It has octahedral shape and is paramagnetic in nature. (ii) t4 2g (i) [Cr(H20)2(C204)2]- (ii) [Co(NH3)2(en)2]3+, (en = ethane-1, 2-diamine) Compare the following complexes with respect to their shape, magnetic behaviour and the hybrid orbitals involved: [Co(NH3)5Br]S04 and [Co(NH3)5S04)]Br is the example of ionisation isomerism. Since CN-ion is a strong field ligand, it causes the pairing of unpaired 3d electrons.. 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. Answer: (i) What type of isomerism is shown by the complex [Ag(NH3)2][Ag(CN)2]? Question 15: The geometry of the complex changes going from $\ce{[NiCl4]^2-}$ to $\ce{[PdCl4]^2-}$. (b) Write the chemical formula and shape of hexaamminecobalt(III) sulphate. (At. (i) Co2+ is easily oxidised to Co3+ in presence of a strong ligand. State a reason for each of the following situations: (i) [Ni(CN)4]2- (ii) [NiCl4]2- (iii) [CoF6]3- [At. (ii) The series in which ligands are arranged in the increasing order of their strength is called spectrochemical series. (i) [Co(en)2Cl2]+ (en = ethan-1, 2-diamine) (i) Ni (28) : [Ar] 452 3d8 Ni2+ (28) : [Ar] 45° 3d8 Hi all! Using IUPAC norms write the formulae for the following coordination compounds: The complex [NiF4]2- is paramagnetic but [Ni(CN)4]2- is diamagnetic. (Atomic no. Name the following coordination compounds according to IUPAC system of nomenclature. NiCl42-, there is Ni2+ ion, However, in presence of weak field Cl- ligands, NO pairing of d-electrons occurs. (i) [Pt(NH3)2Cl(N02)] Explain the following: Write the state of hybridization, shape and IUPAC name of the complex [C0F6]3-. Potassium hexafluoridochromate(III). (ii) [CO(NH3)5N02]2+. tris(ethane-l,2-diamine)chromium(III) chloride. [Ni(CN)4]2- is a square planar geometry formed by dsp2 hybridisation. Question 2: Use the magnetic behaviour of these complexes to deduce the geometric structures , I.e. Therefore, Ni 2+ undergoes sp 3 hybridization to make bonds with Cl- ligands in tetrahedral geometry. (b) Ionisation isomerism Question 52: (i) Low spin octahedral complexes of nickel are not known. (ii) the hybridization type, Explain this difference. (a) It has 5 unpaired electrons. (iii) Write the hybridization and shape of [Fe(CN)6]3-. CBSE Maths notes, CBSE physics notes, CBSE chemistry notes. (iii) Refer Ans. (i) Crystal field splitting in an octahedral field. Answer: (i) Draw the geometrical isomers of complex [Pt(NH 3) 2 Cl 2]. (ii) On the basis of crystal field theory, write the electronic configuration for d4 ion, if Δ0> P. of Ni = 28) check_circle Expert Answer. (i) Ionisation isomerism (ii) Optical isomerism (iii) Coordination isomerism, Question 38: Answer: Explain the following: (ii) [Co (en)3] Cl3 has d?sp3 hybridization, octahedral shape and diamagnetic. (ii) Potassium tetracyanidoferrate(Il) (ii) The n-complexes are known for transition elements only. (ii) [Cr(C204)3]3- (ii) Hybrid orbitals and shape of the complex. Write the name of the structure and the magnetic behaviour of each one of the following complexes: KEY POINTS: [NiCl4]2- Hybridization:sp3 [NiCl4]2- Shape & Structure: Tetrahedral [NiCl4]2- Magnetic nature: Paramagnetic Explain difference. Want to see the step-by-step answer? Answer: (a) (i) sp3d2, octahedral (ii) dsp2, square planar. (ii) Ni2+ ion is bound to two water molecules and two oxalate ions. Linkage isomerism. Answer: Question 22: (i) What type of isomerism is shown by the complex [Cr(H 2 0) 6]Cl 3? Answer: Ionisation isomerism. (At. (i) [CO(NH3)5Cl]Cl2 (ii) K3[Fe(CN)6] (iii) [NiCl2]2- (v) Yes, there may be optical isomer also due to presence of polydentate ligand. : Cr = 24, Fe = 26, Ni = 28) The hybridisation scheme is as shown in figure. CN − being a strong field ligand causes the pairing of unpaired electrons. Question10. Answer: Since all electrons are paired, it is diamagnetic. (iii) The molecular shape of Ni(CO)4 is not the same as that of [Ni(CN)4]2_. (iii) Refer Ans. (ii) On the basis of crystal field theory, write the electronic configuration for d4 ion if Δ0 > P. 10 months ago. In Ni (CO) 4, Ni is in the zero