InterviewSolution
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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.
| 151. |
Study the given figure of structure of TMV (Tobacco Mosaic virus) and select the option that correctly identifies the labellings A and B A. ssDNa CapsomeresB. dsDNA CapsomeresC. ssRNA CapsomeresD. dsRBA Tail fibres |
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Answer» Correct Answer - C TMV (Tobacco mosaic virus) is a ribovirus. It is rod shaped measurning `300 xx 200nm`. It is made up of ssRNA and proteinaceous covering capsid with `2130` capsomeres, which are helically arranged to form a hollow cylinder. |
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| 152. |
Assertion: TMV is a virus which causes mosaic disease. Reason: TMV has RNA as genetic material.A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertion.C. If the assertion is true but reason is false.D. If both the assertion and reason are false. |
| Answer» Correct Answer - A | |
| 153. |
Assertion: The first ionisation energy of `Be` is greater than that of `B`. Reason: 2p-orbital is lower in energy than 2s-orbital.A. If both assertion and reason are true and reason is the correct explanation of assertionB. If the assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If assertion is false but reason is true. |
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Answer» Correct Answer - A The first Ionisation energy of `Be` is greater than Boron because it is difficult to remove electron from `Be` in comparison to boron. It is also true that the `2p`-orbitals. Both assertion and reason are true and reason is correct explanation. |
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| 154. |
The chemistry of lithium is very similar to that to that of magnesium even though they are placed in different groups. Its reason isA. Both are found together in natureB. Both have nearly the same sizeC. Both have similar electronic configurationD. The raito of their charge and size (i.e. charge density) is nearly the same |
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Answer» Correct Answer - D Diagonal relationship. |
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| 155. |
Assertion: Metallic character of first group metals of Mordern periodic table increase on moving down the group. Reason: On moving top to bottom in first group in Modern periodic table, value of ionisation enthalpy continuously decreases.A. If both assertion and reason are true and reason is the correct explanation of assertionB. If the assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If assertion is false but reason is true. |
| Answer» Correct Answer - A | |
| 156. |
Which of the following statement is not correct regarding hydrogen atom?A. It resembles halogens in some propertiesB. it resembles alkali mentals in some propertiesC. it can be placed in 7th group of periodic tableD. it cannot be placed in first group of periodic table |
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Answer» Correct Answer - D Hydrogen resemble alkali metals in some properties so it can be placed in the first group of periodic table. |
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| 157. |
Which of the following properties displays progressive increase with the reise in atomic number across a period in the periodic table?A. ElectronegativityB. Electron affinityC. Ionisation potentialD. Size of the atom |
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Answer» Correct Answer - A Both electronegativity and electron affinity increases in the nuclear charge. But electronegativity increases continuously. |
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| 158. |
Which of the following transitions involves maximum amount of energy?A. `M^(-) (g) rarr M (g)`B. `M(g) rarr M^(+) (g) `C. `M^(+) (g) rarr M^(2+) (g)`D. `M^(2+) (g) rarr M^(3+) (g)` |
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Answer» Correct Answer - D `M^(+2) rarr M^(3+)` after the removel of `2e^(-)` the nuclear charge per `e^(-)` increases,due to which high energy is required to remove `3e^(-)`. |
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| 159. |
`Mg` and `Li` are similar in their properties due toA. Same `e//m` ratioB. same electron affinityC. Same groupD. Same ionic potential |
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Answer» Correct Answer - D `L^(+) and Mg^(+2)` ions have similar polarising power or ionic potential and therefore have similar properties. This type of relationship of the first element of a group with the second of the next group is known as diagonal relationship. |
