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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.
| 951. |
Which of the following is the most suitable description of transition elements?A. Low melting pointsB. No catalytic activityC. Show variable oxidatio statesD. Exhibit inert pair effect |
| Answer» Correct Answer - c | |
| 952. |
Which of the following sentences is not suitable for the capacity of transition metal to form complex compounds ?A. Transition metal ions are small in sizeB. Nuclear charge of transition metal ion 15 comparatively moreC. Co-ordination covalent bond is not directionalD. Transition metal ions possesses different oxidation states |
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Answer» Correct Answer - C Transition metals forms complexes due to small cations, high charge to mass ratio and variable oxidation states |
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| 953. |
` X + H^(+) to Y + Na^(+) + H_2O` `Y+ KCl to K_2Cr_2O_7 + NaCl ` Mention X and Y.A. X=`Na_2Cr_2O_7`, Y=`Na_2CrO_4`B. X=`Na_2Cr_2O_7`, Y=`Na_2Cr_2O_7`C. X=`Na_2CrO_4` , Y=`Na_2Cr_2O_7`D. X=`Na_2Cr_2O_4` , Y=`Na_2CrO_4` |
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Answer» Correct Answer - C `underset"(X)"(2Na_2CrO_4) + underset((H^+))(H_2SO_4) to Na_2Cr_2O_7 + underset((Na^+))(Na_2SO_4) + H_2O` `underset"(Y)"(Na_2Cr_2O_7) + 2KCl to K_2Cr_2O_7 + 2NaCl` |
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| 954. |
`FeCr_2O_4 + Na_2CO_3 + O_2 to ` mention which product is obtained?A. `Na_2CrO_4 + Fe_3O_4 + CO_2`B. `Na_2CrO_4 , Fe_2O_3 + CO_2`C. `Na_2 Cr_2 O_7 + Fe_2O_3 + CO_2`D. `Na_2 Cr_2O_7 + Fe_2O_3 + CO` |
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Answer» Correct Answer - B `4FeCr_2 O_4 + 8Na_2CO_3 + 7O_2 to 2Fe_2O_3 + 8Na_2CrO_4 + 8CO_2` |
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| 955. |
(a). Assertion (A) is true, Reason (R) is also true, Reason (R) is the correct explanation for assertion (A). (b). Assertion (A) is true, Reason (R) is true, Reason (R) is not the correct explanation for Assertion (A). (c). Assertion (A) is true, Reason (R) is false. (d). Assertion (A) is false, Reason (R) is true. Q. Assertion (A): `Zn^(2+)` is diamagnetic. Reason (R): The electrons are lost from 4s-orbital to form `Zn^(2+)`. |
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Answer» Correct Answer - A Both (A) and (R) are correct and reason (R) is the correct explanation for Assertion (A). `Zn^(2+):` the oxidation number of `Zn` is `+2`. `Zn(Z=30)3d^(10)4s^2` `Zn^(2+)=3d^(10)` No unpaired electron is presen hence diamagnetic. |
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| 956. |
(a). Assertion (A) is true, Reason (R) is also true, Reason (R) is the correct explanation for assertion (A). (b). Assertion (A) is true, Reason (R) is true, Reason (R) is not the correct explanation for Assertion (A). (c). Assertion (A) is true, Reason (R) is false. (d). Assertion (A) is false, Reason (R) is true. Q. Assertion (A): to a solution of potassium chromate, if a strong acid is added, it changes its colour from yellow to orange. Reason (R): The colour change is due to the change in oxidation state of potassium chromate. |
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Answer» Correct Answer - C Assertion (A) is true, (R) is false. There is no change in the oxidation state of `Cr` which is `+6` in both cases. `2overset(2+)(Cr)O_4^(2-)+SO_4^(2-)+2H^(o+)tooverset(6+)(C)r_2O_7^(2-)+SO_4^(2-)+SO_4^(2-)+H_2O` The colour of a compound is due to either d-d transition or due to charge transfer transitions. In this case both `CrO_4^(2-)` and `Cr_2O_7^(2-)` are coloured due to charge transfer transition. |
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| 957. |
Atomic radius of Cu is greater than that of Cr but ionic of`Cr^(2+)`is greater than that of `Cu^(2+)`. Give suitable explanation. |
