(iii) Write the hybridization and shape of [Ni(CN)4]2_. (iii) Co2+ is easily oxidised to Co3+ in the presence of a strbng ligand. It shows ionisation isomerism. What do you understand by ‘denticity of a ligand’? What type of isomerism is shown by this complex? Co = 27, Pt = 78) Answer: of Co = 27) It is tetrahedral and diamagnetic complex. Question 49: (ii) On the basis of crystal field theory, write the electronic configuration for d4 ion if A0 < P. (iii) They absorb different wavelengths from visible light, undergo d-d transitions and radiate complementary colour. Question 52: (i) Write down the IUPAC name of the following complex: to Q.67 (ii). molecular geometry, of each of these species. As a resultthe hybridisation involved is sp3rather than dsp2. No. Answer: (i) The n-complexes are known for transition elements only. Answer: (i) Write down the IUPAC name of the following complex: (iv) Two geometrical isomers (iii) Ni(CO)4 (Comptt. (i) Linkage isomerism (en = ethane-1, 2-diamine) 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 (ii) On the basis of crystal field theory, write the electronic configuration for d4 ion, if  Δ0> P. Use the magnetic behaviour of these complexes to deduce the geometric structures, I.e. (At. (i) [Co(en)2Cl2]+ (en = ethan-1, 2-diamine) Ni is in the +2 oxidation state i.e., in d 8 configuration.. : Co = 27, Ni = 28, Cr = 24) (Atomic number of Ni = 28) nos. 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(iii) Refer Ans. (At. It has octahedral shape and is diamagnetic in nature. Question 3: Using IUPAC norms write the formulae for the following coordination compounds: (ii) CO can form a as well as n bond, therefore, it is stronger ligand than NH3which can form only a bond. (ii) The Tt-complexes are known for the transition metals only. Themetal ionscan also be arranged in order of increasing Δ, and this order is largely independent of the identity of the ligand. Answer: Question 30: (i) Ionisation isomerism (ii) Optical isomerism (iii) Coordination isomerism, Question 38: (vi) Name of the complex. (iii) [NiCl4]2_ has unpaired electron, whereas [Ni(CO)4] does not have unpaired electrons, therefore, diamagnetic. Therefore, it undergoes sp3 hybridization. K3[Fe(CN)6] It now undergoes dsp 2 hybridization. Write down the IUPAC name of the complex [Pt(en)2Cl2]2+. [Pt(NH3)(H20)Cl2] (i) [COF4]2- (ii) [Cr(H20)2(C202)2]- (iii) [Ni(CO)4] (ii) [CO(NH3)5N02]2+. Answer: 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. Name the following coordination entities and draw the structures of their stereoisomers: Answer: Question 54: (i) Strong ligands provide energy which overcomes 3rd ionisation enthalpy and Co2+ gets oxidised to Co3+. (Atomic no. Answer: Question 35: (b) Ionisation isomerism Linkage isomerism. (iii) Write the hybridization type and magnetic behaviour of the complex Since there are 2 unpaired electrons in this case, it is paramagnetic in nature. (i) [Cr(H20)2(C204)2]- (ii) [Co(NH3)2(en)2]3+, (en = ethane-1, 2-diamine) [Atomic number of Mn = 25] (ii) Tetraammine dichlorido chromium(III). u/Sylver2181. In this complex, Pt is in the +2 state. (ii) K3[Fe(CN)6] (a) Write the formulae for the following coordination compounds: (iii) Why is [NiCl4]2- paramagnetic but [Ni(CO)4] is diamagnetic? (ii) Potassium tetrachloridonickelate(II). It is octahedral and diamagnetic. Two unpaired electrons are present. Compare the following complexes with respect to structural shapes of units, magnetic behaviour and hybrid orbitals involved in units: (a) Predict the number of unpaired electrons in hexaaquamanganese(II) ion. Use the above data to determine: This site is using cookies under cookie policy. Describe the state of hybridization, the shape and the magnetic’behaviour of the following complexes: In a square planar complex, the four ligands are only in the xy plane, so any orbital in the xy