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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.
| 151. |
With increasing bond order, strength of a bondA. Remain unalteredB. DecreasesC. IncreasesD. None of these |
| Answer» Correct Answer - C | |
| 152. |
The bnumber of nodal planes d orbital hasA. zeroB. oneC. twoD. three |
| Answer» Correct Answer - C | |
| 153. |
The coordinate bond is absent inA. `BH_(4)^(Theta)`B. `CO_(3)^(-2)`C. `H_(3)O^(+)`D. `NH_(4)^(o+)` |
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Answer» Correct Answer - B `.^(-)O-overset(O^(-))overset(|)(C)=O` has covalent bonds only |
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| 154. |
In which of the following `p pi-d pi` bonding is observed ?A. `NO_3^-`B. `SO_3^(2-)`C. `BO_3^(3-)`D. `CO_3^(2-)` |
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Answer» Correct Answer - B In `SO_3^(2-)` d-orbital of S atom overlaps with p-orbital of O atom to form a `pi`-bond. B, C and N do not have d-subshell in the valence shell. |
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| 155. |
A lone pair of electrons in an atom impliesA. A pair of valence electronsB. A pair of electronsC. A pair of electrons involved in bondingD. A pair of valence electrons not involved in bonding |
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Answer» Correct Answer - D A lone pair of electrons is a pair of valence electrons not involved in bonding. |
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| 156. |
The compound containing coordinate bond isA. `SO_(3)^(2-)`B. `CH_(4)`C. `CO_(2)`D. `NH_(3)` |
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Answer» Correct Answer - A `SO_(3)^(2-)` has one coordinate bond `.^(-)(O)-underset(O)underset(darr)(S)-O^(-)` |
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| 157. |
Sulphuric acid provides a simple example ofA. coordinate bondsB. non covalent compoundC. covalent ionD. non-covalent ion |
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Answer» Correct Answer - A `H_2SO_4` has the structure `H-O-underset(O)underset(darr)overset(O)overset(uarr)S-O-H` |
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| 158. |
A coordinate bond is formed when an atom an atom in a molecule hasA. electric charge on itB. All its valency electrons sharedC. A single unshared electronsD. one or more unshared electron pair. |
| Answer» Correct Answer - D | |
| 159. |
Sulphuric acid provides a simple example ofA. co-ordinate bondsB. non-covalent compoundC. covalent and co-orndinate bondD. non-covalent ion |
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Answer» Correct Answer - C Sulpuric acid contain and co-ordinate bond. |
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| 160. |
Which one of the following molecules has a coordinate bondA. `NH_(4)Cl`B. `AlCl_(3)`C. `NaCl`D. `Cl_(2)` |
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Answer» Correct Answer - A `NH_(4)Cl` has a coordinate bond besides covalent and ionic bonds `[H-overset(H)overset(|)underset(H)underset(|)(N)toH]^(+)Cl^(-)` |
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| 161. |
The compound containing coordinate bond isA. `O_(3)`B. `SO_(3)`C. `H_(2)SO_(4)`D. All of these |
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Answer» Correct Answer - D Co-ordinate bond is a special type of covalent bond which is formed by sharing of electrons between two atoms, where both the electrons of the shared pair are contributed by one atom. Since this type of sharing of electrons exists in `O_(3),SO_(3)` and `H_(2)SO_(4)`. therefore all these contains coordinate bond. |
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| 162. |
The bond that exists between `NH_(3)` and `BF_(3)` is calledA. ElectrovalentB. CovalentC. CoordinateD. hydrogen |
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Answer» Correct Answer - C `NH_(3)` has lone pair of electron while `BF_(3)` is electron deficient compound so they form a co-ordinate bond. `NH_(3)to BF_(3)` |
