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.
| 201. |
Which combination is best explained byh the co-ordinate covalent bondA. `H^(+)+H_(2)O`B. `Cl+Cl`C. `Mg+1/2O_(2)`D. `H_(2)+I_(2)` |
|
Answer» Correct Answer - A `H-overset(..)underset(H)underset(|)("O:")+H^(+) to H -overset(..)underset(H)underset(|)(H)to H` |
|
| 202. |
The paramagnetic nature of oxygen molecules is best explained on the basis ofA. Valence bond thoeryB. ResonanceC. Molecular orbital theoryD. Hybridization |
| Answer» Correct Answer - C | |
| 203. |
Which of the following is paramagnetic ?A. `O_(2)^(-)`B. `CN^(-)`C. `CO`D. `NO^(+)` |
| Answer» Correct Answer - 1 | |
| 204. |
Which of the following is paramagnetic?A. `O_(2)^(-)`B. `CN^(-)`C. `CO`D. `NO^(+)` |
|
Answer» Correct Answer - a |
|
| 205. |
Which one is paramagnetic from the followingA. `O_(2)^(-)`B. NOC. Both (a) and (b)D. `CN^(-)` |
|
Answer» Correct Answer - C `O_(2)^(-)` (2x 8 +1=17) has odd number of electrons and hence it is paramagnetic . All the remaining molecules/ions, i.e. `CN^(-)`(6+7+1=14) diamagnetic NO (7+8+1=14) has odd number of electrons and hence it is paramagnetic. |
|
| 206. |
The type of hybrid orbitals used by chlorine atom in `ClO_(2)^(-)` is :A. `sp^(3)`B. `sp^(2)`C. spD. None of these |
|
Answer» Correct Answer - A `H=1/2 (B+M-C+A)` where H=No. of orbitals involved in hybridised (viz. `sp^(2),sp^(3),sp^(3)d,sp^(3)d^(2)`) can be ascertained. V=no. of electrons in valence shell of the central atom, M =no. of monovalent atoms, C=charnge on cation, A=charge on ation, `rArrH=1/2 (7+1)=4` or `sp^(3)` hybridisation as 4 orbitals are involved. |
|
| 207. |
Which of the following compounds has the least tendency to form hydrogen bonds between molecules?A. `NH_3`B. `NH_2 OH`C. HFD. `CH_3F` |
|
Answer» Correct Answer - D A hydrogen bond is formed if H is attached to highly electronegative atoms like F, O, N , etc. |
|
| 208. |
The H-Bonds in solid HF can be best represented as :A. H-F---H-F---H-FB. C. D. |
|
Answer» Correct Answer - C The arrangement of bp and Ip around F is tetrahedral. Therefore, angle should appear around F and not around H atom. |
|
| 209. |
At 300 K and 1.00 atm. pressure, the density of gaseous HF is 3.17 `gL^(-1)`. Which of the following facts, is supported by the given data ?A. Gaseous HF under these conditions is lighter than AlrB. There is no hydrogen bonding in gaseous HF under these conditionsC. There is extensive hydrogen bonding in gaseous HF under these conditionsD. None of these |
|
Answer» Correct Answer - C T=300 K, P=1.00 atm, V=1.00 L `therefore n=(PV)/(RT)` `=(("1.00 atm")("1.00 L"))/(("0.0821 L atm" K^(-1) mol^(-1))("300 K"))` =0.0406 mol Mass of 1 L of HF under these conditions =3.7 g `therefore` Formula mass of HF under these conditions =`(3.17 g)/(0.0406 mol)=78.1 g mol^(-1)` The formula mass of HF =`19 g mol^(-1)` |
|
| 210. |
The two types of bonds present in `B_(2)H_(6)` are covalent and _____. |
|
Answer» Correct Answer - three centre bond-two electrons |
|
| 211. |
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` |
|
