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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.
| 251. |
One mole of complex compound `Co(NH_3)_5Cl_3` gives 3 moles of ions on dissolution in water. One mole of same complex reacts with two moles of `AgNO_3` to yield two moles of `AgCl(s)`. The complex is:A. `[Co(NH_(3))_(3)Cl_(3)].2NH_(3)`B. `[Co(NH_(3))_(4)Cl.NH_(3)]`C. `[Co(NH_(3))_(4)Cl]Cl_(2).NH_(3)`D. `[Co(NH_(3))_(5)Cl]Cl_(2)` |
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Answer» Correct Answer - D The structure of complex is : |
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| 252. |
In the complex `[Fe(H_(2)O)_(6)]^(3+)` `[Fe(CN)_(6)]^(3-)` `[Fe(C_(2)O_(4))_(3)]^(3-)` and `[FeCl_(6)]^(3-)`, that complex that has highest stability isA. `[Fe(H_(2)O)_(6)]^(3+)`B. `[Fe(CN)_(6)]^(3-)`C. `[Fe(C_(2)O_(4))_(3)]^(3-)`D. `[FeCl_(6)]^(3-)` |
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Answer» Correct Answer - C The central metal ion in all complexes is `Fe^(3+)`. `C_(2)O_(4)^(2-)` a negative bidentate ligand will form more stble complex than monodentate neutral or negative ligands. |
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| 253. |
Which of the following has square planar structure ?A. `[Ni(CO)_(4)]`B. `[NiCl_(4)]^(2+)`C. `[Ni(CN)_(4)]^(2+)`D. `All of these. |
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Answer» Correct Answer - C `[Ni(CN)_(4)]^(2+)` with `dsp^(2)`-hybridisation has a square planer structure (see hint to Q. 26 above). |
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| 254. |
Name the metal M which is extracted on the basis of following reactions : `4M + 8CN^(-)+2H_(2)O+O_(2) rarr 4[M(CN)_(2)]^(-)+4OH^(-)` ` 2[M(CN)_(2)]^(-)+Zn rarr [Zn(CN)_(4)]^(2-)+ 2M`A. NickelB. SilverC. CopperD. Mercury |
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Answer» Correct Answer - B Both silver and gold are extracted by this process. |
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| 255. |
The EAN of Ni in `[Ni(CN)_(4)]^(2-)` isA. 34B. 35C. 36D. 28 |
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Answer» Correct Answer - A Oxidation number of Ni in `[Ni(CN)_(4)]^(2-)` is 2. `:.` `EAN = Z - O.N. + 2 xx L` ` = 28 - 2 + 2 xx 4 = 34`. |
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| 256. |
In which of the follwing pair the EAN of central metal atom is not same ?A. `[Fe(CN)_(6)]^(3-) and [Fe(CN)_(6)]^(4-)`B. `[Cr(NH+)3))_(6)]^(3+) and [Cr(CN)_(6)]^(3-)`C. `[FeF_(6)]^(3-) and [Fe(CN)_(6)]^(3-)`D. `[Ni(CO)_(4)] and [Ni(NH_3)_(4)]^` |
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Answer» Correct Answer - A Rerer to A-Level information . |
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| 257. |
Calculate the EAN of the metal in the following complexes and compare it with the atomic number of nearest noble gas.A. `[AuCL_(2)]^(-1)`B. `Fe(C_(2)O_(4))_(3)]^(3-)`C. `[Cr(H_(2)O)_(4)NH_(3)]^(2+)`D. `[Zn(OH)_(4)]^(2-)` |
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Answer» a) `[AuCl_(2)]^(-)` : EAN = `(79-1+2xx2)` = 82,` Xe(Z=86)` b) `[Fe(C_(2)O_(4))_(3)]^(3-)`: EAN = `(26-3 +2xx6)`=35, `Kr(Z=36)` d) `[Cr(H_(2)O)_(4)NH_(3)]^(2+)` : EAN = `(24-2+2 xx 6)` = 34 , `Kr(Z=36)` d) `[Zn(OH)_(4)]^(2-)` : EAN = `(30-2+2xx4)` = 36, Kr(Z=36) `e( [Ni(Py)(en) (NH_(3))]^(2+)` : EAN= `(28 -2+2xx6)`=36, Kr(Z=36) |
