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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.
| 1201. |
Which of the following statement is incorrect regarding lanthanides and actinides?A. Oxidation state +3 is most common in both the seriesB. In both series f-orbitals are progressively filledC. The elements of there series are radioactiveD. Both series show contraction effect |
| Answer» Correct Answer - C | |
| 1202. |
For, both lanthanides and actinides, which of the following statement is false ?A. Both involves the filling of(n - 2) f subshellB. In atoms of both the series, three outer shells are partly filled while the remaining are completely filledC. Some of them are electronegative in natureD. Their cation with unpaired electrons are paramagnetic in nature |
| Answer» Correct Answer - C | |
| 1203. |
`Gd^(+3)` ions has electronic configurationA. `4f^(0) , 5d ^(0), 6s^(0) `B. `f ^(7) , 5d^(1), 6s^(0)`C. `4f ^(7), 5d^(0), 6s^(0)`D. `f^(7), 6d^(0), 7s^(0)` |
| Answer» Correct Answer - C | |
| 1204. |
The Electronic configuration of `Gd^(2+)` is ( atomic no. of Gd is 64)A. `[Xe] 4f^(7)`B. `[Xe] 4f^(7) 5d^(1)`C. `[Xe]4f^(8)`D. `[Xe] 4f^(7)5d^(1) 6s^(2)` |
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Answer» `Gd(64) =[Xe] 4f^(7)5d^(1) 6s^(2)` `:.Gd^(2+) = [ Xe] 4f^(7)5d^(1)` |
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| 1205. |
The outer electronic configuration of Gd (At.No. 64) isA. `4f^(3)5d^(5)6s^(2)`B. `4f^(8)5d^(0)6s^(2)`C. `4f^(4)5d^(4)6s^(2)`D. `4f^(7)5d^(1)6s^(2)` |
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Answer» Correct Answer - D Gd (64) has electronic configuration as `[Xe]_(54)4f^(7)5d^(1)6s^(2) "instead of" [Xe]_(54)4f^(8)6s^(2)` |
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| 1206. |
Compelete and balance the following equations: (i). `Cu(OH)_2+NH_4NO_3+NH_4OHto…+H_2O` (ii). `Au+HCl+HNO_3to…+…+H_2O` (iii). `Ag_2S+2CuCl_2+2"Hg" to…+…S+2Ag` |
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Answer» (i). `Cu(OH)_2+2NH_4NO_3+NH_4OHto` `Cu(NH_3)_4(NO_3)_2+4H_2O` (ii). `Au+4HCl+HNO_3toHAuCl_3+NH+2H_2O` (iii). `Ag_2S+2Hg+2CuCl_2toCu_2Cl_2+Hg_2Cl_2+2Ag+S` |
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| 1207. |
Transition metals and many of their compounds show paramagnetic behaviour where there are unpaired electron or electrons. The magnetic moment arises from the spin and orbital motions in ions or molecule. Magnetic moment of n unpaired electrons is given as `mu=sqrt(n(n+2))` Bohr magneton Magnetic moment increases as the number of unpaired electrons increases. Q. Which among the following ions has maximum value of magnetic momentA. `Cu^(2+)`B. `Mn^(2+)`C. `Cr^(2+)`D. `Ti^(2+)` |
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Answer» Correct Answer - b `Mn(Z=25)implies3d^54s^2` `Mn^(2+)=3d^5(n=5)mu=sqrt(35)BM` `Cr(Z=24)implies3d^54s^1` `Cr^(2+)=3d^4(n=4)mu=sqrt(24)BM` `Ti(Z=22)implies3d^24s^2` `Ti^(2+)=3d^2(n=2),mu=sqrt8BM` |
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| 1208. |
An explosion takes place when conc. `H_2SO_4` is added to `KMnO_4`. Which of the following is formed?A. `Mn_(2)O_(7)`B. `MnO_(2)`C. `MnSO_(4)`D. `Mn_(2)O_(3)` |
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Answer» Correct Answer - A `2KMnO_(4)+underset("conc".)(H_(2)SO_(4))rarrMn_(2)O_(7)+K_(2)SO_(4)+H_(2)O` |
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| 1209. |
An explosion takes place when conc. `H_2SO_4` is added to `KMnO_4`. Which of the following is formed?A. `Mn^2O_7`B. `MnO_2`C. `MnSO_4`D. `Mn_2O_3` |
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Answer» Correct Answer - a `2KMnO_4+H_2SO_4(conc.)toK_2SO_4+H_2O+Mn_2O_7` |
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| 1210. |
If instead of `H_2SO_4`, `HCl` or `HNO_3` of suitable concentration were used, the volume of `KMnO_4` solution used would have beenA. less in case of HCl but more in case of `HNO_3`B. More in case of HCl but less in case of `HNO_3`C. more in both casesD. Less in both cases |
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Answer» Correct Answer - b If HCl were used, the oxygen produced from `KMnO_4` will be partly used up for oxidation of HCl. Hence, HCl required will be more. If `HNO_3` were used, it itself acts as oxidising agent and will oxidise the reductant. Hence `KMnO_4` solution used will be less. |
