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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.
| 801. |
On boiling an aqueous solution of `KClO_(3)` with iodine, the following product is obtainedA. `KIO_(3)`B. `KClO_(4)`C. `KIO_(4)`D. `KCl` |
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Answer» Correct Answer - A `2KClO_(3)+I_(2)rarr2KlO+_(3)+Cl_(2)` |
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| 802. |
Arrange the following in the order of property indicated for each set : (i) `F_(2), Cl_(2) Br_(2),I_(2)`- increasing bond dissociation enthalpy. HF,HCl, HBr, HI- increasing acid strength. `NH_(3), PH_(3), AsH_(3), SbH_(3), BiH_(3)`- increasing base strength. |
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Answer» Correct Answer - (i)Bond dissociation enthalpy should decrease as the bond distance increases from `F_(2)` to `I_(2)` due to the corresponding increase in the size of the atom as we move from `F` to `I`.However, the `F-F` bond dissociation enthalpy is smaller than that of `Cl-Cl` and even smaller than that of `Br-Br`.This is due to the reason that the `F` atom is very small and hence the three lone pairs of electrons on each `F` atom repel the bond pair holding the `F`-atom in `F_(2)` molecule. Thus, the bond dissociation enthalpy increases in the order : `I_(2)ltF_(2)ltBr_(2)ltCl_(2)`. (ii)The relative acid strength of `HF , HCl HBr` and `Hl` depends upon their bond dissociation enthalpies. Since the bond dissociation enthalpy of `H-X` bond decreases from `H-F` to `H-I` as the size of atom increases from `F` to `l`.Therefore, acid strength increases in the opposite order, i.e., the acid strength increases in the order : `HFltHClltHBrltHI`. (iii)Due to the presence of a lone pair of electrons on the central atom in `NH_(3), PH_(3), AsH_(3), SbH_(3)` and `BiH_(3)` all behave as Lewis bases. However, as we move from `NH_(3)` to `BiH_(3)`.the size of the atom increases.As a result, the lone pair of electrons occupies a larger volume. In other words, the electron density on the central atom decreases and hence the basic strength decreases as we move from `NH_(3)` to `BiH_(3)` Thus, the basic strength increases in the order : `BiH_(3)ltSbH_(3)ltAsH_(3)ltPH_(3)ltNH_(3)` |
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| 803. |
Arrange the following in the order of the property indicated for each set. `F_(2),Cl_(2),Br_(2),I_(2)`- increasing bond dissociation enthalpy. |
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Answer» Increasing bond dissociation enthalpy order `I_(2) lt F_(2) lt Br_(2) lt Cl_(2)` |
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| 804. |
What is the colour of `C_(60)` in toluene ? |
| Answer» Correct Answer - Purple colour. | |
| 805. |
Which of the following compound is not hydrolysed?A. `SiCl_(4)`B. `SiF_(4)`C. `C Cl_(4)`D. `PbC l_(4)` |
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Answer» Correct Answer - C `C Cl_(4)` is not hydrolysed. |
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| 806. |
What happens when `K_(4)Fe(CN)_(6)` is heated strongly ? |
| Answer» Correct Answer - `K_(4)Fe(CN)_(6)overset(Delta)rar r4KCN+FeN_(2)` | |
| 807. |
What happens when calcium cynamide is hydrolysed ? |
| Answer» Correct Answer - `CaCN_(2)+H_(2)Orar rCaCO_(3)+NH_(3)+H_(2)O` | |
| 808. |
The species that is not hydrolysed in water isA. `P_5 O_(10)`B. `BaO_2`C. `Mg_(3)N_(2)`D. `CaC_(2)` |
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Answer» Correct Answer - A Chemical reaction occurs when `P_(4)O_(10),Mg_(3)N_2,CaC_2 and BaO_2` reacts with water are as shown below : `P_4O_(10)+6H_2O to 4H_3PO_(4)` `Mg_3N_(2)+6H_(2)O to 3Mg(OH)_(2)+2NH_(3)` `CaC_(2)+2H_2O to Ca(OH)_(2)+C_(2)H_(2)` `BaO_(2)+2H_(2)O to Ba(OH)_(2)+H_(2)O_(2)` Hence , `P_4O_(10)` is not hydrolysed in water . |
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| 809. |
Match the reactions listed in column-I with the products listed in column-II |
| Answer» Correct Answer - A-q,s ; B-s ; C-p ; D-r | |
| 810. |
Match the reactions listed in column-I with the products listed in column-II |
| Answer» Correct Answer - A-p,q ; B-p,r ; C-q,s ; D-p,q,r,s | |
| 811. |
The correct order of acidic strength isA. `HFltHClltHBrltHI`B. `K_(2)Cr_(2)O_(7)+MnO_(2)`C. `Pb_(2)(NO_(3))_(4)+MnO_(2)`D. `K_(2)Cr_(2)O_(7)+HCl` |
