InterviewSolution
Saved Bookmarks
InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 851. |
Using IUPAC norms write the systematic names of the following: (i). `[Co(NH_3)_6]Cl_3` (ii). `[Pt(NH_3)_2Cl(NH_2CH_3)]Cl` (iii). `[Ti(H_2O)_6]^(3+)` (iv). `[Co(NH_3)_4Cl(NO_2)]Cl` (v). `[Mn(H_2O)_6]^(2+)` (vi). `[NiCl_4]^(2-)` (vii). `[Ni(NH_3)_6]Cl_2` (viii). `[Co(en)_3]^(3+)` (ix). `[Ni(CO)_4]` |
|
Answer» (i) Hexaammine cobalt (III) chloride (ii) Diammine chloride (methylamine) platinum (II) chloride (iii) Hexaaquatitanium (III) ion (iv) Tetraamine chloride nitrito N- cobalt (IV) chloride (v) Hexaaquamanganese (II) ion (vi) Tetrachloridonickelate (II) ion (vii) Hexaammine nickel (II) chloride (viii) Tris (ethane - 1,2-diamine) cobalt (III) ion. Tetra carbonyl nickel(0) |
|
| 852. |
Which of the following statement is correct regarding metal carbonyl ?A. a.In `Mn_(2)(CO)_(10)` bond order of `Mn-Mn` is 0 .B. b.In `Fe_(2)(CO)_(9)` number of `Fe -Fe` bonds is 1C. c.In `Ni(CO)_(4)` all bond length are sameD. d.`Fe(CO)_(5)` is diamagnetic |
| Answer» Correct Answer - B::C::D | |
| 853. |
Select the correct `IUPAC` name for `[Ti (pi - C_(5)H_(5))_(2)(O -C_(5)H_(5))_(2)]` .A. `(eta^(5)`-cyclopentadine) bis (cyvclopentadierthyl ) titanate(IV) .B. `(eta^(5)`-cyclopentadine) bis (cyvclopentadierthyl ) titanate(IV) .C. (cyclopentadine) bis ( beta^(5)-cyvclopentadierthyl ) titanate(IV) .D. `(eta^(5)-cyclopentadine)` bis (cyvclopentadierthyl ) titanate(IV) . |
| Answer» Correct Answer - B | |
| 854. |
Select correct statements:A. a.`[Ni(en)_(3)]^(2+)` is less stable than `[Ni(NH_(3))_(6)]^(2+)`B. b.Increase in stability of the complexes due to the presence of multidentate cyclic ligand is called macro-cyclic effect .C. c.`[Ni(en)_(3)]^(2+)` is more stable than `[Ni(NH_(3))_(6)]^(2+)`D. d.For a given ion and ligand the greater the charge on the metal ion the greater is the stability |
| Answer» Correct Answer - B::C::D | |
| 855. |
Which of the following statement(s) is/are correct?A. The stability constant of `[Co(H_(2)O)_(6)]^(3+)` is larger than that of `[Co(H_(2)O)_(6)]^(2+)`B. The cyano complexes more stable than those formed by halide ions .C. The stability of halide complexes follows the order `I^(Θ)gtBr^(Θ)gtCI^(Θ)gtF^(Θ)` .D. The stability constant of `[Co(NH_(3))_(4)]^(2+)` is less than that of `[CuCI_(4)]^(2-)` . |
| Answer» Correct Answer - A::B | |
| 856. |
STATEMENT-1: CO is stronger ligand than `CN^(-)` STATEMENT-2: `CO` and `CN^(-)` both show synergic bonding with metal . STATEMENT-3: `CO and N_(2)` are isoelectronic ligands but `N_(2)` is a weaker ligand than CO but stronger than `NH_(3)` .A. TFTB. TTFC. FFTD. FTT |
| Answer» Correct Answer - 2 | |
| 857. |
IN which of the following cases the synergic bonding takes place at the pi orbital of the ligand .A. `[PtCI_(3)(C_(2)H_(4))]^(Θ)`B. `[Ni(PF_(3))_(4)]`C. `Cr[(C_(6)H_(6))_(2)]`D. `Fe[(pi -C_(5)H_(5))_(2)]` |
|
Answer» Correct Answer - A::B::C The species which show secondary valencies in a complex are Lewis base . |
|
| 858. |
STATEMENT-1: CO is stronger ligand than `CN^(-)` STATEMENT-2: `CO` and `CN^(-)` both show synergic bonding with metal . STATEMENT-3: `CO and N_(2)` are isoelectronic ligands but `N_(2)` is a weaker ligand than CO but stronger than `NH_(3)` .A. TFTB. TTFC. FFTD. FTT |
