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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.
| 301. |
Calculate the `EAN` of `CO` in `[Co(NH_(3))_(6)]^(3+)`(Atomic number of `Co= 27)` . |
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Answer» In `[Co(NH_(3))_(6)]^(3+)=+3` `Y=No` of electrons gained by coordination `=2 xx 6 =12` `EAN` of `Co` in `[Co(NH_(3))_(6)]^(3+)=27 - 3 + 12 =36` which is equal to the atomic number of next higher noble gas which is `(Kr, Z= 36)` Hence Co in `[Co(NH_(3))_(6)]^(3+)` obeys `EAN` rule Futher examples of complexes whose central atom//ion obeys `EAN` rule are given in Table `7.12` . |
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| 302. |
The atomic number of cobalt is 27 .The EAN of cobalt in `Na_(3)[Co(NO_(2)_(4)Cl_(2)]` isA. 24B. 36C. 34D. 35 |
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Answer» Correct Answer - B EAN of CO in `Na_(3) [Co(NO_(2))_(4) Cl_(2)]` = atomic number of Co- oxidation state of Co + 2 `xx` number of electron taken form `NO_(2) ` and Cl ` = 27-3+2xx(4+2)=36` |
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| 303. |
The hypothetical complex triamminediaquachloridocobalt`(III)` chloride can be represented as `:`A. `[Co(NH_(3))_(3)(H_(2)O)_(2)CI]CI_(2)`B. `[Co(NH_(3))_(3)(H_(2)O)CI_(3)]`C. `[Co(NH_(3))_(3)(H_(2)O)_(2)CI]`D. `[Co(NH_(3))_(3)(H_(2)O)_(3)]CI_(3)` |
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Answer» Correct Answer - A While writing the structure of thwe complex the central metal atom/ion is listed first. The ligands are then listed in alphabetical order. The placement of a ligands are then listed does not depend on its chanrge. Thus, the hypothertical complex can be represnted as option (1). |
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| 304. |
Atomic numbers of `Cr` and `Fe` are respectively `24` and `26`. Which of the following is paramagnetic with the spin of the electron?A. `[Cr(CO)_(6)]`B. `[Fe(CO)_(5)]`C. `[Fe(CN)_(6)]^(4-)`D. `[Cr(NH_(3))_(6)]^(3+)` |
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Answer» Correct Answer - D `[Cr(CO)_(6)] implies Cr^(0) (3d^(5) 4s^(2))` Since CO is a strong field ligand, there is resuffle of electron configuration resulting in zero unpaired electrons. `[Fe(CO)_(5)] implies Fe^(0)(3d^(6) 4s^(2))` Since CO is a strong field ligand, there is resuffle electron configuration resulting in zero inpaired electron resulting in zero unpaired electrons. `[Fe(CN)_(6)]^(4-) implies Fe^(2+) (3d^(6))` Since `CN^(-)` is a strong field ligand, there is resulffle of electron configurration resulting in zero unpaired electrons `[Cr(NH_(3))_(6)]^(3+) implies Cr^(3+) (3d^(3))` Than is no resutfle of electron configuration as enough d orbitals are vacant for `d^(2) sp^(3)` hybridization. Thus, there arer 3 unpaired electrons. |
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| 305. |
Which of the following will show optical isomerism? .A. `[ZnCI_(4)]^(2-)`B. `[Cu(NH_(3))_(4)]^(2+)`C. `[Cr(C_(2)O_(4))_(3)]^(3-)`D. `[Co(CN)_(6)]^(3-)` |
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Answer» Correct Answer - C Optical isomerism is very common in otahedral complexes having general formula `[MA_(2)B_(2)C_(2)]^(n-)` `[M(A A)_(3)]^(n-),[M(A A)_(2)b_(2)]^(n-),[M(A A)_(2)AB]^(n-)` and ,. `[M(AB)_(3)]^(n-)` where `A A` is symmetrical bidentate ligand `such as underset(COO^(Θ))overset(COO^(Θ))overset(|)` and `AB` is unsymmetrical bidentate ligand . |
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| 306. |
