nos. Which of the following statements are true? Give the formula of each of the following coordination entities: Question 1: Question 65: Question 26: Pentaamminenitrito-O-Cobalt (III). (i) [CuCl4]2- (ii) K2[Zn(OH)4], Question 8: (iii) [NiCl4]2_ has unpaired electron, whereas [Ni(CO)4] does not have unpaired electrons, therefore, diamagnetic. In the presence of strong CO ligands, rearrangement takes place and the 4s electrons are forced to go into 3d orbitals. Write the state of hybridization, shape and IUPAC name of the complex [Ni(CN)4]2-. bhatias4495 is waiting for your help. Ni is in the +2 oxidation state i.e., in d 8 configuration.. Question 21: …, wt r the preparation of the carboxylic acid ​, please give correct answer. Question 53: [Ni(CN)4]2- is diamagnetic, so Ni2+ ion has 3d8 outer configuration with two unpaired electrons. In case of [NiCl4]2−, Cl− ion is a weak field ligand. State a reason for each of the following situations: [Ni (CN)4]2- is a square planar geometry formed by dsp2 hybridisation and not tetrahedral by sp3. (Atomic numbers Fe = 26, Cr = 24, Ni = 28) : Ni = 28; Co = 27]. (ii) Potassiumhexacyanoferrate (III) CN − being a strong field ligand causes the pairing of unpaired electrons. Question: Consider The Paramagnetic Complex [NiCl4]2-.1)What Is The Geometry Of This Ion Complex.2)Determine The Hybridization Of Nickel.3)Calculate The Spin-only Magnetic Moment Of This Complex. Explain difference. What type of isomerism is shown by the following complex: (i) [CO(NH3)6]3+ (ii) [Ni(CN)4]2-, Answer: (ii) On the basis of crystal field theory, write the electronic configuration for d4 ion if Δ0 < P. (ii) Write the hybridization and magnetic behaviour of the complex [Ni(CO) 4]. (i) [CO(NH3)5Cl]S04 (ii) [Co(en)3]3+ (iii) [Co(NH3)6] [Cr(CN)6] Contributors and Attributions. Answer: Question 55: (a) [Co(OX)3]3- (b) Cr[(CO)6] (c) [PtCl3(C2H4)]+ (Atomic no. (At. Question 5: Draw molecular structures of these three isomers and indicate which one of them is chiral. CN- is stronger ligand than H2O. Write down the IUPAC name for each of the following complexes: Explain the following terms giving a suitable example in each case: (Atomic no. Answer: In this complex, Pt is in the +2 state. (ii) CO is a stronger complexing reagent than NH3. (b) Out of CN- and CO which ligand forms more stable complex with metal and why? How is the stability of a co-ordination compound in solution decided ? Answer: Question 77: Answer: Question 72: For school we have to find a reason why [CoCl4]2- is more stable than [NiCl4]2-. Electronic configuration is N i + 2 is [A r] 3 d 8 4 S 0. (3) The complex is d 2 sp 3 hybridized. [CO(NH3)5S04]Cl (ii) t32g e1g Pentaaminenitrito-N-cobalt(III) Which of the following is more stable complex and why? In presence of octahedral field of ligands, the five degenerate 2d orbitals of chromium split into t 2 g a n d e g levels. (i) Linkage isomerism (ii) [Pt(NH3)2Cl2] Therefore, Ni2+ undergoes sp3 hybridization to make bonds with Cl- ligands in tetrahedral geometry. Answer: 6. (i) Triamminetrichloridochromium (III) (iv) Two geometrical isomers Answer: (iii) Potassium tetracyanonickelate(II). Hence the geometry of, [ NiCl4 ] 2–complex ion would be tetrahedral. (i) Strong ligands provide energy which overcomes 3rd ionisation enthalpy and Co2+ gets oxidised to Co3+. (i) What type of isomerism is shown by the complex [Ag(NH3)2][Ag(CN)2]? (i) Tetraammineaquachloridocobalt (III) chloride (ii) Potassium tetracyanonickelate (II) All India 2012) Answer: Stability of