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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.
| 701. |
Assertion(A) - In case the central atom in a molecule is surrounded only by shared pairs of electrons, the molecule has a regular geometry. Reason(R) -The shared pair of electrons repel each other with equal force so all bonds are equidistant from each other.A. Both A and R true and R is the correct explanation of AB. Both A and R true and R is not a correct explanation of AC. A is true but R is falseD. A is false but R is true |
| Answer» Correct Answer - A | |
| 702. |
A polar covalent bond with positive and negative charge centres at its ends is called a dipole. The polarity of a dipole is measured by its dipole moment. Mathematically it is expressed as dipole moment, `mu=q xx d` where q and d are the net charge and the distance between the two charges respectively. Dipole moment is a vector quantity. The net dipole moment of a polyatomic molecule is the resultant of the various bond moments present in the molecule. The values of dipole moment are expressed in Debye (D) or in SI units in terms of coulomb- metre (Cm). One of the most important applications of dipole moment is in the determination of geometry and shape of molecules besides prediction of a number of properties of the molecules. Debye is equivalent toA. `3.33 xx 10^(-30)` e.s.u. cmB. `1.662xx10^(-27)` e.s.u.cmC. `1 xx 10^(-18)` e.s.u. cmD. `3.33 xx 10^(-12)` e.s.u.cm. |
| Answer» Correct Answer - C | |
| 703. |
A polar covalent bond with positive and negative charge centres at its ends is called a dipole. The polarity of a dipole is measured by its dipole moment. Mathematically it is expressed as dipole moment, `mu=q xx d` where q and d are the net charge and the distance between the two charges respectively. Dipole moment is a vector quantity. The net dipole moment of a polyatomic molecule is the resultant of the various bond moments present in the molecule. The values of dipole moment are expressed in Debye (D) or in SI units in terms of coulomb- metre (Cm). One of the most important applications of dipole moment is in the determination of geometry and shape of molecules besides prediction of a number of properties of the molecules. The molecules `BF_3` and `NH_3` both are covalent compounds but `BF_3` is non-polar while `NF_3` is polar . The reason is thatA. boron is a non-metal and nitrogen is a gas in uncombined stateB. B-F bonds have no dipole moment whereas N-F bonds have dipole momentC. Atomic size of boron is smaller than that of nitrogenD. `BF_3` is planar but NF3 is pyramidal. |
| Answer» Correct Answer - D | |
| 704. |
A polar covalent bond with positive and negative charge centres at its ends is called a dipole. The polarity of a dipole is measured by its dipole moment. Mathematically it is expressed as dipole moment, `mu=q xx d` where q and d are the net charge and the distance between the two charges respectively. Dipole moment is a vector quantity. The net dipole moment of a polyatomic molecule is the resultant of the various bond moments present in the molecule. The values of dipole moment are expressed in Debye (D) or in SI units in terms of coulomb- metre (Cm). One of the most important applications of dipole moment is in the determination of geometry and shape of molecules besides prediction of a number of properties of the molecules. `NH_3` has a net dipole moment, while `BF_3` has zero dipole moment becauseA. `NH_3` is not a planar molecule while `BF_3` is planarB. `NH_3` is a planar, while `BF_3` is non-planarC. Fluorine is more electropositive than nitrogenD. Boron is more electronegative than oxygen. |
| Answer» Correct Answer - A | |
| 705. |
The boiling point of `HCI` is less than that of `HF` . |
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Answer» Correct Answer - T Due to H-bonding in `HF` the boiling point of `HF gt HCI` . |
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| 706. |
Can a non-polar molecule have polar covalent bonds? |
