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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.
| 201. |
Which of the following is not correct about transition metalsA. Their melting and boiling points are highB. Their compounds are generally coluredC. They can form ionic or covalent compoundsD. They do not exhibit variable valency |
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Answer» Correct Answer - D They show variable valency due to presence of vacant d-orbitals. |
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| 202. |
The catalytic of the transition metals and their compounds is described toA. their chemical reactinityB. their magnetic behaviourC. their unfilled d-orbitalsD. their ability to adopt multiple oxidation state and their complexing ability |
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Answer» Correct Answer - D Trasition element on account of their variable (multiply) valrncy have the ability to from intermediate compounds readily, thus acting as good catalyasts. Hence, the catalytic activity of the transition metals and their compounds is described to their ability to adopt multiple oxidation state and their ability to from complexes. |
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| 203. |
A transition element `X` has a configuration `[Ar] 3d^(4)` in its ` + 3` oxidation state. Its atomic number isA. 25B. 26C. 22D. 19 |
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Answer» Correct Answer - A Given, `X=[Ar]d^(4)` `:.` The complete configuration of the ion, `X^(3+)=1s^(2),2s^(2),2p^(6),3s^(2),3p^(6),3d^(4)` `:.X=1s^(2),2s^(2),2p^(6),3s^(2),3p^(6),3d^(5),4s^(2)` ` :. `The atomic number of the element is 25 and the element is Mn. |
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| 204. |
Electronic configuration of a transition element X in `+3` oxidation state is `[Ar] 3d^(5)` . What is its atomic number ?A. 25B. 26C. 27D. 24 |
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Answer» Correct Answer - B `X^(3+) = [Ar]^(18) 3d^(5) = 23 ` electrons . As `X^(3+)` is formed by loss of 3 electrons from X,X will have 26 electrons. `:. `At. No. of X = 26 |
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| 205. |
What may be stable oxidation state of the transition element with the following d electron configuration in the ground state of their atoms ? `3d^(3) , 3d^(5) , 3d^(8) ` and `3d^(4)` |
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Answer» `3d^(3) 4s^(2) = + 5 , 3d^(5) 4s^(1) = + 6 , 3d^(5) 4s^(2) = +2 , + 7 , 3d^(8) 4s^(2) = + 2 , 3d^(4) 4s^*(2) = 3d^(5) 4s^(1) = + 6 ( & + 3)` Note. The maximum oxidation states of reasonable stability correspond to the sum of s and d- electrons upton Mn. After that, there is abrupt decrease in the stability of higher oxidation states. |
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| 206. |
Select the correct order of oxidising power in acidic medium.A. `Cr_(2)O_(7)^(2-)gt MnO_(4)^(-)gtVO_(2)^(-)`B. `MnO_(4)^(-)gtCr_(2)O_(7)^(2-)gtVO_(2)^(+)`C. `VO_(2)^(+)gtMnO_(4)^(-)gtCr_(2)O_(7)^(2-)`D. `MnO_(4)^(-)gtVO_(2)^(+)gtCr_(2)O_(7)^(2-)` |
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Answer» Correct Answer - B Oxidising power depends upon the stability of reduction product obtained. Since, lower oxides of Mn are more stable than that of Cr and V, so the order of oxidising power is `MnO_(4)^(-)gtCr_(2)O_(7)^(2-)gtVO_(2)^(-)` |
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| 207. |
An element of 3d-transition series shows two oxidation states x and y, differing by two units. Then:A. compounds in oxidation state X are ionic if `Xlty`B. compound in oxidation state x are ionic if `xgty`C. compounds in oxidation state y are covalent if `xlty`D. compounds in oxidation state y are covalent if `ygtx` |