Answer» Correct Answer - B Strength of hydrogen bonding depends on the size and electrongativity of the atom. Smaller the size of the atom, greater is the electronegativity and hence stronger is the H - bonding . Thus, the order of strength of H - bonding is `H … F gt H … O gt H ... N`. But each HF molecule is linked only to two other HF molecules while each `H_(2) O` molecule is linked to four other `H_(2)O` molecules through H - bonding . Hence, the decreasing order of boiling points is `H_(2) O gt HF gt NH_(3)`. |
|
| 212. |
Which of the following compounds contain(s) no covalent bond(s)? `KCl,PH_(3),O_(2),B_(2)H_(6),H_(2)SO_(4)`A. `KCl,H_(2)SO_(4)`B. KClC. `KCl,B_(2)H_(6)`D. `KCl,B_(2)H_(6),PH_3` |
|
Answer» Correct Answer - B KCl is not ionic compound |
|
| 213. |
In which of the following compounds does hydrogen bonding occurA. `SiH_(4)`B. LiHC. HlD. `NH_(3)` |
|
Answer» Correct Answer - D Only `NH_(3)` forms H-bonds |
|
| 214. |
Hydrogne bonding is formed in compounds containing hydrogen and _________.A. Highly electronegative atomsB. Highly electropositive atomsC. Metal atoms with d-orbitals occupiedD. Metalloids |
|
Answer» Correct Answer - A Hydrogen bond is formed when hydrogen is attached with the atom which is highly electronegative and having small radius. |
|
| 215. |
Compounds showing hydrogen bonding among `HF,NH_(3), H_(2)Sand PH_(3)` areA. Only `HF,NH_(3)` and `PH_(3)`B. Only HF and `NH_(3)`C. only `NH_(3),H_(2)S` and `PH_(3)`D. All the four |
| Answer» Correct Answer - B | |
| 216. |
Which among the following compounds does not show hydrogen bondingA. ChlorofomB. Ethyl alcoholC. Acetic acidD. Ethyl ether |
| Answer» Correct Answer - D | |
| 217. |
Which of the following compounds in liquid state does not have hydrogen bondingA. `H_2O`B. HFC. `NH_(3)`D. `C_(6)H_(6)` |
| Answer» Correct Answer - D | |
| 218. |
Which structure is linear ?A. `SO_2`B. `CO_2`C. `CO_3^(2-)`D. `SO_4^(2-)` |
|
Answer» Correct Answer - B Out of `SO_2, CO_3^(2-), SO_4^(2-) and CO_2 , CO_2` have sp hybridisation , thus have linear geometry. |
|
| 219. |
strongest hydrogen bonding is shown byA. `H_2O`B. `NH_3`C. `HF`D. `H_2S` |
|
Answer» Correct Answer - C HF have strongest hydrogen bond because the electronegativity of F-atom is high and produce strong electrostatic force of attraction. |
|
| 220. |
The dipole moment of `LiH` is `1.964 xx10^(-29)C-m` and the interatomic diatance between `Li` and `H` in this molceule is `1.596Å` .What is the per cent ionic character in `LiH`.A. `75.0`B. `76.8`C. `79.8`D. 100 |
|
Answer» Correct Answer - B The dipole moment of 100% ionic molecule = (1 electronic charge) `xx` (interatomic distance) ` = 1.602 xx 10^(-19) xx 1.596 xx 10^(-10) = 2.557 xx 10^(-29) ` cm Fractional ionic character = ` "Exp. Value of dipole moment "/"Theoretical value of dipole moment " = (1.964 xx 10^(-29))/(2.557 xx 10^(-29)) = 0.768` The bond in LiH is ` 76.8%` ionic. |
|
| 221. |
The dipole moment of `HBr` is ` 1.6 xx 10^(-30) cm` and interatomic spacing is `1 Å`. The `%` ionic character of `HBr` isA. 7B. 10C. 15D. 27 |
|
Answer» Correct Answer - B Charge of electron=`1.6xx10^(-19)`C Dipole moment HBr=`1.6xx10^(-30)` cm Interiodic spacing =1 Ã… =`1xx10^(-10)` m Considering HBr to be ionic , dipole moment =`1.6xx10^(-19) C xx 10^(-10)m` `=1.6xx10^(-29)` Cm Therefore, percentage ionic character of HBr =`(1.6xx10^(-30))/(1.6xx10^(-29))xx100=10%` |