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| 258. |
The correct IUPAC name of `[Pt(NH_(3))_(2)Cl_(2)]` isA. Diamminedichloridoplatinum (ii)B. diamminedichloridoplatinum (IV)C. Diamminedichloridioplatinum (0)D. dichloridodiammineplatinum (IV) |
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Answer» Correct Answer - A It is the correct IUPAC name. |
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| 259. |
The EAN of iron in `[Fe(CN)_(6)]^(3-)` isA. 34B. 36C. 37D. 35 |
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Answer» Correct Answer - D In `[Fe(N)_(6)]^(3-)` O.S. of Fe = `+23` EAN= 23 (for`Fe^(3+)`)`+6 xx 2` (for6 `CN^(-)` ions ) = `23 + 12 = 35` |
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| 260. |
Write the IUPAC names of i) `[Cr(NH_(3))_(4)Cl_(2)]^(2-)` ii) `[Pt(NH_(3))_(2)Cl_(2)]Cl_(2)` |
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Answer» i) tetraamminedichloridochromium (III) ion. ii) diamminedichloridoplatinum (IV) chloride. |
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| 261. |
Among the following metal carbonyls, the `C-O` bond order is lowest inA. `|Mn(CO)_(6)|^(+)`B. `|Fe(CO)_(5)|`C. `|Cr(CO)_(6)|`D. `|V(CO)_(6)|^(-)` |
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Answer» Correct Answer - D An anionic carbonyl complex can delocalise more electron density to antibonding `pi`-orbital (`dpi-ppi` back bonding ) of CO and thus lowers the bond order , |
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| 262. |
The effective atomic number of iron in `[Fe(CN)_(6)]^(3-)` isA. 34B. 36C. 37D. 35 |
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Answer» Correct Answer - D O.S. of Fe in `[Fe(CN)_(6)]^(3-) is +3` E.A.N. = `(26 - 3)+6 xx 2` = `23+12 = 35` |
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| 263. |
Effective atomic number of Fe in the complex `K_(4)[Fe(CN)_(6)]` isA. 24B. 12C. 36D. 18 |
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Answer» Correct Answer - C In `K_(4)[Fe(CN)_(6)]` E.A.N. of Fe = `Z-O.N. + 2 xx C.N.` = `26-2+2xx6=36` |
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| 264. |
In which of the following compounds the metal is in the lowest oxidation stateA. `[Co(NH_(3))_(5)Br]_(2)SO_(4)`B. `Fe_(3)[Fe(CN)_(6)]_(2)`C. `Mn_(2)(CO)_(10)`D. `K[PtCl_(3)(C_(2)H_(4))]` |
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Answer» Correct Answer - C In `Mn_(2)(CO)_(10)` oxidation number of Mn is zero |
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| 265. |
Assertion (A) Linkage isomerism arises in coordination compounds containing ambidnetate ligand. Reason (R) Ambidentate ligand has two different donor atoms.A. Assertion and reason both are true, reason is correct explanation of assertion.B. Assertion and reason both are true but reason is not the correct explanation of assertion.C. Assertion is true, reason is falseD. Assertion is false, reason is true. |
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Answer» Correct Answer - a a Reason is the correct explanation for assertion. |
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| 266. |
a) Give the electronic configuration of the d-orbitals of Ti in `[Ti(H_(2)O)_(6)]^(3+)` ion and explain why the complex is coloured? [At. No. Of Ti=22] b) Write IUPAC name of `[Cr(NH_(3))_(3)(H_(2)O)_(3)]Cl_(3)` |
