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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.
| 1051. |
Ground state `He_2` does not exist. An electronic excited state of `He_2`A. is more unstable than ground state `He_2`B. is equally unstable as ground state `He_2`C. is more stable towards dissociation than ground state `He_2`D. is less reactive than ground state `He_2` orbital |
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Answer» Correct Answer - C An excited electronic state of `He_2^+` is more stable towards dissociation than ground state `He_2`. This is because excited state must have more bonding than antibonding electrons. |
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| 1052. |
A molecule of chloral hydrate contains two -OH groups attached to a single carbon atom as shown below `Cl-undersetunderset(Cl)(|)oversetoverset(C)(|)(C)-underset(OH)underset(|)oversetoverset(OH)(|)C-H` Its extra stability is due to the presence ofA. hydrogen bonding between H atoms of -OH groups and Cl atomsB. its planar structureC. its symmetrical structureD. none of these |
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Answer» Correct Answer - A It is due to H-bonding between Cl atoms and H atoms to the -OH groups. |
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| 1053. |
a completely filled d orbital `(d^(10))`A. Spherically symmetricalB. Has octahedral symmetricalC. Has octahedral symmetryD. Depends on the atom |
| Answer» Correct Answer - B | |
| 1054. |
Pentagonal bipyramidal structure contains bond angles approximately.A. `120^(@),90^(@),180^(@)`B. `120^(@),72^(@),180^(@)`C. `72^(@),90^(@),120^(@)`D. `72^(@),90^(@),180^(@)` |
| Answer» Correct Answer - D | |
| 1055. |
Which of the following transition metal catios have noble gas core? (a) `Sc^(3+)` (b) `Y^(3+)` (c) `La^(3+)` (d) `Ac^(3+)`A. (i), (ii), (iii)B. (i), (ii), (iii), (iv)C. (i), (ii), (iv)D. (ii), (iii), (iv) |
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Answer» Correct Answer - B After obtaining the electron configuration of the atom (using the building-up principle), remove ns electrons, then `(n-1)d` electrons, until we have the correct positive charge on the ion, e.g., `Sc([Ar]3d^14s^2)rarrSc^(3+)[Ar],Y([Kr]4d^15s^2)rarrY^(3+)([Kr])` |
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| 1056. |
The molecule having one unpaired electron isA. NOB. COC. `CN^(-)`D. `CO_(2)` |
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Answer» (a) NO has 15 electrons in its molecular orbitals which are distributed as follows `NO=sigma 1s^(2)overset(**)sigma1s^(2)sigma2s^(2)overset(**)sigma2s^(2)pi2p_(y)^(2)pi2p_(z)^(2)sigma2p_(x)^(2)overset(**)pi2p_(y)^(1)` As it contains one unpaired electron, hence it is paramagnetic. |
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| 1057. |
How many corners of a cube would be occupied in the case of noble gas?A. 1B. 2C. 4D. 8 |
| Answer» (d) Eight corners of a cube would be occupied in the case of noble gas. | |
| 1058. |
The correct order of the lattice energies of the following ionic compounds isA. `NaClgtMgBr_2gtCaOgtAl_2O_3`B. `NaClgtCaOgtMgBr_2gtAl_2O_3`C. `Al_2O_3gtMgBr_2gtCaOgtNaCl`D. `Al_2O_3gtCaOgtMgBr_2gtNaCl` |
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Answer» Correct Answer - D Lattice energy is directly proportional to the product of the charges while inversely proportional to the size of ions. Thus, the smaller the size of ions and more the charge, the more is the lattice energy. `(Al^(3+))_2(O^(2-))_3gt(Ca^(2+))(O^(2-))gtMg^(2+)(Br^(-))_2gtNa^(+)Cl^(-)` |
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| 1059. |
Which one is configuration of most metallic metal ?A. 2,8,8,1B. 2, 8, 2C. 2, 8, 18, 1D. 2, 8, 1 |
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Answer» Correct Answer - A Metallic character increases down the group. In their respective periods, alkali metals are most metallic |
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| 1060. |
The following compounds have been arranged in order of their increasing thermal statbilties . Identify the correct order . `K_(2)CO_(3) (I)" "MgCO_(3) (II)` `CaCO_(3) (III)" "BeCO_(3) (IV)`A. `I lt II lt III lt IV`B. `IV lt II lt III lt I`C. ` IV lt II lt I lt III`D. ` II lt IV lt III lt I` |
