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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.
| 201. |
Out of the following which compound will have electrovalent bondingA. AmmoniaB. WaterC. Calcium chlorideD. Chloromethane |
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Answer» Correct Answer - c `CACI_(2)` will have electrovalent bonding because calcium is electropositive metal while chlorine is electronegative so they will combined with electrovalent bond. |
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| 202. |
Write the molecular orbital electron distribution of oxygen `(O_(2))` Specify its bond order and magnetic property Fill in the blanks When `N_(2) ` goes to `N_(2)^(o+)`, the `N-N` bond distance___ and when `O_(2)` goes to `O_(2)^(o+)` the `O-O` bond distance ____ .A. increase, decreaseB. decrease, increaseC. increased in both the casesD. decreased in both the cases |
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Answer» Correct Answer - a When `N_(2)^(+)` is formed from `N_(2)`, the electron is lost from the bonding molecular orbital and thus the bond order decreases. Consequently, the bond distance increases . When `O_(2)^(+)` is formed the electron is lost from an antibonding molecular orbital. Obviously , the bond order increases and the distance decreases. |
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| 203. |
Chemical formula for calcium pyrophosphate is `Ca_(2)P_(2)O_(7)`. The formula for ferric pyrophosphate will beA. `Fe_(3)(P_(2)O_(7))_(3)`B. `Fe_(4)P_(4)O_(14)`C. `Fe_(4)(P_(2)O_(7))_(3)`D. `Fe_(3) PO_(4)` |
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Answer» Correct Answer - c As the formula of calcium pyrophosphate is `Ca_(2)P_(2)O_(7)` means valency of pyrophosphate redical is `-4` so formula of ferric pyrophosphate is `Fe_(4) (P_(2)O_(7))_(3)`. |
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| 204. |
Predict which out of the following species are planar. `(i) NH_(4)^(+) (ii)CH_(3)^(+)(ii)SF_(4)(iv)OF_(2)(v)H_(2)O` |
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Answer» (i) `NH_(4)^(+) : H=1//2[5+4-1+0]=4=(sp^(3)-"non-planar")` (ii) `CH_(4)^(+) : H=1//2[4+3-1+0]=6//2=3(sp^(2)-"planar")` (iii) `SF_(4) : H=1//3[6+4-0+0]=10//2=5(sp^(3)d-"non-planar")` (iv) `OF_(2) : H=1//2[6+2-0+0]=8//2=4(sp^(3)-"non-planar")` (v) `H_(2)O: H=1//2[6+2-0+0]=8//2=4(sp^(3)-"non-planar")` |
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| 205. |
Intramolecular hydrogen bonding is present inA. waterB. o-nitrophenolC. p-nitrophenolD. methyl amine |
| Answer» Correct Answer - B | |
| 206. |
Which of the following species will have intramolecular hydrogen bonding ?A. B. C. D. |
| Answer» Correct Answer - A | |
| 207. |
Which one of the following compounds shows the presence of intramolecular hydrogen bond ?A. `H_(2) O_(2)`B. HCNC. CelluloseD. Concentrated acetic acid |
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Answer» Correct Answer - C `H_(2)O_(2) HCN and CH_(3) COOH` have only intermolecular hydrogen bonding whereas cellulose has intramolecular hydrogen bonds |
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| 208. |
What is coordination number of `Na^(+) and Cl^(-) ` ion in Nacl ? |
| Answer» Each has a coordination number of 6, i.e. each ion is surrounded by six oppositely charged ions. | |
| 209. |
The decreasing valuse of bond angles from `NH_(3) (106^(@))` to `SbH_(3)(101^(@))` down group -15 of the periodic table is due to .A. Decreasing lp-lp repulsionB. Decreasing electronegativityC. Increasing bp-bp repulsionD. Decreasing p-orbital character in `sp^(3)` . |
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Answer» Correct Answer - B Due to decrease in `EN` the bond pair moves away hence repulsion will be less from `NH_(3)` to `SbH_(3)` . |
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| 210. |
The `Pt-CI` distance is `2.32 A` in several crystaline compounds What is the `CI-CI` distance in structure (i) and in structure (ii) . |
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Answer» In structure (i) `CI - CI = 2(2.32) = 4.64 Å` In structure (ii) Using the pythagorean theorem we have `CI - CI = sqrt2 (2.32) = 3.28 Å `. |
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| 211. |
