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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.
| 151. |
Which of the following is incorrect regarding `K_(2)Cr_(2)O_(7)` ?A. It oxidezes iodide to iodineB. It oxidizes `H_(2)S` to `SO_(2)`C. It oxidizes `SO_(2)` to `H_(2)SO_(4)`D. It oxidizes ethyl alcohol to acetic acid |
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Answer» Correct Answer - B It oxidized hydrogen sulphide to sulphur. `K_(2)Cr_(2)O_(7) + 4H_(2)SO_(4) + 3H_(2)S rarr K_(2)SO_(4) + Cr_(2)(SO_(4))_(3) + 7H_(2)O + 3S` |
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| 152. |
Manganese achieves its hightest oxidation state in its compoundA. `MnO_(3)`B. `Mn_(3)O_(4)`C. `KMnO_(4)`D. `K_(2)MNO_(4)` |
| Answer» Correct Answer - C | |
| 153. |
What is the hybridisation of Mn in `K_(2)MnO_(4)` ? |
| Answer» Correct Answer - `sp^(3)` | |
| 154. |
Why is `FeI_(3)` not stable ? |
| Answer» Due to more oxidising power of `Fe^(3+)` and less reducing power of `I^(-)` . | |
| 155. |
Which of the following reactions are disproportionation reactions? `(A)" " Cu^(+)rarrCu^(2+)+Cu` `(B)" "3 MnO_(4)^(2-)+4H^(+)rarr2MnO_(4)^(-)+MnO_(2)+2H_(2)O` `(C )" "2KMnO_(4)rarrK_(2)MnO_(4)+MnO_(2)+O_(2)` `(D)" "2MnO_(4)^(-)+3Mn^(2+)+2H_(2)O rarr5MnO_(2)+4H^(+)`A. I,IIB. I,II,IIIC. II,III,IVD. I,IV |
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Answer» Correct Answer - A In (I), `Cu^(+)` is oxidised as well as reduced. In (II), `MnO_4^(-)` ions are oxidised as well as reduced. |
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| 156. |
Which of the following oxide is acidic in nature ?A. `CrO `B. `Cr_(2)O_(3)`C. `CrO_(3)`D. `CrO_(2)` |
| Answer» Correct Answer - C | |
| 157. |
In the dichromate dianion,A. 4 Cr-O bonds are equivalentB. 6 Cr-O bonds are equivalentC. All Cr-O bonds are equivalentD. All Cr-O bonds are non-equivalent |
| Answer» Correct Answer - B | |
| 158. |
Oxidation state of in `Fe_(3)O_(4)` isA. `3//2`B. `4//5`C. `5//4`D. `8//3` |
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Answer» Let the oxidation state of Fe be x, Using the fact that oxidation state of oxygen is usually `-2` and the sum of all possible oxidation numbers is zero, we can proceed as follows: `3x+overset(x-2)overset(Fe_(3)O_(4))(4(-2)=0)rArr 3x - 8 = 0` `rArr 3x = 8 rArr x = 8//3` |
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| 159. |
Which of the following is incorrect ?A. Ionisation enthalpies of the early actionoids are higher than for the early lanthanoidsB. Unlike lanthanoids, actinoids form oxocationsC. Oxides and hydroxides of actinoids are more basis than those of lanthanoidsD. Ion exchange behaviour is exhibited by both actinides and lamnthanides. |
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Answer» Correct Answer - A Because 5d electrons penetrarte less into the inner core of electrons. Therefore, the 5f electrons of actinoids are more effectively shielded form the nuclear charge than the 4f electrons of the are less firmly held, they are easily available for bonding in the actionids. Lanthanoids do not form oxocations while actinoids form oxocations such as `UO_(2)^(2+)`, `UO^(+)`, `UO_(2)^(+)`, etc. These ions decreases in the order. `M^(4+) gt MO_(2)^(2+) gt M^(3+) gt MO_(2)^(+)` This is because higher the charge density (i.e. higher the charge and smaller the ionic size), more is the complex forming ability. The higher tendency of complex formation of `MO_(2)^(2+)` ion as compared to `M^(3+)` is due to higher conc. of charge metal atom M in `MO_(2)^(2+)` . |
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| 160. |
Which of the following actinoids is well known to exhibit `+2` oxidation state ?A. CeB. NoC. LrD. both (2) and (3) |
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Answer» Correct Answer - B The `(+3)` oxidation state is the most stable for all the actinoids but one of the elements with atomic number 96 - 103. The exception is No, which is most stable as No `(+2)`. This is stable because it has a favourable `f^(14)` electronic configuration. Lr exists only in the `(+3)` state and resists both oxidation as well reduction, again illustrating the stability of an `f^(14)` electronic arrangement. Ce is a lanthanoid which is well known to exhibit `+4` oxidation state. |