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Answer» Correct Answer - C The stability of carbonates increases with increasing electropositive character of metal . |
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| 1061. |
`sp^3` has s-characterA. `1//2`B. `1//4`C. `1//8`D. 1 |
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Answer» Correct Answer - B `sp^3` has 1/4 s character. |
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| 1062. |
The `H-O-H` bond angle in water isA. `120^@`B. `109.5^@`C. `107^@`D. `104.5^@` |
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Answer» Correct Answer - D `SN` of `O` in `H_2O` is `2+2=4`. Thus, `O` is `sp^3-`hybridized, has tetrahedral geometry, and angular shape. Due to lone pair-lone pair and lone pair-bond pair repulsion, the bond angle is reduced from `109.5^@` to `104.5^@`. |
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| 1063. |
Maximum hydrogen bonds in water are,A. 4B. 3C. 2D. 8 |
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Answer» Correct Answer - A In water (or ice) oxygen atom can form two H-bonds and each hydrogen atom can form one H-bond. Thus, water in all, can form 4 H-bonds |
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| 1064. |
Which one is most ionic ?A. ` P_2 O_5`B. ` CrO_3`C. ` MnO`D. `Mn_2O_7` |
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Answer» Correct Answer - B Lower oxidation states forms more ionic nature . |
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| 1065. |
Which of the following compounds prossesses the `C-H` bonds with the lowest bond dissociation energy?A. TolueneB. BenzeneC. n-PentaneD. 2, 2-Dimethylpropane |
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Answer» Correct Answer - A The C-H bond of the methyl group in toluene has lowest bond dissociation energy (both for homolytic and heterolytic cleavage) benzyl carbocation as well as benzyl free radical are highly stable due to resonance hyperconjugation |
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| 1066. |
Which of the following molecule is linear?A. `BeCl_2`B. `H_2O`C. `SO_2`D. `CH_4` |
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Answer» Correct Answer - A `SN` of `Be` in `BeCl_2` is `2+0=2` (linear), `SN` of `O` in `H_2O` is `2+2=4` (angular), `SN` of `S` in `SO_2` is `2+1=3` (angular), and `SN` of `C` in `CH_4` is `4+0=4` (tetrahedral). |
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| 1067. |
Number of paired electrons in ` O_2` molecules is :A. ` 7`B. `8`C. ` 16`D. ` 14` |
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Answer» Correct Answer - C In `O_2` paired electrons are ` 14` and two are unpaired . ` |
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| 1068. |
Which of the following is correct?A. The number of electrons present in the valence shell of `S` in `SF_6` is `12`.B. The rate of ionic reaction is very low.C. According to `VSEPR` theory, `SnCl_2` is a linear molecule.D. The correct order of stability to form ionic compounds among `Na^(+)`, `Mg^(2+)`, and `Al^(3+)` is `Al^(3+)gtMg^(2+)gtNa^(+)`. |
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Answer» Correct Answer - A `S` in `SF_6` forms `6` covalent bonds and has no lone pair. Thus, it has 12 valence electrons. Ionic reactions are very fast. `SnCl_2` (`SN` is `2+1=3`) is angular. Stability order to form ionic compounds is `Al^(3+) ltMg^(2+)ltNa^(+)`. |
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| 1069. |
Which is not easily preciptated for a queous soultion ?A. `Cl^-`B. ` SO_4^(2-) `C. `NO_3^-`D. ` CO_3^(2-)` |
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Answer» Correct Answer - C Nitrates of all emtals are water soluble . |
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| 1070. |
Which of the following statements is true?A. The dipole moment of `NF_3` is zero.B. The dipole moment of `NF_3` is less than `NH_3`.C. The dipole moment of `NF_3` is more than `NH_3`.D. The dipole moment of `NF_3` is equal to `NH_3`. |