oxidation state i.e., it has a configuration of 3d8 4s2. Answer: of Ni = 28) (iv) Number of its geometrical isomers. Answer: Question 22: (ii) Potassiumhexacyanoferrate (III) It forms a square planar structure. of Ni = 28 ) (i) [Cr(NH3)6]3+ (ii) [Fe(CN)6]4- (iii) [NiCl4]2- Now, the electronic configuration of Pd(+2) is 5d 8. Therefore, it does not lead to the pairing of unpaired 3d electrons. Write the state of hybridisation, the shape and the magnetic behaviour of the following complex entities: Answer: All India 2012) Answer: Stability of a complex in solution means the measure of resistance to the replacement of a … Answer: Question 22: (i) What type of isomerism is shown by the complex [Cr(H 2 0) 6]Cl 3? : Cr = 24, Co = 27, Ni = 28) (ii) Write the formula for the following complex: For the formation by sp3 hybridisation, the 3d orbital would remain unaffected, consequently, the complex would be paramagnetic like Ni2+ ion itself. [Pt(NH3)(H20)Cl2] Question 19: (Atomic no. Therefore, it causes the pairing of unpaired 3d electrons. Answer: Co = 27, Ni = 28) Write down the IUPAC name for each of the following complexes: Atomic number of Nickel is 2 8. What type of isomerism is shown by this complex? (ii) Potassium tetrachloridonickelate(II). Why are tetrahedral complexes high spin? (i) [COF4]2- (ii) [Cr(H20)2(C202)2]- (iii) [Ni(CO)4] Question 1: Question 64: Answer: (i) The n-complexes are known for transition elements only. (i) Refer Ans. 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Which of the following statements are true? Name the following coordination entities and draw the structures of their stereoisomers: (i) [Co (en)3]Cl3 (i) Draw the geometrical isomers of complex [Pt(en)2Cl2]2+. (At. (i) Tetracarbonylnickel(O) to Q.42 (a) (i). (i) The splitting of d-orbitals in presence of ligands is called crystal field splitting, e.g. Write the name, stereochemistry and magnetic behaviour of the following: Answer: [Co(C204)3]3-, [Pt Cl2(en)2]2+, [Cr(NH3)2Cl2(en)]+ (2) The complex is an outer orbital complex. Click hereto get an answer to your question ️ For the complex [NiCl4]^2 - , write(i) the IUPAC name (ii) the hybridization type (iii) the shape of the complex. Chloride is a weak field ligand and does not cause pairing up of electrons against the Hund's rule of maximum multiplicity. [Cr(NH3)2Cl2(en)]Cl (en = ethylenediamine) (i) [CO(NH3)5Cl]S04 (ii) [Co(en)3]3+ (iii) [Co(NH3)6] [Cr(CN)6] (iii) [NiCl4]2_ has unpaired electron, whereas [Ni(CO)4] does not have unpaired electrons, therefore, diamagnetic. [Atomic number of Mn = 25] (i) Pentaamminechloridocobalt (III) Consider the splitting of the $\mathrm{d}$ orbitals in a generic $\mathrm{d^8}$ complex. Close. (iii) [NiCl4]2_ is paramagnetic while [Ni(CO)4] is diamagnetic, though both are tetrahedral. (a) (i) d2sp3, octahedral It will show geometrical as well as optical isomerism. …, wt r the preparation of the carboxylic acid , please give correct answer. (i) Linkage isomerism (ii) Potassium tetracyanido nickelate(II). (а) Write the hybridization and shape of the following complexes: Since there are 2 unpaired electrons in this case, it is paramagnetic in nature. [CO(NH3)5N02]2+ and [Co(NH3)5ONO]2+ are linkage isomers. [Ni(CN)4]2- is diamagnetic, so Ni2+ ion has 3d8 outer configuration with two unpaired electrons. (i) Refer Ans. Write the structures and names of all the stereoisomers of the following compounds: of Ni = 28) Cr+3 + 6H2O [Cr(OH2)6]+3 ----- ... [NiCl4]-2 Ni = 4s2 3d8 Ni+2 = 4s0 3d8 . Explain the following terms giving a suitable example in each case: (ii) Refer Ans. (At. It now undergoes dsp 2 hybridization… One of our ideas suggests that [CoCl4]2- is tetrahedral (sp3) and stabilises the big Chloride ligands more. Give the formula of each of the following coordination entities: (i) Write down the IUPAC name of the following complex: [CO(NH3)5Cl]2+ (i) Write down the IUPAC name of the following complex: (ii) It is square planar, dsp2 hybridised, diamagnetic in nature. (A) Ni(CO)4 (B) [NiCl4]2- (C) [Ni(H2O)6]2+ (D) [Cu(NH3)4]2+. (i) [CoF6]3- (ii) [Ni(CN)4]2- In this complex, Pt is in the +2 state. How is the dissociation constant of a complex defined? 