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| 161. |
Assertion: In `Cr_2O_7^(2-)` ion, all the `Cr-O` bond lengths are equal. Reason: In `Cr_2O_7^(2-)` ion all the `O-Cr-O` bond angles are equal.A. If both the assertion and reason are ture but the reason is a true explanation of the assertion.B. If both the assertion and reason are true but the reason is not the correct explanation of the assertionC. If the assertion is true but reason is falseD. If assertion is false but reason is true. |
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Answer» Correct Answer - D Correct Assertion : Two `Cr - O` bonds have different bond lengths than the remaining six `Cr - O` bonds Correct Reason : `O - Cr - O` bond in the middle is different from other `O - Cr - O` bond angles. |
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| 162. |
Which of the following is corred ?A. All the actinides resemble the lanthanoids quite closelyB. Actinides do not exhibit similarity with lanthanidesC. The second half of the actinides resembles the lanthanides quite closelyD. The first half of the actinides resembles the lanthanides |
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Answer» Correct Answer - C Because `(+3)` oxidation state is the most stable oxidation state for the later elements (Am `rarr` Lr, excluding No). In spite of similarites, the actinides can be separated form the lanthanides quite easily, as the actinides form complexes more readily. |
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| 163. |
Naturally occurring uranium contains ------- isotopesA. threeB. fourC. twoD. five |
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Answer» Correct Answer - A Three isotopes `99.3%`, `238_(U)`, `0.7%` `235_(U)`, and traces of `234_(U)`. The isotope `235_(U)` is fissile i.e. if it is irradiated with thermal (slow) neutrons the nucleus breaks up into two smaller nuclei. |
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| 164. |
How many actinoids are radioactive ?A. Only oneB. 7C. 14D. 12 |
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Answer» Correct Answer - C All the actinoids are radioactive. All the elements after `82^(Pb)`, that is form `83^(Bi)` onwards, have unstable nuclei and undergo radioactive decay. |
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| 165. |
What would happen when a solution of potassium chromate is treated with an excess of dilute nitric acid ?A. `Cr^(3+) and Cr_2O_7^(2-)` are formedB. `Cr_2O_7^(2-) and H_2O` are formedC. `CrO_4^(2-)` is reduced to +3 state of CrD. `CrO_4^(2-)` is oxidised to +7 state of Cr. |
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Answer» Correct Answer - B Dilute nitruc acid converts chromate into dichromate and `H_2O`. `2K_2CrO_4+2HNO_3 to K_2Cr_2O_7 + 2KNO_3+H_2O` or, `underset("yellow")(2CrO_4^(2-))overset(H^(+))hArrunderset("orange")(Cr_2O_7^(2-))+H_2O` |
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| 166. |
Lanthanum nitrids on hydrolysis producesA. `HN_(3)`B. `N_(2)H_(4)`C. `N_(2)`D. `NH_(3)` |
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Answer» Correct Answer - D At elevated temperatures the lanthanide metals react with N, P, As, Sb and Bi, given Li N, etc. The lithium nitride is hydrolysed by water in a similar way to AIN: `LnN + 3H_(2)O rarr Ln(OH)_(3) + NH_(3)` |
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| 167. |
The general valence shell electronic configuration of actinoids is represented byA. `(n-2)f^(1-14)(n-1)d^(0-1)ns^(2)`B. `(n-2)f^(1-14)(n-1)d^(1-10)ns^(2)`C. `(n-2)f^(0-14)(n-1)d^(0-10)ns^(2)`D. `(n-2)f^(0-14)(n-1)d^(0-2)ns^(2)` |
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Answer» Correct Answer - D Th is `[Rn]6d^(2) 7s^(2)` . Pa, U, Np, Cm and Lr all have `6d^(1)` electron. |