|
| 222. |
Out of o-nitrophenol and p-nitrophenol, which is more volatile ? Explain?A. Hydrogen bondingB. Covalent bondingC. ResonanceD. Conjugation. |
|
Answer» Correct Answer - A o-nitrophenol has intramolecular H-Bonding which makes it more volatile in comparison to nitrophenol which has intermolecular H-bonding |
|
| 223. |
The hybridisation of xenon in `XeF_2` isA. `sp^3`B. `sp^2`C. `sp^3d`D. `sp^2d` |
|
Answer» Correct Answer - C Calculate the value of H (=5) |
|
| 224. |
Which of the following are isolectronic and iso-structural ? `NO_(3)^(Θ)` ,`CO_(3)^(2-)`, `CIO_(3)^(Θ)` ,`SO_(3)` .A. `NO_(3)^(-), CO_(3)^(2-)`B. `SO_(3),NO_(3)^(-)`C. `ClO_(3)^(-),CO_3^(2-)`D. `CO_(3)^(2-),SO_(3)` |
|
Answer» Correct Answer - a |
|
| 225. |
Suppose a gaseous mixtures of `He, Ne Ar ` and `Kr` is irradiated with photons fo frequency appropriate to ionize Ar, What ions will be present in the mixture ? |
| Answer» Initially `Ar^+` will be present , The `EA` value of ` Ar^+` ( 15,76eV) is almost same as `IE_1` for Ar. `IE_1` for `Kr` si ` 14.0eV` Soreaction ` Ar^+ _(-1)e^0 rarr Ar`, will occure exothermally and energy relased will ioise ` Kr` into ` Kr^+` .IE_1` valuer for `He`, Ne are large so tehse stoms will not react . | |
| 226. |
Amongst `LiCl, RbCl, BeCl_2` and `MgCl_2,` the compounds whith the greatrest and the least ionic character respecitely are :A. `LiCl ` and `RbCl`B. `RbCl` and `BeCl_2`C. `RbCl` and `MgCl_2`D. `MgCl_2` and `BeCl_2` |
|
Answer» Correct Answer - B Forllow Fajan`s rule to predict covalent nature . |
|
| 227. |
Which of the follwing forms covalent compound?A. CaB. MgC. SrD. Be |
|
Answer» Correct Answer - D Except for the first elements of groups, IA, IIA, and IIA, the atoms usually form ionic bonds. The number of unpaired electrons in the Lewis symbols for the atoms of elements in these groups equals the group numbers. |
|
| 228. |
Calculate electronegativity of carbon at Pauling scale Given that : ` E_(H-H) =104 .2 kcal "mol"^(-1) E_(C-C) =83.1 kcal "mol"^(-1)` , `E_(C-H) =98.8 kcal "mol"^(-1)`. Electronegativity of hydrogen `=2.1`. |
|
Answer» According to Pauling eqwuation , ` X_C -X-H =0. 208[E_(C-H)-(E_(H-H) xxE_(C-C))^(1//2)]^(1//2)` Where ` X_C` and ` X_H` are electronegativity of `C ` and `H` respectively . ` X_C-2.1 =0.208 [98.8 -(104.2 xx83.1)^(1//2)]^(1//2) 83.1)^(1//2)]^(1//2)` ` X-C -2.1 =0.498 ` or `X_C =2.598`. |
|
| 229. |
Which of the following compound has `mu`= 0 ?A. `C Cl_4`B. `CHCl_3`C. HFD. `NH_3` |
|
Answer» Correct Answer - A In NaCl solid, ions occupy fixed position and are not free to move so it is bad conductor |
|
| 230. |
Which of the following does not conduct electricity ?A. Molten NaOHB. Molten KOHC. Solid NaClD. Aqueous NaCl |
|
Answer» Correct Answer - C In others (`BF_3, CO_2 and CH_4`) cancellation of individual dipoles takes place. |
|
| 231. |
The structure of `IC l_2^-` isA. TrigonalB. Trigonal bipyramidalC. OctahedralD. Square planar. |
|
Answer» Correct Answer - B In `IC l_2^-` iodine is surrounded to 2 surrounding atoms and 3 lone pairs around it. |
|
| 232. |
In piperidine the hybrid state assumed by N isA. spB. `sp^2`C. `sp^3`D. `dsp^2` |
|
Answer» Correct Answer - C In piperidine, the hybrid state of N is `sp^3` |
|
| 233. |