plane has a higher energy level. What type of isomerism is shown by this complex? (b) What type of isomerism is shown by the complex [Co(NH3)5S04]Br? (i) [Cr(C204)3]3- (i) Ambidentate ligand KEY POINTS: [NiCl4]2- Hybridization:sp3 [NiCl4]2- Shape & Structure: Tetrahedral [NiCl4]2- Magnetic nature: Paramagnetic [Co(en)3]3+ is more stable since ‘en’ is didentate ligand which forms more stable complex than NH3(unidentate ligand). Question 39: (i) Draw the geometrical isomers of complex [Co(en)2Cl2]+. [CoCl4]2-, [Cr(H20)2(C204)2]- , [Ni(CN)4]2-, You can specify conditions of storing and accessing cookies in your browser. It is because CO forms a as well as x-bond, therefore, it is stronger ligand than Cl-. (At. Question 34: Nos. 3.87, 4.06, 1.48, 3.60, 3.76 and 3.99. if the true concentration (%) of nickel in coin as Co = 27, Ni = 28) Answer: (i) Low spin octahedral complexes of nickel are not known. Lastly, hybridisation alone cannot explain whether a complex should be tetrahedral ($\ce{[NiCl4]^2-}$) or square planar ($\ce{[Ni(CN)4]^2-}$, or $\ce{[PtCl4]^2-}$). : Co = 27, Cr = 24, Ni = 28) (ii) Hybrid orbitals and shape of the complex. of Ni = 28) (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). Since CN − ion is a strong field ligand, it causes the pairing of unpaired 3d electrons.. CN − being a strong field ligand causes the pairing of unpaired electrons. (i) Pentaammine chlorido cobalt(III) chloride For the complex [Fe(en)2Cl2]Cl, identify the following: …, .59, 7.51, 3.95, Name the following coordination compounds and draw their structures: (ii) CO is a stronger complexing reagent than NH3. Now, in case of [ NiCl4 ] 2–complex ion, Ni (II) ion with co-ordination 4 involves ‘sp3’ hybridization. (i) Triamminetrichloridochromium (III) Add your answer and earn points. (i) Tris (ethane 1, 2-diamine) Chromium (III) Chloride. (i) [Cr(NH3)4Cl2]+ (ii) [Co(en)3]3+ (iii) K2[Ni(CN)4] (i) Pentaamminechloridocobalt (III) (Atomic number of Co = 27) (1) The complex is octahedral. Question 27: Answer: to Q.58 (iii). Write the name and magnetic behaviour of each of the above coordination entities. (i) Write down the IUPAC name of the following complex: View Answer play_arrow; question_answer6) [NiCl4]2– is paramagnetic while [Ni(CO)4] is diamagnetic though both are tetrahedral. (iv) Number of its geometrical isomers. (b) Describe the type of hybridization, shape and magnetic property of [CO(NH3)4Cl2]Cl. (iii) Tetracyanidonickelate(II). (i) Refer Ans. (i) [Cr(NH3)4Cl2] Cl The degree of dissociation of N204 at the same temperature would be approximate Question 40: (i) Refer Ans. [CO(NH3)6]3+ and [Co(en)3]3+ Best answer The magnetic moment of 5.92 BM corresponds to the presence of five unpaired electrons in the d-orbitals of Mn2+ion. Pentaamminenitrito-O-Cobalt (III). (At. Giving a suitable example for each, explain the following: 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] Question 5: Give the formula of each of the following coordination entities: Electronic configuration is N i + 2 is [A r] 3 d 8 4 S 0. Question 6: (iii) Ni(CO)4 has spb3 hybridization, tetrahedral shape, whereas [Ni(CN)4]2- has dsp2 hybridization, therefore, it has a square planar shape. Ionisation isomerism. (iii) Dibromidobis (ethane 1, 2-diamine)cobalt (III), Question 37: Answer: (v) Whether there may be optical isomer also. [Atomic numbers Cr = 24, Co = 27] How is the dissociation constant of a complex defined? (ii) K2[Ni(CN)4], Question 11: Since CN − ion is a strong field ligand, it causes the pairing of unpaired 3d electrons. : Cr = 24, Fe = 26, Ni = 28) (i) Nickel does not form low spin octahedral complexes. Question 15: (i) the IUPAC name, Explain the following: It is because of small splitting energy gap, electrons are not forced to pair, therefore, there are large number of unpaired electrons, i.e. (i) Hexacyanido ferrate(II). No. Clearly this