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| 160. |
Assertion : Fungi may be saprophytic, parasitic or symbiotic. Reason : The cell walls of fungi are composed of chitin and polysaccharide.A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertion.C. If assertion is true but reason is false.D. If both assertion and reason are false. |
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Answer» Correct Answer - B `**` Fungi may be saprophytic, parasitic or symbiotic. `**` The cell walls of fungi are composed of chitin and polysaccharide. |
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| 161. |
Viruses are non-cellular organisms but replicate themselves once they infect the host cell. To which of the following kingdom do viruses belong to ?A. MoneraB. ProtistaC. FungiD. None of the above |
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Answer» Correct Answer - D Viruses are non-cellular organisms but replicate themselves once they infect the host cell. Viruses do not belong to any kingdom. |
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| 162. |
Viruses are non-cellular organisms but replicate themselves once they infect the host cell. To which of the following kingdom do viruses belong to ?A. MoneraB. ProtistaC. FungiD. None of these |
| Answer» Correct Answer - D | |
| 163. |
How many functional groups are present in following compound ? A. 6B. 8C. 7D. 5 |
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Answer» Correct Answer - B |
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| 164. |
Number of functional groups present in the following compounds is: |
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Answer» Correct Answer - 6 |
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| 165. |
How many functional groups are present in the given compound ? |
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Answer» Correct Answer - 7 |
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| 166. |
Degree of unsaturation in the above compound is:A. 7B. 8C. 9D. 10 |
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Answer» Correct Answer - C |
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| 167. |
How many different type of functional groups are present in given compound.A. 6B. 5C. 4D. 3 |
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Answer» Correct Answer - A |
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| 168. |
Degree of unsaturation of this compound is:A. 2B. 3C. 4D. 5 |
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Answer» Correct Answer - C |
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| 169. |
For the given compound answers the following questions: The number of functional group present are (Excluding C=C bonds):A. 2B. 3C. 4D. 5 |
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Answer» Correct Answer - B |
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| 170. |
Which one of the following statements is incorrect in relation to ionisation enthalpy?A. Ionization enthalpy increases for each successive electron.B. The greatest increase in ionization enthalpy is experienced on removal of electron from core noble gas configuration.C. End of valence electrons is marked by a big jump in ionization enthalpy.D. Removal of electron from orbitals bearing lower n value is easier than from orbital having higher n value. |
| Answer» Electrons in orbitals bearing a lower n value are more attracted to the nucleus than electrons in orbitals bearing a higher n value. Hence, the removal of electrons from orbitals bearing a higher n value is easier than the removal of electrons from orbitals having a lower n value. | |
| 171. |
The first ionisation enthalpy of group `13` elements are : Explain this deviation from the general trend. |
| Answer» On moving down a group, ionization enthalpy generally decreases due to an increase in the atomic size and shielding. Thus, on moving down group 13, ionization enthalpy decreases from B to Al. But, Ga has higher ionization enthalpy than Al. Al follows immediately after s–block elements, whereas Ga follows after d–block elements. The shielding provided by d-electrons is not very effective. These electrons do not shield the valence electrons very effectively. As a result, the valence electrons of Ga experience a greater effective nuclear charge than those of Al. Further, moving from Ga to In, the ionization enthalpy decreases due to an increase in the atomic size and shielding. But, on moving from In to Tl, the ionization enthalpy again increases. In the periodic table, Tl follows after 4f and 5d electrons. The shielding provided by the electrons in both these orbitals is not very effective. Therefore, the valence electron is held quite strongly by the nucleus. Hence, the ionization energy of Tl is on the higher side. | |
