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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.
| 101. |
Identify the product and the colour when `MnO_(2)` is fused with solid KOH in the presence of oxygen `(O_(2))`A. `KMnO_(4)`(purple)B. `K_(2)MnO_(4)`(dark green)C. `MnO_(2)` (Colourless)D. `Mn_(2)O_(3)` (brown) |
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Answer» Correct Answer - B `2MnO_(2)to4KOH+O_(2) to underset("(Green)")(2K_(2)MnO_(4))+2H_(2)O` |
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| 102. |
When pyrolusite is fused with KOH, the colour of the product isA. redB. pinkC. blackD. green |
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Answer» Correct Answer - D `MnO_(2)+2KOH+[O]rarrK_(2)MnO_(4)+H_(2)O` `K_(2)MnO_(4)` produced is green in colour. |
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| 103. |
Identify A to E. Pyrolusite on heating with KOH in the presence of air gives a dark green compound (A). The solution of (A) on treatment with `H_2SO_4` gives a purple coloured compound `(B)`, which gives the following reactions: (a). KI on reaction with alkaline solution of (B) changes into a compound (C). (b). The colour of the compoud (B) disappears on treatment with the acidic solution of `FeSO_4`. (c). With conc. `H_2SO_4` compound (B) gives (D) which can compose to yield (E) and oxygen. |
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Answer» `2MnO_2+4KOH+underset((Air))(O_2)tounderset((A)Dark green)(2K_2MnO_4)+2H_2O` `3K_2MnO_4+2H_2SO_4tounderset((B)purpl e)(2KMnO_4)+MnO_2+2K_2SO_4+2H_2O` (a). `2KMnO_4+H_2O+KItoKIO_3(C)+2MnO_2+2KOH` (b). `2KMnO_4+8H_2SO_4+10FeSO_4toK_2SO_42MnSO_4+5Fe_2(SO_4)_3+8H_2O` (d). `2KMnO_4+H_2SO_4toMn_2O_7(D)+K_2SO_4+H_2O` `2MnO_2O_7to4MnO_2(E)+3O_2` |
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| 104. |
Why are Zn, Cd and Hg quite soft and have low melting points ? |
| Answer» These metal atoms have completely filled d-orbitals ( `d^(10)` configuration ). This means that d-electrons are not readily available for metallic bond formation. Quite abviously, the metallic bonds are weak and as a result, these metals are quite soft and also have low melting points. | |
| 105. |
The electronic configuration of actinides Cannot be assigned with degree of certainty because ofA. Samll energy difference between 5f and 6d levelsB. overlapping of inner orbitalsC. Free movement of electrons over all the orbitalsD. None of above |
| Answer» Correct Answer - a | |
| 106. |
(a). Why second asnd third transition series elements show similar size? (b). Why electronic configuration of lanthanoids not known with certainty? (c). The electronic configuration of actinide elements are not known with certainty. Explain. (d). Why there is similarties (horizontal and vertical) in successive members of the transition series? |
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Answer» (a). In the third transition series after lanthanum, there is lanthanoid contraction. Due to this constraction, the size of any atom of the third transition series is almost the same as that of the elements lying just above in the second transition series. This leads to similarity in their properties. (b). In the lanthanoids 4f and 5d subshells are very close in energy. The outermost 6s orbital remains filled with 2 electrons `(6s^2)`. The electron can easily jump from 4f to 5d or vice versa. Hence, their electronic configurations are not known with certainly. (c). In actinides, 5f and 6d subshells are close in energy. Hence, they show a large number of oxidation states. The electrons can jump from 5f to 6d or vice versa. Hence, their electronic configuration are not known with certainty. (d). this is vecause of the fact that along a horizontal row, electrons enter an incomplete inner shell in building up an atom, while the outer level remains almost unchanged. In vertical columns, similarities are due to similar electronic configuration even in the d-subshells. |