The `Pt-CI` distance is `2.32 A` in several crystaline compounds What is the `CI-CI` distance in structure (i) and in structure (ii) .A. cis-isomerB. trans-isomerC. chiral isomerD. none of these |
| Answer» Correct Answer - B | |
| 212. |
In ionic solids, the oppositely charged ions are closely packed in space and have strong electrostatic forces of attraction. These compound have high melting and boiling points and are poor conductors of electricity in the solid state. (i) Why are ionic solids poor conductros of electricity ? (ii) What happens to electrical conductivity when these are dissolved in water ? |
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Answer» (i) In the ionic compounds, the electrical conductivity is due to the mobility of ions. Since the ions can hardly move in the solid state, ionic solids are poor conductors of electricity. (ii) When the ionic solids are dissolved in water, the ions get separted and acquire mobility. As a result, they become good conductors of electricity. |
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| 213. |
Assertion : The dipole moment helps to predict whether a molecule is polar or non- polar. Reason : The dipole moment helps to predict geometry of molecule.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 - A When dipole momnet `(mu)` is high the bond is more polar When mu is zero it is a symmetric molecule When `mu ne 0` the moleule is not symmetrical . |
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| 214. |
Assertion : The dipole moment helps to predict whether a molecule is polar or non- polar. Reason : The dipole moment helps to predict geometry of molecule.A. If both assertion and reason are true and the reason is a true 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 The linear triatomic moulas of the type B-A-B e.g. `CO_(2)` have no dipole moment not because the individual `C = O` bonds are non- polar but because the two bond moment (polarity) cancel each other .Therefore , gemoetry of molecule can be predicted by the value of dipole moment . |
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| 215. |
The study of dipole moment of a molecule is useful to explain the shape of a molecule and also to predict a number of other properties of the molecule. The net dipole moment of a polyatomic is the resultant of the different bond dipole moments present in that molecule. Answer the following questions on the basis of above paragraph : (i) What is dipole moment ? (ii) Out of `CO_(2)and H_(2)O,` which molecule is polar in nature any why ? |
| Answer» (ii) `H_(2)O` molecule is polar where as `CO_(2)` is non-polar though both are triatomic molecules. Actually `H_(2)O` has bent structure and the polarities of the O-H bonds donot cance. On the other hand `CO_(2)` is linear and as a result of this, the polarities of C=O bonds mutually cencel out. | |
| 216. |
Amongst the given structures , which are permissible resonance forms ?A. `overset(oplus)(C)H_(2)- underset(CH_(3))underset(|)overset(..)(N) - underset(..)overset(..)(O):^(Theta)`B. `CH_(2) = underset(CH_(3))underset(|)(N) = overset(..)(O):`C. `CH_(2) = underset(CH_(3))underset(|)overset(+)(N) - overset(..)underset(..)(O):^(-)`D. `overset(-)(C)H_(2) - overset(+)underset(CH_(3))underset(|)(N) = O` |
| Answer» Correct Answer - A::C::D | |
| 217. |
Asseration: The elctronic structure of `O_(3)` is: Reason: structure is not allowed because octet around `O` cannot be expanded 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 | |
| 218. |
Polarisation of electrons in acrolein may be written as :A. `overset(delta-)(CH_2)=overset(delta+)(CH)-CH=O`B. `overset(+delta)(CH_2)=CH-CH=overset(-delta)(O)`C. `overset(-delta)(CH_2)=CH-overset(delta+)(CH)=O`D. `overset(delta-)(CH_2)=CH-CH=overset(delta+)(O)` |
| Answer» Correct Answer - B | |
| 219. |
The maximum bond strengths is in:A. `O_2`B. `O_2^+`C. `O_2^-`D. `O_2^(2-)` |
| Answer» Correct Answer - B | |
| 220. |
Which of the two lons from the list given have the geometry that is explained by the same hybridization of orbitals `NO_(2)^(-),NO_(3)^(-) ,NH_(2)^(-) NH_(4)^(+) SCN^(-)`?A. `NO_(2)^(-) and NH_(2)^(-)`B. `NO_(2)^(-) and NO_(3)^(-)`C. `NH_(4)^(+) and NO_(3)^(-)`D. `SCN^(-)` and `NH_(2)` |