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Answer» a) Oxidation state of Ti : `x+ 6(0) = +3` configuration of `Ti^(3+)` ion = [Ar] `3d^(1)4s^(0)`. Complex is coloured due to the presence of an unpaired electron leading d-d transition. b) Triamminetriaquachromium (III) chloride |
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| 267. |
a) Give the electronic configuration of the d-orbitals of Ti in `[Ti(H_(2)O)_(6)]^(3+)` ion in the otahedral crystal field. b) Why is this complex coloured. Explain on the basis of distribution of electrons in d-orbitals, |
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Answer» a) In the octahedral complex, Ti is n `+3` oxidation state and has `(t_(2g))^1e_(g)^(0)` configuration of `Ti^(3+)` ion. b) The unpaired 3d-electrons is present in the `t_(2g)` level ointhe ground state. The next higher energy level available is `e_(g)`. Thus, there is d-d transition from `t_(2g)` to `e_(g)` `(t_(2_(g))o^1e_(g)^(0) tot_(2g)^0e_(g)^(1))`. This is responsible for the colour of the complex. |
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| 268. |
Which of the following is non-ionizabla ?A. `[Co(NH_(3))_(3)Cl_(3)]`B. `[Co(NH_(3))_(4)Cl_(2)]Cl`C. `[Co(NH_(3))_(5)Cl]Cl_(2)`D. `[Co(NH_(3))_(6)]Cl_(2)` |
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Answer» Correct Answer - A `[Co(NH_(3))_(3)Cl_(3)]` has nothing in ionisation sphere |
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| 269. |
Identity the correct increasing order of crystal field stabilisation energy value for the given complexes .A. `[Ir(nH_(3))_(6)]^(3+) lt [Rh(NH_(3))_(6)]^(3+) lt [Co(NH_(3))_(6)]^(3+)`B. `[Rh(NH_(3))_(6)]^(3+) lt [Co(NH_(3))_(6)]^(3+) lt[Ir(NH_(3))_(6)]^(3+)`C. `[Co(NH_(3))_(6)]^(3+) lt [Ir(NH_(3))_(6)]^(3+) lt [Rh(NH_(3))_(6)]^(3+)`D. `[Co(NH_(3))_(6)]^(3+) lt[Rh(NH_(3))_(6)]^(3+) lt [Ir(NH_(3))_(6)]^(3+)` |
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Answer» Correct Answer - D The order of spitting of the transition metals belonging to different series is : 3d lt 4d lt 5d Therefore , the order of CFSE is : `[Co(NH_3)_6]^(3+)` lt `[Rh(NH_3)_6]^(3+)` lt `[Ir(NH_3)_6)^(3+)`. |
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| 270. |
The pair of coordination complex exhibiting the same kind of isomerism is .A. `[Cr(NH_(3))_(5)Cl]Cl_(2)` and `[Cr(NH_(3))_(4)Cl_(2)]Cl`B. `[Co(NH_(3))_(4)Cl_(2)]^(2+)` and `[Pt(NH_(3))_(2)(H_(2)O)Cl]^(+)`C. `[CoBr_(2)Cl_(2)]^(2-)` and `[PtBr_(2)Cl_(2)]^(2-)`D. `[Pt(NH_(3))_(3)(NO_(3))]Cl` and `[Pt(NH_(3))_(3)Cl]Br` |
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Answer» Correct Answer - b,d (b,d) The complex `[Co(NH_3)_4Cl_2]^(+)` and `[Pt(NH_3)_2(H_(2)O)Cl^(-)` both exhibit geometrical isomerism (b) . The complex `[Pt(NH_3)_3(NO_3)CL] ` and `[Pt(NH_3)Cl]Br` both exhibit ionisation isomerism (d) . |
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| 271. |
The oxidation number of cobalt in `K [Co(CO)_(4)]` isA. `+1`B. `-1`C. `+3`D. `-3` |
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Answer» Correct Answer - C Let O.S. of Co in the complex = x O.S. of K = `+1, O.S. of CN = -1` `1+x + 4(-1)` = 0 x = 3 |
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| 272. |
The oxidation number of cobalt in `K [Co(CO)_(4)]` is |
| Answer» O.N. Of Co : x + 4(0) = - 1 or x = - 1 . Therefore , (c) is the correct answer . | |