| Answer» Correct Answer - B::C | |
| 208. |
A transition element X has a configuration `3d^4` in its `+3` oxidation state. Its atomic number is notA. 25B. 26C. 22D. 19 |
| Answer» Correct Answer - b,c,d | |
| 209. |
Pick out the correct statements from the following. (I) Cobalt (III) is more stable in octahedral complexes. (II) Zinc froms coloured ion or complexes. (III) Most of the d-block elements and their compounds are ferromagnetic. (IV) Osmium show (VIII) oxidation sate.A. I and IIB. I and IIIC. II and IVD. I and IV |
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Answer» Correct Answer - D (I) `Co^(3+)` have higher charge density than `Co^(2+), So Co^(3+)` is more stable in octahedral complexes. (II) Zn exhibits only `+2` oxidation state.So, `Zn^(2+)=[Ar]3d^(10),4s^(0)` Since, iy does not contain any unpaired electron, its compounds are colourless. (III) d-block elements are generally paramagnetic and sometime diamagnetic , but not ferromagnetic. (IV) Osmium and ruthenium are VIII group elements, so can exibit the highest oxidation state`+8` in their oxides, Hence, statements I and IV are correct. |
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| 210. |
How do rou expect the density of transition elements to vary in a given series and why? |
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Answer» In a given series the density of transition elements increases. Eg. : From Titanium to copper there is a significant increase in the density. `rarr` This is due to decrease in metallic radius coupled with increase in atomic mass. |
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| 211. |
Name the element in the 3d series that (i) shows maximum oxidation state (ii) is diamagnetic |
| Answer» (i) Manganese (ii)Zinc | |
| 212. |
Why do transition metals form coloured complexes ? |
| Answer» Most of the compound of transition elements are coloured due to the presence of unpaired electrons in d-subshell of transition metal ions. It can be explained with the help of crystal filed theory. For example, in an octahedral complex `[Ti(H_(2)O)_(6)]^(3+)`, under the influence of ligands (`H_(2)O` molecules ),all the five degenerate d-orbitals of penultimate shell of transition metal ion `Ti^(3+)` split up in to two degenerate sets, one containing `d_(x^(2)-y^(2))` and `d_(z)^(2)` (with higher energy ) and the other containing `d_(xy),d_(yz),d_(zx)` ( with lower energy ). When white light falls, a part of it corresponding to a certain wavelenght ( yellow in this case ) is abosrbed. Due to small energy difference, electronic excitations take place from one set to another (d-d transition ). The remaining colours of white light ( red and blue ) are transmitted and the compound appears purple coloured. | |
| 213. |
How does the ionic `//` covalent character of the compounds of a transition metal vary with its oxidation state? |
| Answer» As the oxidation state increases, more and more valence shell electrons are involved in bonding. The atomic core becomes less shielded resulting in the increased attraction on the electrons. Because ofthis, ionic character of bonds decreases with the increase in oxidation state. | |
| 214. |
The second and third members in each group of the transition elements have similar metallic radii. Assing reason. |
| Answer» This is because of lanthanoid contraction in the members of the lanthanoid family (Z=58 to Z=71) which occupy a position along with Lanthaum (Z=57) in the perodicd table. However, these are placed separately at the bottom of the table in the f-block. The lanthanoid contraction also decrease the size of the elements of the third transition series (Hf onward). As a result, there is a negligible difference in the metallic radii of the second and third members in each group of transition metals. | |