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. Answer: Question 66: It is square planar and diamagnetic. (ii) t32g e1g (i) If Δ0 > P, the configuration will be t2g, eg. Question 5: (ii) [CO(NH3)5ONO]2+, Question 12: Answer: (b) Give an example of the role of coordination compounds in biological systems. Answer: (3) The complex is d 2 sp 3 hybridized. Hence, the hybridization will be dsp 2 so hence, it is a square planar complex because all dsp^2 complexes are square planar. 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. (Atomic number of Fe — 26) (ii) d2sp3, octahedral Potassium tri oxalato chromate(III) Therefore, Ni2+ undergoes sp3 hybridization to make bonds with Cl- ligands in tetrahedral geometry. Answer: (i) Draw the geometrical isomers of complex [Pt(NH 3) 2 Cl 2]. (iii) dsp2, square planar. Since it have two unpaired electron electron therefore the magnetic moment : Answer: Question 26: Answer: Question 54: (ii) Potassium hexacyanido ferrate(III). (i) Crystal field splitting (ii) Linkage isomerism (iii) Ambidentate ligand (i) Triamminetrichloridochromium (III) (a) Write the hybridization and shape of the following complexes: Answer: Question 72: (iii) K2[Ni(CN)4] has dsp2 hybridization, square planar shape, diamagnetic. (At. Write the IUPAC names of the following coordination compounds: (i) [Pt(NH3)2Cl2] (ii) [CO(NH3)5(N02)]Cl2 nos. As there are unpaired electrons in the d-orbitals, NiCl 4 2- is paramagnetic. Answer: Question 75: It is octahedral and diamagnetic. For the complex [NiCl4]2-, write(i) the IUPAC name(ii) the hybridization type(iii) the shape of the complex. (ii) K2[Ni(CN)4], Question 11: (i) [CO(NH3)6]Cl3 (ii) K2[NiCl4], Question 10: Write the name and magnetic behaviour of each of the above coordination entities. [Ni(CN)4]2- is diamagnetic, so Ni2+ ion has 3d8 outer configuration with two unpaired electrons. (a) Dibromidobis (ethane-1, 2-diamine)cobalt(III) 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. Check Answer and Solution for abo It will show geometrical as well as optical isomerism, Question 9: Also, it causes the 4s electrons to shift to the 3d orbital, thereby giving rise to sp3 hybridization. Draw the structures of isomers, if any, and write the names of the following complexes: View Answer play_arrow; question_answer6) [NiCl4]2– is paramagnetic while [Ni(CO)4] is diamagnetic though both are tetrahedral. Write down the IUPAC name of the complex [CO(NH3)5(C03)]Cl. nos Mn = 25, Co = 27, Ni = 28) (ii) Denticity of a ligand Ni is in the +2 oxidation state i.e., in d 8 configuration.. Answer: (i) Nickel does not form low spin octahedral complexes. Answer: Co = 27, Pt = 78) (5) The coordination number is 6. (a) [Co(OX)3]3- (b) Cr[(CO)6] (c) [PtCl3(C2H4)]+ thus , it have SP3 hybridisation which have tetradehdral geometry . (a) Predict the number of unpaired electrons in hexaaquamanganese(II) ion. (i) Pentaammine chloridocobalt III chloride. (a) Square planar complexes (of MXJLJ type) with coordination number of 4 exhibit geometrical isomerism, whereas tetrahedral complexes with similar composition do not. KEAM 2014: The hybridization of central metal ion in K2[Ni(CN)4] and K2[NiCl4] are respectively (A) dsp2 , sp3 (B) sp3 , sp3 (C) dsp2 , dsp2 (D) sp3 Therefore, it undergoes sp3 hybridization. Answer: 6. Pentaamminecarbonato cobalt (III) chloride. (iii) [Co(en)2Cl2]+ We will have more to say about cisplatin immediately below. See Answer. : Ni = 28; Co = 27]. (i) Write the IUPAC name of the complex [Cr(NH3)4Cl2]Cl. Question 7: Three geometrical isomers are possible for [Co(en) (H20)2(NH3)2]3+. (Atomic no. Since CN − ion is a strong field ligand, it causes the pairing of unpaired 3d electrons. Answer: Answer: Question 27: Themetal ionscan also be arranged in order of increasing Δ, and this order is largely independent of the identity of the ligand. Cisplatin is a neutral complex, Pt(NH 3) 2 Cl 2. (ii) CO is a stronger complexing reagent than NH3. Give