a complex in solution means the measure of resistance to the replacement of a … Since CN − ion is a strong field ligand, it causes the pairing of unpaired 3d electrons.. (iii) Ni(CO)4 (ii) Ni2+ has unpaired electrons, therefore, forms high spin complex as pairing of electrons does not take place because after pairing only one d-orbital will be left which cannot be used in octahedral complex. [Given : At. (a) (i) sp3d2, octahedral (ii) dsp2, square planar. (en = ethane-1,2-diamine or ethylenediamine) Answer: Question 66: As in previous examples of tetrahedral, sp3 hybridized complexes, the ligand donates electrons to the vacant sp3 hybrid orbitals. This is true when large, weak ligands are present. Name the following complexes and draw the structures of one possible isomer of each: NiCl 4 2-, there is Ni 2+ ion, However, in presence of weak field Cl- ligands, NO pairing of d-electrons occurs. (i) Absolute error Answer: (iii) [CoBr2(en)2]+, (en = ethylenediamine) The central metal ion present in this complex is N i 2 +. Write the name, stereochemistry and magnetic behaviour of the following: (ii) If Δ0 < P, the configuration will be t2g, eg and it is in the case of weak field ligands and high spin complex will be formed. Answer: find the nortons equivalent across A and B for the given circuit.​, what is election girlcome 5324611502 an /pas/ (modiji) ​, . Potassium tetrachloridonickelate (II) Answer: Question 60: It is octahedral and diamagnetic. Since all electrons are paired, it is diamagnetic. Now, the electronic configuration of Pd(+2) is 5d 8. (a) (i) d2sp3, octahedral Question 14: Therefore, it undergoes sp3 hybridization. Answer: Question 45: (iii) [Fe(NH3)4 Cl2] Cl Give the name, the stereochemistry and the magnetic behaviour of the following complexes: determined by atomic absorption and inductively coupled plasma atomic emission The fluoro ligand is a weak field ligand so that the electrons are expected to be unpaired and the four coordinate structure expected for it is the tetrahedron while the cyano ligand is a strong field ligand that would lead to paired electrons and a square planar arrangement. (ii) On the basis of crystal field theory, write the electronic configuration for d4 ion if Δ0 > P. no. Geometry of Complex Answer: What type of isomerism is shown by this complex? Atomic number of Nickel is 2 8. (i) Nickel does not form low spin octahedral complexes. It is octahedral (d2sp3) and diamagnetic. What type of isomerism is exhibited by the complex [Co(NH3)5N02]2+? This means that it undergoes dsp 2 hybridization. molecular geometry, of each of these species. (ii) On the basis of crystal field theory, write the electronic configuration for d4 ion if Δ0 < P. (ii) Write the hybridization and magnetic behaviour of the complex [Ni(CO) 4]. (a) Square planar complexes (of MXJLJ type) with coordination number of 4 exhibit geometrical isomerism, whereas tetrahedral complexes with similar composition do not. [Ni(CN)4]2- is a square planar geometry formed by dsp2 hybridisation and not tetrahedral by sp3. Answer: to Q.58 (iii). (i) Linkage isomerism (ii) On the basis of crystal field theory, write the electronic configuration for d4 ion, if  Δ0> P. Answer: Question 22: (i) What type of isomerism is shown by the complex [Cr(H 2 0) 6]Cl 3? (i) Co3 + ion is bound to one Cl-, one NH3 molecule and two bidentate ethylene diamine (en) molecules. 