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Answer» Correct Answer - A `HFltHClltHBrltHl` As we go down the group bond energy decreases, hence acidic nature increases. |
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| 812. |
Which of the following statement is correct.A. (A) All interhalogen compounds are gas at room temperature.B. (B) interhalogen are either gas or liquid at room temperature.C. (C) Interhalogens can be solid or liquid or gas at room temperature.D. (D) All interhalogen compounds are liquid at room temperature. |
| Answer» Correct Answer - C | |
| 813. |
Metal halide which is insoluble in water isA. `Agl`B. `KBr`C. `CaCl_(2)`D. `AgF` |
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Answer» Correct Answer - A `Agl` is a covalent compound. |
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| 814. |
The mixture of concentrated `HCl` and `HNO_(3)` made in `3:1` ratio containsA. `ClO_2`B. `NOCl `C. `NCl_3`D. `N_2O_4` |
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Answer» Correct Answer - B The mixture of one part of conc. `HNO_3` and three parts of conc. `HCl` is known as aqua-regia . It contains NOCl. `HNO_3 +3HCl to 2H_2O+NOCl +2[Cl]` |
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| 815. |
In the inter halogen compounds of `AB_3//AB_5` form which is correct:A. (A) A is large size halogenB. (B) B is large size halogenC. (C)B is small size halogenD. (D) Both (A) & (C) |
| Answer» Correct Answer - D | |
| 816. |
Which of the following behaves like pseudohalogen compound:A. (A) NC CNB. (B) `CN^-`C. (C) `N_3^-`D. (D) `l_3^-` |
| Answer» Correct Answer - A | |
| 817. |
Decreasing order of stability of `O_(2), O_(2)^(-), O_(2)^(+)` and `O_(2)^(2-)` isA. `O_(2)^(+) gt O_(2) gt O_(2)^(-) gt O_(2)^(2-)`B. `O_(2)^(2-) gt O_(2)^(-) gt O_(2) gt O_(2)^(-)`C. `O_(2) gt O_(2)^(+) gt O_(2)^(2-) gt O_(2)^(-)`D. `O_(2)^(-) gt O_(2)^(2-) gt O_(2)^(+) gt O_(2)` |
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Answer» Correct Answer - A Order of stability `prop` bond order `:.` Order of th stability of given species `O_(2)^(+) gt O_(2) gt O_(2)^(-) gt O_(2)^(2-)` Bond `2.5 " " 2 " " 1.5 " " 1`. |
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| 818. |
Explain the following (a) `CO_(2)` is a gas whereas `SiO_(2)` is a solid (b) Silicon forms `SiF_(6)^(2-)` ion whereas corresponding fluoro compound of carbon is not known. |
| Answer» (ii) Silicon atoms has vacant 3d orbtials in its valence shell. It is in a position to accept electron pairs from six `F^(-)` ions and expand its co-ordination number upto 6. Therefore, the compound `SiF_(6)^(2-)` can exist. on the other hand, carbon has no vacant 3d orbitals and cannot expand its co-ordination number upto six. tehrefore, the compound `CF_(6)^(2-)` does not exist. | |
| 819. |
Complete the following `C+H_(2)SO_(4)("Conc")to` |
| Answer» `C+2H_(2)SO_(4)("Conc")toCO_(2)+2SO_(2)+2H_(2)O` | |
| 820. |
Complete the following `Cu+H_(2)SO_(4)("Conc")to` |
| Answer» `Cu+2H_(2)SO_(4)("Conc")toCuSO_(4)+SO_(2)+2H_(2)O` | |
| 821. |
Complete the following `"Sucrose"overset("Conc.H"_(2)SO_(4))to` |
| Answer» `C_(12)H_(22)O_(11)overset("Conc.H"_(2)SO_(4))to12C+11H_(2)O` | |
| 822. |
A gas which exists in three allotropic forms `alpha,beta` and `gamma` is :A. `SO_(2)`B. `SO_(3)`C. `CO_(2)`D. NH_(3)` |
| Answer» Correct Answer - B | |
| 823. |
Three allotropes (A),(B) and ( C) of phosphorous in the following changes are respectively. A. White, `beta`-black, redB. `beta`-black, white, redC. Red, `beta`-black, whiteD. Red, violet, `beta`-block |
| Answer» Correct Answer - A | |
| 824. |
A scarlet red compound (X) on treatment with conc. `HNO_(3)` gives compounds (Y) and (Z). (Z) with HCL produces a chloride compond (A) which can also be produced by treating (X) with conc HCL. Compounds (X), (Z) and (A) will beA. `Mn_(3)O_(4),MnO_(2),MnCl_(2)`B. `Pb_(3)O_(4),PbO_(2),PbCl_(2)`C. `Fe_(3)O_(4),Fe_(2)O_(3),FeCl_(2)`D. `Fe_(3)O_(4),Fe_(2)O_(3),FeCl_(3)` |
| Answer» Correct Answer - B | |
| 825. |
Which of the following is not oxidised by `MnO_(2)` ?A. `F^(-)`B. `Cl^(-)`C. `Br^(-)`D. `I^(-)` |
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Answer» Correct Answer - A |