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Answer» Correct Answer - B Because in `NH_3`, the atomic dipole and bond dipole are in the same direction whereas in `NF_3` these are in opposite directions. |
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| 1071. |
Using `MO` theory predict which of the following sepcies has the shortest bond length ?A. `O_(2)^(2+)`B. `O_(2)^(+)`C. `O_(2)^(-)`D. `O_2^(2-)` |
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Answer» Correct Answer - A Bond order of ` O_(2)^(+) , O_2^(-),O_(2)^(2-)` and `O_(2)^(2+)` are ` 2. 5 , 1 . 5 , 1` and 30 respectively . ` :.` Bond order `prop (1)/("Bond length")` . |
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| 1072. |
Which of the following has maximum bond angle?A. `CHF_(3)`B. `CHCl_(3)`C. `CHBr_(3)`D. All have maximum bond angle |
| Answer» Correct Answer - A | |
| 1073. |
Predict the correct order omong the following:A. Lone pair-lone pair gt lone pair -bond pair gtbond pair-bond pairB. Lone pair-lone pair gt bond pair-bond pairgtlone pair bond pairC. Bond pair -bond pair gt lone pair-bond pair gt lone pair-lone pairD. Lone pair-bond pair gt bond pair gt lone pair-lone pair |
| Answer» Correct Answer - A | |
| 1074. |
The decreasing order of bond angle isA. `NH_(3) lt CH_(4) lt C_(2)H_(2) lt H_(2)O`B. `C_(2)H_(2) gt NH_(3) gtH_(2)O lt CH_(4)`C. `NH_(3) gt H_(2)O gt CH_(4) lt C_(2)H_(2)`D. `H_(2)O lt NH_(3) gt CH_(4) lt C_(2)H_(2)` |
| Answer» Correct Answer - B | |
| 1075. |
The enolic form of acetone contains:A. ` 9 sigma , 1` pi bond and `2` lone pairsB. ` 8 sigma , 1` pi bond and `2` lone pairsC. ` 10 sigma , 1` pi bond and `1` lone pairsD. ` 9 sigma , 2` pi bond and `1` lone pairs |
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Answer» Correct Answer - A Acetone shows tautomerism as `underset (("keto form")) underset(O)underset(||) (CH_3 -C-CH_3)hArrCH_2=underset(("enol form "))underset(OH)underset(|)(C-)NH_3`. |
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| 1076. |
CO is practically non-polar sinceA. The `sigma`-electron drift from C to D is almost nullified by the `pi` -electron drift from O to CB. The `sigma`-electron drift from O to C is almost nullified by the `pi` -electron drift from C to DC. the bond moment in lowD. there is a triple bond between C and O |
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Answer» Correct Answer - A The structure of CO is `Coverset(larr)(=)O` |
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| 1077. |
The observed dipole moment of `HCl` is `1.03D`. If the bond length of HCL is `1.3Å`, then the percent ionic character of `H-Cl` bond isA. `17%`B. `34%`C. `40%`D. `10%` |
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Answer» Correct Answer - A We can use dipole moment to calculate the percent ionic character. For this purpose, we calculate the dipole moment of `HCl` assuming it to be 100% ionic. If `HCl` is 100% ionic, then the bonding `e^(-)` pair will be completely in control of `Cl`. As a result, the H atom will carry one unit of positive charge while the Cl atom will carry one unit of negative charge. Assuming that during this polarization the bond length of the molecule does not change, we have `mu_("ionic")=Qxxr` `=(4.8xx10^(-10) "esu")(1.3xx10^(-8)cm)` `=6.24xx10^(-18)"esu" cm` `=6.24D` Thus, `6.24D` implies 100% ionic character. Then `1.3D` will imply % ionic character `=(mu_("observed"))/(mu_("ionic"))xx100%` `=(1.03D)/(6.24D)xx100%=16.5%=17%` |
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| 1078. |
Incorrect order of bond angle is :-A. `NH_(2) gt PH_(3) gt AsH_(3)`B. `H_(2)O gt H_(2)S gt H_(2)Se`C. `BCl_(3) gt AlCl_(3) gt GaCl_(3)`D. `NO_(2)^(+) gt NO_(2) gt NO_(2)^(-)` |
| Answer» Correct Answer - C | |
| 1079. |
Dipole moment is shown byA. `C Cl_(4)`B. `C_(6)H_(6)`C. `BF_(3)`D. HF |
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Answer» Correct Answer - D H-F is polar due to difference of electronegativity of hydrogen and fluorine so it shows positive dipole moment. |
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| 1080. |