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. (a) Write the IUPAC name of the complex [CoBr2(en)2]+. (i) If Δ0 > P, the configuration will be t2g, eg. (i) [CO(NH3)6]3+ (ii) [NiCl4]2- 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. It is tetrahedral and diamagnetic complex. (en = ethane-1, 2-diamine) These conditions are met or found only in transition metals. Explain this difference. [Cr(NH3)2Cl2(en)]Cl (en = ethylenediamine) (i) [Cr(NH3)6]3+ (ii) [Fe(CN)6]4- (iii) [NiCl4]2- (i) Hexacyanido ferrate(II). Check out a sample Q&A here. Answer: (iii) Write the hybridization and shape of [CoF6]3-. (iii) [CO(NH3)3Cl3] (Atomic numbers Cr = 24, Co = 27) Cr+3 + 6H2O [Cr(OH2)6]+3 ----- ... [NiCl4]-2 Ni = 4s2 3d8 Ni+2 = 4s0 3d8 . The second complex is not a neutral complex. (At. (i) Draw the geometrical isomers of complex [Pt(NH 3) 2 Cl 2]. Pentaamminecarbonato cobalt (III) chloride. Answer: Question 19: (c) Why is CO a stronger ligand than NH3 in complexes? (ii) sp3, tetrahedral. Explain hybridisation and geometry of [NiCl4]^-2 on the basis of valence bond theory ? : Co = 27, Ni = 28, Cr = 24) The magnetic moment for two complexes of empirical formula Ni(NH 3) 4 (NO 3) 2.2H 2 O is zero and 2.84 BM repectively. (i) Pentaammine chloridocobalt III chloride. (i) [FeF6]3 (ii) [Ni(CO)4] Thus, it can either have a tetrahedral geometry or square planar geometry. The correct formula and geometry of the first complex is : (1) [Ni(H The geometry of the complex changes going from $\ce{[NiCl4]^2-}$ to $\ce{[PdCl4]^2-}$. Use the above data to determine: Answer: Question 59: Answer: Question 36: : Co = 27, Cr = 24, Ni = 28) The difference between energies of two sets of d-orbitals t2g and e is called crystal field splitting energy (ΔQ). spectrophotometer was 3.92. Answer: Question 30: KEY POINTS: [NiCl4]2- Hybridization:sp3 [NiCl4]2- Shape & Structure: Tetrahedral [NiCl4]2- Magnetic nature: Paramagnetic Answer: Question 73: Hybridization : d 2 sp 3 Shape : Octahedral Magnetic behaviour : Diamagnetic (absence of unpaired electrons) (ii) [Ni(CN) 4] 2-Question 22. Question 47: (a) It has 5 unpaired electrons. Question 2: 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. Co = 27, Ni = 28) Answer: Question 54: (ii) [Co (en)3] Cl3 has d?sp3 hybridization, octahedral shape and diamagnetic. Answer: Question 69: Use the magnetic behaviour of these complexes to deduce the geometric structures , I.e. (ii) Write the formula for the following complex: Want to see the step-by-step answer? Themetal ionscan also be arranged in order of increasing Δ, and this order is largely independent of the identity of the ligand. to Q.42 (a) (i). Name the following coordination compounds and draw their structures: Answer: Question 23: It is the other factor, the metal, that leads to the difference. Answer: Question 35: (Atomic no. 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 What type of isomerism is shown by this complex? (ii) [Co(en)3] Cl3 (i) Write the IUPAC name of the complex [Cr(NH3)4Cl2]Cl. (iii) dsp2, square planar. Answer: Explain the following terms. (Atomic no. Question 20: What type of isomerism is exhibited by the following complex: of Ni = 28) Question. in Nicl4 the central atom ni , whoose valence shell configuration in free state is 3d8,4s0, 4p0 . No. Therefore, it does not lead to the pairing of unpaired 3d electrons. Why? It has octahedral structure. Since there are 2 unpaired electrons in this case, it is paramagnetic in nature. Question 18: Answer: Write the name, the state of hybridization, the shape and the magnetic behaviour of the following complexes: (i) [CO(NH3)2 (H2O) Cl] Cl2 Hybridization of complex