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| 826. |
`F_(2)` is formed by reacting `K_(2) MnF_(6)` withA. `SbF_(5)`B. `MnF_(3)`C. `KSbF_(6)`D. `MnF_(5)` |
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Answer» Correct Answer - A |
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| 827. |
Which of the following is the correct order of acidic strenght ?A. `Cl_(2)O_(7) gt SO_(3) gt P_(4)O_(10`B. `CO_(2) gt N_(2)O_(5) gt SO_(3)`C. `NaO gt MgO gt Al_(2)O_(3)`D. `K_(2)O gt CaO gt MgO` |
| Answer» Correct Answer - A | |
| 828. |
`Ca+C_(2) toCaC_(2)overset(N_(2))toA` Compared `(A)` is as a`//`anA. fertilizerB. dehydration agentC. oxidising agentD. reducing agent |
| Answer» Correct Answer - A | |
| 829. |
`RCIunderset(Si)overset(cu-powder)toR_(2)SiCL_(2)overset(H_(2)O)toR_(2)Si(OH)_(2)overset(condensation)toA` Compound (A) isA. a linear siliconeB. a chlorosilaneC. a linear silaneD. a network silane |
| Answer» Correct Answer - A | |
| 830. |
Noble gases have compleately filled valance shall i.e. `m^(2)sp^(2)` exceps He (i.e) .Noble gases are monoomic under normal conductions .Law bolding point of the ligher noble gases are due to weak van dor wads forces between the atoms and abance of any interature imaractions `Xe` reacts with `F_(2)` so give a sourceof fouoxide mently `XeF_(2),XeF_(4),XeF_(4),XeF_(3)` on complete hydrolyes gives `XeFe_(3)`, `XeF_(4)` and `XeF_(4)` are expected to beA. oxidisingB. reducingC. unreactiveD. strongly basic |
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Answer» Correct Answer - B |
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| 831. |
The noble gases have closed-shell electronic cordigaration and are monatomic gases under normal condition .The low bolling points of the ligher noble gases aree due to the weak dispersion points of the ligher noble gases an due to the weak dispersion forces between the atoms and the alsence of other interalumic interactions. The direct reaction of xenon with flarine loads to a series of compounds with water oxidation number `+2, - 4 and +6, XeF_(4)` reactsviolenatly with water to give `XeO_(2)` .The compound of deduced axbibt nci strouchemistry and their goometries can be deduced considering the total number of electron puirs in the valence shell. The structure of `XeO_(3)` isA. linearB. planarC. pyramidalD. T-shaped |
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Answer» Correct Answer - C |
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| 832. |
The chemical composition of cyolite mineral is:A. `Al_(2)O_(3)`B. `Al_(2)O_(3).12H_(2)O`C. `KAISI_(3)O_(6)`D. `Na_(3)AIF_(6)` |
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Answer» Correct Answer - D Cryolite is `Na_(3)AIF_(6)`. |
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| 833. |
A metal ion `M^(3+)` loses three electrons , its oxidation number will beA. `+3`B. `+6`C. `zero`D. `-3` |
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Answer» Correct Answer - B The metal (M) has lost six electrons in all, its oxidation state is+6. |
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| 834. |
Which of the following is a compound of ruby ?A. `CaCO_(3)`B. `MgCO_(3)`C. `Al_(2)O_(3)`D. `Al(OH)_(3)` |
| Answer» (c ) `Al_(2)O_(3)` is a compound of ruby (`Al_(2)O_(3)+Cr_(2)O_(3)`),red precious stone | |
| 835. |
What is Tincal?A. `Na_(2)CO_(3)cdot10H_(2)O`B. `NaNO_(3)`C. `Na_(2)B_(4)O_(7)cdot10H_(2)O`D. NaCl |
| Answer» (c ) Naturally occuring crude borax is called tincal. Thus, it is chemically `Na_(2)B_(4)O_(7)cdot10H_(2)O`. | |
| 836. |
The hybridisation of the central atom in `SiF_(6)^(2-),[GeCl_(6)]^(2-)` and `[Sn(OH)_(6)]^(6-)` isA. `sp^(3)d`B. `sp^(3)d^(2)`C. `sp^(3)`D. `sp^(3)d^(3)` |
| Answer» (b) The species like `SiF_(6)^(2-),[GeCl_(6)]^(2-)`, `[Sn(OH)_(6)]^(6-)` exist where the hybridastion of the central atom is `sp^(3)d^(2)` | |
| 837. |
which one of the following oxides is neutral?A. `CO`B. `SnO_(2)`C. `ZnO`D. `SiO_(2)` |
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Answer» Correct Answer - A |
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| 838. |
The noble gas mixture is cooled in a coconut bulb at 173K. The gases that are not adsorbs areA. Ne and XeB. He and XeC. Ar and KrD. He and Ne |