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| 1211. |
During titration `H_2SO_4` is preferably used over HCl and `HNO_3` to make the solution acidic becauseA. `H_2SO_4` is a strong oxidising agent and it reacts with `KMnO_4` during titrationB. Some `KMnO_4` is consumed during the reaction with `H_2SO_4`C. `H_2SO_4` does not react with `KMnO_4` or the reducing agent usedD. `H_2SO_4` can turn colourless `KMnO_4` to pink at the end point |
| Answer» Correct Answer - c | |
| 1212. |
Which compound does not dissolve in hot diluted `HNO_3`?A. HgSB. PbSC. CuSD. CdS |
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Answer» Correct Answer - A HgS does not dissolve in hot dil. `HNO_3` |
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| 1213. |
Ammonium dichromate is used in some fireworks. The green-coloured powder blown in the air isA. `Cr_2O_3`B. `Cr_2O_3`C. CrD. `CrO(O_2)` |
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Answer» Correct Answer - B The green coloured compound blown in air is `Cr_2O_3` `(NH_4)_2Cr_2O_7 overset"heat"to Cr_2O_3 + N_2 + 4H_2O` |
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| 1214. |
Ammonium dichromate is used in some fireworks. The green-coloured powder blown in the air isA. `CrO_3`B. `Cr_2O_3`C. `Cr`D. `CrO(O_2)` |
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Answer» Correct Answer - B Ammonium dichromate is used in some firework. The green-coloured powder which blows in air is `Cr_2O_3`. `(NH_4)_2Cr_2O_7overset(Delta)toCr_2O_3+N_2+4H_2O` |
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| 1215. |
The most abundant transition metal in earth crust is :A. FeB. CuC. ZnD. Ag |
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Answer» Correct Answer - A Fe 6.2% by mass in earth crust |
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| 1216. |
Transition elements have greater tendency to form complexes becauseA. they contain partially filled d - orbitalsB. their charge/ size ratio is quite highC. both a and bD. they are metals and all metals form complexes |
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Answer» Correct Answer - C Transition elements forms complex compound because of presence of vacant (n-1) d orbitals and unpaired `e^-` , and high charge/size ratio |
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| 1217. |
The element showing most stable +8 oxidation state in its compounds isA. MnB. FeC. OsD. Sc |
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Answer» Correct Answer - C Os +8 oxidation state is in `OsO_4` `{:(+8,+8),(Os, O_4),(+8,-2):}` |
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| 1218. |
The maximum oxidation state of transition metals are obtained by the following:A. ns electronsB. (n - 1) d - electronsC. ns + (n - 1) d - electronsD. (n + 1) d- electrons |
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Answer» Correct Answer - C Various oxidation states in transition metals is due to participation of `e^-` from ns and (n-1) d orbitals |
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| 1219. |
Which of the following transition metal show variable valency ?A. ScB. FeC. AcD. Zn |
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Answer» Correct Answer - B `Fe to Fe^(2+) , Fe^(3+)` |
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| 1220. |
Pyrolusite is a/anA. oxide oreB. sulphide oreC. carbide oreD. Not an ore |
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Answer» Correct Answer - A Pyrolusite is `MnO_(2)`. Thus, it is an oxide ore. |
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| 1221. |
In the test for nitrate, the composition of brown ring isA. `FeSO_(4).N_(2)O`B. `FeSO_(4).NO`C. `FeSO_(4)NO_(2)`D. `Fe(NO_(3))_(2)` |
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Answer» Correct Answer - B `3Fe^(2+)+NO_(3)^(-)+4H^(+)rarrNO+3Fe^(3+)+2H_(2)O` `Fe^(3+)+NO+5H_(2)Orarrunderset(("Brown complex"))([Fe(H_(2)O)_(5)NO]^(2+)` `FeSO_(4)+NOrarrFeSO_(4).NO` |
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| 1222. |
Which of the following is expected electronic configuration of europium ?A. `6s^(2), 4f^(7), 5d^(0)`B. `6s^(2), 4f^(6)5d^(1)`C. `6s^(2), 4f^(14), 5d^(0)`D. `6s^(2), 4f^(13),5d^(1)` |
