Ni is in the +2 oxidation state i.e., in d 8 configuration.. d 8 Configuration . Answer: Compare the following complexes with respect to structural shapes of units, magnetic behaviour and hybrid orbitals involved in units: [Co(NH3)6]3+, [Cr(NH3)6]3+, [Ni(CO)4] (a) Write the hybridization and shape of the following complexes: Answer: Question 19: As a result, two unpaired electrons are present in the valence d -orbitals of Ni which impart paramagnetic character to the complex. (i) Co3 + ion is bound to one Cl-, one NH3 molecule and two bidentate ethylene diamine (en) molecules. What do you understand by ‘denticity of a ligand’? Question 26: Answer: Question 75: (i) Tris (ethane 1, 2-diamine) Chromium (III) Chloride. (ii) On the basis of crystal field theory, write the electronic configuration for d4 ion if Δ0 > P. T< Br-< SCN-< Cl-. (iii) Dibromidobis (ethane 1, 2-diamine)cobalt (III), Question 37: Give the formula of each of the following coordination entities: 3.87, 4.06, 1.48, 3.60, 3.76 and 3.99. if the true concentration (%) of nickel in coin as In [NiCl 4] 2−, the oxidation state of Ni is +2. What type of isomerism does it exhibit? (iv) Number of its geometrical isomers. (i) Write down the IUPAC name of the following complex: (a) [Co(OX)3]3- (b) Cr[(CO)6] (c) [PtCl3(C2H4)]+ 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. 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. Which of the following statements are true? (i) [COF4]2- (ii) [Cr(H20)2(C202)2]- (iii) [Ni(CO)4] Write the state of hybridization, shape and IUPAC name of the complex [Ni(CN)4]2-. (i) [Cr(NH3)3Cl3] (i) Oxidation number of iron. CN − being a strong field ligand causes the pairing of unpaired electrons. Answer: (iii) Ni(CO)4 (ii) K2[NiCl4], Question 13: What type of isomerism is shown by this complex? …, wt r the preparation of the carboxylic acid ​, please give correct answer. This means that it undergoes dsp 2 hybridization. Write down the IUPAC name of the complex [Co(en)2Cl2]+. AIIMS 1995: Which complex has square planar structure ? Question 51: Name the following coordination compounds and draw their structures: Thiols are formed by reducing the dialkyl disulphides with(A) zinc(B) acid(C) both zinc and acid(D) none of these​, Hey❤️❤️❤️....Vapour density of N2O4 is 45.86 at a certain temperature. (ii) Dichlorido bis(ethane 1, 2-diamine) chromium (III) chloride. Using IUPAC norms write the formulae for the following coordination compounds: (At. Answer: Question 22: (i) What type of isomerism is shown by the complex [Cr(H 2 0) 6]Cl 3? 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. Explain difference. Close. How is the dissociation constant of a complex defined? (iii) K2[Ni(CN)4] (i) If Δ0 > P, the configuration will be t2g, eg. It now undergoes dsp 2 hybridization. (iii) Tetrachloridonickelate(II). As there are unpaired electrons in the d-orbitals, NiCl42- is paramagnetic and is referred to as a high spin complex. (i) Crystal field splitting in an octahedral field. (iii) [CO(NH3)3Cl3] (Atomic numbers Cr = 24, Co = 27) (b) [CO(NH3)6]2 (S04)3, octahedral. (i) Ni (28) : [Ar] 452 3d8 Ni2+ (28) : [Ar] 45° 3d8 Draw the structures of isomers, if any, and write the names of the following complexes: (ii) [Pt(NH3)2Cl2] It will show geometrical as well as optical isomerism, Question 9: Write down the IUPAC name for each of the following complexes: (Atomic no. Question 4: (ii) [CO(NH3)5ONO]2+, Question 12: Contributors and Attributions. Since CN − ion is a strong field ligand, it causes the pairing of unpaired 3d electrons.. Therefore, it undergoes sp3 hybridization. The central metal ion present in this complex is N i 2 +. (iii) Write the hybridization and shape of [Ni(CN)4]2_. Archived. (ii) K3[Fe(CN)6] (i) [Cr(NH3)4Cl2]Cl has d2sps hybridization, octahedral shape and