| 215. |
The atomic sizesof Fe,Co and Ni are nearly same, Explain with reason. Or Atomic size of 3d series elements from chromium to copper is almost the same. Give reason. |
| Answer» As we move from left to right alonga transition series, the nuclear change increases which tends to decrease the size but the addition of electrons in thed-subshellincreases the screening effect which counter balances the effect of increased nuclear charge. | |
| 216. |
Why do chromium group elements have the highest melting point in their respective series ? |
| Answer» The element belonging to this group . (Cr,Mo,W,Sg) have maximum number of unpaired d-electrons ( `d^(5)` configuration). Therefore metallic bonding is the maximum and so are the melting points. | |
| 217. |
Which metal has the lowerst melting point?A. CsB. NaC. HgD. Sn |
| Answer» Correct Answer - c | |
| 218. |
Which of the following is for oxidatin of `SO_(2)` to `SO_(3)` in contact process ?A. `MnO_(2)`B. `CuCl_(2)`C. `V_(2)O_(5)`D. Fe and Mo |
| Answer» Correct Answer - A::C | |
| 219. |
In the iodometric estimation in the laboratory which process is involved?A. `Cr_2O_7^(2-)+H^(o+)+I^(ɵ)to2Cr^(3+)I_2` `I_2+S_2O_3^(2-)toS_4O_6^(4-)+I^(ɵ)`B. `MnO_4^(ɵ)+H^(o+)+I^(ɵ)toMn^(2+)+I_2` `I_2+S_2O_3^(ɵ)toS_4O_6^(2-)I^(ɵ)`C. `MnO_4^(ɵ)+OH^(ɵ)+I^(ɵ)toMnO_2+I_2` `I_2+S_2O_3^(2-)toS_4O_6^(2-)+I^(ɵ)`D. `Cr_2O_7^(2-)+OH^(ɵ)+I^(ɵ)to2Cr^(3+)+I_2` `I_2+S_2O_3^(2-)toS_4O_6^(2-)+I^(ɵ)` |
| Answer» Correct Answer - a,b | |
| 220. |
Which metal has the highest melting point?A. PtB. WC. PdD. Au |
| Answer» Correct Answer - b | |
| 221. |
The element with lowest melting and boiling point isA. TiB. CuC. ZnD. V |
| Answer» Correct Answer - C | |
| 222. |
Which of the following transition metals of 3d series has the lowest melting point ?A. `Ti(Z= 22)`B. `V(Z= 23)`C. `Cr(Z= 24)`D. `Mn(Z=25)` |
| Answer» In the first transition series,Mn has abnormally low melting point. This is due to the fact that Mn has exactly half-filled d-orbitals.As a result, electronic configuration is very stable, i.e., electrons areheldtightly by the nucleus so that the delocatization is less and the metallic bond is much weaker than the elements on its either side. | |
| 223. |
(a) Out of`Ag_(2)SO_(4), CuF_(2), MgF_(2)` and CuCl, which compound will be coloured and why ? (b) Explain `: ` (i)`CrO_(4)^(2-)` is a strong oxidizing agent while `MnO_(4)^(2-)` is not (ii) Zr and Hf have identical sizes. (iii) The lowest oxidation state of manganese is basic while the highest is acidic. (iv) Mn (II) shows maximum paramagnetic character amongst the divalent ions of the first transition series. |
| Answer» (a) `CuF_(2)` (b) (i) Cr in `CrO_(4)^(2-)` is in the highest oxidation state, i.e., `+6` while Mn in `MnO_(4)^(2-)` is in oxidation state`+6` and its most stable oxidation state is`+7` (ii) due to lanthanoil contraction (iv) `Mn^(2+)` has `3d^(5) ` configuration | |
| 224. |
(a) How would you account for the following `:` (i) Actinoid contraction is greater than lanthanoid contraction (ii) Transition metals from coloured compound. (b) Complete the following equation. `2MnO_(4)^(-) + 6H^(+) + 5NO_(2)^(-) rarr` |
| Answer» (b)`2MnO_(4)^(-) + 6H^(+) + 5NO_(2)^(-) rarr 2Mn^(2+) + 5NO_(3)^(-) + 3H_(2)O` | |
| 225. |
In general the melting and boiling points of transition metalsA. increases gradually across the period from left to rightB. decreases gradually across the period from left to rightC. first increases till the middle of the period and then decreases towards the endD. First decreases regulary till the middle of the period and then increases towards the end. |
| Answer» Correct Answer - c | |
| 226. |
Which of the following transition metals of 4d series has the lowest melting point?A. Ti(Z=22)B. V(Z=23)C. Cr(Z=24)D. Mn(Z=25) |
