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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.
| 1001. |
Both `CO_2` and `H_2O` contain polar covalent bonds but `CO_2` is nonpolar while `H_2O` is polar becauseA. H atom is smaller than C atomB. `CO_2` is a linear molecule while `H_2O` is an angular moleculeC. `O-H` bond is more polar than `C-O` bondD. `CO_2` contains multiple bonds while `H_2O` has only single bonds |
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Answer» Correct Answer - B In `CO_2`, the polar bonds are arranged in such a way (due to linear shape) that the bond polarities cancel each other while in `H_2O` molecule, the `H-O` bond polarities do not cancel one another (due to angular shape). |
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| 1002. |
Which of the following shapes of `SF_(4)` is more stable and why? A. (i) , due to 3 lp-bp repulsions at `90^(@)`.B. (ii) , due to 2 lp-bp repulsions.C. Both are equally stable due to 2 lp-bp repulsions.D. Both are unstable since `SF_(4)` has tetrahedral shape. |
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Answer» Correct Answer - B In structure (ii) , lp - bp repulsions are minimum. |
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| 1003. |
The most stable shape of `ClF_(3)` is shown by A. (i) onlyB. (i) and (ii)C. (ii) onlyD. (iii)only |
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Answer» Correct Answer - A In (i), the lp are at equatorial position so there are less lp - bp repulsions as compared to other positions. Hence, T - shape is most stable. |
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| 1004. |
The hydrogen bond is strongest inA. O-H...SB. S-H...OC. F-H...FD. F-H...O. |
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Answer» Correct Answer - C H-bonds are strongest in F-H.....F |
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| 1005. |
The state of hybridisation of S in `SO_2` is sms to that ofA. C in `C_2H_2`B. C in `C_2H_4`C. C in `CH_4`D. C in `CO_2` |
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Answer» Correct Answer - B Both are `sp^2`-hybridised |
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| 1006. |
Which forces are strongest amongst the following ?A. Ion-ion interactionB. ion-dipole forcesC. Dipole-dipole forcesD. Dipole induced dipole forces |
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Answer» Correct Answer - A Electrostatic forces |
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| 1007. |
Which of the following hydrides are ionicA. `CaH_(2)`B. `BaH_(2)`C. `SrH_(2)`D. `BeH_(2)` |
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Answer» Correct Answer - A::B::C `CaH_(2),BaH_(2),SrH_(2)` are ionic hybride. |
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| 1008. |
Pick out the isoelectronic structure from the following `I. CH_(3)^(+) " " II. H_(3)O^(+)` `III. NH_(3) " " IV CH_(3)^(-)`A. I and IIB. III and IVC. I and IIID. II,III and IV |
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Answer» Correct Answer - B::D Among the given structure , `H_(3)O^(+),NH_(3)` and `CH_(3)^(-)` have 10 electrons each and are isoelectronic. `CH_(3)^(+)` has only 8 electrons. So (b) and (d) are correct. |
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| 1009. |
Which species having molecules have same molecular geometry? I. `CH_(4)` II. `BF_(3)` III. `NH_(4)^(+)` IV. `SF_(4)`A. I and IIB. III and IVC. I and IIID. I, III and IV |
| Answer» (c ) `CH_(4)` and `NH_(4)^(+)` both have tetrahedral shape while `BF_(3)` has trigonal planar and `SF_(4)` has see-saw shape. | |
| 1010. |
Consider the molecules `CH_(4),NH_(3)` and `H_(2)O` which of the given statement is false ?A. The H-O-H bond angle in `H_2O` is larger than the H-C-H bond angle in `CH_4`B. The H-O-H bond angle in `H_2O` is smaller than the H-N-H bond angle in `NH_3`C. The H-C-H bond angle in `CH_4` is larger than the H-N-H bond angle in `NH_3`D. The H-C-H bond angle in `CH_4`, the H-N-H bond angle in `NH_3` and the H-O-H bond angle in `H_2O` are all the greater than `90^@` |