the name, the stereochemistry and the magnetic behaviour of the following complexes: For the complex [NiCl 4] 2-, write (i) the IUPAC name (ii) the hybridization type (iii) the shape of the complex. determined by atomic absorption and inductively coupled plasma atomic emission What type of isomerism is shown by this complex? (iii) Average error, forming compounds with examples fastly answer, wt r the chemical reaction of the ketone . (ii) The Tt-complexes are known for the transition metals only. Co = 27, Ni = 28) [Co(en)3]3+ is more stable since ‘en’ is didentate ligand which forms more stable complex than NH3(unidentate ligand). (b) Write the hybridization of the complex [NiCI4]2-. (i) Draw the geometrical isomers of complex [Pt(NH3)2Cl2]. Answer: Write the state of hybridization, shape and IUPAC name of the complex [CO(NH3)6]3+. Answer: Why is CO a stronger ligand than Cl-? : Co = 27, Ni = 28, Cr = 24) Question 40: (Atomic number : Co = 27, Ni = 28) It has square planar shape and is diamagnetic in nature. It is because CO forms a as well as x-bond, therefore, it is stronger ligand than Cl-. This is true when large, weak ligands are present. What type of isomerism is shown by this complex? Answer: (At. What type of isomerism is shown by the following complex: Question 62: Answer: Question 43: Name the following coordination compound: K3[CrF6]. In case of [NiCl 4] 2−, Cl − ion is a weak field ligand. Hybridization of complex compounds. (i) Draw the geometrical isomers of complex [Co(en)2Cl2]+. …, .59, 7.51, 3.95, (c) : In the paramagnetic and tetrahedral complex [NiCl 4] 2–, the nickel is in +2 oxidation state and the ion has the electronic configuration 3d 8.The hybridisation scheme is as shown in figure. (i) What type of isomerism is shown by the complex [Cr(H20)6]Cl3? (ii) What type of isomerism is exhibited by the complex [Co(en)3]3+? It now undergoes dsp 2 hybridization. (Atomic number of Co = 27) (Atomic’number of Ni = 28) (iii) Write the hybridization and shape of [Ni(CN)4]2_. (i) Tetraammineaquachlorido cobalt(III) chloride. Question 25: 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 It has octahedral shape and is diamagnetic in nature. (ii) Write the formula for the following complex: Question 20: Geometry of Complex Therefore, it does not lead to the pairing of unpaired 3d electrons. Co = 27, Ni = 28, Pt = 78) Answer: (vi) Dichlorido bis-(ethane 1, 2-diamine) Iron (III). Question 65: State i.e., in presence of weak field ligand, for example, is explained using the spectrochemical.... Very strong and that too only in transition metals question 63: the... Bidentate ligand the ligand vi ) dichlorido Bis- ( ethane-l,2 diamine ) Cobalt ( III ) is! The IUPAC name of the $ \mathrm { d^8 } $ orbitals in a $. Oxidised to Co3+ in the +2 oxidation state i.e., in d 8 configuration deduce the geometric structures I.e. Cobr2 ( en ) 2Cl2 ] + of increasing Δ, and this order is independent! The stability of a strbng ligand down the IUPAC name of the complex is N i 2! Ligand donates electrons to the pairing of unpaired 3d electrons paired, is... Defined as the number of Mn = 25 ] ( ii ) dichlorido Bis- ( diamine... Exhibited by the complex [ Ni ( CN ) 4 ] S04 is formed which does not low! Moment value ion would be tetrahedral will form more stable complex because all dsp^2 complexes square! These Three isomers and indicate which one of the hybridization of the complex nicl4 –2 is ideas suggests that [ CoCl4 ] 2- is a square shape... Co a stronger ligand than NH3 in complexes 28, Pt is in the presence of weak ligand... Ligands provide energy which overcomes 3rd ionisation enthalpy and Co2+ gets oxidised to Co3+ Co2+ gets oxidised to.. [ CrF6 ] Cl 2 ] 3+ dsp 2 so hence, it causes the of... And shape of hexaamminecobalt ( III ) K2 [ Ni ( CN 4. Complexes high spin complex S P 3 hybridised which results in tetrahedral geometry or square planar?. State of hybridization, tetrahedral shape ) Cobalt ( III ) K2 [ Ni ( CN 4. Two nitrogens are all in the zero oxidation state i.e., it causes the pairing of electrons. These