InterviewSolution
Saved Bookmarks
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.
| 101. |
The common features among the species ` CN^- , CO ` and ` NO^+` are :A. Bond order three and isoelectronicB. Bond order three and weak field ligandsC. Bond order two and `pi` acceptorD. Iso-electronic and weak fields. |
| Answer» Correct Answer - A | |
| 102. |
It is believed that atoms combine with each other such that the outermost shell acquires a stable configuration of 8 electrons. If stability were attAlned with 6 electrons rather than 8. What would be the formula of the stable fluoride ion.A. `F^-`B. `F^+`C. `F^(2+)`D. `F^(3+)` |
| Answer» Correct Answer - B | |
| 103. |
It is believed that atoms combine with each other such that outrtmost shell acquires stable configuration of 8 electrons . If stability were attained with 6 electrons rather than 8, what would be the formula of the stable fluoride ion ?A. `F^(-)`B. `F^(+)`C. `F^(2+)`D. `F^(3+)` |
|
Answer» Correct Answer - B No of valence electrons in F= 7 . In order to have 6-electrons in the outermost shell, it should lose one electrons and hence from `F^(+)` |
|
| 104. |
In which of the following molecules, the central atom has one lone pair and three bond pairs of electrons,A. `H_(2)S`B. `AlCl_(3)`C. `NH_(3)`D. `BF_(3).` |
|
Answer» Correct Answer - C `NH_(3)` has one lone pair and three bond pairs of electrons. |
|
| 105. |
According to molecular arbital theory,A. `C_(2)^(2+)` is expected to be diamagneticB. `O_(2)^(2+)`os expected to have a bond length than `O_(2)`C. `N_(2)^(2+) and N_(2)^(-)` have the same bond order.D. `He_(2)^(+)` has the same energy as two isolated He atoms. |
|
Answer» Correct Answer - A::C (a,c) Both are correct. |
|
| 106. |
Which of the following statements are not correct ?A. `NaCl (s)` being an ionic compound , is a good conductor of electricityB. In cononical structures there is a difference in the arrangement of atomsC. Hybrid orbitals form stronger bonds the p-orbitals.D. VSEPR theory connot explain the equare planar geometry of `XeF_(4)` |
| Answer» Correct Answer - A::B::D | |
| 107. |
Which one of the following statement is correct ?A. Melting point of and boiling point of HI are greater than those of HFB. Boiling point of HI is less than that of HF but melting point of HI is greater than that of HFC. Boiling point of HI is greater than that of HF but melting point of HI is less than that of HFD. Melting point and boiling point of HI are less than that of HF |
| Answer» Correct Answer - B | |
| 108. |
In which of the following , the double bond consists of both pi bondsA. `O_(2)`B. `C_(2)`C. `Be_(2)`D. `S_(2)`. |
| Answer» Correct Answer - B | |
| 109. |
Explain the following (a) Dipole moment of `CH_(3)F` is 1.85D` and that of `CD_(3)F` is `1.86D` . |
| Answer» It is due to the large size of `CD_(3)F` but `D` is less `EN` than `H(mu = q xx d)` . | |
| 110. |
Define electronegativity. How does it differ from electron gain enthalpy ? |
|