Which of the following statement is correct : As the %s-character of a hybrid orbital decreasesA. The bond angle decreasesB. The bond strength increasesC. The bond length decreasesD. Size of orbital decreases |
| Answer» Correct Answer - A | |
| 1081. |
Which of the following is non polar species.A. `PCl_(5)`B. `PCl_(3)`C. `SF_(6)`D. `IF_(7)` |
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Answer» Correct Answer - C `SF_(6)` is symmetrical and hence non polar because its net dipole moment is zero |
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| 1082. |
Which one of the following is least basic in character?A. WaterB. ethanolC. EthaneD. Ether |
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Answer» Correct Answer - C Due to it symmetrical structure. |
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| 1083. |
Among the following, the molecule with the highest dipole moment is :A. `CO_(2)`B. `BF_(3)`C. `SO_(2)`D. Trans-2 -butane |
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Answer» Correct Answer - C `CO_(2) to ` zero dipole moment `BF_(3) to ` zero dipole moment `SO_(2) to ` non-zero dipole moment trans -2 -butene `to `zero dipole moment |
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| 1084. |
In a polar molecule , the ionic charge is `4.8 xx 10^(-10)` esu. If the interatomic distance is `1 Å` unit, then the dipole moment isA. 41.8 debyeB. 4.18 debyeC. 4.8 debyeD. 0.48 debye |
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Answer» Correct Answer - C Given ionic charge `=4.8xx10^(-10)` e.s.u. and ionic distance `=1Å =10^(-8) cm ` we know that dipole moment =ionic charge x ionic distance =`4.8xx10^(-10)xx10^(-8)` `=4.8xx10^(-18)` e.s.u. per cm =4.8 debey |
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| 1085. |
The correct statement for the molecule, `CsI_(3)` isA. it is a covalent moleculeB. it contains `Cs^(+)` and `I_(3)^(-)` ionsC. it contains `Cs^(3+)` and `I^(-)` ionsD. it contains `Cs^(+),I^(-)` and lattice `I_(2)` molecule |
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Answer» Correct Answer - b |
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| 1086. |
Which of the following bonds require the largest amount of bond energy to dissociate the atoms concerned ?A. `H-H` bond in `H_(2)`B. C-C bond in `CH_(4)`C. `N-=N` bond in `N_(2)`D. `O=O` bond in `O_(2)` |
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Answer» Correct Answer - C Due to multiple bonding in `N_(2)` molecule. |
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| 1087. |
Explain why all bonds in `SF_(4)` molecule are not equivalent.A. `89^(@),117^(@)`B. `120^(@),180^(@)`C. `45^(@),118^(@)`D. `117^(@),92^(@)` |
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Answer» Correct Answer - A `SF_(4)` has a See-Saw structure. |
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| 1088. |
Which of the following has `sp^3` hybridisation of central atom ?A. `PCl_(3)`B. `SO_(3)`C. `BF_(3)`D. `NO_(3)^(-)` |
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Answer» Correct Answer - A `PCl_(3)` contain three bonding and one lone pair electron. Hence shows `sp^(3)`-hybridisation |
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| 1089. |
In which one of the following molecules , the central atom said to adopt `sp^2` hybridisation ?A. `BeF_2`B. `BF_2`C. `C_2H_2`D. `NH_3` |
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Answer» Correct Answer - B `BeF_2` -sp hybridisation `BF_3-sp^2` hybridisation `C_2H_2-sp` hybridisation `NH_3-sp^3` hybridisation |
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| 1090. |
In piperdine . `N -H, N` atom has hybridisation :A. spB. `sp^2 `C. `sp^3`D. `dsp^2` |
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Answer» Correct Answer - C N in it has three `sigma`-bonds and one lone pair of electron. |
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| 1091. |
The species which has pyramidal shape isA. `NO_3^(-)`B. `SO_3`C. `PCl_3`D. `CO_3^(2-)` |