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Answer» Correct Answer - B We know that, Europium has atomic number 63 habing expected electronic configuration is `6s^(2), 4f^(6), 5d^(2) and ` observed electronic configuration is `6s^(2), 4f^(7), 5d^(0)` |
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| 1223. |
The lanthanide contraction relates toA. atomic radiiB. atomic as well as `M^(3+)` radiiC. valence electronsD. oxidation states |
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Answer» Correct Answer - B Lanthanoid contaction relates to decrease in atomic as well as ionc size of `M^(3+)` ions. |
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| 1224. |
The atomic size of cerium and promethium is quite close, becauseA. they are in same period in periodic tableB. their electronic configuration is sameC. f-electrons have poor shielding effectD. nuclear charge is higer on certium than promethium |
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Answer» Correct Answer - C The size of cerium and promethium is quite close due to poor shielding effect of f-electrons. |
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| 1225. |
How many unpaired electrons present in 4f-orbital of lutetium ?A. 1B. 2C. 0D. 4 |
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Answer» Correct Answer - C We know that Ltetium has atomic number 71 and electronic configuration `6s^(2), 4f^(14), 5d^(0)` |
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| 1226. |
Neodymium show `+2, +3, and +4` oxidation state of `+4` oxidation state. The most oxidising state known is aqueous solution isA. `+2`B. `+3`C. `+4`D. none of these |
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Answer» Correct Answer - B We know that, +3 oxidation is common in all series. Stable of +3 O.S. depends upon high hydration energy, high lactic energy and high I.P. `Nd^(2+)` - has configuration `4f^(4)` `Nd^(4+)-` has configuration `4f^(2)` Hence `+2,+4` oxidation states are unstable in Nd in aqueous solution. |
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| 1227. |
Which of the following has highest paired 4f electrons ?A. `Yb ^(3+)`B. `Pr ^(3+)`C. `Pm ^(3+)`D. `Sm^(3+)` |
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Answer» Correct Answer - A `Yb^(3+)` has electronic configuration `6s^(0), 4f^(13), 5d^(0),` containg 13 4f-electrons. Six paired electrons and only one unpaired electrons. |
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| 1228. |
In weak field ligand, which one of the following cations has maximum magnetic moment?A. `Fe^(2+)`B. `Cu^(2+)`C. `Ni^(2+)`D. `Co^(2+)` |
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Answer» Correct Answer - A `Fe^(2+)` has maximum number of unpaired electrons. |
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| 1229. |
Cations wit all the paired electrons will have the total magnetic moment ofA. 1.54B. 2.83C. zeroD. 5.92 |
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Answer» Correct Answer - C All paired `e^-` , ion will be dimagnetic |
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| 1230. |
Which is the correct order of ionic sizes ? (Atomic no . : Ce = 58 , Sn = 50 , Yb = 70 and Lu)A. Ce gt Sn gt Yb gt LuB. Sn gt Ce gt Lu gt YbC. Lu gt Yb gt Sn gt CeD. Sn gt Yb gt Ce gt Lu |
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Answer» Correct Answer - A `{:(Ce,gt,Sn,gt,Yb,gt ,Lu),(58,,50,,70,,72):}` |
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| 1231. |
`Yb^(2+)` isA. `reducing agentB. oxidising agentC. redox agentD. one of the above |
| Answer» Correct Answer - A | |
| 1232. |
The `Yb ^(2+) and Lu ^(3+)` ion has electronic configurationA. `4f ^(14) 5d^(1) 6s^(0)`B. `4f^(14) 5d^(0) 6s^(0) and 4f^(14) 5d^(1)6s^(0)` respectivelyC. `4f^(14) 5d^(0)6s^(0)`D. `4f^(14) 5d^(1) 6s^(2)` |
| Answer» Correct Answer - C | |
| 1233. |
The number of unpaired electrons in `Yb^(3+)` is found to beA. zeroB. oneC. twoD. six |
| Answer» Correct Answer - B | |
| 1234. |
The number of unpaired electrons present in Am areA. 2B. 3C. 4D. 7 |
| Answer» Correct Answer - B | |
| 1235. |
Which of the following contains the maximum number of unpaired electrons?A. `TiCl_3`B. `MnCl_3`C. `FeSO_4`D. `CuSO_4` |
| Answer» Correct Answer - b | |
| 1236. |
Which of the following pair of elements are called chemical twins ?A. Zr and HfB. No and TaC. Mo and WD. all of these |
| Answer» Correct Answer - D | |
| 1237. |