In which of the following pairs both the complex show optical isomerism? .A. a)`cis-[Cr(C_(2)O_(4))_(2)CI_(2)]^(3-),cis-[Co(NH_(3))_(4)CI_(2)]`B. b)`[Co(en)_(3)]CI_(3),cis-[Co(en)_(2)CI_(2)]CI`C. c)`[Co(NO_(3))_(3)(NH_(3))_(3)],cis-[Pt(en)_(2)CI_(2)]`D. d)`[PtCI(en)CI],[NiCI_(2)Br_(2)]^(2-)` |
| Answer» Correct Answer - B | |
| 307. |
How many of the given complexes follow E.A.N rule ? (a) `[Fe(CO)_(5)]` (b) `[Co_(2)(CO)_(8)]` ( c) `[Fe(C_(5)H_(5))_(2)]` (d) `K_(3)[Fe(CN)_(6)]` ( e)`[Fe(NO)_(2)9CO)_(2)]` (f)`[CoF_(6)]^(4-)` |
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Answer» Correct Answer - 4(a,b,c,e) (d) `K_(3)[Fe(CN)_(6)]=26-3+2xx6=35` does not follow E.A.N. rule. (f) `[CoF_(6)]^(4-)=27-2+2xx6=37` does not follow E.A.N. rule. |
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| 308. |
For the `[Cr(H_(2)O)_(6)]^(2+)` ion, the mean pairing energy P is found to `23500cm^(-1)`. The magnitude of `Delta_(0)` is `13900cm^(-1)`. Calculate the C.F.S.E. `(cm^(-1))` for this complex ion corresponding to high spin state (x) and low spin state (y). Write your answer as `((y-x)/(100))`. |
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Answer» Correct Answer - 96 For a `d^(4)`ion in a high spin state. CFSE`=-0.6Delta_(0)=-0.6xx(13,900 cm^(-1))=-8340 cm^(-1)` For a `d^(4)` ion in a low spin state, the net CFSE is, `=-1.6Delta_(0)+P=-1.6xx(13,900 cm^(-1))+23500 cm^(-1)= +1,260 cm^(-1)` Since `Delta_(0)(=13,900 cm^(-1)) lt P(=23,500 cm^(-1))`, the high spin configuration would be more stable . |
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| 309. |
Which of the following does not depict the correct , name of the compound ?A. `K_(2) [Zn(OH)_(4)]` : Potassium tetrahydroxozincate (II)B. `[Co(NH_(3))_(5)CO_(3)]Cl:` Pentaammine carbonatochlorocobaltate(III)C. `Na_(3)[Co(NO_(2))_(6)] :` Sodium hexanitrocobaltate (III)D. `K_(3)[Cr(CN)_(6)]`: Potassium hexacyanochromate(III) |
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Answer» Correct Answer - B `[Co(NH_(3))_(5)CO_(3)]Cl` Pentaamminecarbonatocobalt(III) chloride |
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| 310. |
Which of the following is not correctly matched ?A. Coordination compound containing cationic complex ion : `[Fe(H_(2)O)_(2)(C_(2)O_(4))_(2)]_(2) SO_(4)`B. Coordination compound containing anionic complex ion : `[Ag(NH_(3))_(2)] Cl`C. Non-ionic coordination compound : `[Co(NO_(2))_(3)(NH_(3))_(3)]`D. Coordination compound containing cationic and anionic complex ion : `[Pt(NH_(3))_(4)] [CuCl_(4)]` |
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Answer» Correct Answer - B `[Ag(NH_(3))_(2)]Cl` is a coordination compound containing cationic complex ion . |
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| 311. |
How many of the following is correctly matched complex ? |
| Answer» Correct Answer - 3(a,b,d) | |
| 312. |
`Ni^(+2)` form a complex inon in water having the formula `[Ni(H_(2)O)_(6)]^(+2)`. How many of the following statement are true for the complex ion ? (i) The complex is octahedral in shape (ii) The complex is diamagnetic in nature . (iii) `Ni^(+2)` has incompletely filled 3d subshell. (iv). Secodary valency of `Ni^(+2)` is 6 (v) All the bonds (metal-ligand ) are perpendicular to each other. (vi) All the 3d orbitals of `Ni^(+2)` are degenerate (vii) Total spin of the complex is 1. (viii) The hybridisation of `Ni^(+2) is d^(2)sp^(3)` (ix) The complex is more stable than `[Ni(en)_(3)]^(+2)` (x) Effective atomic number of `Ni^(+2)` is 36. |
| Answer» Correct Answer - 4(i, iii, iv, vii) | |
| 313. |
Co-ordination number of Cr in `CrCl_(3^.)5H_(2)O` is six. The volume of 0.1 N `AgNO_(3)` needed to ppt. the chorine in outer sphere in 200 ml of 0.01 M solution of he possible complex es is/are: |