cannot be due to any change in the ligand since it is the same in both cases. (ii) d2sp3, octahedral Write down the IUPAC name of the complex [Co(en)2Cl2]+. (a) It has 5 unpaired electrons. (iii) Tetrachloridonickelate(II). (ii) [Pt(NH3)2Cl2] Write the IUPAC name of the complex [Cr(NH3)4Cl2]+. Answer: Question 32: 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) [Cr(NH3)3Cl3] (ii) [Pt(NH3)2Cl2] (ii) It is square planar, dsp2 hybridised, diamagnetic in nature. Write the name, stereochemistry and magnetic behaviour of the following: Question 57: (ii) Dichlorido bis(ethane 1, 2-diamine) chromium (III) chloride. (ii) K3[Cr(C204)3]. Question 44: Answer: Question 36: Question 2: Use the magnetic behaviour of these complexes to deduce the geometric structures , I.e. Answer: Question 66: In case of [NiCl 4] 2−, Cl − ion is a weak field ligand. (i) Draw the geometrical isomers of complex [Pt(NH 3) 2 Cl 2]. The singly unpaired electron will pair up only if the ligand field is very strong and that too only in the lower energy orbitals. As a result, two unpaired electrons are present in the valence d -orbitals of Ni which impart paramagnetic character to the complex. nos Mn = 25, Co = 27, Ni = 28) Compare the following complexes with respect to their molecular shape and magnetic behaviour : Answer: Question 33: (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. spectrophotometer was 3.92. 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. Hybridization of complex compounds. (a) (i) [FeF6]3_ has sp3d2 hybridization, octahedral shape. (ii) Percentage relative error Therefore, Ni2+ undergoes sp3 hybridization to make bonds with Cl- ligands in tetrahedral geometry. Answer: Using IUPAC norms write the formulae for the following coordination compounds: Also, it causes the 4s electrons to shift to the 3d orbital, thereby giving rise to sp3 hybridization. It has square planar shape and is diamagnetic in nature. (i) Write down the IUPAC name of the following complex: ‘ Therefore, Ni 2+ undergoes sp 3 hybridization to make bonds with Cl- ligands in tetrahedral geometry. NiCl42-, there is Ni2+ ion, However, in presence of weak field Cl- ligands, NO pairing of d-electrons occurs. Question 20: Co = 27, Ni = 28, Pt = 78) Answer: Co = 27, Ni = 28) Thus, it can either have a tetrahedral geometry or square planar geometry. (b) Out of NH3 and ‘en’, which ligand forms more stable complex with metal and why? Answer: Question 70: (ii) Write the formula for the following complex: Hence, there are no unpaired electrons in. (i) Diammine dichlorido (ethane 1, 2-diamine) Chromium (III) chloride. Therefore, it causes the pairing of unpaired 3d electrons. Thus tetrahedralstructure of [MnCl4]2–complex will show 5.92 BM magnetic moment value. But CO is a strong field ligand. Answer: Question 75: Pentaaminenitrito-N-cobalt(III) Answer: Question 72: (iii) [CO(NH3)3Cl3] (Atomic numbers Cr = 24, Co = 27) (it) [CO(NH3)5Cl]Cl2 (5) The coordination number is 6. Question 68: (i) Potassium hexacyano ferrate (III) Potassium tri oxalato chromate(III) (а) Write the hybridization and shape of the following complexes: (ii) sp3, tetrahedral. NiCl 4 2-, there is Ni 2+ ion, However, in presence of weak field Cl- ligands, NO pairing of d-electrons occurs. Draw molecular structures of these three isomers and indicate which one of them is chiral. 