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| 168. |
Assertion : The aqueous solution of `FeCl_(3)` is basic in nature. Reason : The colour changes due to the oxidation of potassium chromate.A. If both the assertion and reason are ture but the reason is a true explanation of the assertion.B. If both the assertion and reason are true but the reason is not the correct explanation of the assertionC. If the assertion is true but reason is falseD. If assertion is false but reason is true. |
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Answer» Correct Answer - D The aqueous solution of `FeCl_(3)` is acidic in nature because `FeCl_(3)` hydrolyses in water to produce acid ion. `FeCl_(3) + 3H_(2)O rarr Fe(OH)_(3) + 3HCl` Therefore, assertion is false but reason is true. |
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| 169. |
Which of the following factor may be regarded as the main cause of lanthanide contraction?A. Poor shielding of `4f`-electrons in compare to other electrons in the sub-shellB. Effective shielding of one of the `4f`-electrons by another in the sub-shellC. Poorer shielding of `5d` electron by `4f`-electrons.D. Greater shielding of `5d` electrons by `5f-electron. |
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Answer» Correct Answer - A This phienomenon is associated with the intervention of the `4f` orbitals which must be filled before the `5d` series of elements begin The filling of `4f` before `5d` orbital results in a regular decrease in atomic radii called is lanthanoid contraction. This is because of poor shielding of one of the `4f`-electrons by another in the sub-shell. |
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| 170. |
Which of the following has `d^(3)` configuration and is quite stable ?A. Fe `( +3)`B. Mn `( +3)`C. V `( +3)`D. Cr `( +3)` |
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Answer» Correct Answer - D Chromium (III) is the important species with this configuration which is stable and well known for complex formation. In other cases this configuration is relatively unimportant. |
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| 171. |
Which of the following has `d^(1)` configuration and is stable ?A. Cr `( +5)`B. Mn `( +6)`C. both (1) and (2)D. V `( +4)` |
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Answer» Correct Answer - D Except `V(+4)`, all other with this configuration are either reducing or undergo disproportionation. For example, disproportionation occurs for `Cr(+5)` and `Mn(+6)` as follows: `3 CrO_(4)^(3-) + 8H^(+) rarr 2 CrO_(4)^(2-) + Cr^(3+) + 4H_(2)O` `3MnO_(4)^(2-) + 4H^(+) rarr MnO_(2) + 2H_(2)O` |
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| 172. |
Transitional elements exhibit variable valencies because they release electrons from the following orbitsA. `ns` orbitB. `ns` and `np` orbitsC. `(n -1) d` orbitD. `(n - 1)d` and `ns` orbits |
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Answer» Correct Answer - D `(n - 1) d` and `ns` orbits. |
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| 173. |
Transitional elements are named transition elements because their characters areA. Like that of p-and-d-block elementsB. In between `s` and p-block elementsC. They are members of `IA` groupD. They are like inactive elements |
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Answer» Correct Answer - B In between `s-` and `p-` block elements. |
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| 174. |
`E_(Mn^(3+)//Mn^(2+))^@`is highly positive than that of `E_(Cr^(3+)//Cr^(2+))^@orE_(Fe^(3+)//Fe^(2+))^@` becauseA. `Mn^(2+)(d^5)` can be easily oxidised to `Mn^(3+)(d^4)` due to low ionisation enthalpyB. third ionisation enthalpy of Mn is much larger due to stable half filled `d^5` electronic configuration of `Mn^(2+)`C. `Mn^(3+)` is more stable than `Mn^(2+)` due to higher oxidation stateD. second ioonisation enthalpy of Mn is higher than third ionisation enthalpy. |
| Answer» Correct Answer - B | |
| 175. |