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Answer» Correct Answer - D The elements Mn has the lowest melting point because due to symmetrical `(d^(5))` configuration, the metallic bond is the weakest. |
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| 227. |
Account for the following `:` (a) ` CuCl_(2) ` is more stable than `Cu_(2)Cl_(2)`. (b) Atomic radii of 4d and 5d series elements are nearly same. (c ) Hydrochloric acid is not used in permanganate titrations. or Account for the following `:` (a) `Eu^(2+)` is a strong reducing agent. (b) Orange colour of dichromate ion changes to yellow in alkaline medium. (c ) `E^(@) ( M^(2+) // M )` values for transiton metals show irregular variation. |
| Answer» (a) `Eu^(2+)`can lose electron to form more stable `Eu^(3+)` Property ) | |
| 228. |
Out of `Cu_(2)Cl_(2)`, which is more stable and why ? |
| Answer» `CuCl_(2)` is more stable than `Cu_(2)Cl_(2)` because `Cu^(2+)`(aq) ion has more negative enthalpy of hydration ( ` Delta_("hyd.")H^(@)`) as compared to `Cu^(+)(aq)` ion due to its smaller size. This compensates the extra ionisation enthalpy involved in the conversion of Cu to `Cu^(2+)` ion. | |
| 229. |
Out of `Cu_(2)Cl_(2)` and `CuCl_(2)` which is more stable and why ? |
| Answer» `CuCl_(2)` is more stable than `Cu_(2)Cl_(2)` . This is because `Cu^(2+) (aq)` has greater negative enthalpy of hydration than `Cu^(+) (aq)`. | |
| 230. |
When `Cu^(2+)` ion is treated with KI,a white precipiate is formed. Explain the reaction with the help of chemical equation . |
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Answer» `i^(-)` ion is a strong reducing agent. It reduces `Cu^(2+)` to `Cu^(+)` `2Cu^(2+) + 4I^(-) rarr underset("(white ppt.)")(Cu_(2)I_(2)) + I_(2)` |
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| 231. |
When `Cu^(2+)` ion is treated with KI, a white precipitate is formed. Explain the rectiaon with the help of chemical equation . |
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Answer» `I^(-)` ion reduces `Cu^(2+)` ion to `Cu^(+)` ion. The chemical equation for the reaction is : `2Cu^(2+)(aq)+4I^(-)(aq) to underset("(White ppt.)") underset("Copper (I) iodide")(Cu_(2)I_(2))+I_(2)` |
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| 232. |
When `Cu^(2+)` ion is treated with KI, a white precipitate is formed. Explain the reaction with the help of chemical equation. |
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Answer» When `Cu^(2+)` ion is treated with KI, it produces `Cu_(2)I_(2)` white precipitate in the final product. `2Cu^(2+)+4I^(-)rarrunderset(("White ppt."))(Cu_(2)I_(2))+I_(2)` (In this reaction, `CuI_(2)` is formed which being unstable, dissociates into `Cu_(2)I_(2)` and `I_(2)`). |
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| 233. |
Write the formula of an oxoanion of manganese(Mn) in which it shows the oxidation state equal to its group number. or Write the formulaof an oxoanion of chromium (Cr) in which it shows the oxidation state eqaul to its group number. |
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Answer» `MnO_(4)^(-)`(permanganate ion). Oxidation state of `Mn= +7` Group number of Mn`=7` `Cr_(2)O_(7)^(2-)` ( dichromate ion) . Oxidation state of `Cr= + 6` Group number of Cr=6 |
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| 234. |
The purple colour of `KMnO_4` is due toA. d - d transitionB. charge transfer transitionC. f - f transitionD. d - f transition |
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Answer» Correct Answer - A Colour is due to presence of unpaired `e^-` in 3d orbitals of Mn the leading to d-d transition giving colour. |
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| 235. |