| Answer» Yes, it is possible in case of mol.ecules having linear and symmetrical shapes. In them, the bonds are polar bup their polarities cancel out resulting in non-polar molecules. For example, in `CO_(2)` (O=C=O) both the bonds, are polar.`C Cl_(4)` molecule, C0Cl bonds are polar but molecule is non polar in nature. | |
| 707. |
Answer the following : (a) `C Cl_(4)` is non-polar but `CH_(3)Cl` is polar. (b) `SiF_(4)` is non-polar although Si-F bonds are polar. (c ) A hetero-diatomic molecule is always polar. |
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Answer» (a) Carbon tetrachloeide `(C Cl_(4))` is a symmetrical molecule with a regular tetrahedral geometry. Therefore, the polarises of the bonds cancel out and the molecule is non-polar. On the other hand, methyl chloride `(CH_(3)Cl)` has irregular geometery due to the presence of one C-Cl bond. The molecule has a specific polarity. (b) Like `C Cl_(4)` the molecule `SiF_(4)` is highly symmetrical and has a tetrahedral geoometry. Therefore, bond polarities cancel out and it is non-polar in nature. ( c) A hetero-diatomic molecule (e.g. HF,HCl, ICl etc.) is dipolar on account of the electronegativity difference in the bonding atoms. However, the same may not be true for polyatomic molecules in which bond polarities may mutually cancel. For example, `BF_(3)` and `C Cl_(4)` molecules. |
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| 708. |
Which of the following bonds is most polar ?A. C-OB. C-FC. O-FD. N-F. |
| Answer» Correct Answer - B | |
| 709. |
The order of increasing polarity is `HCI, CO_(2), H_(2)O` and `HF` molecules isA. `CO_(2), HCI, H_(2)O`B. `HF, H_(2)O, HCI, CO_(2)`C. `CO_(2), HF, H_(2)O`D. `"CC"_(2), HF, H_(2)O, HC` |
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Answer» Correct Answer - a The geometry of `CO_(2)` molecule is linear so th edipole opposite direction and `CO_(2)` becomes non polar. In other molecules order of electromagativity of `CI, O and F` is `CI lt O lt F`. So polarity order is `H - F gt H - Ogt H - CI` bonds |
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| 710. |
Out of the following which has highest dipole moment?A. `2,2`-dimethyl propaneB. trans`-2`- penteneC. cis `-3-`hexenD. `2,2,3, 3-` tertamethyl butane |
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Answer» Correct Answer - c Unsymmetrical molecule has highest dipole moment as observed in `( c)`. |
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| 711. |
The Type of hybrid orbitals used by the chlorine atom in `CIO_(2)^(Theta)` is . |
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Answer» Correct Answer - `sp^(3)` `sp^(3)` (Hybridisation ` = (1)/(2) ( 7 +0 +1) = 4= sp^(3)`) . |
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| 712. |
Which contains both polar and non-polar bonds?A. `NH_(4)CI`B. `HCN`C. `H_(2)O_(2)`D. `CH_(4)` |
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Answer» Correct Answer - C In `H_(2)O_(2),(O -O)` bond is non polar and `(O-H)` bonds are polar . |
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| 713. |
The Type of hybrid orbitals used by the chlorine atom in `CIO_(2)^(Theta)` is .A. `sp^(3)`B. `sp^(2)`C. `sp`D. None of these |
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Answer» Correct Answer - A Hybridisation of `CIO_(2)^(Theta) rArr (1)/(2) (7 +0+1) =4 =sp^(3)` . |
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| 714. |
In which of the following molecules /ions , are all the bonds not equal ?A. `BF_(4)`B. `SF_(4)`C. `SiF_(4)`D. `XeF_(4)` |
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Answer» Correct Answer - b Two axial and two equatorial `S - F` bonds in `SF_(4)`. |
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| 715. |
Which of the following pairs of ions are isoelectronic and also isostructural ?A. `SO_(3)^(2-),NO_(3)^(-)`B. `ClO_(3)^(-),SO_(3)^(2-)`C. `CO_(3)^(2-),SO_(3)^(2-)`D. `ClO_(3)^(-),CO_(3)^(2-)` |