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Answer» Correct Answer - A::D As the number of lone pair of electrons on central element increases, repulsion between those lone pair of electrons increases and therefore, bond angle decreases. Molecules Bond angle `CH_4` (no lone pair of electrons) `109.5^@` `NH_3` (one lone pair of electrons ) `107.5^@` `H_2O` (two lone pair of electrons ) `104.45^@` |
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| 1011. |
Which of the following compound has the highest boiling pointA. HClB. HBrC. `H_(2)SO_(4)`D. `HNO_(3)` |
| Answer» Correct Answer - C | |
| 1012. |
The boiling point of a compound is raised byA. Intramolecular hydrogen bondingB. Intermolecular hydrogen bondingC. covalent bondingD. Ionic covalent |
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Answer» Correct Answer - B Intermolecular hydrogen bonding compound contain more b.p.ll compare the to intramolecular hydrogen bonding compound. |
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| 1013. |
The reason for exceptionally high boiling point of water isA. Its high specific heatB. Its high dielectric constantC. Low ionization of water moleculesD. Hydrogen bonding in the molecules of water |
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Answer» Correct Answer - D Hydrogen bonding increases the boiling point of compound. |
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| 1014. |
The correct order for triple bond energy in `CO, N_2, CN` and `C-=C` is ,A. ` C -= O gt N-= Ngt C-= N gt C-=C`B. `N -=N gt C-= O gt C-= Cgt C -=N`C. `N -=N gt C-= O gtN-= Cgt C -=C`D. `N -=N gt C-= O gtC= Cgt C -=C` |
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Answer» Correct Answer - A The values of bond energy for `CO,N-2,CN` and `-C -=C-`are `107,946, 891` and ` 836 kJ "mol"^(-1)`. The is due to the fact htat `CO` is polar and thus possesses highest bond eenrgy . |
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| 1015. |
The total number of electrons that take part in forming the bond in `N_2` isA. `10`B. `6`C. `4`D. `2` |
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Answer» Correct Answer - B `overset(..)N-=overset(..)N` In `N_2` molecule, the two N atoms are held together by a triple bond which consists of 3 electron pairs or 6 electrons. |
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| 1016. |
Find the bond order of ` N_2` |
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Answer» `N_2 = sigma1s^2,sigma^**1s^2,sigma2s^(2),sigma^**2s^2 [(pi2p_y^1),(pi2p_z^1)]` , `simga 2 p_s^2` `:.` Bond order of ` N_2 = 1/2 xx [ 10 - 4 ] =3`. |
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| 1017. |
The correct order of increasing s character ( in percentage ) in the hybrid orbitals in below molecules `//` ions is ( assume all hybrid orbitals are exactly equivalent) `:` `underset(I)(CO_(3)^(2-))" "underset(II)(XeF_(4))" "underset(III)(I_(3)^(-))" "underset(IV)(NCl_(3))" "underset(V)(BeCl_(2)(g))`A. `II lt III lt IV lt I lt V`B. `II lt IV lt III lt V lt I`C. `III lt II lt I lt V lt IV`D. `II lt IV lt III lt I lt V` |
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Answer» Correct Answer - A `|{:("Species","Hydridisation"),(CO_(3)^(2ɵ),sp^(2)),(XeF_(4),sp^(3)d^(2)),(l_(3)^(ɵ),sp^(3)d),(NCl_(3),sp^(3)),(BeCl_(2),sp):}|` |
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| 1018. |
Find the bond order of ` CO` |
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Answer» `CO= sigma1s^2, sigma^**1s^2, sigma2s^2, sigma2p_x^2 [(pi2p_y^2),(pi2p_z^2)],sigma^**2s^2` `:.` Bond order of ` CO = 1/2 [ 10 - 4 ] = 3` . |
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| 1019. |
In which of the following molecules, bonding is not taking place in excited state `:`A. `CH_(4)`B. `BF_(3)`C. `IF_(7)`D. `PCl_(3)` |
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Answer» Correct Answer - 4 `{:(C^(**)rarr1s^(2)2s^(1)2p^(3),,"4 unpaired electron ":." 4 bonds"),(B^(**)rarr1s^(2)2s^(1)2p^(2),,"3unpaired electron " :." 3 bonds"),(I^(**)rarr5s^(1)5p^(3)5d^(3),,"7 unpaired electron " :." 7 bonds" ),(Prarr3s^(2)3p^(3),,"3 unpaired electron ":." 3 bonds"):}` `.^(**)` represent excited state |
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| 1020. |