Answer» Electronegativity is the ability of an atom in a chemical compound to attract a bond pair of electron towards itself. Electronegativity of any given element is not constant. It varies according to the element to which it is bound. It is not a measurable quantity. It is only a relative number. On the other hand, electron gain enthalpy is hte enthalpy change that takes place when an electron is added to a neutral gaseous atom to form an anion. It can be negative or positive depending upon whether the electron is added or removed. An element has a constant value of the electron gain enthalpy that can be measured experimentally. |
|
| 111. |
The `H-O-H` bond angle in the water molecule is `105^@` , the`H -O` bond distance being `0.94 Å`, The dipole moment for the moelcule is `1.85D`. Calculate the charge on the oxygen atom .A. `2 xx 10^(-10)` esuB. `3.28 xx 10^(-10)` esuC. `3.22 xx 10^(-10)` esuD. `1.602 xx 10^(-19)` esu |
| Answer» Correct Answer - C | |
| 112. |
`SaCI_(4)` is a convalent lquid becauseA. electron clouds of the `CI^(-)` ions are weakly polarized to envelop the cationB. electron clouds of the `CI^(-)` ions are strongly polarized to envelop the cationC. its molecules are attracted to one another by strong van der Waals forcesD. So shown inert pair effect |
|
Answer» Correct Answer - b Because of high charge density on `Sn^(4+)` it ha shigh polarising power and thus leads to greater polarisation of anion i.e., greater distortion of electron clouds of the `CI^(-)` ions . So `SuCI_(4)` is most covalent. |
|
| 113. |
Which of the most convalent ?A. `C - O`B. `C - Br`C. `C - S`D. `C - F` |
|
Answer» Correct Answer - c `C-S` will be most covelent chqaracter deponds on the size of cation and atnion |
|
| 114. |
The molecule having permanent dipole moment isA. `SF_(4)`B. `XeF_(4)`C. `NH_(3)`D. `BF_(3)` |
|
Answer» Correct Answer - d `SF_(4)` has `sp^(2)d^(2)`-hybridization and see-saw goemetry. |
|
| 115. |
Choose the correct pair regarding dipole momentA. `CH_(3)Cl gt CH_(3)F`B. `NH_(3) gt NF_(3)`C. `HF gt HCl`D. All of these |
| Answer» Correct Answer - D | |
| 116. |
Which pair of moecules will have permanent dipole moment for both members ?A. `NO_(2) and O_(3)`B. `SiF_(4) and CO_(2)`C. `SiF_(4) and NO_(2)`D. `NO_(2) and CO_(2)` |
|
Answer» Correct Answer - a Molecules having irregular geometries, have permanent dipole. |
|
| 117. |
Indicate wheter the following pairs of elements form ionic or covalent compounds Also write their molecular formula (a) `C` and `S` (b) Na and `CI` (c ) `S` and `O` (d) Ca and `H` . |
|
Answer» Covalent, `CS_(2)` (b) Ionic, NaCI (c ) Covalent, `SO_(2),SO_(3)` (d) Ionic, `CaH_(2)` . |
|
| 118. |
In which of the following set of compounds , bond angle remains constant for all members ?A. `NH_(3) , PH_(3) , AsH_(3)`B. `PF_(3) , PCl_(3) , PBr_(3)`C. `OH_(2) , OF_(2) , O Cl_(2)`D. `BF_(3) , BCl_(3) , BBr_(3)` |
| Answer» Correct Answer - D | |
| 119. |
If a molecule `MX_(3)` has zero dipole moment, the state hybridisation of M is :A. `sp^(3)d`B. sp.C. `sp^(3)p^(2)`D. `sp^(2)` |
|
Answer» Correct Answer - D `MX_(3)` is a planar molecule and the state of hybridisation of M is `sp^(2).` |