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| 107. |
The electronic configuration of actinides Cannot be assigned with degree of certainty because ofA. overlapping of inner orbirtalsB. small energy difference between 5f and 6d orbitalsC. free movement of electrons over all the orbitalsD. all of these |
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Answer» Correct Answer - B We know that , 5f and 6d orbitals has almost equal energy. Three are some uncertainties regarding the filling of 5f and 6d orbitals. The electron may enter either 5f or 6d lebitals. Hence electronic configuration of actinides known with uncertainties. |
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| 108. |
The electronic configuration of gadolinium ( At. No. 64) isA. `[Xe] 4 f^(8)5d^(0)6s^(2)`B. `[Xe]4f^(7) 5d^(1) 6s^(2)`C. `[Xe]4f^(3) 5d^(5) 6s^(2)`D. `[Xe]4f^(6)5d^(2) 6s^(2)` |
| Answer» Correct Answer - b | |
| 109. |
The electronic configuration of gadolinium (At. No 64) is:A. `[Xe] 4f^8 5d^1 6s^2`B. `[Xe] 4f^7 5d^1 6s^2`C. `[Xe] 4f^3 5d^5 6s^2`D. `[Xe] 4f^6 5d^2 6s^2` |
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Answer» Correct Answer - B `Gd(64) to [Xe] 6s^2 5d^1 4f^7` |
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| 110. |
The `+3` oxidation states of langthanum `( Z= 57)`, gadolinium `(Z= 64)` and lutetium `(Z= 71)` are especially stable. Why ? |
| Answer» This is because in the `+3` oxidation state, they have empty, half-filled and completely filled 4f sub-shell respectively. | |
| 111. |
Explain by giving suitable reason. (a). Yellow coloured aqueous solution of sodium chromate changes to orange-red when `CO_2` under pressure is passed. (b). Green solution of potassium manganate, `K_2MnO_4`, turns purple when `CO_2` is circulated. (c). `Hg^(2+)` and `Hg_2^(2+)` salts are colourless. (d). `Cu^(2+)` salts are paramagnetic while `Cu^(o+)` salts are diamagnetic in nature. |
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Answer» (a). `CO_2` gives `H^(o+)` ions which are responsible for the conversion of chromate into dichromate (orange- red). `CO_2+H_2Ounderset(larr)rarrHCO_3^(ɵ)+H^(o+)` `2CrO_4^(2-)+2H^(o+)toCr_2O_7^(2-)+H_2O` (b). `CO_2` gives `H^(o+)` ions which are responsible for conversion of `MnO_4^(2-)` into `MnO_4^(ɵ)` purple `CO_2+H_2Ounderset(larr)rarrH^(o+)+HCO_3^(ɵ)` `3MnO_4^(2-)+4H^(o+)tounderset(purpl e)(2MnO_4^(ɵ)+MnO_2+2H_2O` (c). `Hg^(2+)` and `Hg_2^(2+)` salts have `5d^(10)` configuration i.e., there is no d-d transition hence colourless. (d). `Cu^(2+)` configuration is `3d^(9)` (one orbital is singly occupied) - Paramagnetic, Cu configuration is `3d^(10)` (all orbitals are doubly occupied)- Diamagnetic. |
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| 112. |
When `H_(2)O_(2)` is shaken with an acidified solution of `K_(2)Cr_(2)O_(7)` in the presence of ether , the ethereal layer turns blue due to the formation ofA. `Cr_(2)O_(3)`B. `CrO_(4)^(-)`C. `Cr_(2)(SO_(3))_(3)`D. `CrO_(5)` |
| Answer» Correct Answer - D | |
| 113. |
`H_(2)S` gas is passed through an acidic solution of `K_(2)Cr_(2)O_(7)` . The solution turns milky, why? |
| Answer» This is due to oxidation of `H_(2)S` to sulphur which is of collodial nature. Therefore, the solution appears to be milky white or pale yellow. | |
| 114. |
To an acidified dichromate solution, a pinch of `Na_(2)O_(2)` is added and shaken. What is observed ?A. blue colourB. Orange colour changing to greenC. Copious evolution of oxygenD. Bluish-green precipitate |