| Answer» In Cu, all the d-electronsarepaired `(3d^(10)4s^(1))`. In Cr , all the d-electrons are unpaired `( 3d^(5) 4s^(1))`. Hence, d-delectron repulsion in Cu are much greater than those in Cr. Therefore ,Cu atom is larger size than Cr. In `Cu^(2+) (3d^(9))`, d-d electron repulsions decrease due to presenceofone unpaired d-electron. Moreover, the electrons are attracted by 29 protons of the nucleus whereas in `Cr^(2+)`, three unpaired electrons are still presenct but they are attractedby only 24 protons of the nucleus . Thus , `Cu^(2+)` is smaller in size than`Cr^(2+)`. | |
| 958. |
Atomic radius of Cu is greater than that of Cr but ionic radius of `Cr^(2+)`. Give suitable explanation. |
| Answer» In Cu, all the d-electrons are paired `(3d^(10)4s^1)` in Cr, all the d-electrons are unpaired `(3d^54s^1)` hence, d-d electron repulsions in Cu are much greater than those in Cr. Hence, Cu atom is larger in size than Cr. In `Cu^(2+)(3d^9)`, d-d electron repulsions decrease due to presence of one unpaired d-electron. moreover, the electrons are attracted. By 29 protons of the nucleus whereas in `Cr^(2+)` three unpaired electrons are still present but they are attracted by only 24 protons of the nucleus. Thus, `Cu^(2+)` is smaller in size than `Cr^(2+)` | |
| 959. |
Fehling solution A consist of an aqueous solution of copper sulphate while Fehling solution B consists of an alkaline solution of ______. |
| Answer» Fehling solution A consists of an aqueous solution of copper sulphate while fehling solution B consists of an alkaline solution of Sodium potassium tartarate. | |
| 960. |
Out of `Fe^(2+)` and `Fe^(3+)` ions, which is more stable ? |
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Answer» `Fe^(3+)` ion is more stable than `Fe^(2+)` ion. This is explained on the basis of the electronic configuration of the two ions `Fe^(3+)` on whit all the five 3d orbitals half filled, is more symmetrical than `Fe^(2+)` on in which four 3d orbitals are half filled and one is filled. Therefore, `Fe^(3+)` on is more stable than `Fe^(2+)` ion. `Fe^(2+):1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(6), Fe^(3+) : 1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(5)` Moreover, `Fe^(3+)` ion has a greater tendency to get hydrated in aqueous solution as compared to `Fe^(2+)` ion. It therefore,expected to be more stable. |
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| 961. |
Calculate equivalent weight of `Cu` in `CuO` and `Cu_(2)O`. At.wt. of `Cu=63.6`. |
| Answer» In `Cu^(+)` ion `(d^(10))` there is no `d-d**` trantion and the ion is colourless. Both `Cu_(2)O` and `Cu_(2)S` are coloured because of charge transfer of electrons from `O_(2-)` or `S^(2-)` to the vacant orbital of `Cu^(+)` ion. | |
| 962. |
Complete the following : `3MnO_(4)^(2-)+4H^(+) to ` |
| Answer» `3MnO_(4)^(2-)+4H^(+) to 2MnO_(4)^(-)+MnO_(2)+2H_(2)O` | |
| 963. |
Why is third ionisation enthalpy of manganese exceptionally high ? |
| Answer» `Mn^(2+)` has the configuration `[Ar]3d^(5)`. With highly symmetrical configuration , the removal of third electron is very difficult. Therefore, third ionisation enthalpy of the metal is exceptionally high ( `A_(i)H_(3)=3200kJ mol^(-1)`). | |
| 964. |
Why are all salts of scandium white ? |
| Answer» In these salts, scandium exists as `Sc^(3+)` ion which is iso-electronic with `[Ar]^(18)`. With completely filled orbitals, the salts of `Sc^(3+)` are white. | |
| 965. |
Traces of `MnO _(4)^(-) ` in conc. `H_(2) SO_(4)` may be changed to `A. `MnO_(4)^(2-)`B. `Mn^(2+)`C. `Mn_(2)O _(7)`D. `MnO_(3)^(-)` |
| Answer» Correct Answer - B | |
| 966. |
The extraction of which of the following metals involves bessemerisation?A. FeB. AgC. AlD. Cu |