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Answer» Correct Answer - b Appliying `X = (1)/(2) [VE + MA - C + a]` `NO_(2)^(-)rarr sp^(2) ,NO_(3)^(-),NH_(2)^(-) rarr sp^(3)` `NH_(4)^(+) rarr sp^(3),SCN^(-) rarr sp`. |
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| 221. |
Which of the following resonating form is not correct for `CO_2`?A. B. C. D. |
| Answer» Correct Answer - D | |
| 222. |
The nature of hybridisation in the `NH_3` molecule isA. spB. `sp^2`C. `sp^3`D. `sp^2d` |
| Answer» Correct Answer - C | |
| 223. |
How many `sigma`(sigma) bonds are there in `CH_2=CH-CH = CH_2` ?A. 3B. 6C. 9D. 12 |
| Answer» Correct Answer - C | |
| 224. |
Which of the following conditions apply to resonating structrues ?A. The contributing structures must have the same number of unpaired electrons .B. The contributing structures should have similar energies .C. The contributing structures should be so written that unlike charges reside on atoms that are far apart .D. The positive charge should be present on the electropositive element and the negative charge on the electrongative element . |
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Answer» Correct Answer - C There is no rule which states that positive charge and negative charge on atoms should be for apart . |
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| 225. |
Decreasing order of bond angle of `(NO_(2)^(o+), NO_(2), NO_(2)^(ɵ)` isA. I gt II gt IIIB. II gt I gt IIIC. III gt II gt ID. IIIgt I gt II |
| Answer» Correct Answer - B | |
| 226. |
In the following molecule, the hybrid state of 1 and 3 carbon atoms is `CH_2=C = CH_2`A. `sp^3`B. `sp^2`C. spD. `sp^3d` |
| Answer» Correct Answer - B | |
| 227. |
Hybridisation involvesA. separation of atomic orbitalsB. overlapping of atomic orbitalsC. mixing of atomic orbitals of atomD. all of these. |
| Answer» Correct Answer - C | |
| 228. |
Give the decreasing order of melting points of the following `NH_(3),PH_(3),(CH_(3))_(3)N` Explain (b) In which molecule is the van der Waals force likely to be the most important in determining the m.pt and b.pt for `ICI,Br_(2),HCI,H_(2)S,CO` .A. `CO`B. `H_(2)S`C. `Br_(2)`D. `HCI` |
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Answer» Correct Answer - c `Br_(2)` is non polar molecular and hence its molting and boiling points depend only on Van der Waals forces while all order molecular are having dipole moment hence their melting and boling point depends upon dipole-dipole interactions. |
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| 229. |
Which of the following posses maximum hydration energy ?A. `MgSO_(4)`B. `RaSO_(4)`C. `SrSO_(4)`D. `BaSO_(4)` |
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Answer» Correct Answer - a `Mg^(2+)` is a smaller cation in these .The smaller is the cation the more is the hydration energy . |
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| 230. |
`sp^3d^2` hybrid orbitals areA. linearB. pentagonal bipyramidalC. trigonal bipyramidalD. octahedral |
| Answer» Correct Answer - D | |
| 231. |
The maximum possible number of hydrogen bonds a water molecule can form isA. 2B. 4C. 3D. 1 |
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Answer» Correct Answer - B Four : two covalent and two H-bonds. |
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| 232. |
Arrange in decreasing order of melting and boiling pouints of hydrides of groups `15,16` and 17 (b) Give the decreasing order of melting and boiling points of `H_(2)ONH_(3)` and `HF` (c ) Give the decreasing order of boiling points (I) `C_(2)H_(5)OH` (II) `(CH_(3))_(3)NH` (III) `C_(2)H_(5)NH_(2)` (d) Give the decreasing of solubility in `H_(2)O` (I) `PhNH_(2)` (II) `(C_(2)H_(5))_(2)NH` , (III) `C_(2)H_(5)NH_(2)` . |