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Answer» Correct Answer - B Species No. of `e^(-)` Hybridisation Shape `{:(SO_(3)^(2-),42,sp^(3),"Pyarmidal"),(ClO_(3)^(-),42,sp^(3),"Pyramidal"),(CO_(3)^(2-),32,sp^(2),"Trigonal planar"),(NO_(3)^(-),32,sp^(2),"Trigonal planar"):}` `SO_(3)^(2-) and ClO_(3)^(-)` ions are iso-electronic as well as iso structural in nature. |
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| 716. |
In which of the following pairs, the two species are not isostructural?A. `CO_(3)^(2+)` and `NO_(3)^(-)`B. `PCI_(4)^(+)` and `SiSI_(4)`C. `PF_(5)` and `BrF_(5)`D. `AIF_(6)^(3-)` and `SF_(6)` |
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Answer» Correct Answer - c (a) `CO_(3)^(2-). NO_(3)^(-)` Triangular planer (b) `PCI_(4)^(+). SiCI_(4)` Terahedral ( c) `PF_(5)` Tringular bipytarminal `BiF_(5)` Square pyramidal (d) `AIF_(6^(3-)). SF_(6)` Octahedral |
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| 717. |
The species in which the cantral atom uses `sp^(2)` hybrid orbital in its bonding is:A. `PH_(3)`B. `NH_(3)`C. `CH_(3)^(o+)`D. `SbH_(3)` |
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Answer» Correct Answer - C Hybridisation of `CH_(3)^(o+) rArr (1)/(2) (4 +3-1)=3 =sp^(2)` . |
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| 718. |
If molecule `MX_3` has zero dipole moment, the sigma bonding orbitals used by M (atomic number `lt21`) areA. pura`P`B. sp-hybridisedC. `sp^(2)`-hybridisedD. `sp^(3)` hybridised |
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Answer» Correct Answer - C Dipole momnet is a vector quantity In trigonal planar geometry (for `sp^(2)` hybridisation) the vector sum of two bond moments is equal and oposite to the dipole moment of third bond . |
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| 719. |
If molecule `MX_3` has zero dipole moment, the sigma bonding orbitals used by M (atomic number `lt21`) areA. Pure pB. sp hybridsC. `sp^(2)` hybridsD. `sp^(3)` hybrids |
| Answer» Correct Answer - C | |
| 720. |
In `BrF_(3)` molecule, the lone pair occupies equatorial position minimizeA. lone pair-bond pair repulsion onlyB. bond pair-bond pair repulsion onlyC. lone pair-bond pair repulsion and lone pair-bond pair repulsionD. lone pair-lone pair repulsion only |
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Answer» Correct Answer - c In `BrF_(3)` molecule, the lone pair occupies equatorial position to minimize lone pair-lone pair repulsion and lone pair bond pair repulsion as per VSEPR theory. |
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| 721. |
Assertion : Crystals of hydrated calcium sulphate gypsure `(CaSO_(4).2H_(2)O)` are soft easilly eleaved. Reason :Crystals anldrous calcium sulphate (anydride: `CaSO_(4))` are hard and very difficult to cleave.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 and the reason is 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 - b Within the `Ca^(2+),SO_(4)^(2-)` layers the ions are hold togather by atrong electrovalent bonds but these separated `Ca^(2)//SO_(4)^(2-)` layers are linked by relative weak H- bond The weak H-bonds `SO_(4)^(2-)` ions is the intermediate region consequenity, the hypsum can be readily eleaved and structure along the layer of water molecules .Anlydride has a complrtly ionic structure involvin g only `Ca^(2+)` and `SO_(4)^(2-)` ions. |
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| 722. |
Which of the following statement is wrong ? .A. A sigma bond is shorter than a pi-bond .B. Bond energies of sigma and pi bonds are of the order of 264 and `347 kJ mo1^(-1)` .C. Free rotation of atoms about sigma bond is allowed but not in case of a-pi bond .D. A sigma-bond determines the direction between C-atoms but a pi-bond has no primary effect which leads to bonding . |
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Answer» Correct Answer - B Bond energy of sigma-bond is greater than that of pi bond . |
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| 723. |