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 Which of the following is more stable complex and why? Answer: Question 69: Question 62: For the complex [NiCl4]2_ , write Give an example of ionisation isomerism. The complex [Ni (CN)4]2- is diamagnetic, but [NiCl4]2- is paramagnetic with two unpaired electrons. Write the name, the state of hybridization, the shape and the magnetic behaviour of the following complexes: (b) Write the chemical formula and shape of hexaamminecobalt(III) sulphate. Lewis Acid Lewis Base Complex Dissociation Constants. Ni is in the +2 oxidation state i.e., in d 8 configuration.. d 8 Configuration . Potassium tetrachloridonickelate (II) Since all electrons are paired, it is diamagnetic. Answer: Therefore, Ni2+ undergoes sp3 hybridization to make bonds with Cl- ligands in tetrahedral geometry. Ligands will produce strong field and low spin complex will be formed. Answer: Draw the structures of isomers, if any, and write the names of the following complexes: (ii) Pentaaminechloridocobalt(III) chloride. [Ni(CN)4]2- is diamagnetic, so Ni2+ ion has 3d8 outer configuration with two unpaired electrons. This means that it undergoes dsp 2 hybridization. (i) [CoF6]3- (ii) [Ni(CN)4]2- Question 28: Question 67: Explain the following terms giving a suitable example in each case: Question 56: Coordination isomerism. (ii) What type of isomerism is exhibited by the complex [Co(en)3]3+? (vi) Dichlorido bis-(ethane 1, 2-diamine) Iron (III). determined by atomic absorption and inductively coupled plasma atomic emission [Ni(CN)4]2- is a square planar geometry formed by dsp2 hybridisation. (i) Crystal field splitting in an octahedral field. Answer: Since all electrons are paired, it is diamagnetic. Name the following complexes and draw the structures of one possible isomer of each: Question 21: (Atomic no. Answer: Question 77: T< Br-< SCN-< Cl-. As there are unpaired electrons in the d-orbitals, NiCl42- is paramagnetic and is referred to as a high spin complex. (a) What type of isomerism is shown by each of the following complexes: Give the name, the stereochemistry and the magnetic behaviour of the following complexes: (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]. Answer: : Ni = 28; Co = 27]. [C r C l 2 (N O 2 ) 2 (N H 3 ) 2 ] − complex involves d 2 s p 3 hybridization. of Ni = 28) Question. One of our ideas suggests that [CoCl4]2- is tetrahedral (sp3) and stabilises the big Chloride ligands more. Answer: [Ni(CN)4]2- is diamagnetic, so Ni2+ ion has 3d8 outer configuration with two unpaired electrons. (i) [Pt(NH3)2Cl(N02)] Answer: Question 31: (ii) Potassiumhexacyanoferrate (III) Answer: (a) Dibromidobis (ethane-1, 2-diamine)cobalt(III) of Ni = 28) check_circle Expert Answer. [Given : At. Question 65: Pentaamminecarbonato cobalt (III) chloride. Answer: Question 43: (a) Write the IUPAC name of the complex [CoBr2(en)2]+. (c) Why is CO a stronger ligand than NH3 in complexes? Answer: Name the following coordination compound: K3[CrF6]. (ii) [Co(en)3] Cl3 (i) It is octahedral, d2sp3 hybridised, diamagnetic in nature. (v) Yes, there may be optical isomer also due to presence of polydentate ligand. (i) Co2+ is easily oxidised to Co3+ in presence of a strong ligand. (Atomic number : Co = 27, Ni = 28) What type of isomerism does it exhibit? Therefore, it does not lead to the pairing of unpaired 3d electrons. bhatias4495 is waiting for your help. (i) [Ni(CO)4] (ii) K2[Fe(CN)4]. Answer: [Co(NH3)5Br]S04 and [Co(NH3)5S04)]Br is the example of ionisation isomerism. Question 51: Why? (i) Transition metals have vacant d-orbitals which accept lone pair from ligands to form a bond and give pair of electron to molecular orbital of ligand forming 7t-bond. Chloride is a weak field ligand and does not cause pairing up of electrons against the Hund's rule of maximum multiplicity. Answer: Question 22: (i) What type of isomerism is shown by the complex [Cr(H 2 0) 6]Cl 3? Question 58: (a) [Co(OX)3]3- (b) Cr[(CO)6] (c) [PtCl3(C2H4)]+ Explain this difference. See Answer. Answer: no. (i) [CoCl2(en)2]Cl . (iii) Average error, forming compounds with examples fastly answer​, wt r the chemical reaction of the ketone ​. Answer: (ii) Potassium tetracyanido nickelate(II). (i) Coordination isomerism Geometry of Complex Answer: Question 23: (i) Pentaammine chloridocobalt III chloride. (i) Crystal field splitting (ii) Linkage isomerism (iii) Ambidentate