| Answer» Correct Answer - 2 | |
| 859. |
`Delta_(0)` increases in the order of `[CrCI_(6)]^(3-)lt[Cr(CN)_(6)]^(3-)lt[Cr(C_(2)O_(4))_(3)]^(3-)` reason The stronger the ligand field the higher will be `Delta_(0)` value .A. If both `(A)` and ` (R )` are correct and `(R )` is the correct explanation of `(A ) ` .B. If both `(A)` and ` (R )` are correct and `(R )` is the correct explanation of `(A ) ` .C. If`(A)` is correct, but `( R )` is incorrect .D. Both `(A)` and `(R ) ` are incorrect . |
|
Answer» Correct Answer - D Accoring to spectrochemical series `CN^(Θ)` is a strong field ligands, so it has more `Delta_(0)` for these complex ions is `[CrCI_(6)]^(3-)lt[Cr(C_(2)O_(4))_(3)]^(2-)lt[Cr(CN)_(6)]^(3-)` . |
|
| 860. |
Assertion A solution of `[Ni(H_(2)O)_(6)]^(2+)` is green but a solution of `[Ni(CN)_(4)]^(2+)` is colourless Reason `[Ni(CN)_(4)]^(2+)` is square planar complex .A. If both `(A)` and ` (R )` are correct and `(R )` is the correct explanation of `(A ) ` .B. If both `(A)` and ` (R )` are correct and `(R )` is the correct explanation of `(A ) ` .C. If`(A)` is correct, but `( R )` is incorrect .D. Both `(A)` and `(R ) ` are incorrect . |
| Answer» Correct Answer - B | |
| 861. |
A solution of `[Ni(H_(2)O)_(6)]^(2+)` is green but a solution of `[Ni(CN)_(4)]^(2-)` is colourless Explain. |
|
Answer» In `[Ni(H_(2)O)_(6)]^(2+) Ni is in +2` oxidation state and having unpaired electrons which do not pair in the presence of the weak `H_(2)O` ligand Hence, it is coloured The d-d transition absorbs red light and the complementary light emitted is green In `[Ni(CN)_(4)]^(2-)` Ni is also in +2 oxidation state and having `3d^(8)` electronic configuration But in presence of strong ligand `CN^(-)` the two unpaired electrons in the 3d orbitals pair up, Thus, there is no unpaired electron present Hence, it is colourless. |
|
| 862. |
What is crystal field splitting energy? How does the magnitude of `triangle_(0)` decide the actual configuration of d orbitals in a coordination entity? |
| Answer» When the ligands approach a transition metal ion, the d-orbitals split into two sets, one with lower energy and the other with higher energy. The difference of energy between the two sets of orbitals is called crystal field splitting energy `(Delta_(0)` for octahedral field) if `Delta_(0) lt P` (pairing energy) the fourth electron enters one of the e.g orbitals giving the configuration `t^(3)2g^(e1)g` thus forming high spin complexes Such ligands for which `Delta_(0) lt P` are called weak field ligands if `Delta_(0) gt P` the fourth electron pairs up in one of the `t_(2g)` thereby forming low spin complexes Such ligands for which `Delta_(0) gtP` are called strong field ligands. | |
| 863. |
The colour of the coordination compounds depends on the crystal field splitting . What will be the correct order of obsorption of wavelength of light in the visible region, for the comolexes, `[Co(NH_(3))_(6)]^(3+)`m and `[Co(H_(2)O)_(6)^(3+)`?A. `[Co(CN)_(6)]^(3-)gt[CO(NH_(3))_(6)]^(3+) gt[Co(H_(2)O)_(6)]^(3+)`B. `[Co(NH_(3))_(6)]^(3+)gt[Co(H_(2)O)_(6)]^(3+)gt[Co(CN)_(6)]^(3-)`C. `[Co(H_(2)O)_(6)]^(3+)gt[Co(NH_(3))_(6)^(3+)gt[Co(CN)_(6)]^(3-)`D. `[Co(CN)_(6)]^(3-)gt[Co(NH_(3))_(6)]^(3+)gt[Co(H_(2)O)_(6)]^(3+)` |
|