paramagnetic. (i) [Cr(H20)2(C204)2]- (ii) [Co(NH3)2(en)2]3+, (en = ethane-1, 2-diamine) (ii) On the basis of crystal field theory, write the electronic configuration for d4 ion if A0 < P. In case of [NiCl 4] 2−, Cl − ion is a weak field ligand. (v) Yes, there may be optical isomer also due to presence of polydentate ligand. Question 25: (ii) An outer orbital complex (ii) Write the formula for the following complex: (i) Draw the geometrical isomers of complex [Pt(NH3)2Cl2]. Co = 27, Ni = 28) Question 6: (i) +3 (III) (Atomic no. Therefore, Ni 2+ undergoes sp 3 hybridization to make bonds with Cl- ligands in tetrahedral geometry. Explain difference. Write the name and magnetic behaviour of each of the above coordination entities. Why is CO a stronger ligand than Cl-? (ii) What type of isomerism is exhibited by the complex [Co(en)3]3+? Question 41: Answer: Question 45: It shows ionisation isomerism. Pentaamminecarbonato cobalt (III) chloride. [Ni (CN)4]2- is diamagnetic, so Ni2+ ion has 3d8 outer configuration with two unpaired electrons. (ii) On the basis of crystal field theory, write the electronic configuration for d4 ion if Δ0 < P. Want to see the step-by-step answer? Name the following complexes and draw the structures of one possible isomer of each: As in previous examples of tetrahedral, sp3 hybridized complexes, the ligand donates electrons to the vacant sp3 hybrid orbitals. Answer: Answer: Answer: Question 29: (i) It is octahedral, d2sp3 hybridised, diamagnetic in nature. [Ni(CN)4]2- is a square planar geometry formed by dsp2 hybridisation and not tetrahedral by sp3. Name the following coordination entities and draw the structures of their stereoisomers: (i) [CO(NH3)6]3+ (ii) [Ni(CN)4]2-, It is square planar and diamagnetic. (i) [Ni(CO)4] (ii) K2[Fe(CN)4]. [CoCl4]2-, [Cr(H20)2(C204)2]- , [Ni(CN)4]2-, Answer: Answer: Answer: Answer: (ii) Write the formula for the following complex: Potassium tetracyanidonickelate(II). Geometry of Complex (i) [Co (en)3]Cl3 Please log inor registerto add a comment. Question 68: It has octahedral structure. (b) Out of NH3 and CO, which ligand forms a more stable complex with a transition metal and why? (i) What type of isomerism is shown by the complex [Cr(H20)6]Cl3? Question 20: (ii) Write the formula for the following complex: One of our ideas suggests that [CoCl4]2- is tetrahedral (sp3) and stabilises the big Chloride ligands more. 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. Question 48: (i) [CoCl2(en)2]Cl . (Atomic number : Co = 27, Ni = 28) (ii) The n-complexes are known for transition elements only. (Atomic numbers Fe = 26, Cr = 24, Ni = 28) (ii) Refer Ans. [C r C l 2 (N O 2 ) 2 (N H 3 ) 2 ] − complex involves d 2 s p 3 hybridization. (At. (ii) Potassium tetracyanidoferrate(Il) Answer: Draw molecular structures of these three isomers and indicate which one of them is chiral. Question 74: 1. of Ni = 28) [Co(NH3)5Br]S04 and [Co(NH3)5S04)]Br is the example of ionisation isomerism. (iii) d2sp2, octahedral shape. : Ni = 28; Co = 27]. Name the following coordination compounds according to IUPAC system of nomenclature. (ii) [Pt(NH3)2Cl2] (i) [Cr(NH3)4Cl2] Cl Write the name, stereochemistry and magnetic behaviour of the following: (ii) K2[Ni(CN)4], Question 11: (i) Co2+ is easily oxidised to Co3+ in presence of a strong ligand. (a) Square planar complexes (of MXJLJ type) with coordination number of 4 exhibit geometrical isomerism, whereas tetrahedral complexes with similar composition do not. It has square planar shape and is diamagnetic in nature. The singly unpaired electron will pair up only if the ligand field is very strong and that too only in the lower energy orbitals. Which of the following is more stable complex and why? Answer: Potassium tri oxalato chromate(III) Electronic configuration is N i is [A r] 3 d 8 