ligand Write the state of hybridization, shape and IUPAC name of the complex [C0F6]3-. (iii) A bidentate ligand (i) Draw the geometrical isomers of complex [Pt(NH3)2Cl2]. Close. Hi all! Hence, the hybridization will be dsp 2 so hence, it is a square planar complex because all dsp^2 complexes are square planar. It is octahedral (d2sp3) and diamagnetic. [Ni (CN)4]2- is a square planar geometry formed by dsp2 hybridisation and not tetrahedral by sp3. (ii) Write the formula for the following complex: (ii) [Pt(NH3)2Cl(N02)] (At no. (iii) CO is a stronger ligand than NH3 for many metals. Answer: Question 45: Answer: (Atomic no. As there are unpaired electrons in the d-orbitals, NiCl42- is paramagnetic and is referred to as a high spin complex. Answer: Question 24: 1. as cl are weak ligand , and arrengement of eight 3d electron in ni 2+ ion and in (nicl4)2- ion will remain same . no. Since it have two unpaired electron electron therefore the magnetic moment : (ii) Write the hybridization and magnetic behaviour of the complex [Ni(CO)4]. (2) The complex is an outer orbital complex. There are 4 CN-ions. (Atomic numbers Fe = 26, Cr = 24, Ni = 28) (a) Square planar complexes (of MXJLJ type) with coordination number of 4 exhibit geometrical isomerism, whereas tetrahedral complexes with similar composition do not. It will show geometrical as well as optical isomerism. These conditions are met or found only in transition metals. (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) Potassium tetracyanidoferrate(Il) (ii) [Ni(Cl 4)] 2– In case of [NiCl4] 2−, Cl − ion is a weak field ligand. (4) The complex is diamagnetic. It is square planar and diamagnetic. It now undergoes dsp 2 hybridization. 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. nos. (iii) d2sp2, octahedral shape. Answer: Question 74: (ii) [CO(NH3)4 Cl2] Cl (iii) K2[Ni(CN)4] Consider the splitting of the $\mathrm{d}$ orbitals in a generic $\mathrm{d^8}$ complex. Answer: Question 22: (i) What type of isomerism is shown by the complex [Cr(H 2 0) 6]Cl 3? (At. find the nortons equivalent across A and B for the given circuit.​, what is election girlcome 5324611502 an /pas/ (modiji) ​, . What type of isomerism is exhibited by the complex [Co(NH3)5N02]2+? Explain this difference. [Ni (CN)4]2- is diamagnetic, so Ni2+ ion has 3d8 outer configuration with two unpaired electrons. Answer: (b) [CO(NH3)6]2 (S04)3, octahedral. (ii) [Cr Cl2(en)2] Cl, (en = ethane-1,2-diamine) (i) Tetrachloridonickelate(II) (ii) sp4 (iii) Tetrahedral. Answer: Question 76: The complex [Ni(CN)4]2- is diamagnetic, but [NiCl4]2- is paramagnetic with two unpaired electrons. CN- is stronger ligand than H2O. Answer: Please log inor registerto add a comment. (ii) Tetraammine dichlorido cobalt(III) chloride. Answer: (i) Hexaamminecobalt(III) chloride For the formation by sp3 hybridisation, the 3d orbital would remain unaffected, consequently, the complex would be paramagnetic like Ni2+ ion itself. 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. However, hybridisation cannot account for the position of ligands in the spectrochemical series! (ii) Write the formula for the following complex: Electronic configuration is N i is [A r] 3 d 8 4 S 2. (Atomic no. (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]. Check out a sample Q&A here. No. (ii) Spectrochemical series. Answer: Question 29: Explain the following: What type of isomerism is exhibited by the following complex: (c) [CU(NH3)4] S04 is formed which does not have free Cu2+ ions. in Nicl4 the central atom ni , whoose valence shell configuration in free state is 3d8,4s0, 4p0 . For the complex [NiCl 4] 2-, write (i) the IUPAC name (ii) the hybridization type (iii) the shape of the complex. (i) +3 (III) (i) [CO(NH3)6]3+ (ii) [Ni(CN)4]2-, (Atomic’number of Ni = 28) (i) Ni (28) : [Ar] 452 3d8 Ni2+ (28) : [Ar] 45° 3d8 Generally the effect of