Out of the all known elements, the percentage of transitional elements is approximatelyA. `30%`B. `50%`C. `60%`D. `75%` |
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Answer» Correct Answer - C `("Transition element + Inner transition element")/("Total element") xx 100` `(33 + 28)/(105) xx 100 = 58.09 ~~ 60%` |
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| 176. |
Which of the following species with `d^(6)` configuration is stable in the presence of strong complexing reagents ?A. Fe `( +2)`B. Co `( +3)`C. Cu `( +1)`D. Ni `( +2)` |
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Answer» Correct Answer - B Iron `(+2)` and `Co(+3)` are important species with this configuration. Iron `(+2)` is quite stable although a mild reducing and `Co(+3)` is stable in the presence of strong complexing agents. `Ni(+2)` is the most important species with `d^(8)` configuration. |
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| 177. |
Which of the following couples has the most positive `E^(Theta)` value ?A. `V^(3+)//V^(2)`B. `Cr^(3=)//Cr^(2+)`C. `Fe^(3+)//Fe^(2+)`D. `Mn^(3+)//Mn^(2+)` |
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Answer» Correct Answer - D The standerd electrode potential for the oxidation of a metal is a composite property that depends on `Delta G^(@)` for the sublimation of the metal, the the ionization enthalpies of the metal atom, and `Delta G^(@)` for the hydrogen of the hydration of the metal ion. `E(Theta)` value for the `Mn^(3+)//Mn^(2+)` cople `(+1.57)` is much more positive than that for others: `V^(3+)//V^(2+)` `(-0.26V)`, `Cx^(3+)//Cr^(2+)` `(-0.41)` and `Fe^(2+)//Fe^(2+)` `(+0.77V)` because of much higher third ionization enthalpy of Mn as the required change is `d^(5)` to `d^(4)` : `Mn^(2+) (3d^(5)) rarr Mn^(3+) (3d^(4))` Thus `Mn^(3+) + bar e rarr Mn^(2+)` is quite favourable and hence `Mn^(3+)` is an oxidizing agent. This clearly explains why the `+3` state of Mn is of little importance. |
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| 178. |
The number of mole of `KMnO_(4)` that will be needed to react completely with one mole of ferrous oxalate in acidic solution is:A. `3//5`B. `2//5`C. `4//5`D. 1 |
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Answer» Correct Answer - A `3 underset(3M)MnO_(4)^(-) + 5 (Fe^(2+) + underset(5M)C_(2) O_(4)^(2-)) + 24 H^(+) rarr` `3Mn_(2) + 5 Fe^(3+) + 10CO_(2) + 12 H_(2) O` Thus `5M` of `Fe_(2)O_(4)` is oxidised by `3M` of `KMnO_(4)` then `1M` of `FeC_(2) O_(4)` is oxidised by `3//5` mole of `KMnO_(4)` |
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| 179. |
Which of the following is the strongest oxidizing agent ?A. `VO_(2)^(+)`B. `Cr_(2)O_(7)^(2-)`C. `CoO_(4)^(2-)`D. `FeO_(4)^(2-)` |
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Answer» On accoumnt of the steady in the effiective nuclear charge along the series, the tendency to exhibit higher oxidation states decreases. For example, the stability of the `(+6)` state decreases as follows : `Cr_(2)O_(7)^(2-) gt MnO_(4)^(2-) gt FeO_(4)^(2-) gt CoO_(4)^(2-)` or `CrO_(4)^(2-)` Therefore, the oxidizing power increases as follows `CoO_(4)^(2-) gt gt FeO_(4)^(2-) gt MnO_(4)^(2-) gt Cr_(2)O_(7)^(2-) gt VO_(2) +` or `CrO_(4)^(2)` This also due to the increasing stability of the lower species to which they are reduced. |
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| 180. |
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 conectration of ironB. Measuring the concentration of `CN-`C. Measuing the solid state magnetic momentD. Thermally decomposing the compound |
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Answer» Correct Answer - C The existence of `Fe^(2+)` and `NO^(+)` in nitroprusside ion `[Fe(CN)_(5) NO]^(2-)` can be established by measuring the magnetic moment of the solid compound which should correspond to `(Fe^(2+) = 3 d^(6))` four un paied electron. |
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| 181. |
How many of oxygen atoms become available form one mole of `KMnO_(4)` as an oxidizing agent in the acidic medium ?A. `2.5`B. 5C. `3.0`D. 6 |