Within each transition series, the oxidation statesA. decreases regularly in moving from left to rightB. first increase till the middle of period and then decreasesC. first decreases till the middle of period and thenD. None of the trend is correct. |
| Answer» Correct Answer - b | |
| 1238. |
Chemical twins are found inA. s - block clementsB. p - block elementsC. d - block elementsD. r - block elements |
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Answer» Correct Answer - C 2nd and 3rd series of transition metals have same ionic radii forming chemical twins |
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| 1239. |
Within each transition series, the oxidation statesA. decreases from left to rightB. first decrease till the middle of the period and then decreasesC. first decreases till the middle of the period and then increasesD. remains same |
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Answer» Correct Answer - B Increases from +1 to +7 and than decreases from +7 to +1 |
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| 1240. |
Metal used for making joints in jewellery isA. CuB. CdC. AgD. Zn |
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Answer» Correct Answer - A Uses of Cu |
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| 1241. |
What is the percentage impurity is 22 carat gold ? |
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Answer» Correct Answer - 8 The percentage purity of 22 carat gold is `=(100)/(24)xx22=91.66~~92.` |
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| 1242. |
The metal which is considered as configuration of neutral titanium atom isA. zincB. cadmiumC. mercuryD. scandium |
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Answer» Correct Answer - D Zinc cadmium and mercury of group 12 have full `d^(10)` configuration in their ground state as well as in their common oxidation states and thus, are not regarded as transition metals. |
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| 1243. |
The ground state electronic configuration of neutral titanium atom isA. `[Ar]4s^(2)4p^(2)`B. `[Ar]3d^(2)4s^(2)`C. `[Ar]4s^(2)p_(x)^(1)p_(y)^(1)`D. `[Ar]3d^(5)` |
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Answer» Correct Answer - B The ground state electronic configuration of Ti is `[Ar] 3d^(2) 4s^(2)` |
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| 1244. |
`Zn` gives `H_(2)` gas with `H_(2)SO_(4)` and `HCl` but not with `HNO_(3)` becauseA. Zn acts as an oxidising agent when reacts with `HNO_(3)`B. `HNO_(3)` is weaker acid than `H_(2)SO_(4)` and HClC. in electrochemical series, Zn is above hydrogenD. `NO_(3)^(-)` ion is reduced in prefence to hydronium ions |
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Answer» Correct Answer - D `Zn+2HClrarrZnCl_(2)+H_(2)` `Zn+H_(2)SO_(4)(dil.)rarrZnSO_(4)+H_(2)` With dil. `HNO_(3)Zn` produce oxide `(N_(2)O)`.`4Zn+10HNO_(3)rarr4Zn(NO_(3))+N_(2)O+5H_(2)O` In this reaction of zinc with HCl or dil. `H_(2)SO_(4),H_(3)O^(+)` ion get reduced to `H_(2)`. In the reaction of Zn with dil. `HNO_(3), NO_(3)^(-)` ions are reduced to `N_(2)O` in preference to hydronium ions . |
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| 1245. |
Describe the preparation of potassium permanganate. |
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Answer» Preparation of `KMnO_(4)` Potassium permanganate is prepared by the fusion of `MnO_(2)` with an alkali metal hydroxide and an oxidising agent like `KNO_(3)` It forms dark green, `K_(2)MnO_(4)` which disproportionates in a neutral or acidic solution to give permanganate. i) `2MnO_(2)+4KOH+O_(2) overset(KNO_(2))rarr2K_(2)MnO_(4)+2H_(2)O` ii) `3MnO_(4)^(2-)+4H^(+) rarr 2MnO_(4)^(-)+MnO_(2)+2H_(2)O` or `3K_(2)MnO_(4)+2H_(2)SO_(4) rarr underset("Potassium permanganate")(2KMnO_(4)+2K_(2)SO_(4)+MnO_(2)+2H_(2)O)` |
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| 1246. |
The equivalent weight of potassium permanganate in acid solution isA. 1/5 of its molecular weightB. 1 /6 of its molecular weightC. 1/10 of its molecular weighD. 1/2 of its molecular weight |
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Answer» Correct Answer - A In acidic medium, `MnO_4^(-) + 8H^(+) + 5e^(-) to Mn^(2+) + 4H_2O` `therefore` Equivalent weight of `KMnO_4 = M/5=158/5`=31.6 |
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| 1247. |