| Answer» Correct Answer - (40+20)ml=60ml | |
| 314. |
Total number of geometrical isomers for the complex `[Rh Cl (CO) (PPh_(3)) (NH_(3))]` is |
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Answer» Correct Answer - 3 |
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| 315. |
Which of the following ligands may be flexidentate ?A. EDTAB. `CO_(3)^(-2)`C. `NO_(3)^(-)`D. All of these |
| Answer» Correct Answer - D | |
| 316. |
Ligands are :A. Lewis acidsB. Lewis basesC. neutralD. none |
| Answer» Correct Answer - B | |
| 317. |
Some salts although containing two different metallic elements give test for one of them in solution Such salts are :A. complex saltB. double saltsC. normal saltD. none |
| Answer» Correct Answer - A | |
| 318. |
Coordination number of `Ni` in `[Ni(C_2O_4)_3]^(4-)` is:A. (a) `3`B. (b) `4`C. (c) `5`D. (d) `6` |
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Answer» Correct Answer - D `C_2O_4^(2-)` is a bidentate group. As the complex contains three bidentate groups, the central metal ion has a coordination number of 6. |
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| 319. |
Write the `IUPAC` name for `[Cr(NH_(3))_(5)CO_(3)]CI` . |
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Answer» `[Cr(NH_(3))_(5)CO_(3)]CI` Pentaamminecarbonatochromium(III) chloride . |
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| 320. |
Write the formulae of the following complexes (a) Pentamminechlorocobalt(III) ion (b) Lithiumtetrahydridoaluminate(III) . |
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Answer» Pentaamminechlorocobalt(III) ion `[Co(NH_(3))5CI]^(2+)` (ii) Lithiumtetrahydrogenaluminate(III) `Li[A1H_(4)]` . |
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| 321. |
Write a balanced equation for the reaction of argentite with `KCN` and name the products in the solution . |
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Answer» Argentite is the sulphide ore of silver `Ag_(2)S + 4KCN rarr 2K[Ag(CN)2]+K_(2)S` . |
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| 322. |
The non -existant metal carbonyl among the following is `A. `Cr(CO)_(6)`B. `Mn(CO)_(5)`C. `Ni(CO)_(4) `D. `Fe(CO)_(5)` |
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Answer» Correct Answer - B By applying the EAN rule , the stability of metal carbonyls can be predicted .All the given metal carbonyls except `Mn(CO)_(5)` follows the EAN rule and thus , it exists as dimer `Mn_(2)(CO)_(10+)` |
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| 323. |
Which of the following can participate in linkage isomerism ?A. `NO_(2)^(-) `B. `H_(2) overset(..)NCH_(2)CH_(2) overset(..)N H_(2) `C. ` H_(2)O`D. `: NH_(3)` |
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Answer» Correct Answer - A `NO_(2)^(-)` can participate in linkage isomerism because it may bond to metal through nitrogen or through oxygen . e.g `[Co(NH_(3))_(5) NO_(2)]Cl_(2) ` pentaminenitro cobalt (III ) chloride and ` underset("pentaminenitrito cobalt (III ) chloride ")([Co(NH_(3))ONO]Cl_(2))` |
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| 324. |
The complex `Na_(2)[Fe(CN)_(5)NO]` is called :A. pentacyanonitroso sodium ferrateB. pentacyanonitroso sodium ferrate (II )C. Sodium pentacyanonitrosonium ferrate (II )D. Sodium pentacyanonitroso ferrate |
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Answer» Correct Answer - C The IUPAC name of `Na_(2) [Fe(CN)_(5) NO]` is sodium pentacyanonitrosonium ferrate (II). |
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| 325. |
Which of the following complexes show linkage isomerism ?A. `[Co (NH_(3))_(6)]Cl_(3) `B. `[Co(H_(2)O)_(5)Co]^(3+)`C. `[Cr(NH_(3))_(5)SCN]^(2+)`D. `[Fe(en)_(2)Cl_(2)]^(+)` |