The noble gases have a particularly stable outer shell electronic configuration represented asA. `ns^(2)np^(6)`B. `ns^(2)np^(5)`C. `ns^(2)np^(4)`D. `ns^(2)np^(8)` |
| Answer» (a) The noble gas have a particular stable outer shell electronic configuration represented as `ns^(2)np^(6)`. | |
| 1021. |
How many line paire of electrons are present in outer shell of ` Cl^-` ? |
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Answer» ` Cl^(-) = 1 s^2 , 2 s^2 2p^6 , 3s^2 3 p^6` No . Of lone pair =4` . |
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| 1022. |
An electronic arrangement is said to be stable if its outer shell consistsA. doublet of electronsB. triplet of electronsC. octet of electronsD. singlet of electron |
| Answer» (c ) Eight electrons (octet) in oter shell show stable electronic arrangement. | |
| 1023. |
The electronic configuration of the outemost shell of the most electronegative element isA. `2s^(2)2p^(5)`B. `3s^(2)3p^(5)`C. `4s^(2)4p^(5)`D. `5s^(2)5p^(5)` |
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Answer» (a) The electronegetivty order of halogens is `F gt Cl gt Br gt l` Flurine is the most electronegeative element of the periodic table. |
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| 1024. |
Which of the following is soluble in waterA. `CS_2`B. `C_2H_5OH`C. `C Cl_4`D. `CHCl_3` |
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Answer» Correct Answer - B Ethyl alcohol is soluble in water because it can form intermolecular hydrogen bonds with water molecules: `CH_3CH_2-underset(H)underset(|)overset(delta-)overset(..)O:* * * * * *overset(delta+)H-underset(H)underset(|)overset(..)O:` |
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| 1025. |
An electrovalent bond is formed betweenA. Good conductor of electricityB. Polar in natureC. Low M.P. and Low B.P.D. Easily available |
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Answer» Correct Answer - B Electrovalent compounds are polar in nature because they are formed by ions |
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| 1026. |
As compared to covalent compounds, electrovalent compounds, generally haveA. high M.P. and Low B.P.B. High dielectric constantC. High M.P. and High B.P.D. high polarity |
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Answer» Correct Answer - A Electrovalent compounds generally have high m.pt and high b.pt due to stronger coulombic forces of attraction. |
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| 1027. |
When `NaCl` is dissolved in water the sodium ion becomesA. oxidizedB. reducedC. hydrolysedD. hydrated |
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Answer» Correct Answer - D When sodium chloride is dissolved in water, the sodium ion is hydrated . |
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| 1028. |
Among the species given below which one is isostere of `N_2` ?A. COB. `O_2`C. `O_2^-`D. `CO^+` |
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Answer» Correct Answer - A CO and `N_2` molecules have same number of atoms as well as same number of electrons. |
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| 1029. |
The hybridization of the central atom will change when `:`A. `NH_(3)` combines with `H^(+)`B. `H_(3)BO_(3)` combines with `OH^(-)`C. `NH_(3_` forms `NH_(2)^(-)`D. `H_(2)O` combines with `H^(+)` |
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Answer» Correct Answer - B `underset(sp^(2))overset(* *)(NH_(3))+H^(+)rarrunderset(sp^(3))(N)H_(4)^(+)` `underset(sp^(3))(2NH_(3))hArrunderset(sp^(3))(NH_(2)^(-))+underset(sp^(3))(NH_(4)^(+))` `underset(sp^(3))(H_(3)BO_(3)+OH^(-)) rarr underset(sp_(+)^(3))([B(OH)_(4)]^(-))` `underset(sp^(3))(H_(2)O)+H^(+)hArr underset(sp^(3))(H_(3)O^(+))` |
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| 1030. |
The hybrid state of B in `BF_4^-` isA. `sp^2`B. spC. `sp^3`D. no specific |
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Answer» Correct Answer - C `H=1/2[3+4-0+1]=1/2xx8=4` so hybridisation is `sp^3` |
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| 1031. |
`C_2H_2` is isostructural withA. `H_2O_2`B. `NO_2`C. `SnCl_2`D. `CO_2` |