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Answer» Correct Answer - C The steric number of central atom of a pyramidal species is four. It has three bonded atoms and one lone pair. In `NO_3^(-)`, due to negative charge, the number of valence electrons in central N atom is six. All are used up to form bonds with oxygens giving zero lone pair. In `SO_3`, the central sulphur atom has 6 valence electrons. All are used up to form 6 bonds with 3 oxygen atoms (each oxygen forms 2 bonds), giving zero lone pair. In `CO_3^(2-)`, the central C atom has 6 valence electrons due to 2 units of negative charge. All are used up to form bonds with 3 O atoms giveng zero lone pair. In `PCI_3`, the central P atom has 5 valence electrons. Three are used to form bonds with `Cl` atoms and two nonbonding electrons form one lone pair. |
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| 1092. |
The total number of valence electrons in `4. 2g` of ` N_3^-` ion are :A. ` 2. 2N`B. ` 4. 2N`C. `16N`D. `3.2 N` |
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Answer» Correct Answer - C ` 42g N_3^(-)` has ` 16N` valence electrons . |
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| 1093. |
Among the following isostructural compounds which one has highest lattice energy ?A. LiClB. MgOC. NaClD. LiF |
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Answer» Correct Answer - B Among the given species, MgO is bi-bivalent ionic compound. Hence it has maximum lattice energy |
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| 1094. |
Which of the following is paramagnetic with bond order `0.5`?A. `F_2`B. `H_2^(+)`C. `N_2`D. `O_2^(-)` |
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Answer» Correct Answer - B `H_2^(+):(sigma_(1s))^1` B.O.=`(1-0)/2=1/2=0.5` `H_2^(+)` is paramagnetic with bond order 0-5 |
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| 1095. |
Oxygen molecule is paramagnetic becauseA. Bonding electrons are more than antibonding electronsB. It contains unpaired electronsC. Bonding electrons are less than antibonding electronsD. Bonding electrons are equal to antibonding electron |
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Answer» Correct Answer - B Paramagnetic nature is due to unpaired electrons in the species. |
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| 1096. |
Which of the following is paramagnetic and also has a bond order equal to 0.5 ?A. `O_2`B. `N_2`C. `He_2`D. `H_2^+` |
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Answer» Correct Answer - D Presence of unpaired electron in any of molecular orbital makes the molecule or ion paramagnetic. |
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| 1097. |
Which is paramagnetic and has bond order `Y_2`?A. `O_2`B. `N_2`C. `F_2`D. `H_2^+` |
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Answer» Correct Answer - D B.O =`1/2` [6-3] =1.5 |
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| 1098. |
Polarisation involves the distortion of the shape of an anion by an adjacently placed cation In this context, which of the following statements is correct ? .A. Maximum polarisation is brought about by a cation of high chargeB. Minimum polarisation is brought about by a cation of low radiusC. A large cation is likely to bring about a large degree of polarisationD. Polarisation power of cation is less than that of anion |
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Answer» Correct Answer - A Fazan rule |
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| 1099. |
The high folowing points and insolution in orgaints solvents of sulphanilic acid are due to its ……..structureA. Simple ionicB. Bipolar ionicC. CubicD. Hexagonal |
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Answer» Correct Answer - B Sulphuric acids have bipolar structure so their mp is high and insoluble in organic solvents. |
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| 1100. |
If the bond enthalpy of `O_(2),N_(2)" and "H_(2)` are 498, 946 and 435.8 kJ `mol^(-1)` respectively. Choose the correct order of decreasing bond strength.A. `H_(2)gtN_(2)gtO_(2)`B. `N_(2)gtO_(2)gtH_(2)`C. `O_(2)gtH_(2)gtN_(2)`D. `H_(2)gtO_(2)gtN_(2)` |
| Answer» (b) Larger the bond dissociation enthalpy, stronger will be the bond in the molecule.So, `{:(N_(2),gt,O_(2),gt,H_(2)),(946 kJ mol^(-1),,498 kJ mol^(-1),,435.8 kJ mol^(-1)):}` | |