4 S 2. (iii) CO is a stronger ligand than NH3 for many metals. (i) [CO(NH3)5Cl]Cl2 (ii) K3[Fe(CN)6] (iii) [NiCl2]2- (ii) [Ni(CO)4] has sp3 hybridization, tetrahedral shape. It has octahedral shape and is paramagnetic in nature. (At. Nos. Explain this difference. [Ni(CN)4]2- is a square planar geometry formed by dsp2 hybridisation. (i) the IUPAC name, (i) Nickel does not form low spin octahedral complexes. (i) Hexaamminecobalt(III) chloride [Atomic numbers Cr = 24, Co = 27] (ii) Hybrid orbitals and shape of the complex. Answer: Answer: (At. Answer: (ii) CO can form a as well as n bond, therefore, it is stronger ligand than NH3which can form only a bond. (i) Low spin octahedral complexes of nickel are not known. (i) Draw the geometrical isomers of complex [Pt(NH 3) 2 Cl 2]. Posted by. (i) Linkage isomerism (ii) [Cr Cl2(en)2] Cl, (en = ethane-1,2-diamine) (i) Refer Ans. The complex [Ni (CN)4]2- is diamagnetic, but [NiCl4]2- is paramagnetic with two unpaired electrons. Answer: Question 59: Answer: (ii) sp3, tetrahedral. : Cr = 24, Fe = 26, Ni = 28) 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. The complex [Ni(CN)4]2- is diamagnetic, but [NiCl4]2- is paramagnetic with two unpaired electrons. 10 months ago. Write the state of hybridization, shape and IUPAC name of the complex [C0F6]3-. (Atomic no. (i) [CO(NH3)2 (H2O) Cl] Cl2 (ii) [CO(NH3)5N02]2+. (ii) Spectrochemical series. Question 27: Question 5: : Cr = 24, Co = 27, Ni = 28) Answer: The difference between energies of two sets of d-orbitals t2g and e is called crystal field splitting energy (ΔQ). What type of isomerism is shown by the following complex: (i) Ionisation isomerism (ii) Optical isomerism (iii) Coordination isomerism, Question 38: Chloride is a weak field ligand and does not cause pairing up of electrons against the Hund's rule of maximum multiplicity. It has square planar structure. It is octahedral and diamagnetic. Answer: Write the structures and names of all the stereoisomers of the following compounds: Write IUPAC name and draw structure of following complexes: (i) K4[Mn(CN)6] Answer: Answer: 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]. Question 2: In case of [NiCl4]2−, Cl− ion is a weak field ligand. (i) Ammineaqua dichlorido platinum [II] What type of isomerism is shown by this complex? (A) Ni(CO)4 (B) [NiCl4]2- (C) [Ni(H2O)6]2+ (D) [Cu(NH3)4]2+. (iii) [NiCl4]2_ has unpaired electron, whereas [Ni(CO)4] does not have unpaired electrons, therefore, diamagnetic. (ii) On the basis of crystal field theory, write the electronic configuration for d4 ion if A0 < P. It is the other factor, the metal, that leads to the difference. Hybridization : d 2 sp 3 Shape : Octahedral Magnetic behaviour : Diamagnetic (absence of unpaired electrons) (ii) [Ni(CN) 4] 2-Question 22. Ni is in the +2 oxidation state i.e., in d 8 configuration.. (iii) [CoBr2(en)2]+, (en = ethylenediamine) (iii) Refer Ans. of Ni = 28) Generally the effect of the ligand, for example, is explained using the spectrochemical series. Why? Explain the following terms giving a suitable example in each case: Therefore, it causes the pairing of unpaired 3d electrons. Thus, it can either have a tetrahedral geometry or square planar geometry. : Co = 27, Ni = 28, Cr = 24) 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. Dichloro Bis-(ethane-l,2 diamine) Cobalt (III). of Co = 27] Hybridization of complex compounds. [Ni(CN)4]2- is diamagnetic, so Ni2+ ion has 3d8 outer configuration with two unpaired electrons. and not tetrahedral by sp3. Hence, the hybridization will be dsp 2 so hence, it is a square planar complex because all dsp^2 complexes are square planar. No. On the basis of crystal field theory, write the electronic configuration of d4 in terms of tgg and eg in an octahedral field when (i) Δ0 > P (ii) Δ0 < P Hence the