Which of the following have identical bond order?A. `CN^(-)`B. `O_(2)^(-)`C. `NO^(+)`D. `CN^(+)` |
|
Answer» Correct Answer - A::C The outer most shells of C,N and O has 4,5 and 6 electrons respectively. Thus `CN^(-)` and `NO^(+)` each has 10 outermost electrons to accommodate in the molecular orbitals. So their bond order is same. `O_(2)^(-)` has 13 and `CN^(+)` has 8 electrons in outermost orbits. |
|
| 234. |
The linear struture is assumed by :A. `SnCl_(2)`B. `NCO^(-)`C. `CS_(2)`D. `NO_(2)^(+)` |
|
Answer» Correct Answer - B::C::D `[O=N=O]^(+), [N-=C-O]^(-),S=C=S` It can be seen from the structure above that `CS_(2)` being sp hybridised has a linear shape and other two molecules are isoelectronic to `CS_(2)`, so they are also linear. |
|
| 235. |
The number of antibonding electron pairs in `O_(2)^(2-)` molecular ion on the basic of molecular orbital theory isA. 4B. 3C. 2D. 5 |
|
Answer» Correct Answer - A `O_(2)^(2-)` consists of four antibonding electron pair [1s and 2s have two antibonding and `2p_(x)2p_(y)` have two antibonding electrons pair]. |
|
| 236. |
The bond order in ` O_(2^(2-))` is-A. 3B. 2C. 1D. `1/2` |
|
Answer» Correct Answer - C `O_(2)^(2-)` have bond order one `B.O. =1/2[10-8]=2/2=1` |
|
| 237. |
Which is false statement about LCAOA. Addition of aromatic orbitals result in molecular orbitalsB. Atomic orbitals of nearly same energy combine to form molecular orbitalsC. Bonding molecular orbitals occupy lower energy than atomic orbitalsD. Each molecular orbital accommodates maximum no. of two electrons |
|
Answer» Correct Answer - C The bonding molecular orbitals have lower energy than the combining atomic orbitals and hence, are much more stable than corresponding antibonding molecular orbitals. |
|
| 238. |
Linear overlapping of two atomic `p`- orbitals leads to a sigma bond. |
|
Answer» Correct Answer - T |
|
| 239. |
The dipole moment of `KCI` is `3.36 xx 10^(-29)Cm` The interatomic distance between `K^(o+)` and `CI^(Θ)` in this unit of `KCI` is `2.3 xx 10^(-10)m` Calculate the percentage ionic character of `KCI` . |
|
Answer» Correct Answer - 0.802 |
|
| 240. |
The correct order in which the ` O-O` bond length increases in the following :A. `H_2O_2 lt O_2gtO_3`B. `O_3 lt H_2O_2 lt O_2`C. ` O_2 gt H_2 O_2 `D. `O_2 lt O_3 lt H_2O_2` |
|
Answer» Correct Answer - D `O-O` bond length. `{:(O_(2),O_(3),H_(2)O_(2)),(120.7 "pm",127.8 "pm",147.5 "pm"):}` |
|
| 241. |
Which of the following is the correct order of stability?A. `H_2gtH_2^(+)gtHe_2gtHe_2^(+)`B. `H_2gtHe_2^(+)gtH_2^(+)gtHe_2`C. `H_2gtH_2^(+)gtHe_2^(+)gtHe_2`D. `H_2gtHe_2gtH_2^(+)gtHe_2^(+)` |
|
Answer» Correct Answer - C Stability of a molecule is directly related to the value of bond order. Thus, the higher the bond order, the greater the stability. If two species have the same bond order `(bo)`, then the stability should be decided by the number of antibonding electrons. The lesser the number of antibonding electrons, the greater the stability. `H_2^(+)` is `(sigma_(1s))^1`. Thus, `bo=(N_b-N_a)/(2)=(1-0)/(2)=0.5` `H_2` is `(sigma_(1s))^2`. Thus, `bo=(2-0)/(2)=1` `He_2^(+)` is `(sigma_(1s))^2(sigma_(1s)^**)^1`. Thus, `bo=(2-1)/(2)=0.5` `He_2` is `(sigma_(1s))^2(sigma_(1s)^**)^2`. Thus, `bo=(2-2)/(2)=0` Both `H_2^(+)` and `H_2^(+)` have equal bond orders but `H_2^(+)` has zero antibonding electrons while `He_2^(+)` has one antibonding electron. Thus, `H_2^(+)` is relatively more stable. The correct order of stability is, thus, `H_2gtH_2^(+)gtHe_2^(+)gtHe_2` Moreover, `H_2^(+)` is somewhat more stable than `He_2^(+)` since there is only one electron in the hydrogen molecule ion and, therefore, it has no electron-electron repulsion. Furthermore, `H_2^(+)` also has less nuclear repulsion than `He_2^(+)`. |
|
| 242. |
The diamagnetic molecules are :A. `B_2, C_2 , N_2`B. ` O_2 , N_2 ,F_2`C. ` C_2, N_2, F_2`D. ` B_2, I_2^(2-) ,N_2` |
|
Answer» Correct Answer - C `C_2, N_2,F_2` has no unpaired electron in their molecualr orbital configuration repersentation. |
|
| 243. |
The correct order of increaisng bond angles is :A. `PF_3 lt PCl_3 lt PBr_3 lt PI_3`B. `PF_3 gt PBr_3 lt PC_3lt gt PI_3`C. `PI_3 gt PBr_3 gt PCl_3 gt PF_3`D. `PF_3lt pCl_3 gt PBr_3 gt PI_3` |
|
Answer» Correct Answer - D `PCl_3 lt PBr_3 lt PI_3` bond angle order is explained in terms of increasing electronegativity fo halogen atom whereas ` PF-3 gt PCl_3` bond angle are explained in terms of ` p pi-d pi` bonding in ` PF_3`, The bond angels are ` 97^@, 100^@, 101.5^@` and ` 102^@` for ` PCl_3` PBr_3` PI_3` and ` PF_3` respectively . |
|
| 244. |
The bond order of a molecule in the excited state can beA. positiveB. negativeC. zeroD. both (2) and (3) |
|
Answer» Correct Answer - B In the excited state, electrons are promoted to higher energy levels. Thus, `N_agtN_b` and bond order `(N_b-N_a)//2` becomes negative. It implies that the molecule is unstable. |
|
| 245. |
Crystal structure of NaCl isA. NaCl moleculesB. `Na^(+)` and `Cl^(-)` ionsC. Na and Cl atomsD. None of the above |
|
Answer» Correct Answer - B NaCl is ionic crystal so it is formed by `Na^(+)` and `Cl^(-)` ions. |
|
| 246. |
The sulphate of a metal has the formula `M_(2)(SO_(4))_(3)`. The formula for its phosphate will beA. `M(HPO_(4))_(2)`B. `M_(3)(PO_(4))_(2)`C. `M_(2)(PO_(4))_(3)`D. `MPO_(4)` |
|
Answer» Correct Answer - D Yet the formula of sulphate of a metal (M) is `M_(2)(SO_(4))_(3)`, it is `M^(3+)` ionso formula of its phosphate would be `MPO_(4)`. |
|
| 247. |
The hybridisation of `P` in phosphate ion `(PO_4^(2-))` is the same as in :A. ` I` in `ICl_4^-`B. `S` in `SO_3`C. `N` in `NO_3^-`D. `S` in `SO_3^(2-)` |
|
Answer» Correct Answer - D `P` in ` PO_4^(3-)` has `sp^3` -hybridsation like S in ` SO_3^(2-)`. |
|
| 248. |
Molecular compounds are usually formed by the combination between :A. High ionisation potential and low electron affinityB. Low ionisation potential and high electron affinityC. High ionisation potential and high electron affinityD. Low ionisation potential and low electron affinity |
| Answer» Correct Answer - B | |
| 249. |
The phosphate of a metal has the formula `MPO_(4)`. The formula of its nitrate will beA. `MNO_3`B. `M_(2)(NO_(3))_(2)`C. `M(NO_(3))_(2)`D. `M(NO_(3))_(2)` |
| Answer» Correct Answer - D | |
| 250. |
Solid NaCl is a bad conductor of electricity becauseA. in Solid NaCl there are no ionsB. Solid NaCl is covalentC. In solid NaCl there is no motion of ionsD. In solids NaCl there are no electrons |
| Answer» Correct Answer - C | |