Answer» (c) As we know that, strong field ligand split the five degenerate energy levers with more energy separation thean weak field ligand, i.e as strenght of ligand increaeses crystal field spilting energy increases. Hence " " `DeltaE=(hc)/(lambda)` `rArr" " DeltaEprop(1)/(lambda)rArrlambdaprop(1)/(DeltaE)` As energy seperation increases, the wavelength decreases/ Thus. the correct order is `[Co(H_(2)O)]^(3+)gt[Co(NH_(3))_(6)]^(3+)gt[Co((CN)_(6)]^(3-)` Hence strength of ligand increaase. `DeltaE` inceases CFSE increaese and `lambda` absored decrease. |
|
| 864. |
What is crystal field splitting energy? How does the magnitude of `triangle_(0)` decide the actual configuration of d orbitals in a coordination entity?A. if `Delta_(0) lt P` , the configuration is `t_(2g)^(3) e_(g) ^(1)` = weak field ligand and high spin complexB. If `Delta_(0) gt P` , the configuration is `t_(2 g)^(3) e_(g) ^(1)` = strong field ligand and low spin complexC. if `Delta_(0) gt P` , the configuration is `t_(2g)^(4) e_(g)^0 `= strong field ligand and high spin complexD. if `Delta_(0) = P` , the configuration is `t_(2g)^(4) e _(g)^(0)` = strong field ligand and high spin complex . |
|
Answer» Correct Answer - A If CFSE `(Delta)_(0) lt P` (Energy required for pairing) , the electrons do not pair up and fourth electron goes to `e_(g)` of higher energy . Hence , high spin complex is formed . Pairing of electrons does not take place in case of weak field ligands . |
|
| 865. |
The colour of the coordination compounds depends on the crystal field splitting . What will be the correct order of obsorption of wavelength of light in the visible region, for the comolexes, `[Co(NH_(3))_(6)]^(3+)`m and `[Co(H_(2)O)_(6)^(3+)`?A. `[Co(CN)_(6)]^(3-) gt [Co(NH_(3))_(6)]^(3+) gt [Co(H_(2)O)_(6)]^(3+)`B. `[Co(NH_(3))_(6)]^(3+) gt [Co(H_(2)O)_(6)]^(3+) gt [Co(CN)_(6)]^(3-)`C. `[Co(H_(2)O)_(6)]^(3+) gt [Co(NH_(3))_(6)]^(3+) gt [Co(CN)_(6)]^(3-)`D. `[Co(CN)_(6)]^(3-) gt [Co(NH_(3))_(6)]^(3+) gt [Co(H_(2)O)_(6)]^(3+)` |
|
Answer» Correct Answer - C As we know that, strong field ligand split the five degenerate energy levels with more energy separation than weak field ligand, i.e., as strength of ligand increases crystal field splitting energy increases. Hence, `DeltaE=(hc)/lambda` `rArr " "DeltaE prop 1/lambda rArr lambda prop 1/(Delta E)` As energy separation increases, the wavelength decreases. Thus, the correct order is `[Co(H_(2)O)_(6)]^(3+) gt [Co(NH_(3))_(6)]^(3+) gt [Co(CN)_(6)]^(3-)` Here, strength of ligand increases, `DeltaE` increases, `CFSE` increases and `lambda` absored decreases. Hence, correct choice is (c). |
|
| 866. |
The value of `Delta_0` for `RhCl_6^(3-)` is `243 kJ//mol` what wavelength of light will promote an electron from. The colour of the complex isA. (a) BlueB. (b) GreenC. (c) YellowD. (d) Orange |
|
Answer» Correct Answer - D `Delta_0RhCl_6^(-3)` is `243 kJ//mol` `E=(hc)/(lambda)` `lambda=(6.626xx10^(-34)xx3xx10^8)/(234xx10^3)` `=511nm`. Nearest color is orange. |
|
| 867. |
Which of the following complex is homoleptic ?A. `H_(2)[PtCl_(6)]` B. `Li[AlH_(4)]`C. `[Ni(CO)_(4)]`D. All of these |
| Answer» Correct Answer - D | |
| 868. |
Which of the following complex is homoleptic ?A. `H_(2)[PtCl_(6)]`B. `Li[AlH_(4)]`C. `[Ni(CO)_(4)]`D. All of these |
| Answer» Correct Answer - D | |