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Answer» Correct Answer - A::C `K_(2)Cr_(2)O_(7)+H_(2) SO_(4)+Na_(2)O_(2)tounderset(("blue colour"))(CrO_(5)+K_(2)SO_(4))+Na_(2)SO_(4)+O_(2)` |
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| 115. |
The colour of the transition metal ions is due toA. d-d transitionB. Charger transferC. Change in the geometryD. None |
| Answer» Correct Answer - a,b | |
| 116. |
The colour of the transition metal ions is/are due to:A. d - s transitionB. d - d transitionC. f - f transitionD. d - f transition |
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Answer» Correct Answer - B Five degenerate d-orbitals splits up into set of two orbitals `t_(2g)` and eg. Transition of `e^-` between these d-orbitals result in d-d transition imparting colour. |
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| 117. |
In which of the following ions , the colour is not due to d-d transition ?A. `Ti(H_(2)O)_(6)^(3+)`B. `CoF_(6)^(3+)`C. `CrO_(4)^(2-)`D. `[Cu(NH_(3))_(4)]^(2+)` |
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Answer» Correct Answer - C In `CrO_(4)^(2-)`, Cr is in `+6` oxidation state.It has `d^(0)` configuration .The coloure is dueto charge transfer and not due to d-d transition |
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| 118. |
The colour of transition metal ion is attributed to:A. small-size of metla ionsB. absorption of light in UV regionC. complete ns sub-shellD. incomplete (n-1) d orbital |
| Answer» Correct Answer - D | |
| 119. |
The colour of the transition metal ions is due toA. d-d transitionB. charge transferC. change in the geometryD. none |
| Answer» In case of `KMnO_(4)` and `K_(2)Cr_(2)O_(7)` , the colour are due to charge transfer. In general , colour is due to d-d transition. | |
| 120. |
Which of the following show oxidation state of `+4` ?A. CeB. AcC. ThD. U |
| Answer» For (b) and for (c ) , `KMnO_(4)` in presence of KOH forms `K_(2)MnO_(4)` and oxygen produced oxidizes HCHO to HCOOH and then to `CO_(2)` and `H_(2)O`. | |
| 121. |
Which of the following statements are correct with reference to the ferrous and ferric ions ?A. `Fe^(3+)` gives brown colour with potassium ferricyanideB. `Fe^(2+)` gives blue precipitate with potassium ferricyanideC. `Fe^(3+)` gives red colour with potassium thiocyanateD. `Fe^(2+)`gives brown colour with ammonium thiocyanate. |
| Answer» `K_(3)[Fe(CN)_(6)]` reacts with `Fe^(2+)` ions to give blue ppt.of`Fe_(4) [Fe(CN)_(6)]_(3)` and potassium thiocyanate reacts with `Fe^(3+)` ions to give a red colour of `Fe(CNS)_(3)` | |
| 122. |
Which of the following statements (s) is (are) correct with reference to the ferrous and ferric ions? (a). `Fe^(3+)` gives brown colour with potassium ferricyanide. (b). `Fe^(2+)` gives blue precipitate with potassium ferricyanide. (c). `Fe^(2+)` gives red colour with potassium thiocyanate. (d). `Fe^(2+)` gives brown colour with ammonium thiocyanate. |
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Answer» b,c `Fe^(2+)+K_3[Fe(CN)_6]toFe_2^(II)[Fe^(II)(CN)_6]darr` `Fe^(3+)+K(SCN)toFe(SCN)_3darr` (blood test) |
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| 123. |
`[Fe(CN)_(4)]^(2-) and [Fe(H_(2)O)_(6)]^(2+)` are of different colours in dilute solutions. Why ? |
| Answer» `rarr` In the given complexes Fe has + 2 oxidation state with `3d^(6)` outer electronic configuration. It has four unpaired electrons in presence of weak ligand `H_(2)O`. But in presence of strong ligand `CN^(-)` the electrons are paired up. Due to the difference in the no of unpaired electrons both complex have different colours in dilute solutions. | |
| 124. |
Reduction of the metal centre in aqueous permanganate ion involvesA. 3 electrons in neutral mediumB. 5 electrons in neutral mediumC. 3 electrons in alkaline mediumD. 5 electrons in acidic medium. |
