central metal ion: Cobalt (Co). Counter ion: Chloride (Cl^-). The complex ion is [Co(NH_3)_5Cl]^{2+}. Oxidation state of Co: x + 5(0) + (-1) = +2 => x = +3. Name: Pentaamminechloridocobalt(III) chloride.

3. Counter ion: Potassium (K^+). Ligands: Six cyano (CN^-) groups. Central metal ion: Iron (Fe). The complex ion is [Fe(CN)_6]^{3-}. Oxidation state of Fe: 3(+1) + x + 6(-1) = 0 => x - 3 = 0 => x = +3. Name: Potassium hexacyanoferrate(III).

4. Counter ion: Potassium (K^+). Ligands: Three oxalato (C_2O_4^{2-}) groups. Central metal ion: Iron (Fe). The complex ion is [Fe(C_2O_4)_3]^{3-}. Oxidation state of Fe: 3(+1) + 3(-2) = 0 => 3 + x - 6 = 0 => x = +3. Name: Potassium trioxalatoferrate(III).

5. Counter ion: Potassium (K^+). Ligands: Four chloro (Cl^-) groups. Central metal ion: Palladium (Pd). The complex ion is [PdCl_4]^{2-}. Oxidation state of Pd: 2(+1) + x + 4(-1) = 0 => 2 + x - 4 = 0 => x = +2. Name: Potassium tetrachloridopalladate(II).

6. Ligands: Two ammine (NH_3) groups, one chloro (Cl) group, and one methylamine (NH_2CH_3) group. Central metal ion: Platinum (Pt). Counter ion: Chloride (Cl^-). The complex ion is [Pt(NH_3)_2Cl(NH_2CH_3)]^+. Oxidation state of Pt: x + 2(0) + (-1) + 0 = +1 => x = +2. Name: Diamminechlorido(methylamine)platinum(II) chloride.

We need to identify the types of isomerism (geometrical, optical, linkage, ionization, etc.) and draw the structures for the isomers of the given coordination complexes.

1. Complex: K[Cr(H_2O)_2(C_2O_4)_2]. This complex has the general formula MA_2B_2, where M is Cr, A is C_2O_4^{2-} (oxalate, a bidentate ligand), and B is H_2O (aqua, a monodentate ligand). It exhibits geometrical isomerism (cis and trans) and the cis isomer exhibits optical isomerism.

   Geometrical Isomerism:

   Trans isomer: The two aqua ligands are opposite to each other.

   \begin{bmatrix}

   Cr(H_2O)_2(C_2O_4)_2

   \end{bmatrix}^-

   (Trans)

   Cis isomer: The two aqua ligands are adjacent to each other.

   \begin{bmatrix}

   Cr(H_2O)_2(C_2O_4)_2

   \end{bmatrix}^-

   (Cis)

   Optical Isomerism:

   The trans isomer is optically inactive because it has a plane of symmetry. The cis isomer is chiral and exists as a pair of non-superimposable mirror images (enantiomers), hence it is optically active.

   cis-isomer (Dextro and Levo forms)

2. Complex: [Co(en)_3]Cl_3. This complex has the general formula M(AA)_3, where M is Co and AA is ethylenediamine (en), a bidentate ligand. This type of complex exhibits optical isomerism but not geometrical isomerism because all the bidentate ligands are identical and arranged symmetrically.

   Optical Isomerism:

   The complex exists as two optical isomers (enantiomers) which are non-superimposable mirror images of each other. Both isomers are optically active.

   \left[ Co(en)_3 \right]^{3+}

   (Optical Isomers)

3. Complex: $[Co(NH\_3)\_5(NO\_2)](NO\_3)\_2$. This complex contains the nitrite ion (NO_2^-), which is an ambidentate ligand. It can coordinate either through the nitrogen atom (nitro complex) or through an oxygen atom (nitrito-O complex). Therefore, it exhibits linkage isomerism.

   Linkage Isomerism:

   Isomer 1 (Nitro): The nitrite ion coordinates through the nitrogen atom.

   $[Co(NH\_3)\_5(NO\_2)](NO\_3)\_2$

   Isomer 2 (Nitrito-O): The nitrite ion coordinates through an oxygen atom.

   $[Co(NH\_3)\_5(ONO)](NO\_3)\_2$

4. Complex: [Pt(NH_3)(H_2O)Cl_2]. This complex has the general formula MA_2BC, where M is Pt, A is Cl (monodentate ligand), B is NH_3 (ammine), and C is H_2O (aqua). It exhibits geometrical isomerism (cis and trans) and potentially optical isomerism for the cis form if the ligands were different, but here it's cis/trans geometrical isomers.

   Geometrical Isomerism:

   Trans isomer: The two chloro ligands are opposite to each other.

   \begin{bmatrix}

   Pt(NH_3)(H_2O)Cl_2

   \end{bmatrix}

   (Trans)

   Cis isomer: The two chloro ligands are adjacent to each other.

   \begin{bmatrix}

   Pt(NH_3)(H_2O)Cl_2

   \end{bmatrix}

   (Cis)

   Optical Isomerism: The cis isomer in this case is not optically active because it has a plane of symmetry passing through Pt, NH_3, H_2O and bisecting the Cl-Pt-Cl angle. The trans isomer is also optically inactive.