| 869. |
Aq. solution of of `KCl * MgCl_(2) * 6H_(2)O` will give test ofA. `K^(+) and Mg^(2+)` onlyB. `K^(+) and Cl^(-)` onlyC. `K^(+) ,Mg^(2+) and Cl^(-)`D. `Mg^(2+) and H_(2)O` only |
| Answer» Correct Answer - C | |
| 870. |
Oxidation number of plantinum in cis-platinA. ZeroB. `+2`C. `+4`D. `+6` |
| Answer» Correct Answer - B | |
| 871. |
Which of the following statement(s) is /are correct with respect to the crystal field theory ?A. It considers only the metal ion d-orbitals and gives no consideration at all to other metal orbitals.B. It cannot account for the `pi` bonding in complexes .C. The ligands are point charge which are either ions or neutral moleculesD. The magnetic properties can be explained in term of splitting of d-orbital in different crystal field. |
| Answer» Correct Answer - A::B::C::D | |
| 872. |
An excess of `AgNO_3` is added to `100mL` of a `0.01M` solution of dichlorotetraaquachromin (III) chloride. The number of moles of `AgCl` precipitated would be:A. (a) `0.002`B. (b) `0.001`C. (c) `0.003`D. (d) `0.01` |
|
Answer» Correct Answer - B `AgNO_3+[Cr(H_2O)_4Cl_2]ClrarrAgCl+[Cr(H_2O)_4Cl_2]NO_3` millimole `{:(100xx0.1=1,,excess,-,-),(0,,,1,1):}` `:.` mole of `AgCl=1xx10^-3=0.001` |
|
| 873. |
Which of the following is not correct for crystal field theory ?A. The crystal field theory (CFT) was originally proposed for explaining the optical properties of crystallise solids.B. Cft assume the ligands to be point charges.C. The interaction between the ligands and the electrons of the central metal is assumed to be covalent in nature.D. The interaction between the ligand and the central metal atom/ion results in splitting of the d orbital energies. |
|
Answer» Correct Answer - C CFT is an electrostatic model which considers the metal-ligand bond to be ionic arising purely form electrostatic interations between the metal ion and the ligand. The five d orbitals in an isolated (gaseous state) metal atom/ion have same energy, i.e., they are degenerate. This degeneracy is maintained if a spherically symmetrical field of negative charges surrounds the metal atom/ion. However, when this negative field is due to ligands (either aninons or the negative ends of dipolar molecules like `NH_(3) H_(2)O`, etc.) in a complex, it becomes asymmetrical and the degeneracy of the d orbitals is lifted. The pattern of splitting depends upon the nature of the crystal field. |
|
| 874. |
An excess of `AgNO_3` is added to `100mL` of a `0.01M` solution of dichlorotetraaquachromin (III) chloride. The number of moles of `AgCl` precipitated would be:A. (a) `0.002`B. (b) `0.003`C. (c) `0.01`D. (d) `0.001` |
|
Answer» Correct Answer - D `AgNO_3+[Cr(H_2O)_4Cl_2]ClrarrAgCl+[Cr(H_2O)_4Cl_2]NO_3` millimole of `AgNO_3` added is `100xx0.1=1` `:.` Mole of `AgCl=1xx10^-3=0.001` (As `AgNO_3` causes ppt of `AgCl` so `AgCl` ppt is `10^-3` moles.) |
|
| 875. |
Which one of the following is an outer orbital complex and exhibits paramagnetic behaviour ?A. (a) `[Cr(NH_3)_6]^(3+)`B. (b) `[Co(NH_3)_6]^(3+)`C. (c) `[Ni(NH_3)_6]^(2+)`D. (d) `[Zn(NH_3)_6]^(2+)` |
|
Answer» Correct Answer - C `._(28)Ni^(2+)` in `[Ni(NH_3)_6]^(2+)` has `1s^2, 2s^2 2p^6, 3s^2 3p^6 3d^8` configuration. It uses 4th orbital to show `sp^3d^2`-hybridizaation to form outer complex with 2 unpaired electrons in 3d-orbital. |