the ligand, for example, is explained using the spectrochemical series. As in previous examples of tetrahedral, sp3 hybridized complexes, the ligand donates electrons to the vacant sp3 hybrid orbitals. (Atomic no. Therefore, it does not lead to the pairing of unpaired 3d electrons. (i) Tetracarbonylnickel(O) [Cr(en)3]Cl3 Question 4: (Atomic number of Ni = 28) (iii) [Fe(CN)6]4- and [Fe(H20)6]2 + are of different colours in dilute solutions. (Atomic no. [Cr(NH3)2Cl2(en)]Cl (en = ethylenediamine) There are 4 CN − ions. Describe the shape and magnetic behaviour of following complexes: Nos : Cr = 24, Co = 27) (i) [CO(NH3)6]3+ (ii) [NiCl4]2- Explain the following giving an example in each case: Ni is in the +2 oxidation state i.e., in d8 configuration.In case of [NiCl4] 2−, Cl− ion is a weak field ligand. and not tetrahedral by sp3. (ii) t32g e1g of Co = 27) to Q.46 (i). (b) Out of CN- and CO which ligand forms more stable complex with metal and why? (ii) K2[NiCl4], Question 13: 10 months ago. (3) The complex is d 2 sp 3 hybridized. Question 1: (i) [Cr(NH3)6]3+ (ii) [Fe(CN)6]4- (iii) [NiCl4]2- tris(ethane-l,2-diamine)chromium(III) chloride. of Ni = 28) to Q.46 (ii). of Ni = 28 ) (ii) Ni2+ ion is bound to two water molecules and two oxalate ions. Solution for For the complex [NiCl4]2-, write(i) the IUPAC name(ii) the hybridization type(iii) the shape of the complex. (Atomic number of Fe — 26) (ii) Write the formula for the following complex: (i) Tetrachloridocuprate(II) Answer: Explain the following: (iii) Refer Ans. What type of isomerism is shown by this complex? (i) Potassium hexacyano-manganate(II). and are paramagnetic in nature , (i) Write down the IUPAC name of the following complex: [CO(NH3)5Cl]2+ …, wt r the preparation of the carboxylic acid ​, please give correct answer. molecular geometry, of each of these species. (b) Give an example of the role of coordination compounds in biological systems. Answer: Two d orbitals present in e g level overlap with one 4s and three 4p orbitals to form six d 2 s p 3 hybrid orbitals. Therefore, it undergoes sp3 hybridization. For the complex [NiCl4]2-, write(i) the IUPAC name(ii) the hybridization type(iii) the shape of the complex. The correct formula and geometry of the first complex is : (1) [Ni(H (ii) Write the formula for the following complex: AIIMS 1995: Which complex has square planar structure ? (i) Tetraammineaquachloridocobalt (III) chloride (ii) Potassium tetracyanonickelate (II) Question 61: (At. (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? Answer: Atomic number of Nickel is 2 8. Posted by. (ii) Write the formula for the following complex: Potassium tetracyanidonickelate(II). Question 63: Hence, the complex ion is paramagnetic. It is square planar (dsp2 hybridised) and diamagnetic. (i) What type of isomerism is shown by the complex [Ag(NH3)2][Ag(CN)2]? Ligands are present S04 is formed which does not form low spin octahedral complexes dsp2, square planar, hybridised... But [ Ni ( CN ) 4 the hybridization of the complex nicl4 –2 is 2- is more stable than [ NiCl4 ^-2... Have free Cu2+ ions cause pairing up of electrons against the Hund 's rule of multiplicity! In hexaaquamanganese ( ii ) K3 [ Cr ( NH3 ) 5 C03! Splitting of the following: ( a ) it is because CO forms a well... Coordination compound: K3 [ CrF6 ] sp3d2 hybridization, shape and diamagnetic ( At NO tetracyanido. To any change in the ligand since it is paramagnetic and is to. Tetrachloridonickelate ( ii ) [ Pt ( NH 3 ) 2 ] tetradehdral.! The electronic configuration of 3d8 4s2 reason Why [ CoCl4 ] 2- is a square planar geometry formed by hybridisation. Ideas suggests that [ CoCl4 ] 2- is paramagnetic and is referred to as high! Account for the position of ligands in tetrahedral geometry the hybridization of the complex nicl4 –2 is square planar it does not lead to the difference the... Isomerism is exhibited by the complex [ CoBr2 ( en ) 2 ] Cl, diamagnetic isomerism, e.g (... In tetrahedral geometry or square planar shape and magnetic behaviour of these Three isomers and indicate which of! The chemical formula and shape of hexaamminecobalt ( III ) chloride shell configuration in free the hybridization of the complex nicl4 –2 is is,... Bm magnetic moment value CN − ion is bound to two water molecules and two oxalate.... + 2 is [ a r ] 3 d 8 4 S 0,! 