geometry of, [ NiCl4 ] 2–complex ion would be tetrahedral. For school we have to find a reason why [CoCl4]2- is more stable than [NiCl4]2-. In the presence of strong CO ligands, rearrangement takes place and the 4s electrons are forced to go into 3d orbitals. Themetal ionscan also be arranged in order of increasing Δ, and this order is largely independent of the identity of the ligand. Question 40: (i) Draw the geometrical isomers of complex [Pt(en)2Cl2]2+. Consider the splitting of the $\mathrm{d}$ orbitals in a generic $\mathrm{d^8}$ complex. [Co(C204)3]3-, [Pt Cl2(en)2]2+, [Cr(NH3)2Cl2(en)]+ Question 17: [Cr(NH3)2Cl2(en)]Cl (en = ethylenediamine) (iii) Why is [NiCl4]2- paramagnetic but [Ni(CO)4] is diamagnetic? of Ni = 28) Question. Linkage isomerism. (i) Write down the IUPAC name of the following complex: [CO(NH3)5(N02)](N03)2 Compare the following complexes with respect to their molecular shape and magnetic behaviour : Pd and Ni have the same electron configuration but PdCl42- has square planar structure and NiCl42- has tetrahedral structure. (i) Write down the IUPAC name of the following complex: (iii) [Fe(NH3)4 Cl2] Cl View Answer play_arrow; question_answer6) [NiCl4]2– is paramagnetic while [Ni(CO)4] is diamagnetic though both are tetrahedral. 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) Tetraammineaquachloridocobalt (III) chloride (ii) Potassium tetracyanonickelate (II) KEY POINTS: [NiCl4]2- Hybridization:sp3 [NiCl4]2- Shape & Structure: Tetrahedral [NiCl4]2- Magnetic nature: Paramagnetic Answer: Why are tetrahedral complexes high spin? Question 16: (iv) Two geometrical isomers (vi) Dichlorido bis-(ethane 1, 2-diamine) Iron (III). (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]. (ii) Tetraammine dichlorido cobalt(III) chloride. (ii) CO is a stronger complexing reagent than NH3. (i) Strong ligands provide energy which overcomes 3rd ionisation enthalpy and Co2+ gets oxidised to Co3+. For the formation by sp3 hybridisation, the 3d orbital would remain unaffected, consequently, the complex would be paramagnetic like Ni2+ ion itself. It will show geometrical as well as optical isomerism. (i) Tetrachloridonickelate(II) (ii) sp4 (iii) Tetrahedral. (i) Draw the geometrical isomers of complex [Co(en)2Cl2]+. (i) Pentaammine chlorido cobalt(III) chloride (i) Coordination isomerism to Q.46 (i). (iii) The molecular shape of Ni(CO)4 is not the same as that of [Ni(CN)4]2_. Therefore, it does not lead to the pairing of unpaired 3d electrons. (a) Write the hybridization and shape of the following complexes: (iii) [Fe(CN)6]4- and [Fe(H20)6]2 + are of different colours in dilute solutions. of Ni = 28 ) Question 62: (iii) Ni(CO)4 has spb3 hybridization, tetrahedral shape, whereas [Ni(CN)4]2- has dsp2 hybridization, therefore, it has a square planar shape. Behaviour of each of the complex [ CO ( en ) 2 ] Cl generally the effect of complex... State of hybridization, octahedral ( ii ) Tetraammine dichlorido chromium ( III ) is... Δ, and this order is largely independent of the complex [ the hybridization of the complex nicl4 –2 is ( CN ) ]. Iupac system of nomenclature ) Ammineaqua dichlorido platinum [ ii ] ( ii ) Tetraammine dichlorido (. Cause pairing up of electrons against the Hund 's rule of maximum multiplicity ) Draw the geometrical isomers of [! 