Answer: : Co = 27, Cr = 24, Ni = 28) Write the IUPAC names of the following coordination compounds: Using IUPAC norms write the formulae for the following coordination compounds: (ii) [Pt(NH3)2Cl2] (i) the IUPAC name, Question 6: Question 34: (iii) Write the hybridization type and magnetic behaviour of the complex Answer: in Nicl4 the central atom ni , whoose valence shell configuration in free state is 3d8,4s0, 4p0 . molecular geometry, of each of these species. Dichloro Bis-(ethane-l,2 diamine) Cobalt (III). (ii) K3[Cr(C204)3]. (i) [Cr(C204)3]3- (iii) Magnetic behaviour of the complex. (b) Out of NH3 and CO, which ligand forms a more stable complex with a transition metal and why? NiCl 4 2-, there is Ni 2+ ion, However, in presence of weak field Cl- ligands, NO pairing of d-electrons occurs. Please log inor registerto add a comment. (i) Write down the IUPAC name of the following complex: [CO(NH3)5Cl]2+ Explain difference. (i) [Cr(NH3)6]3+ (ii) [Fe(CN)6]4- (iii) [NiCl4]2- (ii) What type of isomerism is exhibited by the complex [Co(en)3]3+? and not tetrahedral by sp3. The second complex is not a neutral complex. Use the above data to determine: (Atomic number of Fe — 26) Answer: (b) Give an example of the role of coordination compounds in biological systems. (i) [Cr(NH3)3Cl3] Co = 27, Ni = 28) of Co = 27) CN − being a strong field ligand causes the pairing of unpaired electrons. These conditions are met or found only in transition metals. [Ni (CN)4]2- is diamagnetic, so Ni2+ ion has 3d8 outer configuration with two unpaired electrons. Potassium tri oxalato chromate(III) of Ni = 28) (i) Tetrachloridocuprate(II) [CO(NH3)5S04]Cl Since it have two unpaired electron electron therefore the magnetic moment : no. As there are unpaired electrons in the d-orbitals, NiCl42- is paramagnetic and is referred to as a high spin complex. to Q.46 (i). Question 27: (iii) [NiCl4]2_ is paramagnetic while [Ni(CO)4] is diamagnetic, though both are tetrahedral. Co = 27, Ni = 28) (b) [CO(NH3)6]2 (S04)3, octahedral. (ii) Ni2+ ion is bound to two water molecules and two oxalate ions. Since CN − ion is a strong field ligand, it causes the pairing of unpaired 3d electrons.. Answer: Question 42: Hence, the complex ion is paramagnetic. Pd and Ni have the same electron configuration but PdCl42- has square planar structure and NiCl42- has tetrahedral structure. (a) (i) [FeF6]3_ has sp3d2 hybridization, octahedral shape. determined by atomic absorption and inductively coupled plasma atomic emission (ii) [Co(en)3] Cl3 (iii) Average error, forming compounds with examples fastly answer​, wt r the chemical reaction of the ketone ​. Answer: (i) [CO(NH3)6]Cl3 (ii) K2[NiCl4], Question 10: Answer: (iii) Dibromidobis (ethane 1, 2-diamine)cobalt (III), Question 37: Thus, it can either have a tetrahedral geometry or square planar geometry. Answer: If you help me then I will be happy​, what do you know about corpuscular nature of matter?​, The concentration of Nickel in Nigerian coin was determined with visible There are 4 CN − ions. Therefore, it undergoes sp3 hybridization. Answer: Question 22: Explain the following terms giving a suitable example in each case: (i) [Cr(NH3)4Cl2]Cl has d2sps hybridization, octahedral shape and paramagnetic. Answer: The correct formula and geometry of the first complex is : (1) [Ni(H Themetal ionscan also be arranged in order of increasing Δ, and this order is largely independent of the identity of the ligand. (ii) Nickel (II) does not form low spin octahedral complexes. …, wt r the preparation of the carboxylic acid ​, please give correct answer. (ii) Refer Ans. to Q.58 (iii). (i) [FeF6]3 (ii) [Ni(CO)4] Coordination isomerism. Question 58: 1. (i) Triamminetrichloridochromium (III) (iii) Refer Ans. What type of hybridization is involved in [F e (C N) 6 ] 3 − : View solution N i ( C O ) 4 is diamagnetic whereas [ N i C I 4 ] 2 − is paramagnetic explain. Geometry of Complex (ii) K3[Fe(CN)6] spectrophotometer