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Answer» In neutral medium `MnO_(4)^(-) + 2H_(2)O + 3e^(-) rarrMnO_(2) + 4OH^(-)` In alkaline medium `MnO_(4)^(-) + e^(-) rarr MnO_(4)^(2-)` But `MnO_(4)^(2-)` is further reduced to `MnO_(2)` as shown below `MnO_(4)^(2-) + 2H_(2)O + 2e^(-) rarr MnO_(2) + 4OH^(-)` Complete reaction is `MnO_(4)^(2-) + 2H_(2)O+ 3e^(-) rarr MnO_(2) + 4OH^(-)` which is the same as that for neutral medium. In acidic medium `MnO_(4)^(-) + 8 H^(+) + 5e^(-) rarr Mn^(2+) + 4H_(2)O` Hence, the number of electrons involved in the reduction of metal centre in aqueous `MnO_(4)^(-)` ion in neutral, alkaline and acidic media are 3,3 and 5 respectively. |
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| 125. |
Potassium manganate `(K_2MnO_4)` is formed whenA. Chlorine is passed into aqueous `KMnO_(4)` is formed whenB. manganese dioxide is fused with potassium hydroxide in airC. formaldehyde reacts with potassium permanganate in presence of a strong alkaliD. potassium permanganate reacts with conc. Sulphuric acid. |
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Answer» Correct Answer - B::C (b,c) `2KOH+MnO_(2)+OrarrK_(2)MnO_(4)+H_(2)O` `HCHO+2KMnO_(4)+2KOHrarr2K_(2)MnO_(4)+H_(2)O+O_(2)` |
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| 126. |
Potassium manganate `(K_2MnO_4)` is formed whenA. Chlorine is passed through aqueous `KMnO_4` solutionB. Manganese dioxide is fused with potassium hydroxide in air.C. Formaldehyde reacts with potassium permanganate in the presence of a strong alkali.D. Potassium permanganate reacts with concentrated sulphuric acid. |
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Answer» Correct Answer - B::C Potassium maganate `(K_2MnO_4)` is formed when mangnese dioxide is fused with potassium hydroxide in air of formaldehyde reacts with potassium permanganate in the presence of a strong alkali. `4KOH+2MnO_2+O_2to2K_2MnO_4+2H_2O` `HCHO+2KMnO_4+2KOHtoK_2MnO_4+H_2O+HCOOH` |
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| 127. |
In the preparation of `KMnO_4`, Pyrolusite `(MnO_2)` is first converted to potassium manganate `(K_2MnO_4)` . In this conversion the oxidation state of manganese changes fromA. `+1` to +3B. `+2` to +4C. + 3 to +5D. `+4` to +6 |
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Answer» Correct Answer - D In the pyrolusite `(MnO_2)` , the oxidation state of Mn is +4 , in `K_2MnO_4` , the oxidation state of Mn is +6 |
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| 128. |
When `KMnO_(4)` is reduced with oxalic acid in acidic solution, the oxidation number of `Mn` changes fromA. 4 to 2B. 6 to 4C. `+7` to `+2`D. 7 to 4 |
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Answer» Correct Answer - C `2KMnO_4 + 3H_2SO_4 to K_2SO_4 + 2MnSO_4 + 3H_2O + 5[O]` Hence `Mn^(+7)` in `KMnO_4` is reduced to `Mn^(2+)` in `MnSO_4` |
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| 129. |
In nitroprusside ion, the iron and NO exist as`Fe^(II)` and `NO^(+)` rather than `Fe^(III)` and NO. These forms can be differentiated byA. estimating the concentration of ironB. measuring the concentration of`CN^(-)`C. measuring the solidestate magnetic momentD. thermally decomposig the compound. |
| Answer» Magnetic moment helps to find the number of unpaired electrons. | |
| 130. |
`[Fe(H_(2)O)_(5)NO]^(2+)` is a complex formed in the brown ring test for `NO_(3)^(-)` ion. In this complex.A. NO transfers its electron to `Fe^(2+)` so that we have iron as `Fe(I)` and NO as `NO_^(+)`B. There are three unpaired electron so that its magnetic moment is 3.87 B.M.C. The colour is due to charge transferD. All the above statements are correct. |