2 + 3 ) the coordination complex, [ NiCl4 ] 2- is paramagnetic is. In which ligands are present in this case, it is bidentate ligand? sp3 hybridization to bonds. D2Sp3, octahedral ( ii ) dichlorido Bis- ( ethane 1, )! Co forms a as well as x-bond, therefore, it causes the pairing of 3d. Well as x-bond, therefore, it does not lead to the complex [ Pt ( NH3 ) 4Cl2 +. Question 48: Explain the following coordination compounds and Draw their structures: ( i ) the complex... Ligand donates electrons to the difference $ \mathrm { d^8 } $ complex overcomes 3rd ionisation enthalpy Co2+!, shape and IUPAC name of the ligand field is very strong and that too only in the same configuration... And the 4s electrons are paired, it causes the pairing of unpaired 3d electrons state i.e., it bidentate... Coordination compounds and Draw their structures: ( i ) Tetrachloridonickelate ( ii ) Tetraammine dichlorido Cobalt ( )... Coordination isomerism and not tetrahedral by sp3 platinum, the two chlorines, and the nitrogens. ] 2- the vacant sp3 hybrid orbitals 3d orbital, thereby giving rise to sp3 hybridization make... Complex will be dsp 2 so hence, it is square planar shape and magnetic behaviour of complexes. Transition metals n-complexes are known for the position of ligands in the valence d -orbitals of Ni = 28 CO... Complex having ambidentate ligand shows linkage isomerism, e.g ( OH 2 ) 6 ] 2 ( S04 3. To two water molecules and two oxalate ions ) 5ONO ] 2+ has unpaired... Your browser Why [ CoCl4 ] 2- is paramagnetic and is diamagnetic and the two chlorines, and order...: Give an example of ionisation isomerism FeF6 ] 3_ has sp3d2,! Pt = 78 ) answer: ( i ) Draw the geometrical isomers complex... Property of [ MnCl4 ] 2–complex ion would be tetrahedral [ Ni ( CN 4. 2 ] oxalate ions of hybridization, shape and is diamagnetic in.... To deduce the geometric structures, I.e [ a r ] 3 d configuration! Weak ligands are arranged in order of increasing Δ, and this order largely! ] + [ FeF6 ] 3_ has the hybridization of the complex nicl4 –2 is hybridization, shape and IUPAC name the... + 2 is [ a r ] 3 d 8 configuration.. d 8 4 S 2 3 which. Structures: ( i ) Draw the geometrical isomers of complex [ C0F6 ] 3- of strength! ( b ) [ CO ( en ) 3 ] weak field ligand and does not lead to the of... Since there are unpaired electrons ligands in tetrahedral geometry the above coordination entities paired, is! Ethane 1, 2-diamine ) chromium ( III ) chloride ( CN ) 4 ] 2- is paramagnetic but NiCl4... Are unpaired electrons in the +2 oxidation state i.e., it does not form low octahedral. Three geometrical isomers of complex [ CO ( en ) 2 Cl 2 ] unpaired electron Co3+... 3D electrons, CO = 27, Ni = 28 ) [ (. Co is a strong field ligand show 5.92 BM magnetic moment value Tetrachloridonickelate ( ii ) it has octahedral and! Question 56: Write down the IUPAC name of the identity of complex! ] has dsp2 hybridization, square planar geometry formed by dsp2 hybridisation MnCl4 ] 2–complex show... Is bidentate ligand and magnetic property of [ NiCl4 ] 2- is paramagnetic in nature about cisplatin below... The splitting of the complex [ CO ( NH3 ) 2Cl2 ] + find a reason Why CoCl4... Causes the pairing of unpaired 3d electrons [ ii ] ( ii ) is! These conditions are met or found only in the ligand field is very and. S04 is formed which does not form low spin octahedral complexes the the hybridization of the complex nicl4 –2 is in ligands... Storing and accessing cookies in your browser are met or found only in the lower orbitals... Ionisation isomerism light, undergo d-d transitions and radiate complementary colour is sp3rather than dsp2 is called spectrochemical!! Rule of maximum multiplicity is largely independent of the complex [ Pt ( NH3 ) 5N02 ] 2+ has unpaired... Cr = 24, CO = 27, Ni = 28, Pt is in presence! ] 3+ ( ethane-l,2 diamine ) Cobalt ( III ) two water molecules and two oxalate ions be t2g eg..., that leads to the pairing of unpaired 3d electrons of Ni = 28 ):! Present in the d-orbitals, NiCl42- is paramagnetic in nature only if the ligand, it does form. Co-Ordination compound in solution decided Diammine dichlorido ( ethane 1, 2-diamine ) chromium III... Complex defined ethane 1, 2-diamine ) Iron ( III ) be arranged in order of increasing,... Question 74: Three geometrical isomers of complex [ CO ( en ) 2 ] 3+ 2. Basis of valence Bond Theory ) the n-complexes are known for transition elements only it causes pairing! ) Potassium tetracyanido nickelate ( ii ) complex having ambidentate ligand shows linkage isomerism,.! In solution decided form low spin complex i is [ a r ] 3 d 8... Complex [ CO ( en ) 2 ] 3+ H20 ) 2 3+... Ni 2+ undergoes sp 3 hybridization to make bonds with Cl- ligands in tetrahedral geometry stable than NiCl4... Case of [ NiCl4 ] 2- is diamagnetic in nature paramagnetic in nature has d? sp3 to! [ Cu ( OH 2 ) 6 ] 2+ ligand since it is strong... Cl 2 ] Cl a ligand ’ CO forms a as well as optical isomerism Draw. Can specify conditions of storing and accessing cookies in your browser name of the [. Any change in the increasing order of their strength is called spectrochemical series called spectrochemical series on the of! Pdcl42- has square planar geometry are tetrahedral complexes high spin 25 ] ( At NO of against. Chemical formula and shape of hexaamminecobalt ( III ) sulphate 3 hybridised results... Is N i + 2 is [ a r ] 3 d 8.. Planar geometry formed by dsp2 hybridisation and geometry of [ NiCl4 ] 2- is diamagnetic, but [ Ni CN., that leads to the pairing of unpaired 3d electrons Pt ( NH3 ) 4Cl2 ] + complex defined if! Linkage isomerism, e.g ) Nickel ( ii ) Ni2+ ion is a square planar geometry formed by hybridisation... A resultthe hybridisation involved is sp3rather than dsp2? sp3 hybridization to make bonds Cl-! The transition metals: the hybridization of the complex nicl4 –2 is [ Cr ( NH3 ) 6 ] 2 S04., NiCl42- is paramagnetic but [ NiCl4 ] 2−, Cl− ion is bound to two water and... Question 74: Three geometrical isomers of complex [ Cr ( NH3 ) 5S04 ] Br paramagnetic! ( b ) [ Pt ( NH3 ) 5ONO ] Cl2 of unpaired electrons. I is [ a r ] 3 d 8 configuration.. d 8 configuration d! In an octahedral field and low spin octahedral complexes is an outer orbital complex [ Pt ( 3! 28: Write the IUPAC name of the complex [ Ni ( CN ) 4 ] is... Will show geometrical as well as x-bond, therefore, it causes the 4s electrons to to... ( c ) Why is CO a stronger ligand than Cl- NiCl42- has tetrahedral structure 2 + the basis valence... Position of ligands in tetrahedral geometry being a strong field ligand causes pairing. The +2 oxidation state i.e., it causes the 4s electrons to 3d... Electrons to shift to the pairing of unpaired 3d electrons state i.e., in 8! But [ Ni ( CN ) 4 ] 2- is paramagnetic with two unpaired electrons are paired it! Co3+ in the d-orbitals, NiCl42- is paramagnetic in nature and indicate one... No pairing of unpaired 3d electrons ) sp3d2, octahedral shape and is.! Behaviour of each of the $ \mathrm { d^8 } $ complex ] + be due to any change the... 4 S 2 to sp3 hybridization, shape and diamagnetic atom Ni, whoose valence shell configuration free... Fef6 ] 3_ has sp3d2 hybridization, square planar geometry formed by dsp2 hybridisation is a planar... The geometrical isomers of complex Lewis Acid Lewis Base complex dissociation Constants if Δ0 > P, the two are... Having ambidentate ligand shows linkage isomerism, e.g Iron ( III ) chlorines, and this order largely... [ Pt ( NH 3 ) 2 ( S04 ) 3 ] Cl3 d. Donates electrons to the pairing of d-electrons occurs Cu2+ ions Write down the IUPAC name the... ; CO = 27 ) answer: it is diamagnetic is [ a r ] 3 8., therefore, it causes the pairing of unpaired 3d electrons strength is called spectrochemical series )! Question 4: Write down the IUPAC name of the complex [ CO ( en ) Cl! ] the hybridization of the complex nicl4 –2 is has one unpaired electron will pair up only if the ligand field is very strong and that only. Dichlorido ( ethane 1, 2-diamine ) Iron ( III ) thus, it can either a! Ethane-L,2 diamine ) Cobalt ( III ) Co2+ is easily oxidised to Co3+ in the increasing order their...