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Answer» Correct Answer - D Both `C_2H_2` and `CO_2` have linear geometry |
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| 1032. |
The shape of covalent molecule `AX_3` isA. TriangularB. T-shapeC. PyramidalD. Any of the above three depending upon the number of lone pairs present in A. |
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Answer» Correct Answer - D Shape of the molecule depends upon total no. of electron pairs around the central atom which include I.p as well as b.p. |
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| 1033. |
Why `Al Cl_(3)` is a covalent molecule but is ionized in water to give `Al^(+3)` and `Cl^(-)` ions ? |
| Answer» `AlC_(3)` (anhydrous) is covalent because of very high ionisation energy required for formation of `Al^(+3)` . But in hydrated conditioin , this excessive energy is obtained by hydration energy of smaller `Al^(+3)` ion . | |
| 1034. |
`Na_(2)CO_(3)` does not decompose on heating whereas `CaCO_(3)` decomposes . Why ? |
| Answer» More the carbonate ion is polarised by the cation more is the chance of formation of `CO_(2)` and therefore higher is probability of decomposition . | |
| 1035. |
Which of the following is not true about resonance ?A. The resonating structures are hypothetical.B. The unpaired electrons in various resonating structures are sameC. Hybrid structure is most stable.D. Hybrid structure has maximum energy. |
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Answer» Correct Answer - D Hybrid structure has minimum energy and maximum stability |
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| 1036. |
Resonance occurs due to the :A. delocalization of a lone pair of electronsB. delocalization of a sigma-electronsC. delocalizatioon pi pi - electronsD. migration of protons |
| Answer» Correct Answer - A::C | |
| 1037. |
The correct order of stabilities of the following resonance structures is : (I) `H_2 C = overset (oplus) N = overset (Ө) N` (II) `H_2 overset (oplus) C - N = overset (Ө) N` (III) `H_2 overset (Ө) C - overset (oplus) N -= N` (IV) `H_2 overset (Ө) C - N = overset (oplus) N`.A. `(III) gt (I) gt (IV) gt (II)`B. `(I) gt (III) gt (II) gt (IV)`C. `(I) gt (II) gt (IV) gt (III)`D. `(II) gt (I) gt (III) gt (IV)` |
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Answer» Correct Answer - B Both (I) and (III) have more number of covalent bonds than `(II)` and `(IV)`. But (I) is the most stable as it has negative charge on more electronegative N atom while (III) has negative charge on less electronegative C atom. In both (I) and (III), the octet of every atom is complete. Thus, (III) is the second most stable. Both (II) and (IV) have equal number of covalent bonds but (II) is more stable as it has negative charge on more electronegative N while (IV) has negative charge on less electronegative C atom. Note that in (II), the octet of C is incomplete while in (IV), the octet of N is incomplete. Thus, the correct stability order is `IgtIIIgtIIgtIV` |
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| 1038. |
Which of the following has the shortest bond length?A. `Br_2`B. `F_2`C. `Cl_2`D. `I_2` |
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Answer» Correct Answer - B All halogen molecules have single bond between the atoms. As we move down the group, the bond length increases on account of increase of atomic size: `underset((143 p m))(F-F) lt underset((199 p m))(Cl-Cl) lt underset((288p m))(Br-Br) lt underset((266p m))(I-I)` |
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| 1039. |
Which possess fractional bond order ?A. ` O_2^(+)`B. `O_2^(-)`C. `H_2^(+)`D. `N_2` |
| Answer» Correct Answer - A::B::C | |
| 1040. |
The correct nond order for ` CO` and ` CO^+` are respectively :A. ` 3, 5//2`B. ` 3,2 `C. ` 3,7//2`D. ` 4//2, 3` |
