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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.
| 551. |
Assertion : `N_(2)O` is respented by (i) `N = N = O` and (ii) `N= N rarr O` but the latter is more stable. Reason :From (ii) shown resonance.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 assertion is false but reason is true. |
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Answer» Correct Answer - c From (ii) is more stable due to lesser formed on N- atom. |
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| 552. |
Which of the following statement `(s)` is (are) true ? .A. `CuCI` is more covalent than `NaCI`B. `HF`is more polar than `HBr`C. `HF` is less polar than `HBr`D. Cemical bond formation takes plane when forces of attraction overcome the forces of repulsion . |
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Answer» Correct Answer - A::B::D `HF` is more polar than `HBr` as `EN` of `F` is more than that of `Br` ` CuCI` is more covalent than `NaCI` as 18-electron shell of `Cu^(o+)([Ar]3d^(10))` is more `EN` because inner electrons have poor shielding effect on nucleus increasing the polarising power of nucleus . |
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| 553. |
Assertion : In `IOF_(4)^(-)` a single lone pair is present an iodine atom trans to oxygen to have minimum repulsion between the `I=0` and the lone pair of electrons. Reason : The VSEPR model consider double and triple bonds to have slightly greater repulsive effect then single bonds bonds because of the repulsive effective `pi` electronsA. 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 assertion is false but reason is true. |
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Answer» Correct Answer - a Assertion : To have minimum repulsion between the `I= O` and the lone pair of electron (has greater replusion effect), the lone pair is opposite to the `I= O `bond. Reason : Correct statement. |
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| 554. |
Which among the following are isostructural ? .A. `XeO_(2)F_(2),SF_(4)`B. `CO_(2),I_(3)^(Θ)`C. `SO_(3)^(2-),CO_(3)^(2-)`D. `CIF_(3),XeF_(2)` |
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Answer» Correct Answer - A::B (a) `:SF_(4) implies` Hybridisation `implies (1)/(2) (6 +4) = 5 = sp^(3) d` hybidised with one lp Geometry =Tbp Shape =See-saw `:XeO_(2)F_(2)implies` Hybridisation `implies (1)/(2) (8 +2)= 5 = sp^(3) d` Hybridised and has saem structure and shape as of `SF_(4)` (b) `CO_(2)` and `I_(3)^(Θ)` are sp and `sp^(3)` d hybridised but both of them have linear shape . |
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| 555. |
Out of `NaCl and MgO`, which has higher lattice energy and why ? |
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Answer» MgO has higher lattice energy because each ion carries two unit charge whereas in NaCl each ion carries one unit charge . |
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| 556. |
Which one of the following statements about water is false ?A. Watr is oxidized to oxygen during photo-synthesisB. Water can act both as an acid and as a baseC. There is extensive intramolecular hydrogen bonding in the condensed phaseD. Ice formed by heavy water sinks in normal water |
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Answer» Correct Answer - C In the condensed phase, there is extensive intermolecular hydrogen bonding in water molecules but there is no intramolecular hydrogen bonding |
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| 557. |
Which one of the following molecule (s) is (are) expected to exhibit diamagnetic behaviour ?A. `S_(2)`B. `C_(2)`C. `N_(2)`D. `O_(2)` |
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Answer» Correct Answer - B::C In`C_(2) and N_(2)` , all molecular orbitals are fully filled, i.e., there is no unpaired electron present . Hence both of them are diamagnetc. |
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| 558. |
`CO_(2) ` is isostructural withA. `HgCl_(2)`B. `SnCl_(2)`C. `C_(2) H_(2)`D. `NO_(2)` |
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Answer» Correct Answer - A::C Like `CO_(2),HgCl_(2) and C_(2)H_(2)` have sp hybridization and hence are isostructural ,i.e.,linear |