A solution of `KMnO_4` is reduced to various products depending upon its pH. At pH lt 7 it is reduced to a colourless solution (A), at pH = 7 it forms a brown precipitate (B) and at pH gt 7 it gives a green solution ( C), (A),(B) and( C) areA. (A)-`Mn^(2+) , (B)-`MnO_2` , (C ) -`MnO_4^(2-)`B. `(A)-MnO_2 , (B)-Mn^(2+) , (C ) - MnO_4^(2-)`C. `(A)-Mn^(2+) , (B ) -MnO_4^(2-) , (C ) - MnO_2`D. `(A)-MnO_4^(2-) , (B) - Mn^(2+) , (C ) - MnO_2` |
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Answer» Correct Answer - A At pH lt 7 , in acidic medium `MnO_4^(-) + 8H^(+) + 5e^(-) to underset"(Colourless)"(Mn^(2+)) + 4H_2O` At pH=7 , in neutral medium `MnO_4^(-) + 2H_2O + 3e^(-) to underset"(Brown precipitate )"(MnO_2 )+ 4OH^(-)` At pH gt 7 , in alkaline medium `MnO_4^(-) + e^(-) to underset"(green)"(MnO_4^(2-))` |
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| 236. |
Which is true statement about `KmnO_4`?A. Its solution is unstable in acidic medium.B. It has purple colour.C. `MnO_4^(ɵ)` changes to `Mn^(2+)` in basic solution.D. It is self indicator in `Fe&(2+)` or `C_2O_4^(2-)` titration. |
| Answer» Correct Answer - a,b,d | |
| 237. |
`d_(x)2-_(y)2` orbital is involved in which of the following hybridisation ?A. `sp^3d`B. `sp^3d^2`C. `sp^3d^3`D. None of these |
| Answer» Correct Answer - b,c | |
| 238. |
`d_(xy),d_(yz)` and `d_(xz)` orbitals is involved in which of the following hybridisation?A. `dsp^2`B. `sp^3d`C. `sp^3d^2`D. `d^3sp^3` |
| Answer» Correct Answer - c,d | |
| 239. |
`d_(x^2-y^2)` and `d_(z^2)` orbitals is involved in which of the foollowing hybridisation?A. `sp^3d^2`B. `d^2sp^3`C. `sp^3d^3`D. `d^3sp^3` |
| Answer» Correct Answer - a,b | |
| 240. |
Cerium shows oxidation state of `+4` becauseA. It resembles alkali metalsB. It has very low `IE`C. it has tendency to attain noble has configurationD. It has tendency to attain `f^0` configuration |
| Answer» Correct Answer - d | |
| 241. |
The most common oxidation states of cerium areA. `+2,+4`B. `+3,+4`C. `+3,+4`D. `+3,+5` |
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Answer» Correct Answer - B The electronic configuration of Ce is as `Ce:[Xe]4f^(2)5d^(0)6s^(2)` The most common oxidation state shown by cerium are `+3 and +4` |
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| 242. |
Which of the following has fully filled f-orbitals ?A. Only Eu and YbB. Only Gd and EuC. Only Gd and LuD. Only Yb and Lu |
| Answer» Correct Answer - D | |
| 243. |
The electronic configuration `4f^(0) 5d^(1) 6s^(2)` representsA. actiniumB. lanthanumC. ceriumD. dprasedymium |
| Answer» Correct Answer - B | |
| 244. |
`Eu^(2+)` ion has strong tendency toA. gain electronB. lose electronC. undergo reductionD. none of these |
| Answer» Correct Answer - B | |
| 245. |
`4f^(14)5d^(0)6s^(2)` represents configuration of which of the followingA. LuB. EuC. YbD. Gd |
| Answer» Correct Answer - C | |
| 246. |
The lanthanoide ion with e.c. `4 f^(14) 5d ^(0) 6s^(0)` has tendecny toA. under goes reductionB. gain electronC. lose electronD. all of these |
| Answer» Correct Answer - C | |
| 247. |
The first synthetic element, i.e., element made artificially was `"….................."`. |
| Answer» Correct Answer - technitium `(. _(43)Te)` | |
| 248. |
The lanthanoide which shows all i.e., `+2,+3 and +4` oxidation state isA. europiumB. samariumC. neodymiumD. gadolinium |
| Answer» Correct Answer - C | |
| 249. |
The pair of lanthanoide ions used as raducing agents is/areA. `Eu ^(3+) and Ce ^(4+)`B. `Ce ^(4+) and Tb ^(4+)`C. `Eu ^(2+) and Dy^(4+)`D. `Nd ^(2+) and Sm ^(2+)` |
| Answer» Correct Answer - D | |
| 250. |
What is the correct order of ionic size of lanthanoide ions ?A. `Lu ^(3+) gt Tb ^(3+) gt Sm ^(3+) gt La ^(3+)`B. `Lu ^(3+) lt Tb ^(3+) Sm ^(3+) lt La ^(3+)`C. `Lu^(3+) lt Sm ^(3+) lt Tb ^(3+) lt La ^(3+)`D. `Sm ^(3+) lt Lu ^(3+) lt Tb ^(3+) lt La ^(3+)` |
| Answer» Correct Answer - B | |