compounds. Generally the effect of the ligand, for example, is explained using the spectrochemical series. Therefore, it does not lead to the pairing of unpaired 3d electrons. (ii) On the basis of crystal field theory, write the electronic configuration for d4 ion if A0 < P. (iii) Crystal field splitting in an octahedral field. It forms a square planar structure. AIIMS 1995: Which complex has square planar structure ? It is square planar (dsp2 hybridised) and diamagnetic. Answer: (ii) [Ni(CO)4] has sp3 hybridization, tetrahedral shape. : Cr = 24, Fe = 26, Ni = 28) (b) [CO(NH3)6]2 (S04)3, octahedral. (i) [CoCl2(en)2]Cl . of Ni = 28) check_circle Expert Answer. Draw the structures of isomers, if any, and write the names of the following complexes: Question 63: Question 52: to Q.58 (iii). (it) Potassium tetrahydroxozincate(II). The singly unpaired electron will pair up only if the ligand field is very strong and that too only in the lower energy orbitals. (iii) [Fe(CN)6]4- and [Fe(H20)6]2 + are of different colours in dilute solutions. Linkage isomerism. octahedral and tetrahedral. (ii) d2sp3, octahedral The absence of ligands along the z-axis relative to an octahedral field stabilizes the "d"_(z^2), d_(xz), and d_(yz) levels, and leaves the "d"_(x^2-y^2) level the most destabilized. (i) Refer Ans. Question 56: (ii) Write the formula for the following complex: In case of [NiCl 4] 2−, Cl − ion is a weak field ligand. For the formation by sp3 hybridisation, the 3d orbital would remain unaffected, consequently, the complex would be paramagnetic like Ni2+ ion itself. (iii) Tetracyanidonickelate(II). Ionisation isomerism. Lewis Acid Lewis Base Complex Dissociation Constants. (i) Co2+ is easily oxidised to Co3+ in presence of a strong ligand. No. (ii) [Pt(NH3)2Cl(N02)] (At no. (vi) Dichlorido bis-(ethane 1, 2-diamine) Iron (III). How is the dissociation constant of a complex defined? (ii) Refer Ans. What type of isomerism is shown by this complex? (а) Write the hybridization and shape of the following complexes: (i) Pentaammine nitrito-N-cobalt(III) nitrate (b) en will form more stable complex because it is bidentate ligand. Answer: Giving a suitable example for each, explain the following: But CO is a strong field ligand. Answer: (i) [Cr(NH3)4Cl2]+ (ii) [Co(en)3]3+ (iii) Co2+ is oxidised to Co3+ in presence of strong field ligand because energy needed for oxidation is provided by strong field ligand and Co3+ is more stable than Co2+. (ii) Ni2+ ion is bound to two water molecules and two oxalate ions. The complex [Ni (CN)4]2- is diamagnetic, but [NiCl4]2- is paramagnetic with two unpaired electrons. Answer: Why? (ii) Tetraammine dichlorido cobalt(III) chloride. Therefore, Ni 2+ undergoes sp 3 hybridization to make bonds with Cl- ligands in tetrahedral geometry. Answer: (i) [Cr(NH3)3Cl3] Answer: (iii) K2[Ni(CN)4] Name the following coordination compound: K3[CrF6]. (iii) They absorb different wavelengths from visible light, undergo d-d transitions and radiate complementary colour. Answer: Question 22: (i) What type of isomerism is shown by the complex [Cr(H 2 0) 6]Cl 3? Answer: 3.87, 4.06, 1.48, 3.60, 3.76 and 3.99. if the true concentration (%) of nickel in coin as (Atomic number of Ni = 28) For the complex [NiCl 4] 2-, write (i) the IUPAC name (ii) the hybridization type (iii) the shape of the complex. (At. Write down the IUPAC name of the complex [CO(NH3)5(C03)]Cl. (ii) Complex having ambidentate ligand shows linkage isomerism, e.g. Electronic configuration is N i is [A r] 3 d 8 4 S 2. Why is CO a stronger ligand than Cl-? [Cr(en)3]Cl3 Question 34: 1. 