Which statement is/are correct? (a) Ionic compounds like sulphate and phosphates of Ba and Sr `[e.g BaSO_(4),SrSO_(4),Ba_(3)(PO_(4))_(2)` and `Sr_(3)(PO_(4))_(2)]` are insoluble in water (b) The above compounds are soluble in water (c ) Magnitude of lattice energy `(Delta_(U)H^(Θ))` of the above compounds is greater than their hydration energy `(Delta_(hyd)H^(Θ))` High `Delta_(U)H^(Θ)` of these compounds is due to polyvalent nature of both the cations and the anions (d) In these cases,hydration of ions fails to liberate sufficient energy to offset the lattice energy . |
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Answer» Correct Answer - a,c,d If hydration energy `(Delta_(hyd)H^(Θ)) gt` lattice energy So, statements a,c and d are correct . |
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| 724. |
Write two resonance structure of `N_2 O` that satisfy the octet rule. |
| Answer» `overset(-)overset(* *)(":"N)=overset(+)N=overset(* *)(O":")harr:N-=overset(+)N-overset(-)overset(* *)underset(* *)(O":")` | |
| 725. |
Which of the following has highest visosity?A. GlycerolB. GlycolC. EthanolD. Water |
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Answer» Correct Answer - a Due to `3 O-H` groups glyoerol shows extensive intra-molecular `H-`bonding and hence it has high visocsity. |
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| 726. |
Which of the following is leaser acceptable resonating atructure of `N_(3)^(-)`?A. B. C. D. |
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Answer» Correct Answer - c As per octet rule structure ( c) is correct but it is highly strained resonating atructure. |
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| 727. |
Assertion Bond angles of `NH_(3),PH_(3),AsH_(3)` and `SbH_(3)` decrease in order as mentioned Reasoning The central atom in each possesses a lone pair .A. If both `(A)` and `(R )` are correct and `(R )` is the correct explanation of `(A)`B. If both `(A)` and `(R )` are correct and `(R )` is the correct explanation of `(A)`C. If `(A)` is correct but `(R )` is incorrectD. If `(A)` is incorrect but `(R )` is correct |
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Answer» Correct Answer - B Bond angle of hybrides of group 15 decreases down the group due to increase in size or decrease in `EN` of central atom Thus both `(A)` and `(R )` are correct but `(R )` is not the correct explantion of `(A)` . |
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| 728. |
A list of species having the formula `XZ_(4) ` is given below : `XeF_(4) ,SF_(4), SiF_(4) ,BF_(4)^(-),BrF_(4)^(-),[Cu(NH_(3))_(4)]^(2+) ,[FeCl_(4)]^(2-) ,CoCl_(4)^(2-) and [PtCl_(4)]^(2-)` Defining shape on the basis of the lacation of X and Z atoms , the total number of species having a square planar shape is |
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Answer» Correct Answer - 4 `XeF_(4)` = Square planar , `SF_(4)` = See-saw `SiF_(4)` = Tetrahedral , `BF_(4)^(-)` = Tetrahedral , `BrF_(4)^(-)` = Square planar . `[Cu(NH_(3))_(4)]^(2+)` = Squara planar, `[FeCl_(4)]^(2-)` = Tetrahedral , `[CoCl_(4)]^(2-) `= Tetrahedral , `[PtCl_(4)]^(2-)` = Squara planar. |
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| 729. |
Which one of the following hydrogen halides has the lowest boilling point ? .A. `HF`B. `HCI`C. `HBr`D. `HI` |
| Answer» Correct Answer - B | |
| 730. |
Which statements `(s)` is (are) correct for `AB_(x)` type molecule ? .A. If the `EN` of central atom decreases, the bond angle decreases .B. If the size of central atom increases the bond angle decreases.C. If the `EAN` of atom `B` decreases that bond angle increasesD. If the `EAN` of atom `B` decreases that bond angle decreases . |
| Answer» Correct Answer - A::B::C | |
| 731. |
For `AB_(x)` type molecule which statement (s) si (are) correct about bond angle `(B-A-B)` (I) Bond angel `prop EN` of the central atom `A` (II) Bond angle `prop 1//EN` of the central atom `A` (III) Bond angle `prop` Size of central atom (IV) Bond angle prop 1/Size of central atom .A. I,III,IIIB. II,IVC. I,IVD. II,III |
| Answer» Correct Answer - C | |
| 732. |
