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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 an ionic halide ?A. `UF_(4)`B. `PbCl_(2)`C. `SnCl_(2)`D. `UF_(6)` |
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Answer» Correct Answer - D `UF_6` has essentially covalent character. This is because small highly charged cations polarise the anions more efectively. For a metal with variable oxidation states, the halide with higher oxidation state of metal will be more covalent than the halide with lower oxidation of the metal. For example, `PbCl_6, SnCl_4 and UF_6`, are more covalent in nature than `PBCI_2, SnCl_2 and UF_4` respectively. |
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| 202. |
Which of the statement is true for the above sequence of reactions?A. `Z` is hydrogenB. `Y` is `LiBH_(4)`C. `Z` and `Y` and `F_(2)` and `B_(2)H_(6)` respectivelyD. `Z` is potassium hydroxide |
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Answer» Correct Answer - C `B(s)overset(F_(2) (Z))tounderset((X))(BF_(3))overset(LiH)tounderset((Y))(B_(2)H_(6))+LiBF_(4)` |
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| 203. |
Alumina is an ionic compound but insoluble in water becauseA. its lattice energy is very highB. its hydration energy is very highC. both lattice energy & hydration energy are sameD. none of these |
| Answer» Correct Answer - A | |
| 204. |
The correct sequence of the melting points of 16 elements isA. `H_(2)S gt H_(2)O gt H_(2)Se gt H_(2)Te`B. `H_(2)S gt H_(2)Se gt H_(2)Te gt H_(2)O`C. `H_(2)O gt H_(2)Te gt H_(2)Se gt H_(2)S`D. `H_(2)S gt H_(2)Te gt H_(2)Se gt H_(2)S` |
| Answer» Correct Answer - C | |
| 205. |
Select the incorrect statement of the following.A. Aluminium is often used as reducing agent for libration of other metals from their oxides.B. Anthydrous `AlCl_3` can be prepared by treating `Al_(2)O_(3)` with coke and chlorine gas.C. Aluminium readily dissolves in both dilute as well as in concentrated nitric acid.D. Aluminium hydroxide is soluble in both aqueous alkali and acids indicating its amphoteric nature. |
| Answer» Correct Answer - C | |
| 206. |
Diborane reacts with water to form:A. `HBO_(2)`B. `H_(3)BO_(3)`C. `H_(3)BO_(3)+H_(2)`D. `H_(2)` |
| Answer» Correct Answer - C | |
| 207. |
`Mg+BtoMg_(x)B_(y)overset("HCl")to`Diborane Report your answer as `(x+y)`. |
| Answer» Correct Answer - N//A | |
| 208. |
Which of the following statements is correct for diborane?A. Small amines like `NH_(3), CH_(3)NH_(2)` give unsymmetrical cleavage of diborane.B. Large amines such as `(CH_(3))_(3)N` and pyridine gives symmetrical cleavage of diborane.C. Small as well as large amines both gives symmetrical cleavage of diborane.D. (A) and (B) both |
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Answer» Correct Answer - D `B_(2)H_(6)+2NH_(3)to [H_(2)B(NH_(3))_(2)]^(+)+[BN_(4)]^(-)` `B_(2)H_(6)+2N(CH_(3))_(3)to2H_(3)B larrN(CH_(3))_(3)` |
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| 209. |
Diborane upon hydrolysis givesA. boric anhydride and polymetaborateB. metaboric acid and hydrogenC. orthoboric acid and hydrogenD. boron oxide and metaboric acid |
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Answer» Correct Answer - C `B_(2)H_(6)+6H_(2)O to underset("Orthoboric acid")(2H_(3)BO_(3))+6H_(2)` |
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| 210. |
Which has greater reactivityA. `TeCl_(6)`B. `SF_(6)`C. `TeF_(6)`D. `SeF_(6)` |