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| 163. |
From elementary molecular orbital theory we can deduce the electronic configuration of the singly positive nitrogen molecular ion `N_2^+` asA. `sigma(1s)^(2)sigma^(**)(1s)^(2) sigma(2s)^(2) sigma^(**)(2s)^(2)pi(2p)^(4)sigma(2p)^(1)`B. `sigma(1s)^(2)sigma^(**)(1s)^(2)sigma(2s)^(2)sigma^(**)(2s)^(2)sigma(2p)^(1) pi (2p)^(3)`C. `sigma(1s)^(2)sigma^(**)(1s)^(2)sigma(2s)^(2)sigma^(**)(2p)^(2)pi(2p)^(4)`D. `sigma(1s)^(2)sigma^(**)(1s)^(2)sigma(2s)^(2)sigma^(**)(2s)^(2)sigma(2p)^(2)pi(2p)^(2)` |
| Answer» Correct Answer - A | |
| 164. |
The bond angles of `NH_(3),NH_(4)^(o+)` and `overset(Theta)(NH_(2)` are in the order .A. `NH_(2)^(-)gtNH_(3)gtNH_(4)^(+)`B. `NH_(4)^(+)gtNH_(3)gtNH_(2)^(-)`C. `NH_(3)gtNH_(2)^(-)gtNH_(4)^(+)`D. `NH_(3)gtNH_(4)^(+)gtNH_(2)^(-)` |
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Answer» (b) As the number of lone pairs of electrons increases, bond angle decreases. Therefore, the order of bond angle is `{:(NH_(4)^(+),gt,NH_(3),gt,NH_(2)^(-)),(nolp,,1lp,,2lp):}` |
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| 165. |
Which of the following statement is not correct from the point of view of molecular orbital ?A. `Be_(2)` is not ,a stable moleculeB. `He_(2)` is not stable but `He_(2)^(+)` is expected to existC. Bond strength of `N_(2)` is maximum amongst th homonuclear diatomic moleculesD. The order of energies of molecular orbitals in `F_(2)` molecules is `pi_(2px)=pi_(2py) lt sigma_(2pz)` |
| Answer» Correct Answer - D | |
| 166. |
In compounds of type `ECI_(3)`, where `E=BP`, As or `B`, the angles `CI-E-CI` for different `E` are in the orderA. ` B lt P gt As = Bi`B. ` B gt P gt As gt Bi`C. ` B lt P = As = Bi`D. ` B lt P lt As gt Bi` |
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Answer» Correct Answer - D ` BCl_3` is `sp^2` -hybridised `( 120^@)` , Rest all are ` sp^3` hybridised . Also angle direcases form P to `Bi`. |
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| 167. |
In an octahedral structure , the pair of d orbitals involved in `d^(2)sp^(2)` hybridization isA. `d_(x^2-y^2), d_z^2`B. `d_(xz), d_(x^2-y^2)`C. `d_z^2, d_(xz)`D. `d_(xy),d_(yz)` |
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Answer» Correct Answer - A In the formation of `d^2sp^3` hybrid orbitals , two (n-1) d-orbitals of `e_g` set, i.e. (n-1) `d_z^2` and (n-1)`d_(x^2-y^2)` orbitals, one ns and three np (`np_x, np_y, and np_z)` orbitals combine together. |
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| 168. |
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 Bond angle of water is `104.5^@` (In water, oxygen is `sp^3`-hybridised with two lone pairs and two bond pairs and lp-lp repulsion is more than lp.- b.p.repulsion, which in turn is more than b.p.-b.p. repulsion) |
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| 169. |
The structure of `H_(2)O_(2)` isA. PlanarB. Non-planarC. SphericalD. Linear |
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Answer» Correct Answer - B Structure of `H_(2)O_(2)` is non-polar . It has open book structure,. |
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| 170. |
Which of the following has a linear structure?A. `C Cl_4`B. `SO_2`C. `C_2H_2`D. `C_2H_4` |
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Answer» Correct Answer - C `C_2H_2` has linear structure among given species. |
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| 171. |
Which of the following molecules is not linear?A. `SnCl_(2)`B. HClC. `CO_(2)`D. `HgCl_(2)` |
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Answer» Correct Answer - A `SnCl_(2)` is V-shaped |
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| 172. |
Which one among the following does not have the hydrogen bond?A. PhenolB. Liquid `NH_(3)`C. waterD. liquid HCl |
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Answer» Correct Answer - D Liq. HCl do not have hydrogen bonding because the electronegativity of Cl is not sufficient enough to form H-bonds. |