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Answer» `HgI_(2)overset(Delta)(rarr) Hg_ I_(2) ("violet vapours")` `NO rarr NO^(+) + e^(-)`, `{:(Fe^(2+) ,+,e^(-),rarr,Fe^(+)),((3d^(6)),,,,(3d^(7))),("4 unpaired "e^(-),,,,"3 unpaired "e^(-)):}` `mu` for `Fe^(+) = sqrt(n(n+2))= sqrt(3 xx 5)` BM `= sqrt(15) BM =3.87 BM` |
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| 131. |
Mercury is the only metal which is liquid at `0^(@) C `. This is due to itsA. very high ionization energy and weak metallic bondB. low ionization potentialC. high atomic massD. high vapour pressure |
| Answer» Correct Answer - A | |
| 132. |
Mercury is a liquid metal becauseA. it has a completely filled s-orbitalB. it has a small atomic sizeC. it has a completely filled d- orbital that prevents d-d overlapping of orbitalsD. it has a completely filled d-orbital that causes d-d overlapping |
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Answer» Correct Answer - C The electronic configuration of mercury (80) is , `[Xe]4f^(10),5d^(5),6s^(2)`. Its d-subshell is completely filled, thus it prevents the overlapping of d-oebitals (d-doverlapping). Hence, it is a liquid metal at room temperature. |
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| 133. |
Mercury is the only metal which is liquid at `0^(@) C `. This is due to itsA. very high ionization energy and weak metallic bondB. Low inisation potentialC. High atomic weightD. High vapour pressure |
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Answer» Correct Answer - A Very high ionisation energy and weak metallic bond. |
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| 134. |
A red solid is insoluble in water.However , it becomes solubleif some KI is added to water.Heating the red solidin a test tube results in liberation of someviolet coloured fumesand droplets of a metal appear on the cooler partsof the test tube.The red solid isA. `HgI_(2)`B. `HgO`C. `Pb_(3)O_(4)`D. `(NH_(4))_(2)Cr_(2)O_(7)` |
| Answer» `underset("Red ppt.")(HgI_(2)) +2KI rarrunderset("Soluble")(K_(2)[HgI_(4)])` | |
| 135. |
The correct formula for diamine silver chloride isA. `[Ag(NH_(3))_(2)]Cl`B. `[Ag(NH_(2))_(3)]Cl`C. `AgCl. NH_(3)`D. `[Ag(NH_(4))_(2)Cl` |
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Answer» Correct Answer - A Diammine means `2NH_(3)` in the coordination sphere. |
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| 136. |
Mercury is a liquid metal becauseA. it has a completely filled d-orbital that causes d - d overlappingB. it has completely filled d-orbital that prevents d-d overlappingC. it has a completely filles s-orbitalD. it has a small atomic size |
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Answer» Correct Answer - B The electronic configuration of mercury is `[Xe]4f^(10), 5d^(10), 6s^(2)`. Its d-subshell is completely filled, thus, it prevents the overlapping of d-orbitals (d-d overlapping). Hence, it is liquid at room temperature. |
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| 137. |
A red coloured solid is insoluble in water. However, it becomes soluble if some KI is added to water. Heating red solid in a test tube results in the liberation of some violet coloured fumes and droplets of a metal appear on the cooler parts of the test tube. The red solid isA. `(NH_(4))_(2)Cr_(2)O_(7)`B. `HgI_(2)`C. HgOD. `Pb_(3)O_(4)` |
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Answer» Correct Answer - B The red coloured solid in `Hgl_(2)` `Hgl_(2)+2Kl to underset("Soluble")(K_(2)Hgl_(4))` `Hgl_(2) overset("heat")toH_(g)+underset("Violet fumes")(I_(2))` |
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| 138. |
Write the formulas for the follow co-ordination compounds Potassium trioxalatoaluminate (III) |