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| 559. |
Which of the following are tetrahedral structures ? .A. `[Ni(CN)_(4)]^(2-)`B. `[Ni(CO)_(4)]`C. `[NiCI_(4)]^(2-)`D. `CrO_(4)^(2-)` |
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Answer» Correct Answer - B::C::D (b),(c )and (d) are `sp^(3)` hybridised but (a) is `sp^(2)` hybridised . |
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| 560. |
`AlCl _(3)` is …………. Compound whereas `PCl_(5)` is …………… compound in terms of octer rule. |
| Answer» electron-deficient, hypervalnet | |
| 561. |
Which of the following does not have a coordinate bond ?A. `SO_(2)`B. `HNO_(3)`C. `H_(2)SO_(3)`D. `HNO_(2)` |
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Answer» Correct Answer - d `NHO_(2)` does not have co-ordinate bond Structure is `H - O - H = O`. |
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| 562. |
Some ether is added to anb aqueous soluetion of a mixtrue of `LiCl, NaCl and AlCl_(3)` . Which will be extracted into ether ?A. `LiCl, NaCl`B. `LiCl, AlCl_(3)`C. `NaCl, AlCl_(3)`D. `LiCl, NaCl, AlCl_(3)` |
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Answer» Correct Answer - B LiCl and `AlCl_(3)` have large covalent character and hence extracted into ether. |
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| 563. |
Out of `NO,NO^(o+)` and `CN^(o+)` the paramagnetic species is `NO^(o+)` . |
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Answer» Correct Answer - F NO is paranagnetic |
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| 564. |
The bond between two identical non-metal atoms has a pair of electrons:A. Unqually shared between the twoB. Transferred fully between the twoC. With identical spinsD. Equally shared between them |
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Answer» Correct Answer - D In covlaent bonds between bonds between two identical no-metal atoms share the pair of electrons equally between then e.g `F_(2),O_(2),N_(2)` . |
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| 565. |
Which of the following is a correct statement ? .A. Mobility of `H^(o+)` ions in ice is greater as compared to liquid waterB. Mobility of `H^(o+)` ions in ice is less as compared ti liquid waterC. Mobility of `H^(o+)` ions in ice is equal to that in liquid water .D. Cannot be predicted |
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Answer» Correct Answer - A In liquid water `H^(o+)` ions (obtained from ionisation of `H_(2)O)` are easily hydrated As a result the size of the hybrated `H^(o+)` ions is much larger than in ice Hence their mobility in water is less than in ice . |
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| 566. |
Statement-1. `BF_(3)` molecule is planar while ` NF_(3)` is pyramidal. Statement-2. N atom is smaller than B .A. Statement-1 is Ture , Statement-2 is Ture , Statement-2 is a correct explanation for Statement-1.B. Statement-1 is Ture , Statement-2 is Ture , Statement-2 is not a correct explanation for Statement-1.C. Statement-1 is True , Statement-2 is False .D. Statement-1 is False, Statement-2 is True. |
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Answer» Correct Answer - b Correct explanation . In `BF_(3)` , hybridisation of B atom is `sp^(2)` while iin `NF_(3)` , hybridisation o N atom is `sp^(3)` (of course , N atom is smaller than B because along a period the size decreasses.) |
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| 567. |
Bond angle in `I_(3)^(Θ)` is .A. More than `CIO_(2)`B. `180^(@)`C. Less than `CIO_(2)`D. `gt 109.5^(@)` |
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Answer» Correct Answer - A::B Bond angle in `I_(3)^(Θ)` is `180^(@)` (`sp^(3)` d, linear) whereas bond angle in `CIO_(2) is gt 109.5^(@)` [Refer to IIIustration 1.81 Question 11 Part (m) solution] . |
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| 568. |
Expalin (a) `H_(2)^(o+)` and `H_(2)^(Θ)` ions have same bond order but `H_(2)^(o+)` ions are more stable than `H_(2)^(Θ)` (b) It is possible to have a diatomic molecule with its ground sate `MO` s full with electrons (c ) Why `2p_(x0` or `2p_(y)` orbitals do not combine with 2s orbitals to form `MO` (Taking Z-axis as the internuclear axis) . |
