InterviewSolution
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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.
| 501. |
A sigma `(sigma)` bond is stronger than a pi `(pi)` bond. |
| Answer» Correct Answer - 1 | |
| 502. |
The compound `1,2-`butadiene has `:`A. only Sp hybridized carbon atomsB. only `sp^(2)` hybridized carbon atomsC. both sp and `sp^(2)` hybridized carbon atomsD. `sp,sp^(2)` and `sp^(3)` hybridized carbon atoms |
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Answer» Correct Answer - D `overset(sp^(2))(C)H_(2)=overset(sp)(C)=overset(sp^(2))(C)H-overset(sp^(3))(C)H_(3)` 1, 2-butadiene. |
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| 503. |
The bond order in `O_2^+` is the same as in :A. `N_(2)^(+)`B. `CN^(-)`C. `CO`D. `NO^(+)` |
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Answer» Correct Answer - A `O_(2)^(+) (15e^(-))=K: K^(**)(sigma2s)^(2)(sigma^(**)2s)^(2)(sigma^(**)2s)^(2)(sigma2p_(x))^(2)` `(pi 2 p_(y))^(2)(pi 2p_(z))^(2)(pi^(**)2p_(y))^(1)(pi^(**)2p_(z))^(0)` Hence, bond order =`1/2(10-5) =2.5` `N_(2)^(+) (13e^(-))=KK^(**) (sigma2s)^(2)(sigma^(**)2s)^(2)(sigma2p_(x))^(3)` `(pi 2p_(y))^(2)(pi 2p_(z))^(1)` Hence bond order =`1/2(9-4) =2.5` |
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| 504. |
Bond energies in `NO,NO^(+)` and `NO^(-)` are such asA. `N^(-) gt NO gt NO^(+)`B. `NO gt NO^(-) gt NO^(+)`C. `NO^(+) gt NO gt NO^(-)`D. `NO^(+) gt NO^(-) gt NO` |
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Answer» Correct Answer - C Bond order of `NO^(+)`, No and `NO^(-)` are 3,2.5 and 2 respectively, bond energy `prop` bond order |
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| 505. |
One hybridization of one `s` and one `p` orbital we getA. two mutually perpendicular orbitalsB. two orbitals at `180^(@)`C. four orbitals directed tetrahderallyD. three orbitals in a plane |
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Answer» Correct Answer - b |
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| 506. |
Correct order of stability of species `N_(2),N_(2)^(+),N_(2)^(-)`A. `2.5, 2.5 and 3` respectivelyB. `2,2.5 and 3` respectivelyC. `3,2.5 and 3` respectivelyD. `2.5,2.5 and 2.5` respectively. |
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Answer» Correct Answer - A `N_(2)^(+) (13) : (sigma1s^(2)) (sigma^(**)1s^(2))(sigma2s^(2))(sigma^(**)2s^(2))(pi 2p_(x)^(2) = pi 2p_(y)^(2))(sigma2p_(z)^(1))` B.O. ` = 1/2 xx(9-4) = 2.5` `N_(2)^(-) (15) : (sigma1s^(2))(sigma^(**)1s^(2))(sigma2s^(2))(sigma^(**)2s^(2))(pi2p_(x)^(2) = pi 2p_(y)^(2))(sigma2p_(z))^(2) (pi^(**) 2p_(x)^(1))` B.O. ` = 1/2 xx (10 - 5) = 2.5` `N_(2) (14) : (sigma1s^(2))(sigma^(**)1s^(2))(sigma2s^(2))(sigma^(**)2p_(x)^(2)=pi2p_(y)^(2))(sigma2p_(z)^(2))` B.O. ` = 1/2 xx (10 - 4) = 3.0` |
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| 507. |
The hybridization state of the central atom in `HgCl_2` isA. `sp`B. `sp^2`C. `sp^3`D. `dsp^2` |
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Answer» Correct Answer - A The ground-state electron configuration of `Hg` is `[Xe]4f^(14)5d^(10)6s^2`. Therefore, the `Hg` atom has two valence electrons (the `6s` electrons). Steric number of `Hg` is `2+0=2` which is confirmed by its Lewis structure `Cl-Hg-Cl` There are no lone pairs on the `Hg` atom. The molecule has linear geometry. We conclude that `Hg` is sphybridized because it has the geometry of the two sp hybrid orbitals. |
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| 508. |
In `CaF_(2)`, the number of electron(s) transferred from calcium to fluorine atoms is ______. |
| Answer» Correct Answer - two | |
| 509. |
Covalent bonds in polyatomic molecules are formed by the overlap ofA. pure atomic orbitalsB. hybrid orbitalsC. hybrid orbitals with unhybridized onesD. both (2) and (3) |
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Answer» Correct Answer - D The hybridization bonding scheme is still within the framework of valence bond theory. Electrons in a polyatomic molecule are assumed to occupy hybrid orbitals of the individual atoms. |