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Answer» Correct Answer - D In a bessemer converter, copper pyrites are oxidised to FeO and `Cu_(2)O`. FeO is slagged off. `Cu_(2)O` reacts with `Cu_(2)S` left unoxidised to give Cu. `2Cu_(2)S+3O_(2)rarr2Cu_(2)O+2SO_(2)uarr` `2Cu_(2)O+Cu_(2)Srarr6Cu+SO_(2)uarr` |
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| 967. |
Which of the following is wrong about interstitial compounds ?A. their M.P. is higher than parent metalsB. their density is higher than parent metalsC. their hydrides are powerful reducing agent than parent metals.D. these are neither ionic nor covalent |
| Answer» Correct Answer - B | |
| 968. |
Which one of the following atoms is not involved in the formation of interstitial compounds?A. HydrogenB. CarbonC. NitrogenD. Iodine |
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Answer» Correct Answer - D Only small non-metal atoms can form interstitial compounds, Iodine has a larger size and is unable to occupy the interstitials in the small lattices . |
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| 969. |
Interstitial compounds are formed byA. `Fe, Co`B. `Co, Ni`C. `Fe, Ni `D. all of thes |
| Answer» Correct Answer - D | |
| 970. |
Formation of interstitial compounds makes the transition metalA. more softB. more ductileC. more metallicD. more hard and brittle |
| Answer» Correct Answer - d | |
| 971. |
How would you account for the following ? (i) Many of the transition elements are known to form interstitial compound ? (ii) The metallic radii of third (5d) series of transition metals are virtually the same as those of the corresponding group member of second (4d) series. (iii) Lanthanoids form primarily +3 ions while the actinoids usually have higher oxidation state in their compounds, +4 or even +6 being typical. |
| Answer» (iii) In the elements belonging to Actionids series 5f, 6d and 7s orbitals have comparable enegies and the electrons present in these orbitals can take part in the bond formation to show higher oxidation states. | |
| 972. |
The atomic radii of transition elements in a period are.A. smaller than those of s-block as well as p-block elementsB. greater than those of s-block as well as p-block elements.C. smaloler than toose of s-block but greater than those of p-block elements.D. greater than thiose of s-block but smaller than those of p-block |
| Answer» Correct Answer - C | |
| 973. |
If a compund absorbs orange colour from the white light, then the observed colour of the compound isA. yellowB. orangeC. blueD. violet |
| Answer» Correct Answer - C | |
| 974. |
Which one of the following is transition element ?A. MgB. CaC. NiD. Na |
| Answer» Correct Answer - C | |
| 975. |
Paramagnetism is exhibited by molecules whichA. not attracted in a magnetic fieldB. carrying a positive chargeC. containing unpaired electronsD. containing only paired electrons |
| Answer» Correct Answer - C | |
| 976. |
Which is a stronger reducing agent `Cr^(2+) or Fe^(2+)` and why ? |
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Answer» `Cr^(2+)` is a stronger reducing agent than `Fe^(2+)`. The `E^(@)` values are `E_(Cr^(3+)//Cr^(2+))^(@)=-0.41 V and E_((Fe^(3+)//Fe^(2+))^(0)=0.77 V` Transition `Cr^(2+) rarr Cr^(3+)` is easier than `(3d^(4))" "(3d^(3))` `Fe^(2+) rarr Fe^(3+)` because in latter, an electron is removed from the paired orbital (difficult) `(3d^(6))" "(3d^(5))` Therefore, `Cr^(2+)` is stronger reducing agent (itself gets oxidised easily) than `Fe^(2+)` |
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| 977. |
Which of the following is the softest metals-A. ScB. ZnC. TiD. V |
| Answer» The softest metal is `Zn,` while remaining all other metal are comparatively harder metals. | |
| 978. |
In chromite ore, the oxidation number of iron and chromium are respectively.A. `+ 3, + 2`B. `+3, +6`C. `+2, + 6`D. `+2, +3` |
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Answer» Chromite ore is `FeCr_(2)O_(4)` Ox. State of `Fe = + 2 , Cr=+ 3` |