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Answer» Correct Answer - C `TeCl_6`, is unknown, because Te atom cannot accommodate six large Cl atoms around itself. `SF_6`, is extremely unreactive, because it is coordinately saturated and sterically hindered. `SeF_6`, is slightly more reactive, `TeF_6`, is hydrolysed by water. This is possibly due to larger size of Te, which permits the larger coordination number necessary in the first stage of hydrolysis `TeF_(2) +6H_(2)O to 6HF +H_(6)TeO_(6)` |
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| 211. |
Which of the elements does not show the O.S of +4?A. OB. BC. SeD. Te |
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Answer» Correct Answer - A Because of absence of d-orbital in this valence shell oxygen atom cannot expand its octet. |
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| 212. |
Pick out the incorrect statement.A. The tendency for catenation is marked by shown by sulphur.B. Te =C=Te is unknownC. The + 4 oxidation state is relatively more stable for Se, Te and Po than +6 O.S, but opposite trend holds good for S.D. `S_2` is diamagnetic, but `O_2` is paramagnetic |
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Answer» Correct Answer - D Like `O_(2), S_(2)`, is also paramagnetic, because it has two unpaired electrons a). S-S single bond strength is higher than O-O single bond. After oxygen. M-M single bond strength decreases from S to Po. b). The tendency to form multiple bonds with C, N and O decreases as we descend the group. c). In heavier elements, the tendency of `ns^(2)` electrons of outermost shell to remain paired and not to participate in bond formation increases as we descend the group. |
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| 213. |
Which one of the following has the highest bond energy ?A. O-OB. S-SC. Se-SeD. Te-Te |
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Answer» Correct Answer - B As the size increases from B to Te, the bond dissociation energy decreases O-O bond energy s, however, smallet than that of S-Sbond, because of the repulsive forces between the non-bonding electrons of srmaller sized oxygen atoms. |
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| 214. |
Which of the following shows large number of oxidation stateA. NB. PC. AsD. Bi |
| Answer» Correct Answer - A | |
| 215. |
Which of the following has highest bond dissociation energy?A. `NH_(3)`B. `AsH_(3)`C. `SbH_(3)`D. `BiH_(3)` |
| Answer» Correct Answer - A | |
| 216. |
Of the following elements, the one with maximum electropositive character isA. `Cl_(2)`B. `Br_(2)`C. `I_(2)`D. `F_(2)` |
| Answer» Correct Answer - C | |
| 217. |
Highest hydrogen bonding is possible inA. `BiH_(3)`B. `AsH_(3)`C. `NH_(3)`D. `SbH_(3)` |
| Answer» Correct Answer - C | |
| 218. |
Which of the following element do not form `dpi-p pi`A. NB. PC. AsD. Sb |
| Answer» Correct Answer - A | |
| 219. |
`ppi-ppi` bonding is possible inA. NB. PC. AsD. Sb |
| Answer» Correct Answer - A | |
| 220. |
Which one of the following is planar?A. `ClO_(2)^(-)`B. `ClO_(4)^(-)`C. `ClO_(3)^(-)`D. `IF_(7)` |
| Answer» Correct Answer - A | |
| 221. |
Maximum no. of atonms laying in a same planar in `B_(2)H_(6)` are :-A. 8B. 4C. 6D. 2 |
| Answer» Correct Answer - C | |
| 222. |
In which of the folowing molecules, vacant orbitals do not participate in bonding :-A. `B_(2)H_(6)`B. `Al_(2)Cl_(6)`C. `[H_(3)N.BF_(3)]`D. `Si_(2)H_(6)` |
| Answer» Correct Answer - D | |
| 223. |
Aluminium can be prepared by:A. Electrolytic reduction of aluminia in presence of Cryolite & fluorspar.B. Reduction of `AlCl_3(s)` by potassium amalgam.C. Reduction of aqueous solution of `Al_(2)(SO_(4))_(3) (aq)` by zinc metalD. Thermal decomposition of aluminium oxide. |
| Answer» Correct Answer - A::B | |
| 224. |