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Answer» `MO` Electronic configuration of `H_(2)^(o+)(Z =1) = sigma 1s^(1)` `MO` Electronic configuration of `H_(2)^(Θ) (Z =1) sigma 1s^(2),overset(**)sigma 1s^(1)` `H_(2)^(o+)` ion is more stable due to the presence of one electron in bonding `MO` But `H_(2)^(Θ)` is less stable due to the presence of one electron in the antibonding `overset(**)sigma 1s` `MO` due to which there is some destabilising effect (b) `NO` Because in such case `BO` becomes zero for example in case of `He_(2),Be_(2),Ne_(2)etc` Note In `H_(2),BO` is one `sigma 1s MO` is full while `overset(**)1s` antibonding `MO` is empty (c ) Small `(+)(+)` overlap cancels the `(+)(-)` overlap Hence there is no net overlapping . |
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| 569. |
Which statement `(s)` is (are) wrong for bond angle ?A. `CH-= CH gt BF_(3) gt CH_(4)`B. `H_(2)OgtNCI_(3)`C. `overset(o+)NH_(4) gt NH_(3) gt PCI_(5)`D. `CO_(2) gt NH_(3) gt CH_(4)` |
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Answer» Correct Answer - B::D Statement (b) and (d) are wrong For (a) refer to Section 1.17 (Point3) (a) (v) For (c ) refer to Section 1.17 (Point3) (a) (v) For (d) `CO_(2) (sp, 180^(@)) gt CH_(4) (sp^(3) , 109^(@) 28)` `HH_(3) (sp^(3)` pyramidal `107^(@))` Refer to Section 1.17 (Point 3) (a) (v) . |
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| 570. |
Which of the following statement is wrong ? .A. Multiple bonds are always shorter the corresponding sigle bondsB. The electron-deficient molecules act as Lewis acidsC. Every `AB_(5)` molecule does in fact have square pyramidal structureD. The cenonical structure has no real existnece |
| Answer» Correct Answer - C | |
| 571. |
Which statement `(s)` is (are) wrong for bond angle ?A. `NH_(3) gt NF_(3)`B. `NF_(3)gtNCI_(3)`C. `NO_(2)^(o+)gtNO_(2)`D. `NO_(3)^(Θ) gt NO_(2)^(Θ)` |
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Answer» Correct Answer - A::C::D is wrong`NCI_(3) gt NF_(3)` |
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| 572. |
Which of the following statement about resonance energy is wrong ? .A. The different in energy of the resonance hybrid and the most stable contributing structures (having least energy) is called resonance energy .B. The differnet in energy of the resonane hybrid and the least stable contributing structures (having highest energy) is called resonance energy .C. The differnece in the experimental and calculated enthalpies (bond enthalpy formation or combustion or hydrogenation) is called resonance enrgy .D. Resonance energy is the amount of energy by which the compound is stable . |
| Answer» Correct Answer - B | |
| 573. |
Statement 1: `N_(2) and NO^(+)` are both dia-magnetic. Statement 2: `NO^(+)` is isoelectronic with `N_(2)`A. Statemetn-1 is true, Statrment-2 is also true, Statement -2 is the correct explanation of statement-3B. Statement -1 is true , Statement 2 is also true, Statement-2 is not the correct ezplanation of Statement-3C. Statement-1 is true, Statement-2 is false.D. Statement-1 is false, Statement-2 is true. |
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Answer» Correct Answer - b Correct explanation. Both `NO^(+) and N^(2)` have all paired electrons in their M.O. |
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| 574. |
`N_(2) and O_(2)` are converted into monocations, `N_(2)^(+)` and `O_(2)^(+)` respectively. Which of the following is wrong?A. In `N_(2)^(+),` the N-N bond is weakenedB. In `O_(2)^(+),` the bond order increasesC. In `O_(2)^(+),` paramagnetism decreasesD. `N_(2)^(+)` becomes diamagnetic |
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Answer» Correct Answer - D `N_(2)^(+)` is paramagnetic and not diamagnetic, |
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| 575. |
`N_(2) and O_(2)` are converted into monocations, `N_(2)^(+)` and `O_(2)^(+)` respectively. Which of the following is wrong?A. In `N_(2)^(+), N -N` bond order weakensB. In `O_(2)^(+), the O -O` bond order increasesC. In `O_(2)^(+)` paramagnetism decreasesD. `N_(2)^(+)`becomes diagnetic |
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Answer» Correct Answer - d `B.O = "in" N_(2) = (10-4)//2 = 3` `B.O = "in" N_(2)^(+) = (9-4)//2 = 2.5` `B.O = "in" O_(2) = (10-6)//2 = 2` `B.O = "in" O_(2)^(+) = (10-5)//2 = 2.5` `N_(2)^(+)` contains one unpaired electron and hence it is paramagnetic. Thus option (d) id wrong |