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| 173. |
Which of the following molecules is linear ?A. `C_2H_2`B. `CH_4`C. `H_2O`D. `NH_3` |
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Answer» Correct Answer - A `C_2H_2` is linear, `H-C-=C-H` |
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| 174. |
Which molecules is linearA. `NO_(2)`B. `ClO_(2)`C. `CO_(2)`D. `H_(2)S` |
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Answer» Correct Answer - C `CO_(2)` has sp- hybridization and is linear |
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| 175. |
Carbon tetrachloride has no net dipole moment because ofA. its planar structureB. its regular tetrahedral structureC. similar sizes of carbon ad chlorine atomsD. similar electron affinities of carbon and chlorine |
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Answer» Correct Answer - b |
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| 176. |
Carbon tetrachloride has no net dipole moment because of A. its planar structureB. its regular tetrahedral structureC. Similar sizes of carbon and chlorine atomsD. Similar electron affinities of carbon and chlorine |
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Answer» Correct Answer - D `C Cl_(4)` has no net dipole moment because of its regular tetrahedral structure. |
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| 177. |
The geometry and the type of hybrid orbitals present about the central atom in ` BF_3` is :A. linear ,spB. trigonal planar , `sp^2`C. tetrahedral `sp^3`D. pyramidal , `sp^3` |
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Answer» Correct Answer - A ` B` in ` BF_3` hs ` sp^2` - hybridisation and trigonal planar . |
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| 178. |
Which one of the following orders is not correct in accordance whith the property stated against is ?A. Electronegativity ` : F_2gt Cl_2 gt Br_2 gt I_2`B. Bond dissociation energy `: F_2 gt Cl_2 gt Br _2 gt I_2`C. Oxidsi ing power : ` F_2gt Cl_2 gt Br_2 gt I_2`D. acidic nature in water : `Hl gt HBr gt HC l gt HF` |
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Answer» Correct Answer - B Bond disssociation enrgy order : `{:(,Cl_(2),gt,Br_(2),gt,F_(2),gt,I_(2)),((kJ "mol"^(-1)),242.6,,192.8,,158.8,,151.1):}` |
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| 179. |
An ionic solid is poor conductor of electricity becauseA. ions do not conduct electricityB. charge on the ions is uniformly distributedC. ions have uniform field of influence around itD. ion occupy fixed position in solids |
| Answer» (d) An ionic solid is a good conductor of electricity only in molten or solution state, when ions are free to move. | |
| 180. |
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 radical is -4 so formula of ferrice pyrophosphate is `Fe_(4)(P_(2)O_(7))_(3)`. |
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| 181. |
The correct order of `O-O` bond length in `O_2,H_2 O` and `O_3`.A. `O_2 gt O_3 gt H_2O_2`B. `O_3 gt H_2O_2 gt O_2`C. `H_2O_2 gt O_3 gt O_2`D. `O_2 gt H_2O_2 gt O_3` |
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Answer» Correct Answer - C In `H_2O_2`, O-O bond is pure single bond. In `O_3`. O-O bond has partial double bond character and in `O_2`, O-O bond is double bond so bond length is in the order `H_2O_2 gt O_3 gt O_2` |
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| 182. |
The correct order of `O-O` bond length in `O_2,H_2 O` and `O_3`.A. `O_2 gt O_3 gt H_2O_2`B. `O_3 gt H_2O_2 gt O_2`C. `O_2 gt H_2O_2 gt O_3`D. `H_2O_2 gt O_3 gt O_2` |
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Answer» Correct Answer - D The bond length of O-O in `H_2O_2` is 147.5 pm, in `O_3` is 128 pm and in `O_2` it is 121 pm , so the correct order is `O_2 lt O_3 lt H_2O_2` |
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| 183. |