| 273. |
What will be the correct order for the wavelengths of absorption in the visibleregion for the following: `[Ni(NO_(2))_(6)]^(4-), [Ni(NH_(3))_(6)]^(2+), [Ni(H_(2)O)_(6)]^(2+)` ? |
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Answer» In all the complexes , the metal ion is the same `(Ni^(2+))`. The increasing field strengths of the ligands present at per electrochemical series are in order : `H_(2)O lt NH_(3) lt NO_(2)^(-)` The energies absorbed for exciation will be in the order : `[Ni(H_(2)O)_(6)]^(2+) lt [Ni(NH_(3))_(6)]^(2+) lt [Ni(NO_(2))_(6)]^(4-)` As E = `hc / lambda , the wavelengths absorbed will be in the opposite order . |
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| 274. |
Amongst the following , the most stable complex is : (a) `[Fe(H_(2)O)_(6)]^(3+)` (b) `[Fe(NH_(3))_(6)]^(3+)` (c) `[Fe(C_(2)O_(4))_(3)]^(3-)` (d) `[FeCl_(6)]^(3-)` . |
| Answer» In all the complexes , Fe is in + 3 oxidation state. However, the complex (c) is a chelate because three `C_(2)O_(4)^(2-)` ions act as the chelating ligands. Thus, the most stable complex is (c) . | |
| 275. |
`[Fe(NO_(2))_(3)Cl_(3)]` and `[Fe(O-NO)_(3)Cl_(3)]` showA. linkage isomerismB. Geometrical isomerismC. Optical isomerismD. None of these . |
| Answer» Correct Answer - A | |
| 276. |
In which of the following complex ion, the central metal ions is in a state of `sp^(3)d^(2)` hybridization ?A. `[CoF_(6)]^(3-)`B. `[Co(NH_(3))_(6)]^(3+)`C. `[Fe(CN)_(6)]^(3-)`D. `[Cr(NH_(3))_(6)]^(3+)` |
| Answer» Correct Answer - A | |
| 277. |
A square planar complex is formed by hybridization of which atomic orbitals?A. `s, px, py, d_(yz)`B. `s, px, py, d_(x)^(2)-y^(2)`C. `s, px, py, d_(z)^(2)`D. `s, px, py, d_(xy)` |
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Answer» Correct Answer - B A square planar complex results from `dsp^(2)` hybridisation involving `(n-1)d_(x)^(2)-y^(2),-ns, npx` and `np_(y)` atomic orbitals. |
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| 278. |
Which complex compound possesses `sp^(3)d^(2)` hybridizationA. `|Fe(CN)_(6)|^(4-)`B. `|FeCl_(6)|^(3-)`C. `Fe(NH_(3))_(6)|^(3+)`D. `|Fe(CN)_(6)|^(3-)` |
| Answer» Correct Answer - C | |
| 279. |
Given the IUPAC names of the following complex compounds. ltbr. i) `[CoBr(NH_(3)_(5)]SO_(4)` `[Fe(NH_(3))_(6)][Cr(CN)_(6)]` iii) `Na_(3)[FeCl(CN)_(5)]` iv) `[Fe(OH)(H_(2)O)_(5)]^(2+)` |
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Answer» i) pentamminebromidocobalt (III) sulphate ii) hexammineirono (III) hexacyanochromate (III) iii) pentaamminesulphatocobalt (III) ion iv) pentaaquahydroxoiron (III) ion. |
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| 280. |
Which one of the following complexes? [Atomic numbers, Mn=25, Fe = 26, Co = 27, Ni = 28]A. `[Fe(CN)_(6)]^(4-)`B. `[Mn(CN)_(6)]^(4-)`C. `[Co(NH_(3))_(6)]^(3+)`D. `[Ni)NH_(3))_(6)]^(2+)` |
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Answer» Correct Answer - D See Comprehensive Review. |
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| 281. |
Which of the following options are correct for `[Fe(CN)_(6)]^(3-)` complex ?A. `d^(2)sp^(3)` hybridisationB. `sp^(3)d^(2)` hybridisationC. ParamagneticD. diamagnetic |
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Answer» Correct Answer - A::C a,c are both correct options. |