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Answer» Correct Answer - C M.O. configuration of ` CO:` `sigma1s^2 , sigma^(** )1s^2, sigma^2, sigma2p_x^2,[(pi2p_y^2),(pi2p_x^2)] sigma^(**) 2s^2` M.O. configuration of `CO^+` `sigma1s^2 , sigma^(**) 1s^2, sigma^2, sigma2p_x^2,[(pi2p_y^2),(pi2p_x^2)] sigma^(**) 2s^1` Note that : bnond length of `CO^+` is shorter than `CO` and thus bond order should be greater for `CO^+` Also this will bot be possible if `CO` is supposed to have `M.O` configration : `sigma 1s^2 , sigma^(**) 1s^2 , sigma2s^2`. `sigma^(**) 2s^2 , sigma2p_x^2, [(pi 2p_y^2),(pi 2p_z^2)]`. This that : bond length of `CO^+` is shorter than `CO` and thus bond order should be greater for `CO^+` to `3` and ` 5//2` . The problem have been explained by concluding that 2s-orbital of oxygen has lower energy thabn 2-srobital of carbon . Thus on mixing ` sigma 2^(**)` attatins so high energy that it goes up ` sigma2p_x` as well as ` pi 2 p_x` and ` pi2p_y ` level . |
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| 1041. |
Which of the following has fractional nond order ?A. ` O_2^(2+)`B. `O_2^(2-)`C. `F_(2^(2-)`D. `H_2^-` |
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Answer» Correct Answer - D Bond order for ` H_(2)^(-) = + 1//2` |
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| 1042. |
Which of the following has //have identical nond order ?A. `CN^-`B. `O_2^-`C. `NO^+`D. `CN^+` |
| Answer» Correct Answer - A::C | |
| 1043. |
`RbO_2` isA. peroxide and diamagneticB. superoxide and paramagneticC. peroxide and paramangeticD. superoxide and diamagnetic |
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Answer» Correct Answer - B `RbO_2` is an ionic compound consisting of `Rb^(+)` (alkali metal ion) and `O_2^(-)` (superoxide ion). The superoxide ion contains an odd number of electrons `(17 e^(-))` and is, thus, paramangnetic. |
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| 1044. |
The `C -C` bond length is `1.54 Å C =C` bond length is `1.33 Å` What is the circumference of benzene ring ? Bond length between single and double bonds `= 1.4 Å` .A. `(C - C) = (C = C) = (C -=C)`B. `C -=C lt C = C lt C - C`C. `C-C lt C = C lt C -= C`D. `C = C lt C -= C lt C- C` |
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Answer» Correct Answer - B Bond order ` propto 1/("Bond length")` |
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| 1045. |
The incorrect order of decreasing boiling point isA. `HF gt HI gt HBr gt HCl`B. `H_2O gt H_2Te gt H_2Se gt H_2S`C. `Br_2 gt Cl_2 gt F_2`D. `CH_4 gt GeH_4 gt SiH_4` |
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Answer» Correct Answer - D Among hydrides of carbon family , the b.p. increases down the group due to increase in van der Waal forces. |
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| 1046. |
The bond angle formed by different hybrid orbitals are in the orderA. `sp^2gt sp^3gt sp`B. `sp^3 gt sp gt sp^2`C. `sp^3 gt sp^2 gt sp`D. `sp gt sp^2 gt sp^3` |
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Answer» Correct Answer - D sp(`180^(@)`)gt `sp^2 (120^(@))` gt `sp^3(109.5^(@))`. |
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| 1047. |
The number of bonding and anti-bonding electrons respectively in CO molecule is :A. 8, 2B. 2,8C. 4,2D. 2,4 |
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Answer» Correct Answer - A `CO : KK (sigma_(2s))^2 (sigma_(2s)^**)^2 (pi_(2px))^2(pi_(2py))^2 (sigma_(2pz))^2` No. of bonding electrons=8 No. of anti-bonding electrons = 2 |
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| 1048. |
What is the hybrid state of carbon in ethyne, graphite and diamond ?A. `sp^(2), sp, sp^(3)`B. `sp, sp^(2), sp^(3)`C. `sp^(3) , sp^(2) , sp`D. `sp^(2), sp^(3), sp` |
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Answer» Correct Answer - B In ethyne, ` H - C -= C - H`, C is sp hybridised . In graphite , one C atom is attached to 3 other C atoms which are `sp^(2)` hybridised. In diamond , one C atom is attached to 4 other C atoms which are `sp^(3)` hybridised . |
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| 1049. |
The bond order in `He_2^(+)` ions is :A. 0.5B. `1.0`C. 1.5D. `2.0` |
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Answer» Correct Answer - A `He_2^(+)=(sigma_(1s))^2 (sigma_(1s))^1 : BO =1/2 (2-1) =1/2 =0.5` |
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| 1050. |
Why carbon exists as graphite but silicon does not exist in any such form ? |
| Answer» Silicon has lowest bond energy than carbon , so less tendency of catenation is present . | |