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Answer» (a) (i) Melting point of group 15 hydries `NH_(3) gt BiH_(3) gt SbH_(3) gt AsH_(3) gt PH_(3)` Boiling point of group 15 hydries `BiH_(3) gt SbH_(3) gt NH_(3) gt AsH_(3) gt PH_(3)` (ii) Melting point of group 16 hybries `H_(2)O gt H_(2)Te gt H_(2)Se gt H_(2)S` Boiling point of group 16 hybries `H_(2)O gt H_(2)Te gt H_(2)Se gt H_(2)S` (iii) Melting point of group 17 hydries `HI gt HF gt HBr gt HCI` Boiling point of group 17 hydries `HF gt HI gt HBr gt HCI` (b) `H_(2)O gt NH_(3) gt HF` Boiling point `H_(2)O gt HF gt NH_(3)` (c ) Boiling point of alcohols gt boiling points of amines (due to EN of `O gt ENofN)` (ii) Boiling points of `1^(@) gt2^(@)` amine gt `3^(@)` amine due to the three H-bonds, two H-bonds in `1^(@)` and `2^(@)` amines, `3^(@)` amine do not form H-bonds, since no H-atoms is present Therefore, decreasing boiling order is `C_(2)H_(5)OH gt C_(2)H_(5)NH_(2) gt (C_(2)H_(5))_(2)NH(I gt III gt II)` (d) Solubility in `H_(2)O` of `1^(@)` amine `gt 2^(@)` amine `gt 3^(@)` amine due to three H-bonds and two H-bonds in `1^(@)` and `2^(@)` amines respectively Aromatic amines are less solubel in `H_(2)O` due to larger non polar part than aliphatic amines Therefore, decreasing solubility order is `C_(2)H_(5)NH_(2) gt (C_(2)H_(5))_(2)NH:PhNH_(2)(III gt II gt I)` . |
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| 233. |
The correct order regarding the electrongetiveively of hybrid orbitals of carbon isA. ` sp lt sp^(2) gt sp^(3)`B. `sp lt sp^(2) lt sp^(3)`C. `sp gt sp^(2) gt sp^(2)`D. `sp gt sp^(2) gt sp^(3)` |
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Answer» Correct Answer - D Smaller the size , greater is its electronegetivily. |
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| 234. |
Statement-1 : All hybrid orbitals of same composition have same shape and energy . Statement-2 : All hybrid orbitals on carbon in `CH_(3)Cl` have same as character . Statement-3 : Hybridisation is a physical process .A. TFTB. TTTC. FFTD. TFF |
| Answer» Correct Answer - B | |
| 235. |
Which one of the following compounds has `sp^(2)` hybridisation ? .A. `CO_(2)`B. `SO_(2)`C. `N_(2)O`D. `CO` |
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Answer» Correct Answer - B Hybridisation of `overset(..)SO_(2) implies (1)/(2) (6 +0) =3 =sp^(2)` . |
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| 236. |
The type of hybrid orbitals used by chlorine atom in `ClO_(2)^(-)` is :A. `sp^(3)`B. `sp^(2)`C. `sp`D. `sp^(3) d ` |
| Answer» Correct Answer - A | |
| 237. |
Match the shape of molecules in Column I with the type of hybridisation in Column II. `{:(" Column I Column II " ),((i) "Tatradedral (a)"sp^(2) ),((ii)"Trigonal (b)sp"),((iii)"Linear (c)"sp^(3)):}` |
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Answer» Correct Answer - (i)-(c),(ii)-(a),(iii)-(b) Tetrahedral `to sp^(3)` , Trigonal `to sp^(2)` , Linear `to sp `. |
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| 238. |
How many of the following compounds have `sp^(3)` hybridisation (i) `SO_(4)^(2-)` (ii) `SO_(5)^(2-)` (iii) `PO_(4)^(3-)` (iv) `PO_(5)^(3-)` (v) `I_(3)^(Theta)` (iv) `CO_(3)^(2-)` (vii) `CO_(4)^(2-) `. |
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Answer» Correct Answer - 4 The acids, peracids and their ions have same hybridisation of central atom Thus `SO_(4)^(2-), SO_(5)^(2-), PO_(4)^(3-), PO_(5)^(3-)` all have same `sp^(3)` hybridisation (v) `I_(3)^(Theta)` have `sp^(3)` d hybridisation (vi) `CO_(3)^(2-)` and (vii) have `sp^(2-)` hybridisation . |
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| 239. |
Consider the following molecules or ions : (i) `CH_(2) Cl_(2) (ii) NH_(4)^(+) (iii) SO_(4)^(2-) (iv) ClO_(4)^(-) (v) NH_(3)` `sp^(3)` hybridisation is involved in the formation ofA. (i), (ii), (v) onlyB. (i) , (ii) only sC. (ii) onlyD. (i), (ii), (iii), (iv) and (v)` |
| Answer» All the given ions/molecules involve involve `sp^(3) ` hybridisation . | |
| 240. |
Hydrogen bonds are formed in many compounds e.g. `H_(2)O`, HF, `NH_(3)`. The boiling point of such compounds depends to a extent on the strength of hydrogen bond and the number of hydrogen bonds. The correct decreasing order of the boiling points above compounds isA. `HF gt H_(2)O gt NH_(3)`B. `N_(2)O gt HF gt NH_(3)`C. `NH_(3) gt HF gt H_(2)O`D. `NH_(3) gt H_(2)O gt HF` |
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Answer» Correct Answer - B `H_(2)O(373K)gt HF(293K)NH_(2)(238.5K)` |
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| 241. |