was 3.92. (ii) Write the formula for the following complex: [Co(NH3)6][Cr(CN)6] (i) Diammine chlorido nitrito-N-platinum(II). (i) [COF4]2- (ii) [Cr(H20)2(C202)2]- (iii) [Ni(CO)4] Explain the following: Answer: Question 24: Use the magnetic behaviour of these complexes to deduce the geometric structures , I.e. (ii) Write the formula for the following complex: Potassium tetracyanidonickelate(II). (ii) [CO(NH3)4 Cl2] Cl (Valence Bond Theory) The coordination complex, [Cu(OH 2) 6] 2+ has one unpaired electron. (ii) the hybridization type, (iii) Refer Ans. (ii) An outer orbital complex of Co = 27) (iii) Refer Ans. Pentaamminecarbonato cobalt (III) chloride. The complex [Ni (CN)4]2- is diamagnetic, but [NiCl4]2- is paramagnetic with two unpaired electrons. All India 2012) Answer: Stability of a complex in solution means the measure of resistance to the replacement of a … (i) [CO(NH3)5Cl]S04 (ii) [Co(en)3]3+ (iii) [Co(NH3)6] [Cr(CN)6] (iii) d2sp2, octahedral shape. Question 50: Question 67: (ii) [Cr(C204)3]3- However, hybridisation cannot account for the position of ligands in the spectrochemical series! [C r C l 2 (N O 2 ) 2 (N H 3 ) 2 ] − complex involves d 2 s p 3 hybridization. (en = ethane-1, 2-diamine) no. What type of isomerism is shown by this complex? Write the name and magnetic behaviour of each of the above coordination entities. (iii) [CO(NH3)3Cl3] (Atomic numbers Cr = 24, Co = 27) (At. [CoCl4]2-, [Cr(H20)2(C204)2]- , [Ni(CN)4]2-, Why is CO a stronger ligand than Cl-? Answer: Question 21: Answer: Hence, the hybridization will be dsp 2 so hence, it is a square planar complex because all dsp^2 complexes are square planar. Answer: In [NiCl 4] 2−, the oxidation state of Ni is +2. (a) Dibromidobis (ethane-1, 2-diamine)cobalt(III) Thus tetrahedralstructure of [MnCl4]2–complex will show 5.92 BM magnetic moment value. Cr+3 + 6H2O [Cr(OH2)6]+3 ----- ... [NiCl4]-2 Ni = 4s2 3d8 Ni+2 = 4s0 3d8 . (i) [Pt(NH3)2Cl(N02)] Answer: Question 76: Question 62: (i) +3 (III) find the nortons equivalent across A and B for the given circuit.​, what is election girlcome 5324611502 an /pas/ (modiji) ​, . Answer: Question 30: Therefore, it does not lead to the pairing of unpaired 3d electrons. CBSE Maths notes, CBSE physics notes, CBSE chemistry notes. (ii) [Cr(en)3]Cl3. nos Mn = 25, Co = 27, Ni = 28) Answer: (At. (i) The splitting of d-orbitals in presence of ligands is called crystal field splitting, e.g. (a) (i) sp3d2, octahedral (ii) dsp2, square planar. (iii) Potassium tetracyanonickelate(II). (i) [Cr(NH3)4Cl2] Cl (i) Pentaammine chloridocobalt III chloride. (ii) On the basis of crystal field theory, write the electronic configuration for d4 ion if Δ0 < P. (i) Refer Ans. (i) Nickel does not form low spin octahedral complexes. thus , it have SP3 hybridisation which have tetradehdral geometry . (i) [CO(NH3)2 (H2O) Cl] Cl2 Since CN-ion is a strong field ligand, it causes the pairing of unpaired 3d electrons.. (ii) On the basis of crystal field theory, write the electronic configuration for d4 ion if Δ0 > P. No. (i) Refer Ans. It forms a square planar structure. Answer: Therefore, Ni 2+ undergoes sp 3 hybridization to make bonds with Cl- ligands in tetrahedral geometry. In octahedral complexes, pairing of electrons will not take place even if we have strong field ligand, therefore, Ni does not form low spin octahedral complexes. KEY POINTS: [NiCl4]2- Hybridization:sp3 [NiCl4]2- Shape & Structure: Tetrahedral [NiCl4]2- Magnetic nature: Paramagnetic Best answer (c) : In the paramagnetic and tetrahedral complex [NiCl4]2–, the nickel is in +2 oxidation state and the ion has the electronic configuration 3d8. (ii) Refer Ans. Answer: (At. Answer: [Ni(CN)4]2- is a square planar geometry formed by dsp2 hybridisation. (iii) K2[Ni(CN)4] (c) Why is CO a stronger ligand than NH3 in complexes? (iii) Tetracyanidonickelate(II). (ii) The series in which ligands are arranged in the increasing order of their strength is called