The colours of the transition metal are due to d-d excitation. The energy required for d-d electron axcitation is available in the visible range. Transition metal ions have the tendency to absorb certain rediations from the visible region and exhibit the complementary colour. The transition metal ions which have completely filled d-orbitals are colourless as the excitation of electron or electrons is not possible within d-orbitals. The transition metal ions which have completely empthy d-orbitals are also colourless. In `KMnO_4` and `K_2Cr_2O_7`, there are no unpaired electrons at the central atom but they are dep in colour. The colour of these compounds is due to charge transfer spectrum. for example in `MnO_4` electron is momentrily transferred from O to the metal and thuys oxygen changes from `O^(2-)` and `O^(ɵ)` maganese from `Mn^(7+)` to `Mn^(6+)`. Q. `K_2Cr_2O_7` gives coloured solution in water. The colour is due toA. d-d transition in Cr-atomsB. Presence of unpaired electron in d-orbital of oxygenC. charge transfer from 0 to CrD. none of the above |
| Answer» Correct Answer - c | |
| 1248. |
The colours of the transition metal are due to d-d excitation. The energy required for d-d electron axcitation is available in the visible range. Transition metal ions have the tendency to absorb certain rediations from the visible region and exhibit the complementary colour. The transition metal ions which have completely filled d-orbitals are colourless as the excitation of electron or electrons is not possible within d-orbitals. The transition metal ions which have completely empthy d-orbitals are also colourless. In `KMnO_4` and `K_2Cr_2O_7`, there are no unpaired electrons at the central atom but they are dep in colour. The colour of these compounds is due to charge transfer spectrum. for example in `MnO_4` electron is momentrily transferred from O to the metal and thuys oxygen changes from `O^(2-)` and `O^(ɵ)` maganese from `Mn^(7+)` to `Mn^(6+)`. Q. Select the correct statement:A. colour of the transition metal ion arises due to d-d transition.B. colour of certain oxysalts of transition metals is due to charge tranfer.C. Both are correct.D. None is correct. |
| Answer» Correct Answer - c | |
| 1249. |
The colours of the transition metal are due to d-d excitation. The energy required for d-d electron axcitation is available in the visible range. Transition metal ions have the tendency to absorb certain rediations from the visible region and exhibit the complementary colour. The transition metal ions which have completely filled d-orbitals are colourless as the excitation of electron or electrons is not possible within d-orbitals. The transition metal ions which have completely empthy d-orbitals are also colourless. In `KMnO_4` and `K_2Cr_2O_7`, there are no unpaired electrons at the central atom but they are dep in colour. The colour of these compounds is due to charge transfer spectrum. for example in `MnO_4` electron is momentrily transferred from O to the metal and thuys oxygen changes from `O^(2-)` and `O^(ɵ)` maganese from `Mn^(7+)` to `Mn^(6+)`. Q. Which of the following compounds is (are) coloured due to charge transfer spectra and not due to d-d transition?A. `KMnO_4`B. `K_2CrO_4`C. `CrO_3`D. all of these |
| Answer» Correct Answer - d | |
| 1250. |
The colours of the transition metal are due to d-d excitation. The energy required for d-d electron axcitation is available in the visible range. Transition metal ions have the tendency to absorb certain rediations from the visible region and exhibit the complementary colour. The transition metal ions which have completely filled d-orbitals are colourless as the excitation of electron or electrons is not possible within d-orbitals. The transition metal ions which have completely empthy d-orbitals are also colourless. In `KMnO_4` and `K_2Cr_2O_7`, there are no unpaired electrons at the central atom but they are dep in colour. The colour of these compounds is due to charge transfer spectrum. for example in `MnO_4` electron is momentrily transferred from O to the metal and thuys oxygen changes from `O^(2-)` and `O^(ɵ)` maganese from `Mn^(7+)` to `Mn^(6+)`. Q. Which is a coloured ion?A. `[Cr(H_2O)_6]^(3+)`B. `[Cu(CN)_4]^(3-)`C. `[Ti(H_2O)_6]^(4+)`D. `[Sc(H_2O)_6]^(3+)` |
| Answer» Correct Answer - a | |