Which one among the following does not have the hydrogen bond?A. PhenolB. Liquid`NH_(3)`C. WaterD. `HCI` |
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Answer» Correct Answer - D `HCI` does not form hydrogen bonds because chlorine is not electronegative enough and it has larger size so hydrogen bonds are not formed They are generally formed with flurine oxygen, and nitrogen . |
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| 733. |
One hybridization of one `s` and one `p` orbital we getA. Two mutually perpendicualr orbitalsB. Two orbitals at `180^(@)` .C. Two orbitals directed tetrahedrallyD. Three orbitals in a plane |
| Answer» Correct Answer - B | |
| 734. |
Which of the following phenomena will occur when two atoms of the elements having same spin of electron approach for bonding ?A. Orbital overlap will not occurB. Bonding will not occurC. Both (A) and (B) are correctD. None of the above are correct |
| Answer» Correct Answer - C | |
| 735. |
The molecule having one unpaired electrons is .A. `NO`B. `CO`C. `CN^(Theta)`D. `O_(2)` |
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Answer» Correct Answer - A In `NO` there are `7 +8 =15` electrons. Definitely it has an unpaired electron . |
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| 736. |
Which connot be expained by `VBT`?A. OverlappingB. Bond formationC. Paramagenatic nature of oxygenD. Shapes of molencules |
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Answer» Correct Answer - c Paramagnetic nature of oxigen can be esplained by `MOT`. |
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| 737. |
Which of the following leads to bonding?A. B. C. D. |
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Answer» Correct Answer - b Only symmetrical overlap leads to bonding |
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| 738. |
The hybridisation of sulphur in sulphur dioxide isA. `sp`B. `sp^(3)`C. `sp^(2)`D. `dsp^(2)` |
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Answer» Correct Answer - C Hybridisation ` = (1)/(2) (6 + 0) =3 =sp^(2),V` shaped molecule . |
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| 739. |
Statement 1: `H_(2)` molecule is more stable than `H_(2)-HE` molecule Statement 2: The antibonding electeon in `H_(2)-He` molecule destabilies. It.A. Statemetn-1 is true, Statrment-2 is also true, Statement -2 is the correct explanation of statement-2B. Statement -1 is true , Statement 2 is also true, Statement-2 is not the correct ezplanation of Statement-2C. Statement-1 is true, Statement-2 is false.D. Statement-1 is false, Statement-2 is true. |
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Answer» Correct Answer - a Statement 2 is the correct explanation for Statement 1. |
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| 740. |
Statement-1 `H_(2)` molecule is more stable than HeH molecule . Statement -2 . The antibonding electron in the molecular destabilises it .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 . `H_(2)` molecular is more stable than HeH because bond or der of `H_(2) = 1` , that of ` HeH = (1)/(2)` |
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| 741. |
Assertion . `H_(2)` molecule is more stable than HeH molecule. Reason . The antibonding elelctron in the molecule destabilises is .A. If both assertion and reason are correct, and reason is the correct explanation of the assertion.B. If both assertion and reason are correct , but reason is not the correct explanation of the assertion.C. If assertion is correct, but reason is incorrect .D. If both assertion and reason are incorrect. |
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Answer» Correct Answer - b Correct explanation . `H_(2)` molecular is more stable than Heh because bond order of `H_(2)` = 1 and that of Heh ` = (1)/(2)` (Higher the bond order , more is the stability). |
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| 742. |
Which one of the following pairs of species have the same bond order ?A. `CO, NO`B. `O_(2), NO^(+)`C. `CN^(-) , CO`D. `N_(2), O_(2)^(-)` |