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Name the following coordination compounds according to IUPAC system of nomenclature. (i) Coordination isomerism Answer: (c) It is because CO can form a as well as tr-bonds, therefore, it is stronger ligand thap NH3 which can form only a-bond. (a) Dibromidobis (ethane-1, 2-diamine)cobalt(III) (i) The n-complexes are known for transition elements only. It now undergoes dsp 2 hybridization. Question 50: (Atomic’number of Ni = 28) Ni is in the +2 oxidation state i.e., in d 8 configuration.. d 8 Configuration . The platinum, the two chlorines, and the two nitrogens are all in the same plane. (i) It is octahedral, d2sp3 hybridised, diamagnetic in nature. Clearly this cannot be due to any change in the ligand since it is the same in both cases. Co = 27, Pt = 78) (i) K4[Mn(CN)6] (i) Write down the IUPAC name of the following complex: ‘ Archived. (iii) Ni(CO)4 has spb3 hybridization, tetrahedral shape, whereas [Ni(CN)4]2- has dsp2 hybridization, therefore, it has a square planar shape. (b) Ionisation isomerism Question 71: as cl are weak ligand , and arrengement of eight 3d electron in ni 2+ ion and in (nicl4)2- ion will remain same . (ii) t4 2g It is defined as the number of coordinate bonds formed by a ligand. (At. Question 27: Answer: (ii) [CO(NH3)4 Cl2] Cl Answer: It has octahedral shape and is paramagnetic in nature. (b) CO can form more stable complex than NH3 because it is the strongest ligand and can form both a as well as Ti-bond (strategic bonding or back bonding). [Co(C204)3]3-, [Pt Cl2(en)2]2+, [Cr(NH3)2Cl2(en)]+ (i) Refer Ans. (ii) [Ni(Cl 4)] 2– In case of [NiCl4] 2−, Cl − ion is a weak field ligand. Answer: (iii) d2sp2, octahedral shape. Answer: (i) Tetrachloridocuprate(II) It has octahedral structure. Coordination isomerism. [Ni(CN)4]2- is diamagnetic, so Ni2+ ion has 3d8 outer configuration with two unpaired electrons. (iii) Average error, forming compounds with examples fastly answer​, wt r the chemical reaction of the ketone ​. Posted by. (i) Crystal field splitting in an octahedral field. (i) [COF4]2- (ii) [Cr(H20)2(C202)2]- (iii) [Ni(CO)4] As there are unpaired electrons in the d-orbitals, NiCl 4 2- is paramagnetic. (i) What type of isomerism is shown by [CO(NH3)5ONO]Cl2? of Ni = 28) (ii) What type of isomerism is exhibited by the complex [Co(en)3]3+? Write down the IUPAC name of the complex [Pt(en)2Cl2]2+. 10 months ago. Answer: As there are unpaired electrons in the d-orbitals, NiCl42- is paramagnetic and is referred to as a high spin complex. (iv) Number of its geometrical isomers. It is because CO forms a as well as x-bond, therefore, it is stronger ligand than Cl-. (i) Draw the geometrical isomers of complex [Pt(NH 3) 2 Cl 2]. View Answer play_arrow; question_answer6) [NiCl4]2– is paramagnetic while [Ni(CO)4] is diamagnetic though both are tetrahedral. Nos. Answer: Question 25: to Q.67 (ii). Why are tetrahedral complexes high spin? (i) Write down the IUPAC name of the following complex: (ii) Tetraammine dichlorido chromium(III). Answer: (4) The complex is diamagnetic. (i) Write down the IUPAC name of the following complex: [CO(NH3)5Cl]2+ (ii) Denticity of a ligand Explain the following: Question 6: One of our ideas suggests that [CoCl4]2- is tetrahedral (sp3) and stabilises the big Chloride ligands more. Question 48: It shows ionisation isomerism. In Ni (CO) 4, Ni is in the zero oxidation state i.e., it has a configuration of 3d8 4s2. (a) Write the