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Answer» Correct Answer - C `[Cr (NH_(3))_(5)SCN]^(2+)` shows linkage isomerism . |
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| 326. |
Which of the following compounds exhibits linkage isomerism?A. `[Co(en)_(3)]Cl_(3)`B. `[Co(NH_(3))_(6)] [Cr(en)_(3)]`C. `[Co(en)_(2)(NO_(2))Cl]Br`D. `[Co(NH_(3))_(5)Cl]Br_(2)` |
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Answer» Correct Answer - C `[Co(en)_(2) (NO_(2))Cl]Br` and `[Co(en)_(2) (ONO) Cl]Br` are linkage isomers . |
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| 327. |
When aqueous solution of potassium fluoride is added to the blue coloured aqueous `CuSO_(4)` solution , a green precipitate is formed . This observation can be explained as follows .A. On adding `KF , H_(2)O` being weak field ligand is replaced by `F^(-)` ions forming `[CuF_(4)]^(2-)` which is green in colour .B. Potassium is coordinated to `[Cu(H_(2)O)_(4)]^(2+)` ion present in `CuSO_(4)` and gives green colour .C. On adding `KF , Cu^(2+)` are replaced by `K^(+)` forming a green complex .D. Blue colour of `CuSO_(4)` and yellow colour of KI form green colour on mixing . |
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Answer» Correct Answer - A Aqueous `CuSO_(4)` solution contains `[Cu(H_(2)O)_(4)]^(2+)` ions which are blue in colour . When aqueous solution of KF is added , `H_(2)O` being weak field ligand can be replaced by `F^(-)` ions forming `[CuF_(4)]^(2-)` which is green in colour . `[Cu(H_(2)O)_(4)]^(2+) underset(( "from" KF)) (+4F^(-)) to underset("Green")([CuF_(4)]^(2-)) + 4 H_(2)O` |
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| 328. |
The denticity of the ligand `N(CH_(2)CH_(2)NH_(2))_(3)` isA. bidentateB. tridentateC. tetradentateD. pentadentate. |
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Answer» Correct Answer - C Number of donor atoms in `N(CH_(2)CH_(2)NH_(2))_(3)` is four hence it is a tetradentate ligand . |
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| 329. |
Find the number of ligand (s) which is//are polydentate ligand `en, dmg dien `EDTA` . |
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Answer» Correct Answer - 2 dien,`EDTA` |
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| 330. |
Calculate the oxidation number of Ni in `K_(2)[Ni(CN)_(4)]` . |
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Answer» Let the charge on Ni be x Charge on `CN^( ᶱ)=-1` `K_(2)[Ni(CN)_(4)]rarr2K^(o+)+[Ni(CN)_(4)]^(2-)` Charge on complex ion , `[Ni (CN)_(4))_(4)]^(2) = -2` `implies x + 4 x (-1) = -2` `implies x = +2` Oxidation number of Ni is II . |
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| 331. |
Calculate the oxidation number of Ni ion `[Ni (H_(2)O_(6))_(6)]^(2+)]` . |
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Answer» Oxidation number is calculated as follows Charge on complex ino` = +2` Let charge on `Ni =x` Charge on `H_(2)O=O` `implies x + 6 xx (0) = + 2` `implies x = 2+` Hence oxidation number of `Ni is II . |
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| 332. |
Why `NH_(3)` form coordinate complex, while the isoelectronic species `CH_(4)` does not . |
| Answer» `NH_(3)` has a lone pair of electron on `N` hence it can act as a ligand and form coordinate complex whereas in `CH_(4)` C` has no lone pair of electrons present on it hence it cannot form complex . | |
| 333. |
Which one of the following complex is not expected to exhibit isomerism?A. `[Ni(NH_(3))_(4)(H_(2)O_(2)]^(2+)`B. `[Pt(NH_(3))_(2)CI_(2)]`C. `[Ni(NH_(3))_(2)CI_(2)]`D. `[Ni(en)_(3)]^(2+)` |