The species which is not paramagnetic among the following isA. `Cl^-`B. `Be`C. `Ne^(2+)`D. `As^+` |
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Answer» Correct Answer - A Paramagnetic character is based upon presence of unpaired electron. `.17Cl^(-)=1s^2 , 2s^2 2p^6, 3s^2 3p_x^2 3p_y^2 3p_z^2` In `Cl^-` no unpaired electron , so it is in nature diamagnetic `._4Be=1s^2 , 2s^1 2p_x^1` `._10Ne^(2+)=1s^2 , 2s^2 2p_x^2 2p_y^1 2p_z^1` `._33As^(+)=1s^2, 2s^2 2p^6 , 3s^2 3p^6 3d^10 , 4s^2 4p_x^1 4p_y^1 4p_z^0` While all others have unpaired electron, so they are paramagnetic in nature. |
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| 184. |
The charge/size ratio of a cation determines its polarizing power. Which one of the following sequences represents the increasing order of the polarizing power of the cationic species, `K^+,Ca^(2+), Mg^(2+), Be^(2+)`A. `Mg^(2+) lt Be^(2+) lt K^(+) lt Ca^(2+)`B. `Be^(2+) lt K^(+) lt Ca^(2+) lt Mg^(2+)`C. `K^(+) lt Ca^(2+) lt Mg^(2+) lt Be^(2+)`D. `Ca^(2+) lt Mg^(2+) lt Be^(2+) lt K^(+)` |
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Answer» Correct Answer - C Higher the charge/size ratio, more is the polarizing power. `:. K^(+) lt Ca^(2+) lt Mg^(2+) lt Be^(2+)` |
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| 185. |
Which of the following pairs of molecules have bond order three and are isoelectronic?A. `CN^(-),CO`B. `NO^(+),CO^(+)`C. `CN^(-),O_(2)^(+)`D. `CO,O_(2)^(+)` |
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Answer» Correct Answer - A `CN^(-)=14 `electrons, CO=14 electrons `B.O.=1/2 [10-4] =6/2 =3` |
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| 186. |
In which of the following pairs, the two species are isostructural :A. `SF_(4)` and `XeF_(4)`B. `SO_(3)^(2-)` and `NO_(3)^(-)`C. `BF_(3)` and `NF_(3)`D. `BrO_(3)^(-)` and `XeO_(3)` |
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Answer» Correct Answer - D `BrO_(3)^(-)` Electron pair over the central the atom =`3+1/2[7-6+1] rArr =4` Electron pair =3 No. of atom bonded to the central atom =4 Geometry will be pyramidal. `XeO_(3)` Electron pair =`3+1/2 [8-6] rArr 4` Electron pair =3 No.of atom bonded to the central atom =4 Geometry will be pyramidal |
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| 187. |
Planar structure is shown byA. `CO_(3)^(2-)`B. `BCl_(3)`C. `N(SiH_(3))_(3)`D. All of these |
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Answer» Correct Answer - D C in `CO_(3)^(2-)` and B in `BCl_(3)` are `sp^(2)` hybridised thus acquire planar geometry. Although N in `N(SiH_(3))_(3)` is `sp^(3)` hybridised but the steric hinderence caused by three bulky groups make it planar for higher stability. |
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| 188. |
The isoelectronic pair isA. `Cl_(2)O,Icl_(2)^(-)`B. `I Cl_(2)^(-),ClO_2`C. `IF_(2)^(+),I_(3)^(-)`D. `ClO_(2)^(-),ClF_(2)^(+)` |
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Answer» Correct Answer - D Number of electrons in `ClO_(2)^(-)` =17+8+8+1=34 Number of electrons in `ClF_(2)^(+) =17+9+9-1=34` |
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| 189. |
Assertion :- Bond order of `O_(2)` and BN is same . Reason `O_(2) and BN` are isoelectronicA. If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion.B. If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion.C. If Assertion is True, but the Reason is False.D. If both Assertion & Reason are false |
| Answer» Correct Answer - C | |
| 190. |
Which of the following is a correct set ?A. `H_(2)O,sp^(3)` angularB. `BCl_(3),sp^(3)`, angularC. `NH_(4)^(+)`,dsp^(2),square planarD. `CH_(4),dsp^(2)`, tetrahedral |
| Answer» Correct Answer - (a) | |
| 191. |
The isoelectronic pair isA. `Cl_(2)O,ICl_(2)^(-)`B. `Cl_(2)^(-),ClO_(2)`C. `IF_(2)^(+),I_(3)^(-)`D. `ClO_(3)^(-), CIF_(2)^(+)` |
| Answer» (d) `ClO_(3)^(-)"and"CIF_(2)^(+)` contain 34 electrons each hence the are isoelectronic. | |
| 192. |