|
| 876. |
Low spin complex of `d^6`-cation in an octahedral field will have the following energy:A. (a) `(-12)/(5)Delta_0+P`B. (b) `(-12)/(5)Delta_0+3P`C. (c) `(-2)/(5)Delta_0+2P`D. (d) `(-2)/(5)Delta_0+P` |
|
Answer» Correct Answer - B `d^6`-cation with low spin has electronic configuration `t_(2g)^6e_g^0`. Total energy`=(-0.4Delta_0per e^(-)xx6)` `+(e^-` pairing eenrgy of 3 pairs) `=-2.4 Delta_0+3P=-(12)/(5)Delta_0+3P` |
|
| 877. |
Which of the following is a low spin (spin-paired) complex ?A. `[Co(NH_(3))_(6)]^(3+)`B. `[Ni(NH_(3))_(6)]^(2+)`C. `[Co(NH_(3))_(6)]^(2+)`D. `[Fe(C_(2)O_(4))_(3)]^(3-)` |
|
Answer» Correct Answer - A `[Co(NH_(3))_(6)]^(3+) rarr d^(2) sp^(3)` hybridization `rarr` zero unpaired electrons `[Ni(NH_(3))_(6)]^(2+) rarr sp^(3) d^(2)` hybridization `rarr` Two unpaired electrons `[Co(NH_(3))_(6)]^(2+) rarr sp^(3) d^(2)` hybridization `rarr` Three unpaired electrons A low spin (or spin-paired) complex, such as `[Co(NH_(3))_(6)]^(3+)` is one in which the electrons are paired up to give a maximum number of doubly occupied d orbitals and a minimum number of unpaired electrons. Usually inner orbital complexes `(d^(2)sp^(3))` are low-spin (or spin paired) complexes. `Fe^(3+)` has a `d^(5)` electronic configuration. Thus complexes with weak field ligands (such as halide ions) will have a high spin arrangement with five unpaired electrons. On the other hand, strong field ligands such as `CN^(-) SCN^(-)` and oxalate form complexes with `Fe^(3+)` which have a spin paired arrangement. |
|
| 878. |
Which among the following will be named as dibromidobis (ethylene diamine) chromium (III) bromide?A. `[Cr(en)_(2) Br_(2) ]Br`B. `[Cr(en) Br_(4)]^(-)`C. `[Cr(en) Br_(2)]Br`D. `[Cr(en)_(3)]Br_(3)` |
| Answer» Correct Answer - A | |
| 879. |
Which among the following will be named as dibromidobis (ethylene diamine ) chromium (III) bromide ?A. `[Cr(en)_(3)]Br_(3)`B. `[Cr(en)_(2)Br_(2)]Br`C. `[Cr(en)Br_(4)]^(-)`D. `[Cr(en)Br_(2)]Br` |
| Answer» Correct Answer - B | |
| 880. |
The magnetic moment (spin only) of `[NiCl_4]^(2-)` isA. 1.82 BMB. 5.46BMC. 2.82 BMD. 1.41 BM |
| Answer» Correct Answer - C | |
| 881. |
`CN^(-)` is a strong field ligand. This is due to the fact thatA. it carries a negative chargeB. it is a pseudohalideC. it can accept electrons form metal speciesD. it form high spin complexes with metal species |
|
Answer» Correct Answer - C Ligands such as CO, `CN^(-)` and `NO^(+)` have empty `pi` orbitals with the sorrect symmetry to overlap with the metal `t_(2g)` orbitals, forming `pi` bonds. This is often described as back bonding. Normally the `pi` orbitals on the ligands are higher energy than th emetal `t_(2g)` orbitals. No more electrons are added to the scheme as the ligand `pi` orbitals are empty, but the `pi` interaction increases the value of `Delta_(o)` . This accounts for th eposition of these ligands as strong field ligands, at the right of the spectrochemical series. |
|
| 882. |
`CN^(-)` is a strong field ligand. This is due to the fact thatA. (a) it can accept electron from metal speciesB. (b) it forms high spin complexes with metal speciesC. (c) it carries negative charge.D. (d) it is a pseudohalide |
|