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Answer» Correct Answer - A `[Ni(NH_(3))_(2)C1_(2)` is a tetrahedral complex. Tetrahedral complexes do not show geometrical isomerism because the relative positions of the unidentate ligands bonded to the central metal are the same with respect to each other. `[Ni(NH_(3))_(2)C1_(2)]` is a square planar complex. It exhibits geometrical isomerism. In cis isomer two `H_(2)O` ligands are at `90^(@)` while in trans isomer two `H_(2)O` ligands are at `180^(@)` angle. `[Pt(NH_(3))_(2)C1_(2)]` is a square planar complex. It also exhibits geometrical isomerism as two `NH_(3)` (or two C1) could be at `90^(@)` or at 180: `[Ni(en)_(3)]^(3+)` , an octachedral complex exhibits optical isomerism |
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| 334. |
When 0.1 mol `CoCl_(3)(NH_(3))_(5)` is treated with excess of `AgNO_(3)`, 0.2 mole of AgCl are obtained. The conductivity of solution will correspond toA. `1 : 3` electrolyteB. `1 : 2` electrolyteC. `1 : 1` electrolyteD. `3 : 1` electrolyte |
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Answer» Correct Answer - B 0.2 mole of AgCl are obtained when 0.1 mol `CoCl_(3) (NH_(3))_(5)` is treated with excess of `AgNO_(3)` which show that one molecule of the complex gives two `Cl^(-)`ions in solution . Thus , the formula of the complex is `[Co(NH_(3))_(5) Cl]Cl_(2)` i.e., `1 : 2` electrolyte . |
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| 335. |
Crystal field stabilization energy for high spin `d^4` octahedral complex isA. `-1.8 Delta_(0)`B. `-1.2 Delta_(0)`C. `-1.6 Delta_(0)`D. `-0.8 Delta_(0)` |
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Answer» Correct Answer - D The electronic configuration of `d^(4)` complex is `t_(2g)^(3) e_(g)^(1)`. The CFSE is given by CFSE `= (-0.4 x + 0.6 y) Delta_(0)` where, x is the number of electrons in `t_(2)` and y is the number of electrons in `e_(g)`. Therefore, the CFSE is given by `CFSE = [0.4 xx 3] + [0.6 xx 1] Delta_(0) = [-1.2 + 0.6]Delta_(0) = 0.6Delta_(0)` |
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| 336. |
What is the EAN of central metal in `[Ni(gly)_2]` (At. No. of `Ni=28`)A. (a) `30`B. (b) `34`C. (c) `36`D. (d) `32` |
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Answer» Correct Answer - B `EAN=28-2+8=34`. |
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| 337. |
As per IUPAC nomenclature, the name of the complex `[Co(H_(2)O)_(4) (NH_(3))_(2)] Cl_(3)` isA. tetraaquadiaminecobalt (III) chlorideB. tetraaquadiaminecobalt (III) chlorideC. Diaminetetraaquacobalt III) chlorideD. Diaminetetraaquacobalt III) chloride |
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Answer» Correct Answer - D |
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| 338. |
As per IUPAC nomenclature, the name of the complex `[Co(H_(2)O)_(4) (NH_(3))_(2)] Cl_(3)` isA. Tetraaquadiaminecobalt(III) chlorideB. Tetraaquadiamminecobalt(III) chlorideC. Diaminetetraaquacobalt (III) chlorideD. Diamminetetraaquacobalt (III) chloride |
| Answer» Correct Answer - D | |
| 339. |
Which of the following lighands leads to minimum value of `Delta_(@)` ?A. `NO_(2)^(-)`B. `CN^(-)`C. `F^(-)`D. `OH^(-)` |