| 172. |
Energy of an electron in the ground state of the hydrogen atom is `-2.18xx10^(-18)J`. Calculate the ionisation enthalpy of atomic hydrogen in terms of `J mol^(-1)`. Hint: Apply the idea of mole concept to derive the answer. |
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Answer» It is given that the energy of an electron in the ground state of the hydrogen atom is `-2.18 xx 10^(-18) J`.Therefore, the energy required to remove that electron from the ground state of hydrogen atom is `2.18 xx 10^(-18) J.` `therefore` Ionization enthalpy of atomic hydrogen `= 2.18 xx 10^(-18) J` Hence, ionization enthalpy of atomic hydrogen in terms of `J "mol"^(-1) = 2.18 xx 10^(-18) xx 6.02 xx 10^(23) J "mol"^(-1)` `= 1.31 xx 10^(6) J "mol"^(-1)` |
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| 173. |
Chemical periodicity |
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Answer» The concept of chemical periodicity is central to the study of inorganic chemistry. There is no single best form of the periodic table because the choice depends on the purpose for which the table is used. Some forms emphasize chemical relations and valence, whereas others stress the electronic configuration of the elements or the dependence of the periods on the shells and subshells of the atomic structure. This chapter discusses the periodic law in terms of the electronic structure of atoms. It discusses the use of the periodic table and graphs to systematize trends in physical and chemical properties and to detect possible errors, anomalies, and inconsistencies. The chapter also describes the use of the periodic table to predict new elements and compounds and to suggest new areas of research. The most characteristic chemical property of an element is its valence. There are numerous measures of valence, each with its own area of usefulness and applicability. The chapter illustrates the trends in atomic, physical, and chemical properties of the elements in each group and along each period. |
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| 174. |
Which one of the following is the smallest in size?A. `N^(3-)`B. `O^(2-)`C. `F^(ɵ)`D. `Na^(ɵ)` |
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Answer» Correct Answer - D All of these are isoelectronic ions (10 electrons). In isolectronic species, as the number of protons (atomic number) goes on increasing, size goes on decreasing due to stronger attraction on the electrons (Z for N, O, F and Na=7, 8, 9 and 11) |
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| 175. |
The size of species `I, I^(+)` and `I^(ɵ)` decrease in the order.A. `I^(o+) gt I^(ɵ) gt I`B. `I^(ɵ) gt I gt I^(o+)`C. `I^(ɵ) gt I^(o+) gt I`D. `I gt I^(o+) gt I^(ɵ)` |
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Answer» Correct Answer - B Size of negative ion `gt` neutral atom `gt` positive ion of the parent element. |
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| 176. |
Which one of the following is the smallest in size?A. `Na^(o+)`B. `N^(3-)`C. `O^(2-)`D. `F^(ɵ)` |
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Answer» Correct Answer - A For the isoelectronic species, positive ion species have smallest size. |
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| 177. |
The major component of fungal cell wall is a polymer calledA. ChitinB. XylanC. CelluloseD. Peptidoglycan |
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Answer» Correct Answer - A The major component of fungal cell wall is a polymer called Chitin. |
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| 178. |
ChlorophyII-a is present inA. ArchaebacteriaB. BacteriaC. CyanobacteriaD. All of the above |
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Answer» Correct Answer - C Chlorophyll-a is present in Cyanobacteria. |
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| 179. |
Which of the following is used as a bioweapon ?A. Bacillus subtilisB. Bacillus licheniformisC. Bacillus thuringiensisD. Bacillus anthracis |
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Answer» Correct Answer - D Bacillus anthracis is used as a bioweapon. |
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| 180. |
Fungi and bacteria, which feed on dead organic matter are calledA. SymbioticB. ParasiticC. SaprophyticD. All of the above |
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Answer» Correct Answer - C Fungi and bacteria, which feed on dead organic matter are called Saprophytic. |
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| 181. |