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| 282. |
Atomic number of `Mn, Fe, Co` and `Ni` are 25, 26, 27 and 28 respectively. Which of the following outer orbital octahedral complexes have same number of unpaired electrons ?A. `[MnCl_(6)]^(3-)`B. `[FeF_(6)]^(3-)`C. `[CoF_(6)]^(3-)`D. `[Ni(NH_(3))_(6)]^(2+)` |
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Answer» Correct Answer - A::C a,c are the correct answers. |
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| 283. |
`[Fe(H_2O)_6]^(3+)` is strongly paramagnetic whereas `[Fe(CN)_6]^(3-)` is weakly paramagnetic. Explain. |
| Answer» For answer, consult section 10. | |
| 284. |
Consider the follwing complexes ion `P,Q` and `R` `P =[FeF_(6)]^(3-), Q=[V(H_(2)O)_(6)]^(2+)` and `R=[Fe(H_(2)O)_(6)]^(2+)` The correct order of the complex ions, according to their spin only magnetic moment values (inBM) is .A. `R lt Q lt P`B. `Q lt R lt P`C. `R lt P lt Q`D. `Q lt P lt R` |
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Answer» Correct Answer - B P = `|FeF_(6)|^(3-)` has `Fe^(3+)` , i.e., `3d^(5)` (5 unpaired electrons) Q = `|V(H_(2)O)_(6)|^(2+)` has `V^(2+)` , i.e., `3d^(3)` (3unpaired electrons ) R = `|Fe((H_(2)O)_(6)|^(2+)` has `Fe^(2+)` i.e., `3d^(6)` (4 unpaired electrons) `:.` the order of spin only magnetic moment is `Q lt R lt P` . |
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| 285. |
The number of isomers exhibited by `[Cr(NH_(3))_(3)Cl_(3)]` isA. 2B. 3C. 4D. 5 |
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Answer» Correct Answer - A Octahedral complexes of `MA_(3)B_(3)` type exist only in two isomeric forms. |
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| 286. |
Draw the structure of isomers if any and write the names of the following complexes? i) `[Cr(NH_(3))_(4)Cl_(2)]^(+)` ii) `[Co(en)_(3)]^(3+)` |
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Answer» i) tetraamminechloridochromium (III) ion. ii) tris (ethane-1, 2-diamine) cobalt (III) ion. |
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| 287. |
What is the coordination entity formed when excess of aqueous KCN isadded to an aqueous solution of copper sulphate? Why is it that no precipitateof copper sulphide is obtained when `H_(2)S` (g) is passed through this solution? |
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Answer» On mixing the aqueous solutions of KCN and `CuSO_(4)` the complex formed is potassium tetracyanocuprate(II), since `CN^(-)` ions are strong ligands, the complex is quite stable. It is evident from the stability constant value `(K = 2.0 xx 10^(27))`. It is not cleaved by `H_(2)S` gas when passed through the aqueous solution and no precipitate of CuS is formed. `CuSO_(4)(aq) + 4KCN(aq) rarr K_(2)[Cu(CN)_(4)] + K_(2)SO_(4)(aq)` (soluble) |
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| 288. |
how many geometrical isomers are posssible in the following coordination entitles ? (a) `[Cr(C_(2)O_(4))_(3)]^(3-) (b) [Co(NH_(3))_(3)CL_(3)]` |
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Answer» (a) `[Cr(C_(2)O_(4))_(3)]^(3-)` : No geometrical isomerism is possible . (b) `[Co(NH_(3))_(3)CL_(3)]` : Two geometrical isomers : fac and mer For details, consult section 7 . |
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| 289. |