Boron can be obtained by:A. reduction of `B_(2)O_(3)` by `C`.B. reduction of `BCl_(3)` with `H_(2)` at `1270k`.C. thermal decomposition of boron halides at `1173 k`.D. electrolytic reduction of `KBF_(4)` in `KF` at `1073 k`. |
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Answer» Correct Answer - B::C::D (B)`2BCl_(3)+3H_(2)overset(1270 K)to 2B+6HCl` ( C)`2BCl_(3)overset(1173 K)toB+3Cl_(2)` (D)`K^(+)[BF_(4)]^(-) overset(1073 K) hArr K^(+)+[BF_(4)]^(-)` `[BF_(4)]^(-) + e^(-) to B +2F_(2)` (At anode) `K^(+) + e^(-) to K` (At cathode) |
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| 225. |
Which reaction is not valid ?A. `HCl +F_(2) to HF+Cl_(2)`B. `HF+Cl_(2) to F_(2)+HCl`C. `ZN+HCl to ZnCl_(2)+H_(2)`D. `Al+HCl to AlCl_(3)+H_(2)` |
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Answer» Correct Answer - B Fluorine is more electronegative than chlorine hence, it can displace Cl from Cl from HCl while chlorine cannot displace fluroine from HF. Therefore the following reaction is not valid. `HF+Cl_(2) to F_(2)+HCl` |
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| 226. |
The electrolysis of a certain liquid resulted in the formation of hydrogen at the cathode and chlorine at the anode. The liquid is (EAMCET 1979)A. Pure waterB. `H_2SO_4` solutionC. NaCl solution in waterD. `CuCl_2` solution in water |
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Answer» Correct Answer - C `NaCl+2H_2O to 2NaOH+ underset(("anode"))(Cl_(2))+underset(("Cathode"))(H_(2)` |
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| 227. |
What products are expected from the reaction between colemanite powder and sodium carbonate solution, when they are heated?A. `CaCO_(3)`B. `Na_(2)B_(4)O_(7)`C. `NaBO_(2)`D. `CaO` |
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Answer» Correct Answer - A::B::C `Ca_(2)B_(6)O_(11)+2Na_(2)CO_(3)to CaCO_(3)darr+Na_(2)B_(4)O_(7)+NaBO_(2)` |
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| 228. |
chlorine can be manufactring fromA. Electrolysis of NaClB. Electrolysis of brineC. Electrolysis of bleaching powderD. All of these |
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Answer» Correct Answer - B `2NaCl+2H_2O overset("Electrolysis")to underset(("aq."))(2NaOH)+underset((g))(Cl_(2))+underset((g))(H_(2))` |
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| 229. |
Sodium bismuthate isA. a powerful reducing-agentB. used in the estimation of `Mn^(2+)` ionsC. a non-toichiometric compoundD. obtaine3d by treating `Bi_2O_3 "with conc. "NaOH` |
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Answer» Correct Answer - B Oxidation of `Mn^(2+)` ions is carried out quantitively to `MnO_(4)^(-)` by warming with sodium bismuthate in excess of `HNO_(3)` `2Mn^(2+)+5NaBiO_(3)+14H^(+) to 2MnO_(4)^(-)+5Na^(-)+5Bi^(3+)+7H_(2)O` (a). It is a strong oxidizing agent as shown in (b) (c) It is a stoichiometric compound (d). `BiO_(3)+6NaOH to 2Na_(3)Bi_3+3H_(2)O` |
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| 230. |
Large0Scale manufactring of nitric acid by Ostwald process utilizes the reactionA. `2NaNO_(3)+H_(2)SO_(4) to Na_(2)SO_(4)+2HNO_(3)`B. `4NH_(3)+5O_(2) to 4NO+6H_(2)O`C. `NO_(2)^(+)+NO_(3)^(-)+H_(2)O to 2HNO_(3)`D. `2NO+O_(2)+H_(2)O to HNO_(3)+HNO_(2)` |
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Answer» Correct Answer - B `4NH_(3)+5O_(2) underset("Catalyst")overset(Pt//Rh)to 4NO+6H_(2)O` `2NO+O_(2) overset(1120K)to 2NO_(2)` `3NO_(2)+H_(2)O to 2HNO_(3)+NO` |
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| 231. |
Which one of the following has highest enthalpy of hydration?A. `F_(2)`B. `Cl_(2)`C. `Br_(2)`D. `I_(2)` |
| Answer» Correct Answer - A | |
| 232. |