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Answer» Correct Answer - C The octahedral crystal field splitting `(Delta_(o))` depends upon the field produced by the lignds and charge on the metal ion. For given metal ion, some ligands are able to produce strong fields in which case the splitting will be larger whereas other produces weak fields and consequently result in small splitting of d orbitals. In general, ligands can be arranged in a series in a series in the order of increasing field strength. A spectrochemical series, is a list of ligands arranged in increasing field strength. A spectrochemical series, is a list of ligands arranged in increasing of their abilities to split the d orbital energies. `I^(-) lt Br^(-) lt C1^(-) lt F^(-) lt OH^(-) lt H_(2)O lt NH_(3) lt en lt NO_(2)^(-) lt CN^(-) lt CO` Note that the order in the spectrochemical series is the same regardless of what metal atom (or ion) is present. `NO_(2)^(-)` , `CN^(-)` and CO are strong called strong field ligands, because they cause a large splitting of the d orbital energy levels. On the other hand, the halide ions and hydroxide ion are week field ligands because they split the d orbitals to a lesser extent. |
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| 340. |
The crystal field theory attributes the colour of cooedination compounds toA. d-d transitionB. charge transfer spectrumC. polarisationD. All of these |
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Answer» Correct Answer - A One of the most distinctive properties of transitio metal complexes is their wide range of colours. Colours of transtition metal ions in aqueous solution correspond to the presence of their aqua complexes. The color of a substance is due to the absorption of light of a specific wavelength in the visible part of the electromagnetic spectrum and transmission or reflection of the rest of the wavelengths. An object that absorbs all visible light appars balck. This implies that some of the visible spectrum is being removed form light as it passess through the sample of transition metal complex, so teh light that emerges is no longer white. The colour of the complex is complementary to that which is absorbed. Teh complementary colour is the colour generated form the wavelength left over, if green light is absorbed by the complex, it appears red. The color in the coordination compounds can be readily explained in terms of teh cryptal field theory. Consider the complex ion, `[Ti(H_(2)O)_(6)]^(3+)` which is violet in colour. This is an octahedral complex ion where the single 3d electron (`Ti^(3+)` is a `3d^(1)` system) in the metal is in the `t_(2g)` level in teh ground state of the complex. The next higher state available for the electron is the empty `e_(g)` level. When light is absorbed by the complex it excites the electron form `t_(2g)` level to the `e_(g)` level: `t_(2g)^(1)e_(g)^(0) rarr t_(2g)^(0)e_(g_^(1)` Consequently, the complex appears violet in colour. Since this excitation involves electron,s jump form one degenerate set of d orbitals to another set of d orbitals, it is commonly called d-d transition. It is important to note that in the absence of ligands, crystal field splitting to note that in the absence of ligands, crystal field splitting does not occur and hence the substance is colourless as there is no d-d transition. For example, removal of `H_(2)O` Ligands form `[Ti(H_(2)O)_(6)]C1_(3)` on heating renders it colourless. Similarly anhydrous `CuSO_(4)` is with, but `CuSO_(4).5H_(2)O` iws blue in colour. |
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| 341. |
Which would exhibit ionisation isomerism?A. (a) `[Cr(NH_3)_6]Cl_3`B. (b) `[Co(NH_3)_5Br]SO_4`C. (c) `[Cr(en)_2Cl_2]`D. (d) `[Cr(en)_3Cl_3]` |
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Answer» Correct Answer - B The compound which has same composition but give different ions in solution, show ionization isomerism. So `[Co(NH_3)_5Br]SO_4` is ionization isomer. `[CO(NH_3)_5Br]SO_4hArr[Co(NH_3)_5Br]^(2+)+SO_4^(2-)` `[Co(NH_3)_5SO_4]Br hArr [Co(NH_3)SO_4]^+ +Br^-` |