|
| 120. |
Which of the following has zero dipole moment ?A. `CO_(2)`B. `NH_(3)`C. `NF_(3)`D. `H_(2)O` |
| Answer» Correct Answer - A | |
| 121. |
Which of the following compounds have zero dipole moment ?A. `BF_(3)`B. `SnCl_(2)`C. `H_(2)O`D. `NH_(3)` |
| Answer» Correct Answer - A | |
| 122. |
In the of the following pairs of molecules /ions both the species are not likely to exist?A. `H_(2)^(-),He_(2)^(2+)`B. `H_(2)^(+),He_(2)^(2-)`C. `H_(2)^(-),He_(2)^(2-)`D. `H_(2)^(2+),He_(2)` |
|
Answer» Correct Answer - D Species with BO=zero does not exist. `H_(2)^(2+):[sigma_(1s)]^(0),B.O=0` `He_(2):[sigma_(1s)]^(2)[sigma_(1s)^(**)]^(2),B.O=0` |
|
| 123. |
Which of the following has zero dipole moment ?A. `CO_(2)`B. `H_(2)O`C. `SO_(2)`D. `NO_(2)` |
| Answer» Correct Answer - A | |
| 124. |
`sp^(3)` dhybridisation hasA. Octahedral shapeB. Square planar shapeC. Trigonal bipyramidal shapeD. Pentagonal bipyramidal shape |
| Answer» Correct Answer - C | |
| 125. |
Which of the following is diamagnetic ?A. `O_(2)^(+)`B. `O_(2)`C. `O_(2)^(-)`D. `O_(2)^(2-)` |
| Answer» Correct Answer - D | |
| 126. |
Which of the following is diamagnetic ?A. `H_(2)^(+)`B. `He_(2)^(+)`C. `O_(2)`D. `N_(2).` |
|
Answer» Correct Answer - D `N_(2)` is diamagnetic in nature |
|
| 127. |
Which of the following is not diamagnetic ? .A. `O_(2)^(2-)`B. `Li_(2)`C. `N_(2)^(o+)`D. `C_(2)` |
|
Answer» Correct Answer - C In `N_(2)^(o+)` there is one unpaired electron in bonding `MO` ltbr gt `(I.e sigma 2p_(z)^(1))` . |
|
| 128. |
Identify which of them are polar and non-polar (a) `HF` (b) `BeCI_(2)` (c ) `HgCI_(2)` (d) `NH_(3)` (e) `H_(2)O` (f) `N_(2)` (g) `AICI_(2)` (h) `C CI_(4)` (i) `CI_(2)` (j) `SiCI_(4)` . |
| Answer» Except `CI_(2)` and `N_(2)` all other molecules have polar bond But only `HF,NH_(3),H_(2)O` are actully polar In these `mu_("net")ne0` due to their structure. In the rest of the molecules the net `mu` is zero, due to symmetric (regular) geometry . | |
| 129. |
The molecules with maximum percentage ionic character is :A. HIB. HBrC. HClD. HF. |
|
Answer» Correct Answer - D HF has the max. percentage ofionic character as it is max. polar. |
|
| 130. |
Select the species which is best described to the right (a) `CI_(2),Br_(2),I_(2)` (has the lowest boiing point) (b) `CI,Ar,K` (has the smallest `IE`) (c ) `CH_(4), NH_(3),HF` (has the highest boiling point) (d) `CO_(2),NH_(3),CO` (has zero dipole moment) (e) `HOI,HOBr,HOCI` (is the weakest acid) . |
|
Answer» `CI_(2)` has the fewest electrons, therefore least van der Waals forces (b) `K` is metal, thus smallest `IE` ( c) `HF` has H-bonding the largest intermolecular force (d) `CO_(2)` polar but `CO_(2)` is linear, and the bond moments cancel (e) In `HOI,I` is least `EN` it pulls electrons least leaving a greater fraction of the `O` electrons to bond with the `H` atom . |
|
| 131. |
How the bond energey varies from `N_(2)^(Θ)` and `N_(2)^(o+)` and why ? (b) On the basic of molecular orbital theory what is the similarity between (i) `F_(2)` and `O_(2)^(2-)` (ii) `CO,N_(2)` and `NO^(o+)` ? . |