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| 979. |
The electronic configuration of Eu (Atomic No. 63), Gd (Atomic No. 64) and Tb (Atomic No. 65) are:A. `"["Xe"]"4f^(6)5d^(1) 6s^(1), "["Xe"]"4f^(7)5d^(1) 6s^(2) and "["Xe"]"4f^(8)5d^(1)6s^(2)`B. `"["Xe"]"4f^(7)5d^(1) 6s^(2), "["Xe"]"4f^(7)5d^(1) 6s^(2) and "["Xe"]"4f^(9)6s^(2)`C. `"["Xe"]"4f^(7)5d^(1) 6s^(2), "["Xe"]"4f^(8) 6s^(2) and "["Xe"]"4f^(8)5d^(1)6s^(2)`D. `"["Xe"]"4f^(6)5d^(1) 6s^(2), "["Xe"]"4f^(7)5d^(1) 6s^(2) and "["Xe"]"4f^(9)6s^(2)` |
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Answer» Correct Answer - B B It is the correct answer |
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| 980. |
Why hydrogen peroxide is added to acidified potassium dichromate, a blue colour is produced due to formation ofA. `CrO_(3)`B. `Cr_(2)O_(3)`C. `CrO_(5)`D. `CrO_(4)^(2-)` |
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Answer» Blue coloure is due to the formation of `CrO_(5)` as follows `:` `Cr_(2)O_(7)^(2-) + 2H^(+) + 4H_(2)O_(2) rarr 2CrO_(5) + 5H_(2)O` |
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| 981. |
Which one of the following statements is correct when `SO_(2)` is passed through acidified `K_(2)Cr_(2)O_(7)` solution?A. `SO_(2)` is reducedB. Green `Cr_(2)(SO_(4))_(3)` is formedC. The solution turns blueD. The solution is decolourlesed. |
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Answer» Correct Answer - B `{:(K_(2)Cr_(2)O_(7),+,H_(2)SO_(4),+,3SO_(2)),(" "darr,,,,),(K_(2)SO_(4),+,Cr_(2)(SO_(4))_(3),+,H_(2)):}` |
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| 982. |
Which of the following statements is correct when `SO_(2)` is passed through acidified `K_(2)Cr_(2)O_(7)` solution ?A. Green `Cr_(2)(SO_(4))` is formed.B. The solution turns blue.C. The solution is decolourisedD. `SO_(2)` is reduced. |
| Answer» `K_(2)Cr_(2)O_(7) + H_(2)SO_(4) + 3SO_(2) rarr K_(2)SO_(4) + underset("(Green)")(Cr_(2)(SO_(4))_(3))+ H_(2)O` | |
| 983. |
When sulphur dioxide is passed through an acidified `K_(2)Cr_(2)O_(7)` solution, the oxidation state of sulphur is changed fromA. `+4` to `+6`B. `+6` to `+4`C. `+4` to 0D. `+4` to `+2` |
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Answer» `ul(K_(2) Cr_(2)O_(7) +4H(2) SO_(4)rarr K_(2)SO_(4) + Cr_(2)(SO_(4))_(3) + 7H_(2)O+ 3(O))` `SO_(2) + (O) + H_(2) O rarr H_(2) SO_(4) ] xx3` `K_(2)Cr_(2)O_(7) + H_(2)SO_(4) + 3SO_(2) rarr K_(2)SO_(4) + Cr_(2)(SO_(4))_(3) + H_(2)O` In `SO_(2)` , ox. state of `S= + 4 `. In `SO_(4)^(2-)`,ox.state of `S= +6` |
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| 984. |
When `SO_(2)` is passed through acidified `K_(2)Cr_(2)O_(7)` solutionA. If both assertion and reason are correct and reason is correct explanation for assertion.B. If both assertion and reason are correct but reason is not correct not correct explanation for assertion.C. If assertion is correct but reason is incorrect.D. If both assertion and reason are incorrect. |
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Answer» Correct Answer - C (c) Correct reason. Sulphur dioxide acts as a reducing agent in the reaction. |
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| 985. |
Explain why mercury (I) ion exists as `Hg_(2)^(2+)` ions while copper (I)ion exists as `Cu^(+)`ion. |
| Answer» The electronic configuration of Hg(I), i.e., `Hg^(+) ` is `[Xe] 4f^(14)5d^(10) 6s^(1)` and thus has one electron in the valence 6s-orbital. If this were so, all Hg(I) compounds should be paramagnetic butactually they are diamagnetic. This behaviour can be explained if we assume that the singly filled 6s-orbitals of the two `Hg^(+)` ions overlap to form a Hg-Hg covalent bond. Thus, `Hg^(+)` ions exist as dimeric species, i.e., `Hg_(2)^(2+)` . In contrast, the electronic configuration of Cu(I) ion, i.e., `Cu^(+)` is `[Ar] 3d^(10)` . Therefore, it has no unpaired electrons to form dimeric species, i.e., `Cu_(2)^(2+)` and hence it always exists as `Cu^(+)` ion. | |