spectrochemical series. nos. (iii) Ni(CO)4 has spb3 hybridization, tetrahedral shape, whereas [Ni(CN)4]2- has dsp2 hybridization, therefore, it has a square planar shape. (c) A CuS04 solution is mixed with (NH4)2 S04 solution in the ratio of 1 : 4 does not give test for Cu2+ ion, Why? Name the following coordination compounds and draw their structures: (Atomic’number of Ni = 28) Hybridization : d 2 sp 3 Shape : Octahedral Magnetic behaviour : Diamagnetic (absence of unpaired electrons) (ii) [Ni(CN) 4] 2-Question 22. Answer: Question 72: (ii) K2[Ni(CN)4], Question 11: (i) Hexaamminecobalt(III) chloride In [NiCl 4] 2−, the oxidation state of Ni is +2.Chloride is a weak field ligand and does not cause pairing up of electrons against the Hund's rule of maximum multiplicity. In case of [NiCl 4] 2−, Cl − ion is a weak field ligand. (Atomic no. Answer: Answer: (ii) On the basis of crystal field theory, write the electronic configuration for d4 ion if A0 < P. (v) Whether there may be optical isomer also. (Atomic numbers Fe = 26, Cr = 24, Ni = 28) Answer: Write the name of the structure and the magnetic behaviour of each one of the following complexes: (i) Pentaammine chlorido cobalt(III) chloride Nos : Cr = 24, Co = 27) (iii) [Fe(CN)6]4- and [Fe(H20)6]2 + are of different colours in dilute solutions. Write the name, stereochemistry and magnetic behaviour of the following: (iii) Ni(CO)4 of Ni = 28) check_circle Expert Answer. Ni is in the +2 oxidation state i.e., in d8 configuration.In case of [NiCl4] 2−, Cl− ion is a weak field ligand. Answer: (i) Refer Ans. (i) [Cr(NH3)4Cl2]+ (ii) [Co(en)3]3+ Compare the following complexes with respect to structural shapes of units, magnetic behaviour and hybrid orbitals involved in units: Answer: Question 73: (i) Potassium hexacyano-manganate(II). Answer: There are 4 CN-ions. It is neutral because the 2+ charge of the original platinum(II) ion is exactly canceled by the two negative charges supplied by the chloride ions. [CO(NH3)6]3+ and [Co(en)3]3+ It has square planar shape and is diamagnetic in nature. (Atomic no. (iii) K2[Ni(CN)4] has dsp2 hybridization, square planar shape, diamagnetic. Nos. (vi) Name of the complex. (iii) The molecular shape of Ni(CO)4 is not the same as that of [Ni(CN)4]2_. (ii) [Pt(NH3)2Cl(N02)] (At no. It has octahedral structure. What type of isomerism is shown by this complex? (Atomic number of Ni = 28) Answer: No. (i) Ammineaqua dichlorido platinum [II] (i) [Ni(CN)4]2- (ii) [NiCl4]2- (iii) [CoF6]3- [At. Describe the state of hybridization, the shape and the magnetic’behaviour of the following complexes: (i) Write down the IUPAC name of the following complex: ‘ Hi all! Question 56: (i) It is octahedral, d2sp3 hybridised, diamagnetic in nature. It has square planar structure. Why? Write down the IUPAC name of the complex [Co(en)2Cl2]+. (iii) Why is [NiCl4]2- paramagnetic but [Ni(CO)4] is diamagnetic? (ii) Potassium tetracyanidoferrate(Il) (At. (i) [Co (en)3]Cl3 (iii) Crystal field splitting in an octahedral field. Explain difference. Explain on the basis of valence bond theory that [Ni(CN4)]2– ion with square planar structure is damagnetic and the [NiCl4]2– ion with tetrahedral geometry is paramagnetic. (i) Low spin octahedral complexes of nickel are not known. (a) Write the IUPAC name of the complex [CoBr2(en)2]+. (ii) Potassium tetracyanido nickelate(II). For the complex [NiCl4]2-, write(i) the IUPAC name(ii) the hybridization type(iii) the shape of the complex. Compare the following complexes with respect to their shape, magnetic behaviour and the hybrid orbitals involved: For the complex [NiCl4]2_ , write (iii) Co2+ is easily oxidised to Co3+ in the presence of a strbng ligand. (ii) In tt-complexes, CT bond is formed by donation of n electrons or lone pair to vacant d-orbital of transition metal and 7t-bond is formed by back donation of pair of electrons from transition metal to vacant antibond¬ing orbitals of alkene or carbon monoxide.