The state of hybridization of `C_2, C_3, C_5`, and `C_6` of the hydrocarbon `CH_3-overset(CH_3)overset(|)underset(CH_3)underset(|6)C-underset(5)CH=overset(CH_3)overset(|)underset(4)CH-underset(3)CH-underset(2)C-=underset(1)CH` is in the following sequence:A. `sp^3, sp^2, sp^2,` and `sp`B. `sp, sp^2, sp^2,` and `sp^3`C. `sp, sp^2, sp^3,` and `sp^2`D. `sp, sp^3, sp^2`, and `sp^3` |
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Answer» Correct Answer - D `C_2` is `sp`-hybridized as it forms `2pi` bonds. `C_3` is `sp^3-`hybridized as it forms `4sigma` bonds. `C_5` is `sp^2`-hybridized as it forms `1pi` bond. `C_6` is `sp^3`-hybridized as it forms `4 sigma` bonds. |
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| 193. |
Draw the Lewis structure of `BeCl_2`. |
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Answer» Step 1: The number of valence electrons is `16:(2xx7)=14` from two Cl atoms and 2 from `Be` (which is in group IIA). Step 2: Since Cl is more electronegative than `Be`, chlorine cannot form more than one covalent bond with `Be`. Thus, the only possible skeleton structure is `Cl-Be-Cl` Note that Cl forms more than one covalent bond only when it interacts with highly electronegative elements like O or F which help in the excitation of its valence electrons. Step 3: Since two single covalent bonds are involved in the skeleton structure, `(16-4=)12` electrons are left to be accounted for Step 4: Since each Cl forms one bond and Be forms two bonds, a total of `6+4+6=16` electrons is needed to complete all octets. Only 12 valence electrons are available. Step 5: Since the number of remaining electrons is four less than the number of electrons needed to complete all octets, two more bonds are required, i.e., two extra bonds would be added because of the shortage of four electrons. Since Cl can form only one bond, this solution is not possible. Thus, `BeCl_2` molecule must have an incomplete octet. The Be atom, which is less electronegative than Cl, has incomplete octet, giving the Lewis structure: `:overset(..)underset(..)Cl-Be-overset(..)underset(..)Cl:` Beryllium is a group IIA (Gr. 2) and a second period element. The electron configuration of `Be` is `1s^2 2s^2`. Thus, it has only two valence electrons in the `2s` orbital. In the gas phase, beryllium hydride `(BeH_2)` exists as discrete molecules. The Lewis structure of `BeH_2` is `H-Be-H` As we can see, only four valence electrons surrounded the Be atom, and there is no way to satisfy the octet rule for Be in this molecule. We might expect a similar situation for the compounds of the other group IIA elements: `Mg, Ca, Sr, Ba`, and `Ra`. However, these metallic elements have lower ionization enthalpies on account of larger radii, so they usually form ions (with complete octet) by losing two electrons, e.g., in `MgCl_2`, the `Mg^(2+)` ions has 8 valence electrons (octet). Lithium metal `(1s^2 2s^1)` of group IA (alkali metals) forms mainly ionic compounds by losing its valence electron. The `Li^(+)` ion so formed has He configuration and has just two valence electrons in its valence shell. Thus, in lithium chloride (LICI), chlorine has complete octet while Li has incomplete octet. Elements in group IIIA (Gr. 13), particularly B and Al, also tend to form compounds such as `BCl_3`, `AlCl_3`, etc., in which less than eight electrons surround the central atom. Compounds such as `BeCl_2`, `BCl_3`, and `AlCl_3` in which the central atom shares less than 8 electrons are sometimes referred to as electron-deficient compounds. This deficiency refers only to satisfying the octet rule for the central atom. The term does not imply that there are less electrons than there are protons in the nuclei (as in the case of a cation) because the molecule is neutral. |
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| 194. |
Write the Lewis dot structure of the nitrite ion `(NO_(2)^(Θ))` . |