Among the following oxides how many of them are suboxides ? (a) `C_(3)O_(2)` , (b) `N_(2)O` , (c ) `NO_(2)` , (d) `CO` (e) `Fe_(3)O_(4)` , (f) `KO_(2)` |
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Answer» Correct Answer - 2 `C_(3)O_(2)` and `N_(2)O` |
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| 182. |
Which of the following are isoelectronic? `NO_(3)^(ɵ), CO_(3)^(2-). ClO_(3)^(ɵ), SO_(2)`A. `NO_(3)^(ɵ), CO_(3)^(2-), ClO_(3)^(ɵ)`B. `NO_(3)^(ɵ), CO_(3)^(2-), SO_(2)`C. `CO_(3)^(2-), ClO_(3)^(ɵ), SO_(2)`D. None of these |
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Answer» Correct Answer - B `{:("No. of electrons in" NO_(3)^(ɵ) = 7+8 xx 3 +1 = 32),("No. of electrons in" CO_(3)^(2-) = 6+8 xx 3 + 2 = 32),("No. of electrons in" SO_(2) = 16 + 8 xx 2 = 32):}]` Hence iso-electronic |
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| 183. |
Magnetic moment of `V(Z = 23),Cr(Z = 24),and Mn(Z= 25)` are ` x,y,z `repectively henceA. `x = y = z`B. `x lt y lt z`C. `x lt z lt y`D. `z lt y lt x` |
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Answer» Correct Answer - C `V = (Z = 23) rArr 3d^(3) 4s^(2) (n = 3)` (`n` is the number of unpaired electrons) `(x)`. `Cr(Z=24) rArr 3d^(5) 4s^(1) (n = 6) (y)` `Mn(Z = 25) rArr 3d^(5) 4s^(2) (n=5) (z)` The more the number of unpaired electrons, the higher is the magnetic moment. So the answer is `x lt z lt y`. |
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| 184. |
Inert pair effect is shown byA. `s`-blockB. `p`-blockC. `d`-blockD. `f`-block |
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Answer» Correct Answer - B Group `13, 14` and `15` show inert pair effect. |
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| 185. |
`EA` is positive whenA. `O^(ɵ)` is formed from `O`B. `O^(o+)` is formed from `O`C. `O^(2-)` is formed from `O`D. `EA` is always a negative value |
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Answer» Correct Answer - C `EA` is positive due to electron-electron repulsion. |
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| 186. |
Which of the following molecule is theoretically not possible ?A. `OF_(4)`B. `O_(2)F_(2)`C. `OF_(2)`D. `SF_(4)` |
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Answer» Correct Answer - A Oxygen atom generally has `-2` oxidation state `(OS)` But `O` has `-1 OS` (in peroxides and peracids), `+1 OS` in `OF_(4), OS` of `O` is `+4` which is not possible. |
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| 187. |
Ionization enthalpies areA. all positive quantitiesB. all negative quantitiesC. either positive or negative quantitiesD. all netural quantities |
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Answer» Correct Answer - A Ionization is always an endothermic process. By convetion, enegry absoerbed by atoms (or ions) in the ionization process has a positive value. |
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| 188. |
Ionization enthalpies tend to decrease going down any column of main group elements because______going down the column.A. nuclear charge increasesB. number of shilding electrons increasesC. atomic size increasesD. effective nuclear charge increases |
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Answer» Correct Answer - C Elements in the same group have similar valence shell electron configuration. As we go down the group, both the nuclear charge `(Z)` and the number of inner shielding electrons `(sigma)` increase. As a result, `Z_(eff) (Z - sigma)` for the outermost electron,of each atom will be essentially the same. For example, compare `Li` and `Na`. `Al`though the number of protons has increased from `3` in `Li` to `11`in `Na`, teh latter has `10` shielding inner electrons while the former has only two. However, as the principle quantum number `n` increases, so does the average distance of a valence electron from the nucleus. A greater separation between the electron and the nucleus means a weakes attraction, so that it become increasibgly easier to remove the electron as we go from element to element down a group. |
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| 189. |
The basic nature of the hydroxides of group `13` decreases progessively down the group. |
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Answer» Correct Answer - F The basic nature of the hydroxides increases progressively down the group. |
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| 190. |
Second and successive electron affinity of an elementA. is always successive (energy is released)B. is always positive (energy is absorbed)C. can be positive or negativeD. is always zero |