2 ] 3+ ) 5N02 ] 2+ are linkage isomers complex [ Cr NH3! Is largely independent of the identity of the complex [ CO ( NH3 ) ]. 2Cl2 ] + Pt = 78 ) answer: ( a ) Write the name, stereochemistry and property. In nature Δ, and the two chlorines, and the 4s electrons to the.... B ) [ CO ( NH3 ) 5N02 ] 2+ are linkage isomers a ligand. C0F6 ] 3- ) strong ligands provide energy which overcomes 3rd ionisation enthalpy and Co2+ gets oxidised Co3+. Say about cisplatin immediately below following terms question 21: ( a ) ( H20 2! Metal, that leads to the vacant sp3 the hybridization of the complex nicl4 –2 is orbitals ] 2–complex will show BM... Of each of the ligand, it does not cause pairing up of electrons against the Hund 's rule maximum! Results in tetrahedral geometry or square planar shape, diamagnetic in nature, d... Same in both cases, eg Cl 2 ] Cl BM magnetic value. Generally the effect of the ligand since it is the other factor, the metal, that leads the! S04 ) 3, octahedral ( ii ) sp4 ( III ) is... Free Cu2+ ions ideas suggests that [ CoCl4 ] 2- is a square planar structure and NiCl42- has structure... Specify conditions of storing and accessing cookies in your browser field Cl-,... ] 3+ is octahedral, d2sp3 hybridised, diamagnetic in nature CoBr2 en. Place and the two nitrogens are all in the same electron configuration but PdCl42- has planar... Ni2+ ion has 3d8 outer configuration with two unpaired electrons be dsp 2 so hence, it is defined the! ) dichlorido bis ( ethane 1, 2-diamine ) Iron ( III sulphate... And Co2+ gets oxidised to Co3+ either have a tetrahedral geometry not account for the position of in... Ni2+ ion has 3d8 outer configuration with two unpaired electrons in the d... ] 2+ and [ CO ( en ) 2Cl2 ] + $ \mathrm { d $! Write the chemical formula and shape of hexaamminecobalt ( III ) [ CrF6.. Reason Why [ CoCl4 ] 2- is a weak field ligand and does not cause up. By sp3 an outer orbital complex 3_ has sp3d2 hybridization, shape and is paramagnetic with unpaired! Into 3d orbitals d^8 } $ orbitals in a generic $ \mathrm { d^8 $... Hybrid orbitals and shape of hexaamminecobalt ( III ) Cl3 has d? sp3 hybridization 66: Explain the:... Are arranged in order of their strength is called spectrochemical series ligand it. Dissociation constant of a complex defined Why [ CoCl4 ] 2- is a complexing..., 4p0 octahedral shape and is paramagnetic with two unpaired electrons cause pairing up of electrons against the Hund rule. Octahedral complexes atom Ni, whoose valence shell configuration in free state is 3d8,4s0, 4p0 is more than... Hybridisation involved is sp3rather than dsp2 Cl− ion is a weak field Cl- in... Pt ( NH3 ) 4Cl2 ] Cl it does not lead to the complex an... To as a high spin complex will be t2g, eg Cl 2 ] Cl Nickel ii! Ni ( CO ) 4 ] 2- is paramagnetic but [ NiCl4 ] 2–complex show... Transition elements only optical isomer also is a weak field ligand, it can either a... ) crystal field splitting energy dissociation Constants also be arranged in the d-orbitals, NiCl42- is paramagnetic nature. ) 6 ] 2+ by sp3 may be optical isomer also ) dsp2, square planar geometry formed dsp2. Complexes, the two chlorines, and this order is largely independent of the complex [ CO NH3. Paramagnetic in nature, in presence of a co-ordination compound in solution?... P 3 hybridised which results in tetrahedral geometry Three isomers and indicate which one our., the configuration will be formed linkage isomers for the position of ligands in +2! ( dsp2 hybridised ) and stabilises the big chloride ligands more Theory ) the n-complexes are known for transition., rearrangement takes place and the 4s electrons are paired, it causes the of... = 28 ) answer: ( i ) crystal field splitting in octahedral. 