When aqueous KCN is added to a solution of `CuSO_(4)`, a whitie precipitate solube in RCN is formed. But no precipate to formed when `H_(2)S` gas is bubbled through this solution. Explain. |
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Answer» The white precipate formed dissolves in excess of KCN to form a soluble complex. ltrbgt `CuSO_(4) + 2KCN to underset("white ppt")Cu(CN)_(2) + K_(2)SO_(4)` `Cu(CN)_(2) + 2KCN to underset("Soluble complex")(K_(2)[Cu(CN)_(4)])` This complex is quite stable and `H_(2)S` gas is not in a position to break this. That is why, no precipitate is formed. |
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| 290. |
Aqueous copper sulphate solution (blue in colour) gives: (i). A green precipitate with aqueous potassium fluoride and (ii). A bright green solution with aqueous potassium chloride. Explain these experimental results. |
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Answer» Aqueous solution of copper sulphate which is blue in colour exists as `[Cu(H_(2)O))_(4)]SO_(4)` and gives `[Cu(H_(2)O))_(4)]^(2+)` in solution. It is a labile complex entity in which the ligands `H_(2)O` get easily replaced by `F^(-)` ions of KF and by `Cl^(-)` ions of KCl. `[Cu(H_(2)O)_(4)]^(2+)+4F^(-)underset("tetrafluoriodocuprate")([CuF_(4)]^(2-)+4H_(2)O)` `overset(("blue"))([Cu(H_(2)O)_(4)]^(+))+4CI^(-)rarr underset(("bright green solution"))underset("tetrachloridocuprate"(II))([CuCI_(4)]^(2-)+4H_(2)O)` |
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| 291. |
When crystals of `CuSO_(4)` are dissolved in water, there is hardly any evidence for the presence of `Cu^(2+)` ions or ammonia molecules. A new ion `[Cu(NH_(3))_(4)]^(2+)` is furnished in which ammonia molecules are directly linked with the metal ion . Similarly the aqueous solution of `Fe(CN)_(2).4KCN` does not give tests of `Fe^(2+)` and `CN^(-)` ions but gives test of a new ion, `[Fe(CN)_(6)]^(4-)` also called complex ions. Which one of the following statements is correct ?A. `[Cu(NH_(3))_(4)]^(2+)` is paramagnetic while `[Fe(CN)_(6)]^(4-)` is paramagnetic .B. `[Cu(NH_(3))_(4)]^(2+)` is paramagnetic while `[Fe(CN)_(6)]^(4-)` is diamagnetic .C. Both are paramagneticD. Both are diamagnetic |
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Answer» Correct Answer - b b) it is the correct answer. |
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| 292. |
Write equations to represent the following observations: When NaOH is added to aqueous `CuSO_(4)` solutin, a pale blue precipitate is formed. On adding aqeous `NH_(3)` solution, When NaOH is added to give a deep blue solution. If the solution is made acidic with dilute `HNO_(3)`, the colour is converted back to pale blue solution. |
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Answer» `Cu^(2+)(aq) + 2OH^(-)(aq) to Cu(OH)_(2)(g)` `Cu(OH)_(2)(s) + 4NH_(3)(aq) to [Cu(NH_(3))_(4)]^(2+)(aq) + 2OH^(-) (aq)` `[Cu(NH_(3))_(4)]^(2+) (aq) + 4H_(3)O^(+) to [Cu(OH)_(4)]^(2+) + 4NH_(4)^(+) (aq)` |
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| 293. |
When crystals of `CuSO_(4)` are dissolved in water, there is hardly any evidence for the presence of `Cu^(2+)` ions or ammonia molecules. A new ion `[Cu(NH_(3))_(4)]^(2+)` is furnished in which ammonia molecules are directly linked with the metal ion . Similarly the aqueous solution of `Fe(CN)_(2).4KCN` does not give tests of `Fe^(2+)` and `CN^(-)` ions but gives test of a new ion, `[Fe(CN)_(6)]^(4-)` also called complex ions. 15. The hybrid state of `Cu[Cu(NH_(3))_(4)]^(2+)` isA. `sp^(3)`B. `sp^(3)`dC. `sp^(3)d^(2)`D. `dsp^(2)` |