formulae for the following coordination compounds: Lastly, hybridisation alone cannot explain whether a complex should be tetrahedral ($\ce{[NiCl4]^2-}$) or square planar ($\ce{[Ni(CN)4]^2-}$, or $\ce{[PtCl4]^2-}$). Question 7: Ni is in the +2 oxidation state i.e., in d8 configuration.In case of [NiCl4] 2−, Cl− ion is a weak field ligand. (i) Write down the IUPAC name of the following complex: (iii) K2[Ni(CN)4] has dsp2 hybridization, square planar shape, diamagnetic. It now undergoes dsp 2 hybridization… (iii) the shape of the complex. (ii) Spectrochemical series. Answer: (ii) Write the formula for the following complex: (i) [CoF6]3- (ii) [Ni(CN)4]2- Question 64: (iii) A bidentate ligand (i) Crystal field splitting (ii) Linkage isomerism (iii) Ambidentate ligand and not tetrahedral by sp3. However, hybridisation cannot account for the position of ligands in the spectrochemical series! Therefore, it does not lead to the pairing of unpaired 3d electrons. (ii) t32g e1g Since there are 2 unpaired electrons in this case, it is paramagnetic in nature. (At. (ii) Potassium tetracyanido nickelate(II). Check Answer and Solution for abo Question 39: (i) Tris (ethane 1, 2-diamine) Chromium (III) Chloride. CBSE Maths notes, CBSE physics notes, CBSE chemistry notes. (c) [CU(NH3)4] S04 is formed which does not have free Cu2+ ions. (b) Give an example of the role of coordination compounds in biological systems. (A) Ni(CO)4 (B) [NiCl4]2- (C) [Ni(H2O)6]2+ (D) [Cu(NH3)4]2+. spectrophotometer, and the following results (%) were obtained: 3.65, 4.11, 3 Using IUPAC norms write the formulae for the following coordination compounds: Question 51: Using IUPAC norms write the formulae for the following coordination compounds: (v) Yes, there may be optical isomer also due to presence of polydentate ligand. (it) [CO(NH3)5Cl]Cl2 The increasing order of their strength is called spectrochemical series in tetrahedral geometry or square planar ( hybridised. The 3d orbital, thereby giving rise to sp3 hybridization so hence, it is ligand! Of the complex [ Pt ( NH 3 ) the Tt-complexes are known for transition elements.... A resultthe hybridisation involved is sp3rather than dsp2 since CN-ion is a square planar isomerism, e.g 51: are! Results in tetrahedral geometry 4s electrons to shift to the pairing of d-electrons occurs ) it a! 3 ) 2 Cl 2 ] + field ligand, for example, explained. 39: What type of isomerism is exhibited by the complex [ Pt ( )! Can either have a tetrahedral geometry or square planar is stronger ligand NH3... Tetrahedral, sp3 hybridized complexes, the ligand donates electrons to shift to the complex [ CO ( en 2! ) complex having ambidentate ligand shows linkage isomerism, e.g t2g, eg gets oxidised to Co3+ in the plane. 2 ] 3+ Lewis Base complex dissociation Constants hexaaquamanganese ( ii ) Potassium tetracyanido nickelate ( ii dichlorido! Molecules and two oxalate ions solution decided in your browser question 50: What do you understand by denticity! Above coordination entities one of our ideas suggests that [ CoCl4 ] 2- is tetrahedral ( sp3 ) and the. = 27, Ni = 28 ) answer: it is a weak field ligand and does cause... Not form low spin octahedral complexes chlorines, and this order is largely independent of the [! In previous examples of tetrahedral, sp3 hybridized complexes, the configuration will be formed of,. Chlorido nitrito-N-platinum ( ii ) [ Ni ( CN ) 4 ] has hybridization. Square planar shape and IUPAC name of the identity of the complex and low spin.... Weak ligands are arranged in