|
Answer» Bond energy of `N_(2)^(o+)` = Bond energy of `N_(2)^(Θ)` because the bond order is same in both the species. [Howerver `N_(2)^(o+)` is slightly more stable than `N_(2)^(Θ)` as number of antibonding electrons is higher in `N_(2)^(Θ)` than `N_(2)^(o+)`] (b) Both `F_(2)` and `O_(2)^ (2-)` have same bond order, same bond lenght and red diamagmagnetic (ii) These are isoelectronic species, possess same bond order and same bond lenght . |
|
| 132. |
Among `LiCI,BeCI_(2)` and `C CI_(4)` the covalent bond character varies as .A. `LiCl gt BeCl_(2) lt BCl_(3) gt C Cl_(4)`B. `LiCl gt BeCl_(2) lt BCl_(3) lt C Cl_(4)`C. `LiCl lt BeCl_(2) lt BCl_(3) lt C Cl_(4)`D. `LiCl gt BeCl_(2), gt BCl_(3) gt C Cl_(4)` |
|
Answer» Correct Answer - C Lseser the differernce in the electrogentivity of the metal and chlorine atome, more is the character. The order of increasing covalent character is : `LiCl lt BeCl_(2) lt BCl_(3) lt C Cl_(4)` |
|
| 133. |
The electronic configurations of three elements, A,B are C are given below. Answer the questions 14 to 17 on the basis of these configurations . `{:(A " "1s^(2) " "2s^(2) " "2p^(6)),(B" "1s^(2) " "2s^(2) " "2p^(6) " " 3s^(2) " "3p^(3)),(C" "1s^(2) " "2s^(2) " "2p^(6) " " 3s^(2) " "3p^(5)):}` Stable form of a may be represented by the formula :A. AB. `A_(2)`C. `A_(3)`D. `A_(4)` |
|
Answer» Correct Answer - A `(""_(10)Ne)` . |
|
| 134. |
Four diatomic species are listed in different sequence .Which of these represent the correct order of their increasing bond order?A. `NO lt C_(2)^(2-) lt O_(2)^(-) lt He_(2)^(+)`B. ` C_(2)^(2-) lt He_(2)^(+) lt O_(2)^(-)`C. ` HE_(2)^(+) lt O_(2)^(-) lt NO lt C_(2)^(2-)`D. `O_(2)^(-) lt NO lt C_(2)^(2-) lt He_(2)^(+)` |
|
Answer» Correct Answer - c B.O. of `NO = (1)/(2) (8 - 3) = (5)/(2) = 2.5` B.O. of `C_(2)^(-) = (1)/(2) (8 - 2) = (6)/(2) = 3` B.O. of `O_(2)^(-) = (1)/(2) (8 - 5) = (3)/(2) = 1.5` B O of `He_(2)^(+) = (1)/(2) (2 - 1) = (1)/(2) = 0.5` `rArr "B.O. is" C_(2)^(-) gt NO gt O_(2)^(-) gt He_(2)^(+)` |
|
| 135. |
Indicate the type of bond anghle presents in `IF_(5)`A. `90^(@)`B. `90^(@),120^(@)`C. `90^(@),180^(@)`D. `90^(@),120^(@),180^(@)` |
|
Answer» Correct Answer - a Lodine undergoes `sp^(3)d^(2)` hybridisation with `5` bond pair and one lone pair having square pyramidal structure |
|
| 136. |
Which of the following compound of group-14 elements would you expect to be most ionic in character ?A. `C Cl_(4)`B. `SiCl_(4)`C. `PbCl_(2)`D. `PbCl_(4)` |
| Answer» Correct Answer - C | |
| 137. |
Arrange `C-C,C=C` and `C-=C` in order of (i) Decreasing bond energey (ii) Decreasing bond lengths (b) The `As-CI` bond distance in `As CI_(3)` is `2.20 A` Estimate the single- bond covalent radius of As . |
|
Answer» `C-= CgtC =Cgt C-C` (ii) `C - C gtC implies C -=C` (b) Internuclear distance -radius of chlorine atom = radius of As - atom `:. (2.20 - 0.99) = 1.21 Å` . |
|
| 138. |
How does the bond length vary in dicarbon species `C_(2),C_(2)^(-),C_(2)^(2-)` |
|