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Answer» Correct Answer - A The equivalent mass of `KMnO_(4)` in acidic medium is M/5, thus its one mole contains five equivalents. According to law to equivalents, five equivalents of `KMnO_(4)` will release five equivalents of oxygen atoms, `5xx1//2 = 2.5` mole oxygen atoms become available. |
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| 182. |
The magnetic moment of metal ion of first transition series is `2.83 BM`. Therefore, it will have unpaied electronsA. 2B. 4C. 3D. 6 |
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Answer» Correct Answer - A `mu = sqrt(n (n + 2)) (mu =` magnitic moment) `(n =` no. of unparied electrons.). `2.8 = sqrt(n (n + 2))` `n (n + 2) = 8` `n^(2) + 2n - 8 = 0` `n = + 2`. |
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| 183. |
The pair of compounds that can exist together is:A. `FeCl_(3), SnCl_(2)`B. `HgCl_(2), SnCl_(2)`C. `FeCl_(2), SnCl_(2)`D. `FeCl_(3), KI` |
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Answer» Correct Answer - C Both `FeCI_(2)` and `SnCl_(2)` are reducing agent and therefore can exist together. |
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| 184. |
Which of the following is used in medicine as a powerful laxative ?A. `Hg_(2)CI_(2)`B. `HgCI_(2)`C. `ZnCI_(2)`D. `CdCI_(2)` |
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Answer» Correct Answer - A Calomel is used as a purgative in medicine. |
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| 185. |
The lanthanide contraction is responsible for the fact thatA. `Zr` and `Y` have about the same radiusB. `Zr` and `N//b` have similar oxidation stateC. `Zr` and `Hf` have about the same radiusD. `Zr` and `Ce` ahave same oxidation state |
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Answer» Correct Answer - C Due to poor shielding of `(n - 2) f`- electrons, the size of `Zr, and `Hf` remains the same. |
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| 186. |
In 3d series highest oxidation state is shown byA. MnB. FeC. CuD. V |
| Answer» Correct Answer - A | |
| 187. |
The most common oxidation states of 3d series elementsA. `+2`B. `+3`C. `+4`D. `+7` |
| Answer» Correct Answer - A | |
| 188. |
The general valence shell electronic configuration of lanthanoides is represented byA. `(n-2)f^(1-14)(n-1)d^(0-1)ns^(2)`B. `(n-2)f^(1-14)(n-1)d^(10)ns^(2)`C. `(n-2)f^(0-14)(n-1)d^(0-2)ns^(2)`D. `(n-2)f^(1-14)(n-2)d^(0-5)ns^(2)` |
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Answer» Correct Answer - A Lanthanum, the element preceding the lanthanides in the periodic table, has the electronic configuration `[Xe] 5d^(1)6s^(2)`. Like lanthanum, the lanthanides also exhibit the stable oxidation state of `+3`. It is expected that in these elements the successive electrons will be filled in the 4f subshell, thereby the elements may have the electronic configuration form `[Xe]4f^(1) 5d^(1)6s^(2)` to `[Xe] 4f^(14) 5d^(1) 6s^(2)`. However, the actual ground state electronic configuration of lanthanides (determined by atomic spectroscopy) shows that there is an electron in 5d subshell only in Ce, Gd and Lu while in all other elements this electron is shifted to the 4f subshell. This type of shuttling of electrons can be understood in terms of the comparable energies of the 4f and 5d subshells. Whether there is an electron in 5d orbital or not, is of little importance because the lanthanides mostly form configuration of `M^(3+)` ions varies in a regular manner form `[Xe] 4f^(1)` for `Ce^(3+)` to `[Xe] 4f^(14)` for `Lu^(3+)` . |
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| 189. |
`V_(2) O_(5)` is red or orange in colour. It is `a//an`….oxideA. AcidicB. BasicC. AmphotericD. Neutral |
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Answer» Correct Answer - A `V_(2) O_(5)` is red or orange in color and it is an amphoteric oxide. |
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| 190. |
The valence shell electronic configuration of `Cr^(2+)` ion isA. `3 p^(2) 4 s^(2)`B. `4 s^(2) 3 d^(2)`C. `4 s^(2) 3 d^(0)`D. `4 s^(0) 3 d^(0)` |