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Answer» Step 1: The valence shell electron configurations of N and O atoms are `2s^2 2p^3` and `2s^2 2p^4`, respectively, and the ion itself has one negative charge. Thus, the total number of valence electrons is `5+(2xx6)+1=18`. Step 2: The best skeleton structure is `O-N-O` which avoids `O-O` bond and ring structure. Step 3: Since the skeleton structure involves two single bonds, the number of valence electrons remaining is `18-4=14`. Step 4: Since each O atom is forming one bond and N atom is forming two bonds, the number of valence electrons needed to complete all octets is 6 for each O atom and 4 for N atom, i.e., 16 electrons. Step 5: Since the number of remaining electrons is two less than the number of electrons needed to complete all octets, one additional bond is required. It is placed between the N atom and either of the O atoms: `O=underset(I)N-O` or `O-underset(II)N=O` Completing the octets by placing the dots around the appropriate symbols in Lewis structure gives `underset(I) [[underset(..)overset(..)O=N-overset(..)underset(..)O:]]^(-)` or `underset(II)[[:underset(..)overset(..)O-N=overset(..)underset(..)O]]^(-)` Step 6: The Lewis structure drawn for nitrite ion accounts for the 18 valence electrons and indicates the negative charge. |
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| 195. |
Statement I The electronic structure of `O_(3)` is Statement II structure is not allowed because octet around `O` cannot be expanded.A. Statement 1: is true, statement 2 is true : statement 2 is a correct explanation for statement 1B. Statement 1 is true, statement 2 is true , statement 2 is not a correct explanation for statement 1C. Statement 1 is true, statement 2 is falseD. Statement 1 is false, statement 2 is true |
| Answer» Correct Answer - B | |
| 196. |
Assertion: `o-`and `p`-nitrophenol can be separated by steam distillation. Reason: `o-`nitrophenols have intramolecular hydorgen bonding while `p`-nitrophenols exists as associated molecules.A. If both assertion and reason are true and the reason is the correct explanation of the assertionB. if both assertion and reason are true but reason is not the correct explanation of the assertionC. if assertion is true but reason is falseD. if the assertion and reason both are false |
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Answer» Correct Answer - A o & p-nitrophenols can be separated by steam distillation because o-nitrophenol is steam volatile.here, both assertion & reason are correct & reason is correct explanation of assertion. |
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| 197. |
Among the following , the compound that contains ionic, covalent and coordinate linkage isA. `NH_(4)Cl`B. `NaCl`C. CaOCD. `NH_(3)` |
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Answer» Correct Answer - A In `NH_(4)Cl` , there are `3N -H` covalent bonds `[H larr overset(H)overset(|)underset(H)underset(|)(N)-H]^(+)Cl^(-)` 1N `to`H coordinate bond and 1 `NH_(4)^(+)-Cl^(-)` ionic bond. |
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| 198. |
The electronegativity of an element is low. The bond formed between two identical atoms of the above element is most likely to beA. CovalentB. MetallicC. IonicD. Co-ordinate covalent |
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Answer» Correct Answer - B Low electronegativity elements mean metals which form metallic bond in between |
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| 199. |
Which one of the following contains ionic, covalent and co-ordinate and coordinate linkage isA. NaOHB. NaClC. NaCND. NaNC |
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Answer» Correct Answer - D `Na^(+)[NC]^(-)` |
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| 200. |
The hybrid state of postively charged carbon in vinyl `(CH_2 = CH^+ )` cation is :A. `sp^2`B. `sp`C. `sp^3`D. unpredictable |
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Answer» Correct Answer - B It coniyins ` 2 sigma` and `1 pi`-bond . |
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