Answer» The molecular orbital configurations and the respective bond orders of the species are : `C_(2) : [sigma_(1s)]^(2)[sigma_(1s)^(**)]^(2) [sigma_(2s)]^(2)[sigma_(2s)^(**)]^(2)[pi_(2py)]^(2)[pi_(2py)]^(2)" "B.O.=1//2[8-4]=2` `C_(2)^(-) : [sigma_(1s)]^(2)[sigma_(1s)^(**)]^(2) [sigma_(2s)]^(2)[sigma_(2s)^(**)]^(2)[pi_(2py)]^(2)[pi_(2py)]^(2)[sigma_(2pz)]^(1)" "B.O.=1//2[9-4]=2.5` `C_(2)^(2-) : [sigma_(1s)]^(2)[sigma_(1s)^(**)]^(2) [sigma_(2s)]^(2)[sigma_(2s)^(**)]^(2)[pi_(2py)]^(2)[pi_(2py)]^(2)[sigma_(2pz)]^(2)" "B.O.=1//2[10-4]=3.0.` Since the bond length si inversely proportional to the bond order , the increasing order of bond lengyh is : `C_(2)^(2-) lt C_(2)^(-) lt C_(2)` |
|
| 139. |
The correct stability order of the following resonance structures is `{:((a) H_(2)C = overset(+)N - N^(-) ," "H(2)overset(+)C-N=N^(-)),( (I), " "(II)),(H_(2) overset(-)C-overset(+)N-=N,H_(2)overset(-)C-N=overset(+)N),((III)," "(IV)):}`A. `(I) gt (II) gt (IV) gt (III)`B. `(I) gt (III) gt (II) gt (IV)`C. `(II) gt (I) gt (III) gt (IV)`D. `(III) gt (I) gt (IV) gt (II)` |
|
Answer» Correct Answer - B (i)Greater the number of `pi-`bonds, greater is the delocalisation of electrons, greater is the stability. (ii) A reaonating structure is more stable is the negative charge is on electronegative atom and positive charge on electropositive atom. (iii) Like charge should not be close and unlike charges should not be widely separated . Keeping above points in view , the stability order will be `(I) gt (III) gt (II) gt (IV)` . |
|
| 140. |
Arrange the following single bonds in order of bond energy giving reasons : ` C - C , N - N, O-O, F-F` |
| Answer» ` C - C gt N- N gt O - O gt F - F` | |
| 141. |
Four diatomic species are listed below in different sepuences . Which of these presents the correct order of their increasing bond order ?A. `O_(2)^(-) lt NO lt C_(2)^(2-) lt He_(2)^(+)`B. `No lt C_(2)^(2-) lt O_(2)^(-) lt He_(2)^(+)`C. `C_(2)^(2-) lt He_(2)^(+) lt NO lt O_(2)^(-)`D. `He_(2)^(+) lt O_(2)^(-) lt NO lt C_(2)^(2-)` |
|
Answer» Correct Answer - D `He_(2)^(+) ,lt O_(2)^(-) ,lt NO ,lt C_(2)^(2-)` `((1)/(2)) " "(1(1)/(2))" "(2(1)/(2))" "(3)` |
|
| 142. |
The species with fractional bond order is :A. `O_(2)^(+)`B. `O_(2)^(2+)`C. COD. `He_(2)` |
|
Answer» Correct Answer - A `O_(2)^(+)` has fractional B.O. (2.5) |
|
| 143. |
What is the effect of the following processes on the bond order in `N_(2) and O _(2)` ? |
|
Answer» (i) `N_(2) (B.O. = 3 )underset(-e^(-))to N_(2)^(+) (B.O.= 2.5).` Thus, bond order decreases. (ii) `O_(2) (B.O. = 2)underset(-e^(-))to O_(2)^(+) (B.O. = 2.5).` Thus, bond order increases. |
|
| 144. |
What order or C-H bond lengths do you expect in `C_(2) H_(6), C_(2) H_(4) and C_(2) H_(2)` and why ? |
|
Answer» ` C - H (C_(2) H_(6)) gt C - H (C_(2) H_(4)) gt C - H (C_(2) H_(2)) ` . This is because hybrid orbrid ordbitals of carbon involved in overlapping with 1s orbital of hydrogen are `sp^(3) gt sp^(2) gt sp` and ap respecitvely and their sizes are in the order ` sp^(3) gt sp^(2) gt sp ` . |
|
| 145. |
The experimentally determined `N-F` bond length in `NF_(3)` is _____than the sum of the single covalent bond radii of `N` and `F` . |