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| 576. |
Which of the following bonds have lowest bond energy?A. C-CB. N-NC. H-HD. O-O |
| Answer» Correct Answer - D | |
| 577. |
Which of the following best defines a crystal ?A. A coloured substance soluble in waterB. A simple lattice containing ions, atoms or moleculesC. A clear or coloured substance which can transmit lightD. A salt which has been grown from its saturated solution. |
| Answer» Correct Answer - B | |
| 578. |
The chemical inertness of `N_2` is attributed toA. the presence of large no. of bonding electrons in comparison to antibonding electronB. its high heat of dissociationC. the presence of triple bonds between nitrogen atoms which make the molecule quite stableD. all the statements are correct |
| Answer» Correct Answer - D | |
| 579. |
Give the change in bond order in the following ionisation process? i. `O_(2) rarr O_(2)^(o+)+e^(-)` , ii.`N_(2) rarr N_(2)^(o+)+e^(-)` |
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Answer» According to molecular orbital theory, electronic configuration and bond order of `N_(2), N_(2)^(+), O_(2)` and `O_(2)^(+)` species are as follows. `N_(2)(14e^(-))=sigma1s^(2),sigma^(star)1s^(2), sigma2s^(2), sigma^(star)2s^(2), (pi2p_(x)^(2)=pi2p_(y)^(2)).sigma2p_(z)^(2))` Bond order =`1/2[N_(b)-N_(a)]=1/2[10-4]=3` `N_(2)^(+)(13e^(-))=sigma1s^(2), sigma^(star)1s^(2),sigma2s^(2),sigma^(star)2s^(2),(pi2p_(x)^(2)=pi2p_(y)^(2))sigma2p_(z)^(1)` Bond order =`1/2[N_(b)-N_(a)]=1/2(9-4)=2.5` `O_(2)16e^(-)=sigma1s^(2),sigma^(star)1s^(2),sigma2s^(2),sigma2p_(z)^(2), (pi2p_(x)^(2)=pi2p_(y)^(2)),(pi^(star)2p_(x)^(1)=pi^(star)2p_(y)^(1))` Bond order =`1/2[N_(b)-N_(a)]=1/2(10-6)=2` `O_(2)^(+)(15e^(-)= sigma1s^(2), sigma^(star)1s^(2), sigma2s^(2), sigma^(star)2s^(2), sigma2p_(z)^(2), (pi2p_(x)^(2)=pi2p_(y)^(2)), (pi^(star)2p_(x)^(1)=pi^(star)2p_(y))` Bond order =`1/2[N_(b)-N_(a)]=1/2[10-5]=2.5` a) `underset(B.O=3)N_(2) to underset(B.O=2.5)(N_(2)^(+)+e^(-)` Thus, bond order of decreases. b) `underset(B.O=2)O_(2) to underset(B.O=2.5)(O_(2)^(+) + e^(-)` Thus, bond order increases. |
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| 580. |
The shape of `ClO_(3)^(-)` ion according to VSEPR theory is:A. Triangle planarB. PyramidalC. TetrahedralD. square planar. |
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Answer» Correct Answer - B is the correct answer. |
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| 581. |
The bond between B and C will beA. IonicB. ColvalentC. HydrogenD. Coordinate |
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Answer» Correct Answer - B The bonds formed between B and C i.e. in `PCl_(5)` will be covalent in nature. |
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| 582. |
Which of the following statement is not correct for sigma and pi- bonds formed between two carbon atoms ?A. Sigma -bond determines the direction between carbon atoms but a pi-bond has no primary in this regardB. Sigma -bond is stronger than a pi- bondC. Bond energies of sigma and pi-bond are of the order of `264 kj//"mol"` and `347 kJ//"mol"` ,respectivelyD. Free rotation of atoms about a sigme`-`bond is allowed but not in case of a pi`-`bond. |
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Answer» Correct Answer - c As sigma bond is stronger that the `pi("pi")` bond , so it most be having higher bond energy than `pi("pi")` bond. |
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| 583. |
During the formation of a molecular orbital from atomic orbital , the electron density is :A. not zero in the nodal planeB. maximum in the nodal planeC. zero in the nodal planeD. zero on the surface of the lobe |
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Answer» Correct Answer - c The electron density is zero in the modal plane the formation of a molecular orbital from atomic orbitals of the same atom |
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| 584. |
During the formation of a molecular orbital from atomic orbital , the electron density is :A. Minimum in the nodal planeB. Maximum in the nodal planeC. Zero in the nodal planeD. Zero on the surface of the lobe |