Answer» Correct Answer - D Cyanide ion is strong field ligand because it is a pseudohalide ion. Pseudohalide ions are stronger coordinating ligand and they have the ability to form `sigma` bond (from the pseudohalide to the metal) and `pi` bond (from the metal to pseudohalide). |
|
| 883. |
`CN^(-)` is a strong field ligand. This is due to the fact thatA. it carries negative chargeB. it is a pseudohalideC. it can accept electrons from metal speciesD. it forms high spin complexes with metal species |
|
Answer» Correct Answer - B `CN^(-)` is a strong field ligand because it is an example uf pseudohalide Pseudohalide ions are stronger coordinating ligands and they have the ability to form `sigma" and "pi`-bonds |
|
| 884. |
Considering `H_2O` as a weak field ligand, the number of unpaired electrons in `[Mn(H_2O)_6]^(2+)` will be (At. no. of `Mn=25`)A. (a) twoB. (b) fourC. (c) threeD. (d) five |
|
Answer» Correct Answer - D `Mn^(25)rarr3d^5+4s^2` ` Mn^(2+) to 3d^(6) In presence of weak ligand field, there will be no pairing of electrons. So it will form a high spin complex. i.e. the number of unpaired electrons `=5`. |
|
| 885. |
`CuSO_4.5H_2O` is represented asA. (a) `[Cu(H_2O)_5]SO_4`B. (b) `[Cu(H_2O)_3SO_4].2H_2O`C. (c) `[Cu(H_2O)_4]SO_4.H_2O`D. (d) All of these |
|
Answer» Correct Answer - C In `CuSO_4` only four `H_2O` act as ligand or coordination number for `Cu^(2+)` is four. |
|
| 886. |
According of Lewis, the ligands areA. acidic in natureB. basic in natureC. neither acidic nor basicD. some are bais and others are cidic |
| Answer» (b) All the ligands are Lewis bases. | |
| 887. |
Assertion Chelates are relatively more stable than non-cheltated complexes Reason Complexes containing ligands which can be easily replaced by other ligands are called labile complexes .A. If both `(A)` and ` (R )` are correct and `(R )` is the correct explanation of `(A ) ` .B. If both `(A)` and ` (R )` are correct and `(R )` is the correct explanation of `(A ) ` .C. If`(A)` is correct, but `( R )` is incorrect .D. Both `(A)` and `(R ) ` are incorrect . |
| Answer» Correct Answer - B | |
| 888. |
Which of the following complexes will be dehydrated to relatively minimum extent by conc. `H_(2)SO_(4)` under identical condition .A. `[Cr(H_(2)O)_(5)Cl]Cl_(2).H_(2)O`B. `[Cr(H_(2)O)_(4)Cl_(2)]Cl.2H_(2)O`C. `[Cr(H_(2)O)_(6)]Cl_(3)`D. all of these |
| Answer» Correct Answer - C | |
| 889. |
Which of the following energy diagrams shows the electron distribution according to the crystal field model of the 3d-electrons in `[CoCl_(4)]^(2-)` ?A. B. C. D. |
| Answer» Correct Answer - D | |
| 890. |
The figure represents the synergic bonding interaction in metal carbonyl complex. On the basis of the explain the following: (i) Strength of Metal-ligand bond (ii) Bond order of CO in carbonyl complex as compared to bond order in carbon monoxide. |
| Answer» Correct Answer - (i) Increases (i) Decreases | |
| 891. |
Assertion `Ni(CO)_(4)` is tetrahedral in shape Reason `Ni` atom is in zero oxidation state and undergoes `sp^(3)` -hybridisation in `Ni(CO)_(4)` .A. If both `(A)` and ` (R )` are correct and `(R )` is the correct explanation of `(A ) ` .B. If both `(A)` and ` (R )` are correct and `(R )` is the correct explanation of `(A ) ` .C. If`(A)` is correct, but `( R )` is incorrect .D. Both `(A)` and `(R ) ` are incorrect . |