Standard electrode potential is highest forA. `1/2 F_(2)(g)+e^(-) Leftrightarrow F^(-)(aq)`B. `1/2 Cl_(2)(g)+e^(-) Leftrightarrow Cl^(-)(aq)`C. `1/2 Br_(2)(g)+e^(-) Leftrightarrow Br^(-)(aq)`D. `1/2 I_(2)(g)+e^(-) Leftrightarrow I^(-)(aq)` |
| Answer» Correct Answer - A | |
| 233. |
Highest oxidation state of oxygen isA. `+2`B. `+4`C. `+5`D. `+6` |
| Answer» Correct Answer - A | |
| 234. |
Potassium nitrate on thermal decomposition givesA. `N_2`B. `O_(2)`C. `H_(2)`D. `O_(3)` |
| Answer» Correct Answer - B | |
| 235. |
Milk of sulphur is obtained byA. passing `H_(2)S` through `HNO_(3)`B. the reaction of `Na_(2)S_(2)O_(3)" with "HCl`C. melting sulphur in a dishD. boiling milk of lime with sulphur and then with HCl. |
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Answer» Correct Answer - D Milk of lime is boiled with sulphur and then calcium penta sulphide so-formed is decomposed with HCl to get white amorphous precipitate of sulphur. `underset(("Milk of lime"))(3Ca(OH)_(2))+12S to underset(("Milk of sulphur"))(2CaS_(5)+CaS_(2)O_(3)+3H_(2)O)` |
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| 236. |
Borax is:A. `Na_(2)B_(4)O_(7)`B. `Na_(2)B_(4)O_(7).4H_(2)O`C. `Na_(2)B_(4)O_(7).7H_(2)O`D. `Na_(2)B_(4)O_(7).10H_(2)O` |
| Answer» Correct Answer - D | |
| 237. |
When borax is dissolved in water:A. `B(OH)_(3)` is formed onlyB. `[BH_(2)(NH_(3))_(2)]^(+) [BH_(4)]^(-)` is formed onlyC. both `B(OH)_(3)` and `[B(OH)_(4)]^(-)` are formedD. `[B_(3)O_(3)(OH)_(4)]^(-)` is formed only |
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Answer» Correct Answer - C `[B_(4)O_(5)(OH)_(4)]^(2-)+5H_(2)OhArr2B(OH)_(3)+2[B(OH)_(4)]^(-)` or `[B_(4)O_(7)]^(2-)+7H_(2)Oto2B(OH)_(3)+2[B(OH)_(4)]^(-)` |
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| 238. |
How will you obtain? Borazole from sodium borohydride (in three steps only) |
| Answer» Correct Answer - A::B::C::D | |
| 239. |
What is inorganic benzene ? Why is it so called ? How will you get it from diborame ? |
| Answer» Correct Answer - A::B::C::D | |
| 240. |
Which of the following industrial chemicals is produced in the greatest amount annually ?A. `HNO_(3)`B. `H_(3)PO_(4)`C. `H_(2)`D. `H_(2)SO_(4)` |
| Answer» Correct Answer - D | |
| 241. |
About half of `H_3SO_4` produced in world is used toA. Manufacture of soapB. Manufacture of plasticsC. Manufacture of paintsD. Manufacture of fertilizers |
| Answer» Correct Answer - D | |
| 242. |
`H_2O_2` can be usedA. both an oxidizing and as a reducing agentB. only as an oxidizing agentC. only as a reducing agentD. neither as an oxidizing agent nor as a reducing agent |
| Answer» Correct Answer - A | |
| 243. |
Which one of the following is used for drying of ammonia ?A. CaOB. Anhydrous `CaCl_2`C. `P_2O_5`D. `"Conc. "H_SO_4` |
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Answer» Correct Answer - A Except CaO, all react with `NH_3` Anhy.. `CaCl_2," forms "[Ca(NH_3)_8] Cl_2," with "NH_3. P_2O_5`, is acidic in nature and `H_3SO_4` is a strong acid and therefore, these react with `NH_3.` |
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| 244. |
`H_2O_2` cannot act asA. oxidizing agent onlyB. reducing agent onlyC. both oxidizing and reducing agentD. nitrating agent |
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Answer» Correct Answer - C Derivatives of nitrogen (III) have both oxidizing and reducing properties. `underset(("Oxidant"))(2Naoverset(+3)(NO_(2)))+2KI+2H_(2)SO_(4) to I_(2)+2overset(+2)(NO)+K_(2)SO_(4)+Na_(2)SO_(4)+2H_(2)O` `2KMnO_(4)+5Naoverset(+3)(NO_2)+3H_(2)SO_(4) to 2MnSO_(4)+5Na overset(+5)(NO_(3))+K_2SO_(4)+3H_(2)O` |
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| 245. |