| Answer» Correct Answer - C | |
| 585. |
Predict the correct order of repulsions among the following :A. bond pair-bond pair `gt ` lone pair-bond pair `gt` lone pair -lone pairB. lone pair-bond pari `gt` bond pair-bond pair `gt` lone pair-lone pairC. lone pair -lone pair `gt` lone pair-bond pair `gt` bond pair-bond pairD. lone pair-lone pair `gt` bond pair-bond pair `gt` lone pair-bond pair |
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Answer» Correct Answer - C According to VSEPR theory, lone pair-lone pair `gt` lone pair-bond pair ` gt ` bond pair-bond pair. |
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| 586. |
Why axis bonds of `PCl_(5)` are longer than equatorial bonds ? |
| Answer» This is due to greater repulsion on the axial bond pairy by the equatorial bond paris of electrons . | |
| 587. |
Which of the following have been arranged in increasing bond order as well as bond dissociation energy ? .A. `O_(2)^(-2) lt O_(2)^(Theta) lt O_(2)^(o+) lt O_(2)`B. `O_(2)^(-2) lt O_(2)^(Theta) lt O_(2) lt O_(2)^(o+)`C. `O_(2)ltO_(2)^(o+) lt O_(2)^(2-)lt O_(2)^(Theta)`D. `O_(2)^(o+) lt O_(2)^(2-)lt O_(2)^(Theta)ltO_(2)` |
| Answer» Correct Answer - B | |
| 588. |
Which combination will give the strongest ionic bond?A. `Na^(+) and CI^(-)`B. `Mg^(2+) and CI^(-)`C. `Na^(+) and O^(2-)`D. `Mg^(2+) and O^(2-)` |
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Answer» Correct Answer - d Ionic bond strength depends on the force of attraction between cation and anion |
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| 589. |
Molecular axis is Z axis , then which of the following combination of orbitals will result in formation of `sigma` molecular orbitals ?A. `p_(x) - p_(x)`B. `s -s `C. `p_(z) - p_(z)`D. `p_(y) - p_(y)` |
| Answer» Correct Answer - B::C | |
| 590. |
Which of the following order is correct for the bond dissociation energy of `O_(2), O_(2), O_(2)^(-) and O_(2)^(2-)`?A. `O_(2)^(+) gt O_(2) gt O_(2) gt O_(2)^(2-)`B. `O_(2)^(+) gt O_(2) lt O_(2) lt O_(2)^(2-)`C. `O_(2)^(+) lt O_(2) lt O_(2) lt O_(2)^(2-)`D. `O_(2)^(+) gt O_(2) gt O_(2)^(-) gt O_(2)^(2-)` |
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Answer» Correct Answer - a The bond are as follows, `O_(2) = 2, O_(2)^(+) = 2.5, O_(2)^(-) = 1.5, O_(2)^(2-) = 1`, |
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| 591. |
Number of right angle bonds in `PCl_(5)` is _______. |
| Answer» Correct Answer - 6 | |
| 592. |
Which of the following can show variable valency ?A. FB. ClC. BrD. I |
| Answer» Correct Answer - B::C::D | |
| 593. |
Maximum valency that is possible for P with Cl is _________. |
| Answer» Correct Answer - 6 | |
| 594. |
How many P-O-P bonds are present in `P_(4)O_(10)` ? |
| Answer» Correct Answer - 6 | |
| 595. |
What will be the bond order between O - O in peroxide of sodium ? |
| Answer» Correct Answer - 1 | |
| 596. |
How many hybrid orbitals are used for bond formation in `SF_(4)` ? |
| Answer» Correct Answer - 4 | |
| 597. |
`BH_(4)^(-) and NH_(4)^(+)` are isolobal . Explain. |
| Answer» Both have tetrahedral shapes, i.e., four lobes of `sp^(3)` hybridized orbitals . | |
| 598. |
State the type of hybrid orbitals associated with (i) P in `PCl_(5)` and (ii) S in `SF_(6)` . |
| Answer» (i) `sp^(3) d `of P in `PCl_(5) (ii) sp^(3) d^(2)` of S in `SF_(6)` . | |
| 599. |
In a moleculte A - B, electronegativities of atom A and B are 2.0 and 4.0 respectively. Calculate the percent ionic character of A- B bond using (i) Pauling equation (ii) Hannay and Smith equation. |
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Answer» (i) According to Pauling equation % ionic character = `18 (chi _(A) - chi_(B))^(1-4) = 18 (4 - 2)^(1-4) = 18 xx 2 ^(1-4) = x (say)` ` long x = log 18 + 1.4 log 2 = 1.2553 + 1.4 xx 0.3010 = 1.6767` x = Antilog 1 .` 6767 = 47. 50 %` (ii) According to Smith and Hannay equation ` %` ionic character = `16 (chi _(A) - chi_(B)) + 3.5 (chi_(A) - chi_(B))^(2) = 16 xx 2 + 3.5 xx 2^(2)` ` = 32 + 14 = 46 %` |
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| 600. |
One DEBYE (D) is equal toA. `1 xx 10^(-4)`esu-cmB. `1 xx 10^(-18)`esu-cmC. `1 xx 10^(-10)`esu-cmD. `1 xx 10^(-16)`esu-cm |
| Answer» Correct Answer - B | |