| 986. |
In which of the following there is a change in oxidation number ? (a) An aqueous solution of `CrO_(4)^(2-)` is acidified. (b) `SO_(2)` gas is passed through acidified `Cr_(2)O_(7)^(2-)` solution. (c) `Cr_(2)O_(7)^(2-)` solution is made alkaline. (d)` CrO_(2)Cl_(2)` is dissolved in NaOH. |
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Answer» (a) `overset(+6)(2[CrO_(4)]^(2-))+2H^(+) to overset(+6)([Cr_(2)O_(7)]^(2-))+H_(2)O` : no change in O.N (b) `overset(+6)([Cr_(2)O_(7)]^(2-))+overset(+4)(3SO_(2))+2H^(+) to overset(+3)(2[Cr]^(3+))+overset(+6)(3[SO_(4)]^(2-))+H_(2)O` :Change in O.N (c) `overset(+6)([Cr_(2)O_(7)]^(2-))+2OH^(-) tooverset(+6)(2[CrO_(4)]^(2-))+H_(2)O` :No Change in O.N. (d) `overset(+6)(CrO_(2))Cl_(2)+2NaOH to Na_(2) overset(+6)(CrO_(4))` :No change in O.N |
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| 987. |
Photography is based on the nature of silver halides. Except AgF, the silver halides are photosensitive. These undergo decomposition in light and turn black due to formation of free silver. `2AgBroverset(light)to2Ag+Br_2` The photography films are prepared by adding `20%` aqueous solution of `AgNO_3` to `NH_4Br` solution containing gelatic. When such a film is exposed, emulsion gets affected and a latent image is formed on the film. When this exposed film or plate is dipped ina developer which contains a reducing agent, the part affected most during exposure are reduced to the maximum. The image becomes visible. It is called a negative. The remaining sensitive emulsion on the negative is removed by dissolving it in hypo solution (fixer). Fihnally, a positive of the nagative already prepared is made on silver bromide paper. Q. Silver bromide dissolves in hypo solution forming:A. `Ag_2S_2O_3`B. `Ag_2S`C. `Na_3[Ag(SO_3)_2)]`D. `NaAgS_2O_3` |
| Answer» Correct Answer - c | |
| 988. |
Assertion : Cuprous salts are diamagnetic in nature. Reason : `Cu^(+)` ion has filled 3d-orbitals.A. If both assertion and reason are correct and reason is correct explanation for assertion.B. If both assertion and reason are correct but reason is not correct not correct explanation for assertion.C. If assertion is correct but reason is incorrect.D. If both assertion and reason are incorrect. |
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Answer» Correct Answer - A (a) Reason is the correct explanation for assertion . |
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| 989. |
Cuprous ion is colourless, while cupric ion is colured becauseA. both have half filled p and d-orbitalsB. coprous ion has incomplete d-orbital and cupric ion has complete d-orbitalsC. both have unpaired electrons in d-orbitalsD. cuprous ion has a complete d-orbital and cupric ion has an incomplete d-orbitals |
| Answer» Correct Answer - D | |
| 990. |
Cupric compounds are more stable than their cuprous counterparts in solid state. This is becauseA. the endothermic character of the 2nd I P of Cu is not so highB. size of `Cu^(2+)` is less than `Cu^(+)`C. `Cu^(2+)` has stabler electronic configuration as compared to `Cu^(+)`D. the lattice energy released for cupric compounds is much higher than `Cu^(+)`. |
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Answer» Correct Answer - B::D (b,d) These are both correct options. |
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| 991. |
Mercurous ion is written as `Hg_(2)^(2+)` whereas cuprous ion is written as `Cu^(+)` . Explain. |