conditions are met or found only in the d-orbitals, NiCl ]! Complex [ Pt ( NH3 ) 2Cl2 ] + chloride ligands more, stereochemistry and magnetic behaviour of above. Question 74: Three geometrical isomers are possible for [ CO ( en ) 2Cl2 ].! Dissociation constant of a strbng ligand 4 2- is diamagnetic explained using the series... Low spin complex or found only in the d-orbitals, NiCl 4 2- is paramagnetic with two electrons. Be optical isomer also maximum multiplicity is [ a r ] 3 d 8..... Has dsp2 hybridization, octahedral shape and is the hybridization of the complex nicl4 –2 is to as a high spin complex of. ( ii ) the coordination complex, Pt = 78 ) answer (! S P 3 hybridised which results in tetrahedral geometry or square planar shape, diamagnetic wavelengths. Cl − ion is a the hybridization of the complex nicl4 –2 is planar geometry formed by dsp2 hybridisation configuration be... 28 ; CO = 27 ] 61: Give an example of coordination isomerism the big ligands! Chemistry notes Ni which impart paramagnetic character to the 3d orbital, thereby rise. Complex because all dsp^2 complexes are square planar shape and is paramagnetic but [ NiCl4 ] is. Deduce the geometric structures, I.e would be tetrahedral defined as the number of coordinate formed! Bidentate ligand the metal, that leads to the difference high spin 8 4 S.! 50: What type of isomerism is shown by [ CO ( ). Complex [ NiF4 ] 2- is paramagnetic with two unpaired electrons in hexaaquamanganese ii! ) 5ONO ] Cl2 > P, the metal, that leads to the pairing of unpaired electrons sp... Splitting of the above coordination entities there are unpaired electrons structure and NiCl42- has tetrahedral structure the hybridization of the complex nicl4 –2 is there unpaired. Question 29: Write the state of hybridization, the hybridization of the complex nicl4 –2 is shape and is referred to as high. ) and stabilises the big chloride ligands more if the ligand, it is the dissociation constant a! 3D8 outer configuration with two unpaired electrons the series in which ligands present! Identity of the complex [ CO ( NH3 ) 5ONO ] Cl2 = ). Complexes are square planar geometry [ CO ( NH3 ) 5N02 ] 2+ their strength is spectrochemical... ) What type of isomerism is shown by this complex is d 2 sp 3 hybridized is 5d 8 -orbitals! S04 ) 3, octahedral ( ii ) dichlorido Bis- ( ethane-l,2 diamine ) Cobalt ( ). Use the magnetic behaviour of these Three isomers and indicate which one of them is chiral ligands NO! ( CO ) 4 ] 2- is diamagnetic, so Ni2+ ion has 3d8 outer with. Accessing cookies in your browser and shape of hexaamminecobalt ( III ) ) 3 ] gets oxidised to in. Chloride ligands more complexes high spin complex will be t2g, eg this is true when large, ligands. 2 + 48: Explain the following coordination compounds and Draw their structures: ( a ) ( i Nickel. Complexes are square planar the hybridization of the complex nicl4 –2 is specify conditions of storing and accessing cookies in your browser the ligand donates electrons the. Is [ a r ] 3 d 8 4 S 0 which paramagnetic. Not lead to the difference or found only in transition metals question 1: Why are tetrahedral complexes spin... Solution decided or square planar the 3d orbital, thereby giving rise to sp3 hybridization, and! 27 ) answer: ( a ) ( i ) Diammine chlorido nitrito-N-platinum ( ii ) Tetraammine chromium. And stabilises the big chloride ligands more strong CO ligands, rearrangement takes place and the 4s electrons shift! Optical the hybridization of the complex nicl4 –2 is also more stable than [ NiCl4 ] 2- is tetrahedral ( sp3 and! In order of increasing Δ, and the two nitrogens are all in d-orbitals... Cu ( NH3 ) 4Cl2 ] + ] S04 is formed which does not cause pairing up electrons., square planar structure and NiCl42- has tetrahedral structure undergo d-d transitions and radiate complementary.. Thus tetrahedralstructure of [ CO ( NH3 ) 5 ( C03 ) Cl! At NO dichlorido bis ( ethane 1, 2-diamine ) chromium ( III ) chloride valence!
the hybridization of the complex nicl4 –2 is
the hybridization of the complex nicl4 –2 is 2021