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Answer» Correct Answer - C M.O. configuration of `CN^(-)` ion `(14e^(-))` is `KK sigma (2s)^(2) sigma^(**)(2s)^(2) pi(2p_(x))^(2) pi (2p_(y))^(2) pi (2p_(z))^(2)` B.O. ` = (8-2)/(2) = (6)/(2) = 3 ` M.O. cofiguration of CO `(14r^(-))` : `KK sigma (2s)^(2) pi (2p_(y))^(2) pi (2p_(y))^(2) sigma (2p_(z))^(2) sigma^(**) (2s)^(2) ` B.O. ` = (8-2)/(2) = (6)/(2) = 3 ` Thus , `CN^(-)` and CO have same bond order . |
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| 743. |
Which of the following species have the same shape and same bond order ? (i) `CO_(2) (ii) N_(3)^(-) (iii) O_(3) (iv) NO_(2)^(-)`A. (i) and (ii)B. (iii) and (iv)C. (i) and (iii)D. (ii) and (iv) |
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Answer» Correct Answer - A::B `CO_(2) and N_(3)^(-)` are linear with bond order 2 in each case `O_(3) and NO_(2)^(-)` are V-shaped with bond order 1.5 in each case . |
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| 744. |
`LiF` a least soluble among the fluorides of alkali metals, becauseA. smaller size `Li` impart significant covalent character in `LiF`B. the hydration energies of `Li^(+)` and `F^(-)` are quit higherC. lattice energy of `LiF` is quite higher due to the smaller size of `Li^(+) and F^(-)`D. `LiF` has strong polymeric network in solid |
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Answer» Correct Answer - c The small size of both `Li^(+)` and `F` ion leads to a very high value of lattice energy and thus crystale of `LiF` is very diffcult to break . |
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| 745. |
In the molecular orbital diagram for `O_2^(+)` ion the highest occupied orbital isA. `sigmaMO` orbitalB. `piMO` orbitalC. `pi^** MO` orbitalD. `sigma^**` MO orbital |
| Answer» Correct Answer - C | |
| 746. |
Assertion : `B_(2)` molecule is paramagnetic. Reason :The highest occupied molecular orbital is of `sigma` type.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 `B_(2)` is `sigma 1s^(2) , sigma^(**)1s^(2), sigma2s^(2), sigma^(**)2s^(2) [{:(pi 2 py^(1)),(pi 2 pz^(1)):}]` |
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| 747. |
Statement-1 : `O_(2)^(-)` and `O_(2)^(+)` are paramagnetic . and Statement-2 : Bond order of `O_(2)^(+)` and `O_(2)^(-)` is same .A. Statement-1 is True , Statement-2 is True , Statement-2 is a correct explanation for Statement-10B. Statement-1 is True , Statement-2 is True , Statement-2 is NOT a correct explanation for Statement-10C. Statement-1 is True , Statement-2 is FalseD. Statement-1 is False , Statement-2 is True |
| Answer» Correct Answer - C | |
| 748. |
Assertion(A) - The bond order of NO molecule is 2.5. Reason( R )-NO molecule is paramagnetic in nature.A. Both A and R true and R is the correct explanation of AB. Both A and R true and R is not a correct explanation of AC. A is true but R is falseD. A is false but R is true |
| Answer» Correct Answer - B | |
| 749. |
Find the pair that has the same bond order with diamagnetic and paramagnetic properties respectively.A. `F_(2) and O_(2)`B. `N_(2) and O_(2)^(2-)`C. `Li_(2) and B_(2)`D. `B_(2) and O_(2)` |
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Answer» Correct Answer - C `{:("Specoes","Bond order","Magnetic Behaviour"),(F_(2),1,"Diamagnetic"),(O_(2),2,"Paramagnetic"),(N_(2),3,"Diamagnetic"),(O_(2)^(2-),1,"Diamagnetic"),(Li_(2),1,"Diamagnetic"),(B_(2),1,"Paramagnetic"):}` `Li_(2)(B.O.1)` is diamagnetic while `B_(2)(B.O=1)` is paramagnetic in nature. |
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| 750. |
Considering X-axis as the internucler axis, which out of the following atomic orbitals will from a sigma bond ? `(a)1sand 1s" " (b) 1s and 2p_(x)" "(c )2p_(y)and2p_(y)" "(d)1s and2s.` |
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Answer» `2p_(y)`and `2p_(y)` atomic orbitals are involved in the sidewise overlap leading to the formation of `pi`-bond. `(a)1sand 1s" " (b) 1s and 2p_(x)" "(d)1s and2s.` |
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