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Answer» Correct Answer - B `O_((g))+e^(-) underset(-EA_(1))(rarr)O_((g))^(ɵ) + e^(-) underset(+EA_(2))(rarr) O_((g))^(2-)` Energy is required to add an eelctron to the negatively charge species due to electron-electron repulsion. |
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| 191. |
The decreasing order of electron affinity of `F,Cl` and `Br` is `F gt Cl gt Br`. |
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Answer» Correct Answer - F False. The decreasing order of electron affinity of F, Cl and Br is `Cl gt F gt Br`. In general, the electron affinity decreases down the group but the electron affinity for chlorine is more than that for fluorine, and similarly the electron affinity for sulphur is more than that for oxygen because in F ans S, due to very small size of the atom. There are strong electron-electron repulsion in the relatively small`2p` orbitals of F and thus, the incoming electron does not experience much attraction. Hence, the incoming electron is not added easily as it is added in large `3p` orbitals of Cl. Consequently, the `Delta_(eg)H^(ɵ)` or EA of F is less negative than that of Cl. |
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| 192. |
Which of the following pairs of elements have almost similar atomic radii ?A. `Zr, Hf`B. `Cu, Ag`C. `Sc, Ti`D. `Pd, Pt` |
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Answer» Correct Answer - A Due to consequence of lanthanide contraction Size of `Zr =` size of `Hf` Size of `Nb=` size of `Ta` Size of `Mo=` size of `W` Size of `Co=` size of `Ni` (due to imperfect screening effect of `3d` orbitals) `|{:("Group" .rArr,4,5,6,),("Period" rArr 5,Zr,Nb,Mo,),("Period" rArr 6,Hf,Ta,W,):}|` |
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| 193. |
The number of the followinjg pairs contains elements with similar atomic radii a. `Co, Ni` , b. `Zn, Mo` , c. `Rh, Ir` , d. `Hf, Ti` |
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Answer» Correct Answer - 6 `CO` and `Ni` have similar atomic radii. |
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| 194. |
Chloride of an element `A` gives neutral solution in watt . In the periodic table, the elements `A` belong toA. First groupB. Third groupC. Fifth groupD. First transition group |
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Answer» Correct Answer - A Salts of strong acid and strong base (e.g. `NaCl`) do not hydrolyse, and thus they have `pH = 7` (neutral). Group I forms salts of strong acid `(S_(A))` and strong base `(S_(B))`. |
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| 195. |
Lattice energy of `BeCO_(3)(I), MgCO_(3)(II)` and `CaCO_(3)(III)` is in order.A. `I gt II gt III`B. `III gt II gt I`C. `II gt III gt I`D. `III gt I gt II` |
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Answer» Correct Answer - A All of the carbonates are of group `2` elements. Lattice energy decreases down the group `(darr)`. |
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| 196. |
Among `BECO_(3), CaCO_(3), SrCO_(3)` and `BaCO_(3)`, how many number than of these compounds are thermally more stable than `MgCO_(3)` ? |
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Answer» Correct Answer - 3 Only `BeCO_(3)` is less stable than `MgCO_(3)`. Since stability of the carbonates of group 2 increases down the group. |
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| 197. |
A potentiometer wire is 5m long and a potential difference of 6 V is maintained between its ends. Find the emf of a cell which balances against a length of 180 cm of the potentiometer wire. |
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Answer» Length of potentiometer wire L = 5m Potential difference V = 6 Volt Potential gradient ϕ = V/L = 6/5 = 1.2V/M Balancing length l = 180cm = 1.80m Emf of the cell E = ϕl = 1.2 × 1.8 = 2.16V |
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| 198. |
Prokaryotes are included in the kingdomA. ProtistaB. MoneraC. ProtozoaD. Plantae |
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Answer» Correct Answer - B Prokaryotes are included in the Monera kingdom. |
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| 199. |
Name the important groups of organisms included in prokaryotes. |
| Answer» The prokaryotes includes Bacteria, Cyanobacteria, Mycoplasma, Rickettsias, Actinomycetes and Archaebacteria. | |
| 200. |
Which human pathogen is a flagellate protozoan ?A. It is autotrophicB. It is filamentousC. It is microscopicD. It is prokaryotic |
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Answer» Correct Answer - C Nostoc is microscopic. |
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