| Answer» Correct Answer - A | |
| 892. |
In metal carbonyl having and genral formula `M(CO)_(x)`, where, M = metal, x = 4 and the metal is bonded toA. carbon and oxygenB. oxygenC. carbonD. `C -= O` |
|
Answer» Correct Answer - C In metal carbonyls such as `[M(CO)_(x)]`, the metal is bonded to the carbony1 ligand (CO) through C atom. The metal-carbon bond in metal carbonyls possesses both `sigma` and `pi` character. |
|
| 893. |
In metal carbonyl having and genral formula `M(CO)_(x)`, where, M = metal, x = 4 and the metal is bonded toA. carbon and oxygenB. `CequivO`C. oxygenD. carbon |
|
Answer» Correct Answer - D In metal carbonyl `M(CO)_(4)` , metal is bonded to the ligand CO through carbon atoms. These compounds contam both `sigma" and "pi`-bonded complexes. `sigma`-bond between metal and carbon atom is formed when a vacant hybrid orbital of the metal atom overlaps with an orbital on C-atom of carbon rnonooxide containing a low pair of electron Formation of `pi`-bond is occurs when a filled orbital of the metal atom overlaps with a vacant antibonding `pi`-orbital of C-atom of Co. |
|
| 894. |
Which of the following complex ion`(s)` is/are not expected to absorb visible light?A. (a) `[Zn(NH_3)_6]^(2+)`B. (b) `[Sc(H_2O)_3(NH_3)_3]^(3+)`C. (c) `[Ti(en)_2(NH_3)_2]^(4+)`D. (d) `[Cr(NH_3)_6]^(3+)` |
|
Answer» Correct Answer - D `{:(Ti^(4+),:3d^(0),,,),(Cr^(3+),:3d^3,,,),(Zn^2+,:3d^(10),,,),(Sc^(3+),:3d^(0),,,):}}` Completely filled or empty d-orbitals are colourless. |
|
| 895. |
Copper sulphate dissolves in ammonia due to the formulation ofA. `Cu_(2)O`B. `[Cu(NH_(3))_(4)]SO_(4)`C. `[Cu(NH_(3))_(4)]OH`D. `[Cu(H_(2)O)_(4)]SO_(4)` |
|
Answer» Correct Answer - B `CuSO_(4) + 4 NH_(3) to [Cu(NH_(3))_(4) ] SO_(4)` |
|
| 896. |
Which of the following has the regular tetrahedral structure?A. (a) `[Ni(CN)_4]^(2-)`B. (b) `SF_4`C. (c) `BF_4^(-)`D. (d) `XeF_4` |
|
Answer» Correct Answer - C B in `BF_4^(-)` has `sp^3`-hybridisation leading to tetrahedral geometry. |
|
| 897. |
The species having tetrahedral shape isA. `[PdCl_(4)]^(2-)`B. `[Ni(CN)_(4)]^(2-)`C. `[Pd(CN)_(4)]^(2-)`D. `[Ni(Cl)_(4)]^(2-)` |
| Answer» Correct Answer - D | |
| 898. |
A co-ordination complex, compound of cobalt has the molecular formulae containing five ammonia molecules, one nitro group and two chloride atoms for one cobalt atom. One mole of this compound produces three mole ions in an aqueous solution on reacting with excess of `AgNO_3`, `AgCl` precipitate. The ionic formula for this complex would be:A. (a) `[Co(NH_3)_5(NO_2)Cl_2`B. (b) `[Co(NH_3)_5Cl][Cl(NO_2)]`C. (c) `[Co(NH_3)_4(No_2)Cl][(NH_3)Cl]`D. (d) `[CO(NH_3)_5][(NO_2)_2Cl_2]` |
|
Answer» Correct Answer - A The most probable complex which gives three moles ions in aqueous solution may be `[Co(NH_3)_5NO_2]Cl_2` because it gives two chlorine atoms on ionisation. `[Co(NH_3)_5NO_2]Cl_2rarr[Co(NH_3)_5NO_2]^(2+)+2Cl^(-)` |
|
| 899. |
In the complex `[CoCl_(2)(en)_(2)]Br`, the co-ordination number and oxidation state of cobalt are :A. 6 and +3B. 3 and +3C. 4 and +2D. 6 and +1 |
| Answer» Correct Answer - A | |
| 900. |
Which would exhibit coordination isomerism `[Cr(NH_(3))_(6)][Co(CN)_(6)]` (b) `[Cr(en)_(2)CI_(2)]^(o+)` (c ) `[Cr(NH_(3))_(6)]CI_(3)` `[Cr(edta)]^(-1)` . |
| Answer» Correct Answer - a | |