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Answer» Correct Answer - d d) it is the correct answer. |
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| 294. |
Blue solution of `CuSO_(4)` becomes darker when treated with ammonia becauseA. Ammonia molecule replaces water molecule in solutionB. Ammonia is stronger ligand than waterC. Ammonia is highly soluble in waterD. None of the above. |
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Answer» Correct Answer - B `Cu^(2+) + 4NH_(3) rarrunderset(("Deep blue"))([Cu(NH_(3))_(4)]^(2+)` |
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| 295. |
Which of the following has a square planar geometry? .A. `[PtCl_(4)]^(2-)`B. `[CoCl_(4)]^(2-)`C. `[FeCl_(4)]^(2-)`D. `[NiCl_(4)]^(2-)` |
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Answer» Correct Answer - A The complex with Pt in +2 oxidation state and C.N. 4 has a square geomatery |
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| 296. |
Hexafluorocbaltate(III) ion is found to be high spin complex, the probable hybrid state of cobalt in it isA. `d^(2)sp^(3)`B. `sp^(3)`C. `sp^(3)d`D. `sp^(3)d^(2)` |
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Answer» Correct Answer - D `sp^(3)d` |
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| 297. |
Hexafluorocobaltate (III) ion is a high spin complex. The hybrid state of cobalt isA. `d^(2)sp^(3)`B. `sp^(3)d^(2)`C. `dsp^(2)`D. `sp^(3)d` |
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Answer» Correct Answer - B The formula of the ion is `[CoF_(6)]^(3-)` . It is an octahedral complex with high spin . As such the hybrid state should be `sp^(3)d^(2)` . |
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| 298. |
Which of the following hybrid state is associated with low spin complex?A. `sp^(3)`B. `sp^(3)d^(2)`C. `d^(2)sp^(3)`D. `sp^(3)d` |
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Answer» Correct Answer - C Low spin complex refers to the involvement of inne d-orbitals in hybridisation |
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| 299. |
(a) Define a ligand . (b) Discuss the magnetic behaviour , nature and geometry of `(NiCl_(4))^(2-)` ion on the basis of valence bond theory (VBT) |
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Answer» (a) For the definiton of ligand , consult section 3 . (b) Magnetic behaviour : paramagnetic Geometry of the complex : `sp^(3)` hybridised Nature of the complex : tetrahedral For details , consult section 10 . |
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| 300. |
Which of the following compounds is square planar and does not have any unpaired electron ?A. `Ni(CO)_(4)`B. `[Ni(H_(2)O)_(6)]^(2+)`C. `[NiCl_(4)]^(2-)`D. `[Ni(CN)_(4)]^(2-)` |
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Answer» Correct Answer - D `Ni(CO)_(4)` : Tetrahedral. No unpaired electron. `[Ni(H_(2)O)_(6)]^(2+)` : Octahedral . Two unpaired electrons. `[NiCl_(4)]^(2+)` : Tetrahedral. Two unpaired electrons `[Ni(CN)_(4)]^(2-)` : Square planar. No unpaired electrons |
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