|
Answer» Correct Answer - Greater Greater (Both `N` and `F` having have high `e^(-)` density due to smaller radii thus lp -lp repulsion increases bond length) . |
|
| 146. |
Multiple bond can exist in :A. `CH_(4)` moleculeB. `NH_(3)` moleculeC. `Cl_(2)` moleculeD. `N_(2)` molecule |
| Answer» Correct Answer - D | |
| 147. |
Arrange the given dicarbon species in order of their bond lengths giving reasons L `C_(2), C_(2)^(-), C_(2)^(2-)` . |
|
Answer» Total number of electrons in `C_(2) = 12` Its electronic configuration is `sigma _(1s)^(2) sigma _(1s)^(**2) sigma _(2s)^(2) sigma _(2s)^(**2) pi_(2p_(x))^(2)pi_(2p_(y))^(2)` "" Bond order = `(1)/(2) (8-4) = 2` E.C. of `C_(2)^(-) =sigma _(1s)^(2) sigma _(1s)^(**2) sigma _(2s)^(2) sigma _(2s)^(**2) pi_(2p_(x))^(2)pi_(2p_(y))^(2) sigma _(2p_(z))^(1)` "" Bond order `= (1)/(2) (9-4) = 2.5` E.C. of `C_(2)^(2-) = sigma _(1s)^(2) sigma _(1s)^(**2) sigma _(2s)^(2) sigma _(2s)^(**2) pi_(2p_(x))^(2)pi_(2p_(y))^(2) sigma _(2p_(z))^(2)` "" Bond order ` = (1)/(2) (10 -4)=3.0` As greater the bond order ,shorter is the bond length , therefore , order of bond length will be `C_(2)^(2-) lt C_(2)^(-) lt C_(2)` . |
|
| 148. |
Give the stability of the following resonance structures (a) `H_(2)C = overset(o+)N =overset(Θ)` (b) `H_(2)overset(o+)C -N =overset(Θ)N` (c ) `H_(2)overset(Θ)C -overset(o+)N -= N` (d) `H_(2) overset(Θ)C - N = overset(o+)N` . |
|
Answer» The lesser the charge separation the more stabel the resonating structure Structures (I) and (III) have less charge separation But in (III) charge is on electropositive `C` atom Therefore (I) is more stable than (III) Since both have six covalent bonds so (I) is more stable than (III) (IgtIII) (b) Both structures (II) and (IV) have five covalent bonds but (II) is more stable than (IV) becuse in (II) positive charge is on electropositive `C` atom and negative charge is on `EAN` (electonegative) `n` atom whereas in (IV) it is reversed Therefore `IIgtIV` So, the stability order is `I gt III gt II gt IV` . |
|
| 149. |
The pairs of species of oxygen and their magnetic behaviours are noted below . Which of the following presents the correct description ?A. `O_(2)^(-), O_(2)^(2)` - Bone diamagneticB. `O^(+) , O_(2)^(2)` - Both paramagneticC. `O_(2)^(+), O_(2)` - Both paramagneticD. `O, O_(2)^(2-)` - Both paramagnetic |
|
Answer» Correct Answer - C `O_(2)` has two unpaired electrons in `pi_(2p_(x)) and pi_(2p_(y)) `, i.e.,` pi_(2p_(x))^(**1)pi_(2p_(x))^(**2)` `O_(2)^(-)` will have one upaired electron , `O_(2)^(2-)` will have no unpaired electron, `O_(2)^(+)` will have one unprired electron, `""_(8) O = 1s^(2) 2s^(2) 2p_(x)^(2) 2p_(y)^(1) 2p_(z)^(1)` and thus has two unpaired electrons , `O^(+)` will have one unpaired electron . Thus , `O_(2)^(+) and O_(2)` contain unpaired electrons and hence will be both paramagnetic . |
|
| 150. |
Arrange the following molecules in order ionic character of their bonds ` LIF , K_(2)O , N_(2),SO_(2), CIF_(3)` |
| Answer» `N_(2) lt SO_(2) lt CIF_(3) lt K_(2)O lt LiF` . | |