| Answer» `K_(3)[Al(C_(2)O_(4))_(3)]` | |
| 139. |
Write the formulas for the follow co-ordination compounds Tetracarbonylnickel (0) |
| Answer» Correct Answer - `[Ni(CO)_(4)]` | |
| 140. |
Write the formulas for the follow co-ordination compounds Potassium tetrahydroxozincate (II) |
| Answer» Correct Answer - `K_(2)[Zn(OH)_(4)]` | |
| 141. |
On the basis of the following observations made with aqueous solutions, assign secondary valencies to metals in the following compounds. |
| Answer» i) Secondary 4 ii) Secondary 6 iii) Secondary 6 iv) Secondary 6 v) Secondary 4 | |
| 142. |
Give the formula the follow complexes. Potassium hexacyano ferrate (III) |
| Answer» Correct Answer - `K_(3)[Fe(CN)_(6)]` | |
| 143. |
Write the formulas for the follow co-ordination compounds Tetraammineaquachloro cobalt (III) chloride |
| Answer» `[Co(NH_(3))_(4)(H_(2)O)Cl]Cl_(2)` | |
| 144. |
What is the oxidation state of cobalt in `[Co(NH_(3))_(6)]^(3+)` ? |
| Answer» `[Co(NH_(3))_(6)]^(3+): x+6(0)=+3,x=+3` | |
| 145. |
How would you account for the Cobalt (II) is stable in aqueous solution but in the presence of complexing reagents is easily oxidised. |
| Answer» Cobalt (III) ion has greater tendencr to form complexes than cobalt (II) ion. Therefore, Co (II) ion, being stable in aqueous solution, changes to Co (III) ion, in the presence of complexing reagents and gets oxidised. | |
| 146. |
How would you account for the The `d_(1)` configuration is very unstable in ions. |
| Answer» Ions of transition metals with `d^(1)` configuration tend to lose one electron to acquire `d^(0)` configuration that is quite stable. Therefore, such ions (with `d^(1)`) undergo either oxidation or disproportionation hence unstable. | |
| 147. |
What are co-ordination compounds? Give two examples. |
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Answer» Co-ordination compounds : Transition inetal atoms or ions form a large number of compounds in which anions or neutral groups are bound to metal atom or ion through coordinate covalent bonds. Such compounds are called co-ordination compounds ( or) complex compounds. Eg. : `[Fe(CN)_(6)]^(4-), [Co(NH_(3))_(6)]^(+3)`. |
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| 148. |
Explain the catalytic action of Iron(III) in the reaction between `I and S_(2)O_(8)^(2-)` ions. |
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Answer» Transition metal ions can change their oxidation states and become more effective catalysts. Iron (III) Catalyses the reaction between iodide and per sulphate ions `2I^(-)+S_(2)O_(8)^(-2)overset(Fe^(+3))rarrI_(2)+2SO_(4)^(-2)` Catalytic action explained as follows `2Fe^(+3)+2I^(-)rarr2Fe^(+2)+I_(2)` `2Fe^(+2)+S_(2)O_(8)^(-2)rarr2Fe^(+3)+2SO_(4)^(-2)` |
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| 149. |
Transition elements have high melting points. Why ? |
| Answer» Transition elements have high melting points because of involvement of greater number of electrons from (n - 1)d iri addition to ns electrons in the interatomic metallic bonding of these metals. | |
| 150. |
What are interstitial compounds ? How are they formed ? Give two examples. |
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Answer» The compounds which are formed when small atoms like H, C or N are trapped inside the crystal lattices of metal are called interstitial compounds. Eg : TiC, `Fe_(3)H, VH_(0.56)` etc. `rarr` These are non stoichiometric and are neither typically ionic nor covalent. `rarr` They have high melting points, higher than of pure metals. `rarr` They are very hard, some borides approach diamond in hardness. `rarr` They retain metallic conductivity. `rarr` They are chemically inert. |
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