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| 342. |
The equation which is balanced and represents the correct product(s) is .A. `[Co(NH_(3))_(5)CI]^(+) + 5H^(+) rarr 5NH_(4)^(+) + CI^(-)`B. `Li_(2)O + 2KCI rarr 2LiCI + K_(2)O`C. `CuSO_(4) + 4KCN rarr K_(2) [Cu(CN)_(4)] + K_(2)SO_(4)`D. |
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Answer» Correct Answer - A Equation (4) is not balanced w.r.t charge: `[Mg(EDTA)]^(2-)` Equation (3) gives `K_(3) [Cu(CN)_(4)]` as the product Equation (2) is unfavourable in the forward direction as `K_(2)O` is unstable while `Li_(2)O` is stable. Only Equalition (4) is correct and balanced. |
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| 343. |
Identify the colourless complexesA. `Ti(NO_(3))_(4)`B. `[Cu(NC CH_(3))_(4)]^(+) BF_(4)^(-)`C. `[Cr(NH_(3))_(6)]Cl_(3)`D. `K_(3)[VF_(6)]` |
| Answer» Correct Answer - A::B | |
| 344. |
Which of the following complex has higher `DeltA_(0)` VALUE?A. `[Fe(H_(2)O)_(6)]^(2+)`B. `[FeCl_(6)]^(3-)`C. `[Fe(H_(2)O)_(6)]^(3+)`D. All have equal |
| Answer» Correct Answer - C | |
| 345. |
The colour of `Cu^(o+)` compounds is .A. WhiteB. BlueC. OrangeD. Yellow |
| Answer» Correct Answer - A | |
| 346. |
Identify the incorrect statements about `[Cu(NH_(3))_(4)]^(2+)`A. The complex is tetrahedralB. The complex is square planarC. `Cu^(2+)` in the complex is `dsp^(2)` hybridisedD. `Cu^(2+)` in the complex is `sp^(2)` hybridised |
| Answer» Correct Answer - A::C | |
| 347. |
In quantitative analysis of second group in laboratory, `H_2S` gas is passed in acidic medium for precipitation. When `Cu^(2+)` and `Cd^(2+)` react with `KCN`, then for product true statement isA. (a) `K_2[Cu(CN)_4]` more solubleB. (b) `K_2[Cd(CN)_4]` less stableC. (c) `K_2[Cd(CN)_3]` more stableD. (d) `K_3[Cu(CN)_2]` less stable |
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Answer» Correct Answer - D In `K_3[Cu(CN)_2]` copper is in `+1` oxidation state while in `K_2[Cd(CN)_2]` cadmium is in `+2` oxidation state. The more the oxidation state of the central metal atom, the more is the stability of the complex. |
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| 348. |
Diethylenetriamine is :A. chelating agentB. tridentate neutral moleculeC. tridentatemonoanionD. (A) and (B) both |
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Answer» Correct Answer - D Diethylenetriamine is a tridentate neutral molecules with three donor nitrogen atoms. Polydentate ligand which uses its two or more donor atoms to bind a single metal ion producing a ring is called as chelating ligand. |
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| 349. |
A square planar complex is formed by hybridisation of which atomic oritals?A. `s,p_(x),p_(y),d_(yz)`B. `s,p_(x),p_(y),d_(x2-y2)`C. `s,p_(y),p_(z),d_(xy)`D. `s,p_(x),p_(y),d_(x2)` |
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Answer» Correct Answer - B `dsp^(2)` hybrisation gives square planar complex with `d_(x2-y2)` orbitals involved in hybridisation forming angles of `90^(@)` . |
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| 350. |
The colour of a complex compound is due to .A. Promotion of 3d-electrons of the central atom//ion to 4p-orbitals.B. Promation of 3d-electrons of the central atom//ion to 4s-orbitalsC. Promation of 3d-electrons of the central atom//ion within d-orbitalsD. Promation of 3d-electrons of the central atom//ion to 4s-orbitals |
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Answer» Correct Answer - C The colour of the complex compound is due to `d-d` electronic transition . |
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