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Answer» Correct Answer - D Electronic configuration of chromium `Cr rarr [Ar] 3d^(5) 4 s^(0)` `Cr^(2 +) rarr [Ar] 3d^(4) 4 s^(0)` |
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| 191. |
Which of d-block elements can exhibit the highest possible oxidation state ?A. FeB. RhC. OsD. both (2) and (3) |
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Answer» Correct Answer - D Under the old nomenclature system, Group 8, 9 and 10 were collectively called Group VIII. In some respects this made chemical sense, because the three groups have some interesting relationships. In particular, the first row elements, Fe, Co and Ni have similarities in their chemistry while Ru, Os, Rh, Ir, Pd, and Pt have enough common chemistry that they have the collective name of the platinum metals. Rh an Os are stable to atomospheric attack whilst Fe is subject to corrosion in the rusting. Whereas Fe fails to attain the group oxidation state `(+8)`, Ru and Os attain the oxidation state of `+8`, through they are only elements to do so in the entire d-block. The lower oxidation states of `+2` and `+3`, which are so well known for Fe are particularly unknown for Ru and Os. |
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| 192. |
The `f`-block elements of the periodic table contains those element in whichA. Only `4f` orbitals are progressively filled in 6th period.B. only `5f` orbtials are progressively filled in 7th period.C. `4f` and `5f` orbitals are progressively filled in 6th and 7th periods respectively.D. None of these |
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Answer» Correct Answer - C Electronic configuration of `f`-block element `(n - 2) f^(1-14)`, `(n - 1) d^(10) , ns^(2) f`-block starts with 6th period `(n = 6)`. |
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| 193. |
Which of the element is not a typical transition element ?A. FeB. PdC. CrD. Zn |
| Answer» Correct Answer - D | |
| 194. |
Which of the following does not have abnormal electronic configuration ?A. CrB. PdC. PtD. Hg |
| Answer» Correct Answer - D | |
| 195. |
A transition element `X` has a configuration `[Ar] 3d^(4)` in its ` + 3` oxidation state. Its atomic number isA. 19B. 26C. 22D. 25 |
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Answer» Correct Answer - D Number of electrons in excited state |
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| 196. |
Of the ions `Zn^(2), Ni^(2 +)` and `Cr^(3 +)` [atomic number of `Zn = 30, Ni = 28, Cr = 24`]A. Only `Ni^(2+)` is coloured and `Zn^(2+)` and `Cr^(3+)` are colourlessB. All three are colourlessC. All three are colouredD. Only `Zn^(2+)` is colourless and `Ni^(2+)` and `Cr^(3+)` are coloured |
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Answer» Correct Answer - D `Ni^(2 +)` and `Cr^(3 +)` are coloured. But `Zn^(2 +)` is colourless because of absence of unpaired `e^(-2)` |
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| 197. |
Which is coloured because of d-d transition ?A. `KMnO_(4)`B. `K_(2)CrO_(4)`C. `CoCl_(3)`D. All of these |
| Answer» Correct Answer - C | |
| 198. |
There are 14 elements in actinoid series. Which of the following elements does not belong to this series?A. U, NpB. NpC. TmD. Fm |
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Answer» Correct Answer - C Actinoid series has elements from atomic no. 90 to 103. Thulium ( Tm) has atomic no. 69. |
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| 199. |
Which of the following elements does not belong to the first transition series?A. `Fe`B. `Ag`C. `V`D. `Cu` |
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Answer» Correct Answer - B Ag belongs to second `(4d)` transition seriess remaining all are in first transition series. |
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| 200. |
`MnO_(4)^(2-)+H^(+) to "Product"` Product is formedA. `MnO_(4)^(-)`B. `MnO_(2)`C. MnD. Both(1) & (2) |
| Answer» Correct Answer - D | |