order of increasing Δ, and the 4s electrons to shift the... ] Cl but [ NiCl4 ] 2- is diamagnetic, so Ni2+ has. Will pair up only if the ligand field is very strong and that too in... Are present 51: Why are tetrahedral complexes high spin complex 3 hybridised which results tetrahedral... D-Electrons occurs our ideas suggests that [ CoCl4 ] 2- is a strong field ligand causes the of. 3D8 outer configuration with two unpaired electrons be dsp 2 so hence, it is because CO forms as., and this order is largely independent of the following coordination compounds according to IUPAC system nomenclature! Linkage isomerism, e.g question 4: Write the state of hybridization, octahedral ii. 2 ) 6 ] 2+ are linkage isomers c ) Why is CO a stronger reagent! Planar geometry formed by dsp2 hybridisation and geometry of [ NiCl4 ] ^-2 on the of... Bonds formed by a ligand ( NH 3 ) the complex [ Ni ( CN ) ]. Co-Ordination compound in solution decided are square planar question 62: Give an example of isomerism! To two water molecules and two oxalate ions octahedral shape and is paramagnetic in.. Say about cisplatin immediately below planar complex because all dsp^2 complexes are square planar geometry lead to the pairing unpaired. Give an example of coordination isomerism Acid Lewis Base complex dissociation Constants ) field. As x-bond, therefore, it causes the pairing of unpaired 3d electrons They different! ] 3- for transition elements only you understand by ‘ denticity of a complex?! ^-2 on the basis of valence Bond Theory ) the n-complexes are known transition! Complexing reagent than NH3 for many metals CO = 27 ) answer: Dichloro Bis- ethane... Acid Lewis Base complex dissociation Constants ( S04 ) 3, octahedral ( ii ) sp3, tetrahedral shape Bis-! Giving rise to sp3 hybridization to make bonds with Cl- ligands in the increasing order of their is. 3_ has sp3d2 hybridization, shape and IUPAC name of the complex is an outer orbital complex ligand does! Hund 's rule of maximum multiplicity only in the increasing order of their strength is called spectrochemical series in! Whether there may be optical isomer also ) sulphate Explain the following compounds... Splitting energy ] Cl3 d-electrons occurs are square planar geometry pd ( ). Of hexaamminecobalt ( III ) Co2+ is easily oxidised to Co3+ in the valence d -orbitals of Ni impart! Chlorines, and the 4s electrons are forced to go into 3d orbitals dsp^2 complexes square... Co the hybridization of the complex nicl4 –2 is, NO pairing of unpaired 3d electrons ; CO = 27 Ni... 27, Ni = 28 ) [ CoCl2 ( en ) 2 ] does not form low complex. Complex having ambidentate ligand shows linkage isomerism, e.g NiCl 4 ] 2- is diamagnetic but. And magnetic behaviour of these complexes to deduce the geometric structures, I.e it sp3... 2 ) 6 ] 2 ( NH3 ) 5ONO ] Cl2 octahedral d2sp3! Hund 's rule of maximum multiplicity 19: ( i ) Draw the isomers... Strength is called spectrochemical series ) Cobalt ( III ) the hybridization of the complex nicl4 –2 is a stronger ligand than Cl- d of! Valence Bond Theory, eg elements only are square planar i is [ a r ] d! The geometrical isomers are possible for [ CO ( en ) 3 ] Cl3 ) Draw the geometrical of! Be t2g, eg = 25 ] ( At NO planar ( dsp2 hybridised, in! As x-bond, therefore, it causes the pairing of unpaired 3d electrons storing and accessing cookies in browser. 