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Answer» `""_(80)Hg:[Xe]4f^(14)5d^(10)6s^(2),Hg^(+)[Xe]4f^(14)5d^(10)6s^(1)` In `Hg^(+)` ion, there is one unpaired electron and `Hg^(+)` (mercurous) salts are expected to be paramagnetic. But the magnetic moment of mercurous salts is zero which indicates that they are diamagnetic in nature. This is possible in case 6s electron has neenn used by two Hg atoms in bonding. Therefore, mercurous ion is `Hg^(2)^(2+)`. `""_(29)Cu:[Ar]3d^(10)4s^(1),Cu^(+):[Ar]3d^(10)`. In `Cu^(+)` ion, there is no unpaired electrons and `Cu^(+)` (cuprous) salts are expected to be diamagnetic. These are actually so. Therefore, cuprous ion is written as `Cu^(+)`. |
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| 992. |
Which of the following cuprous compounds is not stable-A. `CuCl_(2)`B. `Cu_(2)(CNS)_(2)`C. `Cu_(2)Cl_(2)`D. `Cu_(2)SO_(4)` |
| Answer» The compound `Cu_(2)SO_(4)` is not stable because `Cu` is stabilized only in `Cu^(+2)` state in its compounds with `SO_(4)^(-2)` ions hence `Cu_(2)SO_(4)` is not stable while `CuSO_(4)` is stable. | |
| 993. |
A developer used in photography is-A. A weak acidB. A weak baseC. A mild reducing agentD. Anoxidizing agent |
| Answer» A developer is a weak reducing agent, `e.g.`, Ferrous oxalate, the parts affected by light on photographic plate are reduced to the maximum extent whereas part not affected by light remains unaffected. | |
| 994. |
Give one example each for ionic and neutral ligands. |
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Answer» `rarr` Examples for ionic ligands - `CN^(-), I^(-), Cl^(-)` ` rarr` Examples for Neutral ligands - `NH_(3),H_(2)O` |
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| 995. |
Explain the terms Ligand |
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Answer» Ligand : The ions or molecules bound to the central atom/ion in the coordination entity are called ligands. These may be a) simple ions such as `Cl^(-)` b) small molecules such as `H_(2)O" or "NH_(3)` c) large molecules such as `H_(2)NCH_(2)CH_(2)NH_(2)" or "N(CH_(2)CH_(2)NH_(2))_(3)` or even (d) macro-molecules, such as proteins. On the basis of the number of donor atoms available for coordination, the ligands can be classified as : a) Unidentate : One donor atom, `Eg. : overset(..)NH_(3), :overset(..)underset(..)Cl:` etc. b) Bidentate : Two donor atoms, `Eg. : H_(2)NCH_(2)CH_(2)NH_(2)` (ethane-1,2-diamine or en), `C_(2)O_(4)^(2-)` (oxalate), etc. c) Polydentate : More than two donor atoms Eg. `:N(CH_(2)CH_(2)NH_(2))_(3))` EDTA. etc. |
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| 996. |
How many moles of AgCl is precipitated when I mole of `CoCl_(3)` is treated with `AgNO_(3)` solution? |
| Answer» 3 moles of AgCl is precipitated by the reaction of 1 mole of `CoCl_(3)` with `AgNO_(3)` solution 3 `AgNO_(3) +CoCl_(3) rarr Co(NO_(3))_(3)+3 AgCl darr` | |
| 997. |
What is an ambidentatc ligand ? Give example. |
| Answer» A unidentate ligand containing two possible donor atoms can co-ordinate through either of donor atoms. Such ligands are called ambidentate ligands. `Eg : NO_(2)^(-)` | |
| 998. |
Explain the terms polydentate ligand |
| Answer» Polydentate ligands : Ligands having more than one donor atom in the co-ordinating group and are capable of forming two or more coordinate bonds with same central atom simultaneously are called poly dentate ligands. Eg : `C_(2)O_(4)^(-2)`. | |
| 999. |
What is a chelate ligand ? Give example. |
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Answer» The ligands which can form two co-ordinate covalent bonds through two donar atoms are called bidentate ligands. These bidentate ligands are also called chelate ligands. `Eg. : C_(2)O_(4)^(-2), CO_(3)^(-2)` etc. |
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| 1000. |
Out of `TiF_(6)^(2-),CoF_(6)^(3-) , Cu_(2)Cl_(2)` and `NiCl_(4)^(2-)`(Z ofTi `=22, Co=27 , Cu =29 , Ni = 28) ` , the colourless species areA. `Cu_(2)Cl_(2)` and `NiCl_(4)^(2-)`B. `TiF_(6)^(2-) ` and `Cu_(2)Cl_(2)`C. `CoF_(6)^(3-)` and `NiCl_(4)^(2-)`D. `TiF_(6)^(2-) ` and `CoF_(6)^(3-)` |
| Answer» `Cu_(2)Cl_(2)(Cu^(+)= 3d^(10)),TiF_(6)^(2-)(Ti^(4+)= 3d^(0))` | |