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Answer» Correct Answer - C Among the C-X bond ( where , X=Cl, Br, I), the correct decreasing order of bond energy is `C-Cl gt C -Br gt C-I` . |
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| 839. |
Which of the following elements from both neutral as well as acidic oxides ?A. SnB. SiC. CD. P |
| Answer» (c ) CO is neutral and `CO_(2)` is acidic. | |
| 840. |
`H_(2)O` is neutral while `H_(2)S` is acidic - explain. |
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Answer» `H_(2)O` is neutral while `H_(2)S` is acidic. Reason : The O-H bond dissociation Enthalpy is greater than the S - H bond dissociation Enthalpy. |
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| 841. |
`SO_(2)` does not act as a/anA. bleaching agentB. oxidising agentC. reducing agentD. dehydrating agent |
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Answer» Correct Answer - D `SO_2` acts as reducing agent , oxidising agent and as a bleaching agent . It does not act as dehydrating agent . |
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| 842. |
Why is `H_(2)O` a liquid while `H_(2)S` is a gas ? |
| Answer» `H_(2)O` is liquid due to the presence of intermolecular hydrogen bonding. While `H_(2)S` is gas because it is not having such type of bonding. | |
| 843. |
Oxygen molecule has the formula `O_(2)` while sulphur has `S_(8)` - explain. |
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Answer» `to` Due to small atomic size and high electronegativity oxygen forms `P pi-P pi` multiple bond and exists as `O_(2)` molecule. `to` Due to large atomic size and less electronegativity sulphur forms strong S - S single bonds and exists `S_(8)` molecule. |
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| 844. |
Sulphur does not exist as `S_2` molecule becauseA. it is less electronegativeB. it is not able to constitute `p pi- p pi `bondsC. it has ability to exhibit catenationD. of tendency to show variable oxidation states |
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Answer» Correct Answer - B Sulphur does not form `p pi - p pi ` bond due to its larger size , hence does not exists as `S_2` molecules. |
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| 845. |
Sulphur molecule is converted into sulphur ion, when itA. Gains two electronsB. Loses two electronsC. Gains two protonsD. Shares two electrons |
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Answer» Correct Answer - A `(1)/(8)S_(8)+2_(e)^(-)rarrS^(2-)` |
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| 846. |
The most stable allotropic form of sulphur is:A. rhombicB. monoclinicC. plasticD. milk of sulphur |
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Answer» Correct Answer - A `The most stable allotropic of sulphur is rhombic |
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| 847. |
Which of following elements is highest electronegative?A. `O`B. `S`C. `Te`D. `Se` |
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Answer» Correct Answer - A `{:("Element",O,S,Sc,Te,Po),("Electronegativity",3.5,2.5,2.4,2.1,2.0):}` |
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| 848. |
Which of following elements is highest electronegative?A. `S`B. `Se`C. `Te`D. `O` |
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Answer» Correct Answer - D `O` is highest electronegative element among `S`, `Se`, `Te` and `O`. |
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| 849. |
The hybridization of C in diamond, graphite and ethyne is in the orderA. `sp^(3), sp , sp^(2)`B. `sp^(3), sp^(2)`, spC. sp, `sp^(2), sp^(3)`D. `sp^(2), sp^(3), sp` |
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Answer» Correct Answer - B `sp^(3)` (diamond), `sp^(2)` (graphite), sp( ethyne). |
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| 850. |
Which is true regarding nitrogen ?A. Less electronegativityB. Has low ionisation enthalpyC. d-orbitals are availableD. Ability to forom `ppi-ppi` bonds with iteslf |
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Answer» Correct Answer - D Because of its samll size and high electronegativity N has a strong tendency to form `ppi-ppi` bonds with itself. It has high ionisation energy and does not have d-orbitals. |
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