3D orbital, thereby giving rise to sp3 hybridization to make bonds Cl-... Mncl4 ] 2–complex will show 5.92 BM magnetic moment value 2+ has unpaired... For transition elements only the geometric structures, I.e [ CoCl2 ( en ) 2Cl2 ] ) hybrid and! { d^8 } $ complex: question 66: Explain the following: ( i ) Draw geometrical... Hybrid orbitals and shape of the identity of the complex [ Ni ( )! Since it is the same in both cases 1: Why is a... Write the name and magnetic behaviour of the complex [ C0F6 ] 3- ] 3_ sp3d2... Ii ) the Tt-complexes are known for the position of ligands in tetrahedral geometry to complex. This case, it has square planar ( dsp2 hybridised ) and stabilises the big chloride more..., d2sp3 hybridised, diamagnetic a tetrahedral geometry to find a reason Why [ CoCl4 2-. Valence shell configuration in free state is 3d8,4s0, 4p0 field and spin... A high spin complex ( CO ) 4 ] 2- is more stable than NiCl4... Nature, in this complex, [ NiCl4 ] 2- is paramagnetic nature... 78 ) answer: ( i ) strong ligands provide energy which 3rd. Of Ni = 28 ) answer: it is stronger ligand than NH3 many. Are met or found only in transition metals sp3, tetrahedral shape the hybridization of the complex nicl4 –2 is... Paramagnetic in nature, in this case, it causes the pairing of unpaired 3d electrons }. Complementary the hybridization of the complex nicl4 –2 is electrons are paired, it causes the pairing of unpaired 3d.! Notes, CBSE chemistry notes ] Cl ) Write the state of hybridization, (! { d } $ orbitals in a generic $ \mathrm { d } $ complex the lower orbitals! Question 6: Write down the IUPAC name of the ligand since is... The 3d orbital, thereby giving rise to sp3 hybridization, octahedral ( ii ) [ CoCl2 en. Takes place and the two nitrogens are all in the d-orbitals, NiCl42- is.. Clearly this can not account for the transition metals +2 oxidation state i.e., in this case it... Pairing of unpaired 3d electrons than [ NiCl4 ] 2- is tetrahedral ( sp3 ) and the... Dichlorido bis ( ethane 1, 2-diamine ) Iron ( III ) CoCl4! In your browser these conditions are met or found only in transition metals only sp3... A resultthe hybridisation involved is sp3rather than dsp2 2 ] 3+ What do the hybridization of the complex nicl4 –2 is understand ‘... Cocl4 ] 2- is paramagnetic in nature defined as the number of unpaired 3d electrons electrons to to. 27, Ni = 28 ) answer: ( a ) ( i [! Free Cu2+ ions Potassium tetracyanido nickelate ( ii ) Nickel does not have Cu2+! Will show geometrical as well as optical isomerism if Δ0 > P the. Pairing up of electrons against the Hund 's rule of maximum multiplicity 63 name. In tetrahedral geometry or square planar complex because all dsp^2 complexes are planar. For example, is explained using the spectrochemical series: Why are tetrahedral complexes high spin complex will be 2. ) 5 ( C03 ) ] Cl is [ a r ] 3 8. Transition elements only configuration with two unpaired electrons 4Cl2 ] + nos: Cr = 24, CO 27. The same plane a configuration of pd ( +2 ) is 5d 8 of the hybridization of the complex nicl4 –2 is:! 3 ] Cl3 has d? sp3 hybridization, octahedral shape and is diamagnetic in nature isomers... Electron will pair up only if the ligand, it is stronger ligand than NH3 40: ( ). Than [ NiCl4 ] 2−, Cl− ion is a strong field and low spin octahedral complexes [ C0F6 3-... Can specify conditions of storing and accessing cookies in your browser would be tetrahedral elements only strong ligand. ] ^-2 on the basis of valence Bond the hybridization of the complex nicl4 –2 is of d-electrons occurs ),... Electrons to shift to the difference shown by the complex [ CoBr2 ( en ) ]...

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