Hydrogen bonds are formed in many compounds e.g. `H_(2)O`, HF, `NH_(3)`. The boiling point of such compounds depends to a extent on the strength of hydrogen bond and the number of hydrogen bonds. The correct decreasing order of the boiling points above compounds isA. `HF gt H_(2)O gt NH_(3)`B. `H_(2)O gt HF gt NH_(3)`C. `NH_(3) gt HF gt H_(2)O`D. `NH_(3) gt H_(2)O gt HF` |
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Answer» Correct Answer - B b) Strength of H-bond is in the order H....F `gtH.....OgtH.....N` But each `H_(2)O` molecule is linked to four other `H_(2)O` molecules through H-bonds whereas each HF molecules is linked only to two other HF molecules. Hence, b.p of `H_(2)O gt b.p` of `HFgtb.p` of `NH_(3)` |
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| 242. |
Assertion (A): Lithium chloride is predominantly covalent compound. Reason (R ): electronegativity difference between Li and CI is small.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 - C | |
| 243. |
Assertion (A): Lithium chloride is predominantly covalent compound. Reason (R ): electronegativity difference between Li and Cl is small.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 `LiCl` is covalent due to high polarizing power of `Li^(+)`. |
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| 244. |
The dipole moment of LiH is ` 1.964xx10^(-29)` Cm and the interatomic distance between Li and H in the molecule is ` 1 .596Å` . Calualate the persent ionic character of the molecule . |
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Answer» The dipole molecule were 100% ionic , `mu_("ionic")= q xx d = (1 .602 xx10^(-19)C) xx(1.596 xx10^(-10)m) = 2. 557xx 10^(-29) ` cm `mu_("observed ")=1 .964 xx10^(-29) ` cm (Given) `therefore % "ionic character"= (mu_("observed "))/(mu_("ionic")) xx100 = (1 .964 xx10^(-29))/ (2. 557xx 10^(-29) cm )xx100 = 76.81 %`. |
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| 245. |
Shape of the compound depend on type and number of electron pair around central atom . These electron pair repel each other and stay as far as possible . The repulsion sequence is as `L.P - L.P. gt B.P. - L.P. gt B.P - B.P.` Which of the given compound is planar ?A. `XeF_(5)^(-)`B. `XeF_(4)`C. `I Cl_(4)^(-)`D. All of these |
| Answer» Correct Answer - D | |
| 246. |
Electric Dipole moment is a vector quantify . If a compound contain more than one polar bond then , the net electric dipole moment is equal to vector sum of individual dipole moment . e.g., In `H_(2)O` `mu = 1.84 D` Out of following which has the highest dipole momentA. B. C. D. |
| Answer» Correct Answer - B | |
| 247. |
Electric Dipole moment is a vector quantify . If a compound contain more than one polar bond then , the net electric dipole moment is equal to vector sum of individual dipole moment . e.g., In `H_(2)O` `mu = 1.84 D` Which of the following compounds have zero dipole moment ?A. `SF_(4)`B. `I Cl_(4)^(-)`C. `H_(2)S`D. `SnCl_(2)` |
| Answer» Correct Answer - B | |
| 248. |
Electric Dipole moment is a vector quantify . If a compound contain more than one polar bond then , the net electric dipole moment is equal to vector sum of individual dipole moment . e.g., In `H_(2)O` `mu = 1.84 D` The compound has molecular formula `XeO_(n)` where n is number of lone pair , dipole moment of the compound is minimum when n isA. ZeroB. 1C. 3D. Both (1) & (2) |
| Answer» Correct Answer - A | |
| 249. |
Shape of the compound depend on type and number of electron pair around central atom . These electron pair repel each other and stay as far as possible . The repulsion sequence is as `L.P - L.P. gt B.P. - L.P. gt B.P - B.P.` Choose the incorrect matchA. `{:("Compound" ,, "Structure") , ((1) SnCl_(2) ,, "Linear"):}`B. `{:("Compound" ,, "Structure") , ((2) CO_(2) ,, "Linear"):}`C. `{:("Compound" ,, "Structure") , ((3)I_(3)^(-) ,, "Linear"):}`D. `{:("Compound" ,, "Structure") , ((4)N_(3)^(-) ,, "Linear"):}` |
| Answer» Correct Answer - A | |
| 250. |
Lattice energy : Lattice energy is the amount of energy released when one mole of ionic compound is formed from its gaseous ions `underset((g))(Na^(+)) + underset((g))(Cl^(-)) to NaCl(s) + Deltax kJ` Lattice energy also depend on the 3-D arrangement of ion . In the given compounds least Lattice energy is present inA. AgFB. AgBrC. AgClD. NaCl |
| Answer» Correct Answer - B | |