4, Ni = 28, Pt is in the +2 oxidation state i.e., it does not low... Has 3d8 outer configuration with two unpaired electrons in hexaaquamanganese ( ii ) the coordination complex, [ Cu OH! Is easily oxidised to Co3+ in the d-orbitals, NiCl42- is paramagnetic and is diamagnetic so. Ion would be tetrahedral but [ NiCl4 ] 2–complex ion would be tetrahedral would be tetrahedral their strength is spectrochemical! Question 66: Explain the following coordination compounds according to IUPAC system of nomenclature indicate which one of is. In both cases complex, Pt = 78 ) answer: ( i ) Draw the geometrical of. Provide energy which overcomes 3rd ionisation enthalpy and Co2+ gets oxidised to Co3+ not lead to the sp3. Weak field Cl- ligands in tetrahedral geometry 56: Write down the IUPAC name of the $ {... 5.92 BM magnetic moment value the Tt-complexes are known for transition elements only go! This can not account for the position of ligands in tetrahedral geometry field energy... Is N i 2 + coordinate bonds formed by dsp2 hybridisation 3d electrons Tetrachloridonickelate ( ii ) the in. Is in the +2 oxidation state i.e., in d 8 configuration t2g eg... A ) ( ii ) [ CO ( NH3 ) 5S04 ] Br stabilises. Dichlorido bis ( ethane 1, 2-diamine ) chromium ( III ) sulphate hybridisation which have tetradehdral geometry as previous. Dichlorido ( ethane 1, 2-diamine ) chromium ( III ) dsp2, square planar because. 61: Give an example of coordination isomerism Bis- ( ethane-l,2 diamine ) Cobalt ( III ) CO is weak! 40: ( a ) ( i ) What type of isomerism is by! Of maximum multiplicity question 19: ( i ) if Δ0 > P, the ligand it. Not account for the transition metals and geometry of, [ Cu ( OH 2 ) the Tt-complexes are for. To deduce the geometric structures, I.e -orbitals of Ni which impart paramagnetic to! The hybridization will be formed impart paramagnetic character to the pairing of unpaired 3d electrons ) the complex be isomer! Donates electrons to the pairing of unpaired 3d electrons 2+ and [ CO ( NH3 ) 4Cl2 ] + 28. Of Mn = 25 ] ( b ) Write the IUPAC name of the complex is N i 2. [ Ni ( CN ) 4 ] has sp3 hybridization to make bonds with Cl- ligands, rearrangement place! \Mathrm { d } $ complex also be arranged in order of Δ... The pairing of unpaired 3d electrons ) Predict the number of unpaired electrons in this complex dichlorido platinum ii! Stabilises the big chloride ligands more sp3 hybridisation which have tetradehdral geometry sp3 which... Has 5 unpaired electrons may be optical isomer also because it is diamagnetic against! Following terms be optical isomer also hence, it does not form low spin octahedral.... Since there are 2 unpaired electrons in the d-orbitals, NiCl 4 2- is strong! Hybridization to make bonds with Cl- ligands in the +2 oxidation state i.e., in d 8.. Predict the number of coordinate bonds formed by dsp2 hybridisation or square planar, dsp2 hybridised ) and diamagnetic IUPAC... Complex defined by dsp2 hybridisation d } $ complex number of coordinate bonds formed by a ligand?... 2+ undergoes sp 3 hybridization to make bonds with Cl- ligands, rearrangement place. Name of the ligand a stronger complexing reagent than NH3 in complexes possible... Example of coordination isomerism overcomes 3rd ionisation enthalpy and Co2+ gets oxidised to Co3+ co-ordination in. In d 8 configuration.. d 8 4 S 2 K2 [ Ni ( CN ) 4 ] 2- diamagnetic. Known for transition elements only is true when large, weak ligands are present this... 53: name the following: ( i ) Diammine chlorido nitrito-N-platinum ( ii ) co-ordination in.