Pick out the incorrect statement w.r.t. `NH_3`A. It contains a lone pair of electrons, which can bonds to a proton to form tetrahedra `NH_(4)^(+)` ionsB. `N_2` is formed, when `NH,_3` is passed over heated copper (II) oxideC. `NH_3` burns in air to form `N_2` and steamD. In Ostwald process for the manufacture of `HNO_3, NH_3,` is oxidized in presence of Pt/Rh catalyst to give `NO and H_2O` |
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Answer» Correct Answer - C `NH_3`, shows no reaction towards air. It burns with a pale green flame in oxygen. `4NH_3+3O_(2) to 2N_(2)+6H_(2)` a) NH, molecule is trigonal pyramidal is shape with a lone pair of electron in one of the four `sp^3` hybridized orbitals. When it is bonded to `H^+`, it forms tetrahedral `NH_(4^(+))` ions, because all the four sp hybrid orbitals possess four bonded pairs of electrons. b). `2NH_(3)+3CuO to N_(2)+3H_(2)O+Cu` d). `4NH_(3)+5O_(2) overset(Pt//Rh)to 4NO+6H_(2)O` |
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| 246. |
Arrange the oxides of group `15` elements in decreasing order of their acidityA. `N_(2)O_(5) gt P_(2)O_(5) gt As_(2)O_(5) gt Sb_(2)O_(5) gt Bi_(2)O_(5)`B. `Bi_(2)O_(5) gt Sb_(2)O_(5) gt As_(2)O_(5) gt P_(2)O_(5) gt N_(2)O_(2)`C. `P_(2) O_(5) gt N_(2)O_(5) gt As_(2)O_(5) gt Sb_(2)O_(5) gt Bi_(2)O_(5)`D. `N_(2)O_(5) gt Bi_(2)O_(5) gt P_(2)O_(5) gt As_(2)O_(5) gt Sb_(2)O_(5)` |
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Answer» Correct Answer - A All the oxides of nitrogen (expect NO and `N_(2)O "neutral oxides")` and phosphorus are strongly acidic in nature , oxides of arsenic are weakly acidicl oxides of antimony are amphoteric and those of bismuth are weakly basic. As the the size of nitrogen atom is small and it has a more strong positive field, so it interacts with water, more strongly pulling the electron pairs between O-H and thus helps in hte release of `H^(+)` ions. However, this tendency diminishes with the increases in size and therefore, decreases the acidic character or conversely increases the basic character. |
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| 247. |
The most efficient agent for the absorption of `SO_(3)` isA. `80% H_(2)SO_(4)`B. `98% H_(2)SO_(4)`C. `50% H_(2)SO_(4)`D. `20% H_(2)S_(2)O_(7)` |
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Answer» Correct Answer - B 98% `H_(2)SO_(4)` is used for absorbing dense fog of acid which is formed by dissolving `SO_(3)` in water. Hence 98% `H_(2)SO_(4)` is the most efficient agent for the absorption of `SO_(3)`. |
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| 248. |
The oxidation state of `Pt` in `Xe^(+)[Ptf_(6)]^(-)` isA. `+4`B. `+5`C. `+6`D. `+7` |
| Answer» Correct Answer - B | |
| 249. |
When orthoboric acid `(H_(3)BO_(3))` is heated, the residue isA. boronB. metaboric acid and hydrogenC. boric anhydrideD. borax |
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Answer» Correct Answer - C `2H_(3)BO_(3)overset(Delta ("red hot"))toB_(2)O_(3) + 3H_(2)O` |
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| 250. |
Colloidal sulphur is obtained whenA. sulphuris heated gradually to a high temperatureB. sulphur is heated with `Ca(OH)_2`C. hydrogen sulphide gas is passed through an aqueous solution of nitric acidD. sulphur is warned with `CS_(2)` |
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Answer» Correct Answer - C Sulphur is precipitated in the colloidal form, when HS is bubbled through as aq. solution of `HNO_3` or (Milk of lime) when `Na_2SO_3`, is treated with HCI. `2HNO_(3) to 2NO_(2)+H_(2)O+O` `H_(2)S+O to S+H_(2)O` =`2HNO_(3)+H_(2)S to S+2NO_(2)+2H_(2)O` `or Na_(2)S_(2)O_(3)+2HCl to 2NaCl+SO_(2)+S+H_(2)O` |
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