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| 510. |
The structure of `IF_4` can be best desribed by :A. B. C. D. none of these |
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Answer» Correct Answer - C `IF_5` shows `sp^3 d^(2)` -hybridisation with one lone pair on `I` atom. |
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| 511. |
Propyne molecule containsA. 6 sigma and 2 pi bondsB. 5 sigma bondsC. 5 pi bonds and 1 sigma bondD. 2 sigma and 3 pi bonds . |
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Answer» Correct Answer - A `H - C-= C - underset(H)underset(|) overset(H) overset(|) (C) - H`, No. of sigma bonds = 6, No. of pi bonds = 2 . |
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| 512. |
The solublity of `KCl` is relatively more in :A. `C_6 H_6 (D =0)`B. `(CH-3)_2 CO(D=2)`C. ` CH_3OH(D=32)`D. `C Cl_4(D=0)` |
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Answer» Correct Answer - C Polar solutes are more soluble in polar solvent . |
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| 513. |
Which of the following is incorrect about hybridization?A. The concept of hybridization is not applied to isolated atoms.B. Hybridization is the mixing of at least two nonequivalent atomic orbitals.C. The number of hybrid orbitals generated is more than the number of pure atomic orbitals that participate in the hybridization process.D. Hybridization requires an input of energy. |
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Answer» Correct Answer - C Hybrid orbitals can be formed from various numbers of atomic orbitals. The number of hybrid orbitals formed always equal the number of atomic orbitals used. For example, if we combine an s orbital and two p orbitals to get a set of equivalent orbitals, we get three hybrid orbitals called `sp^2` hybrid orbitals. `(1)implies` Hybridization is used only to explain the bonding scheme in a molecule. `(2)implies` Therefore, a hybrid orbitals is not a pure (that is, native) atomic orbitals. Hybrid orbitals have very different shape from pure atomic orbitals. `(4)implies` As it involves the promotions of electrons. However, the system more than recovers this energy during extra bond formation. |
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| 514. |
The molecule having one unpaired electron isA. NOB. COC. `CN^(-)`D. `O_(2)` |
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Answer» Correct Answer - A `NO-sigma 1s^(2)sigma^(**)1s^(2)sigma2s^(2)sigma^(**)2s^(2)sigma2p_(z)^(2)pi2p_(x)^(2)=pi 2p_(y)^(2)pi^(**)2p_(x)^(1)`. 1 unpaired electron `CO-sigma 1s^(2)sigma^(**)1s^(2)sigma2s^(2)sigma^(**)2s^(2)sigma2p_(z)^(2)pi2p_(x)^(2)=pi2p_(y)^(2)` `CN^(-)` similar to that of CO `O_2 -sigma 1 s^(2) sigma^(**)1s^(2)sigma2s^(2)sigma^(**)2s^(2) sigma 2p_(z)^(2)= pi 2p_(x)^(2)=pi2p_(y)^(2)pi^(**)2p_(y)^(2)pi^(**)2p_(y)^(1)` |
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| 515. |
The H-bond is strongest inA. O-H...SB. S-H....OC. F-H....FD. S-H....O |
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Answer» Correct Answer - C The strongest hydrogen bond is in hydrogen fluoride because the power of hydrogen bond `prop` electronegativity of atom and electronegativity `prop1/("atomic size")` So fluorine has maximum electronegativity and minimum atomic size. |
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| 516. |
Which of the following contains a coordinate covalent bond?A. `N_(2)O_(5)`B. `BaCl_(2)`C. HClD. `H_(2)O` |
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Answer» Correct Answer - A Structure of `N_(2)O_(5)` is `O=underset(O)underset(uarr)(N)-O-underset(O)underset(uarr)(N)=O` |
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| 517. |
Which of the following contains a coordinate covalent bond?A. `CH_3NC`B. `CH_(3)OH`C. `CH_(3)Cl`D. `NH_(3)` |
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Answer» Correct Answer - A `CH_(3)Nvec(=)(C)` contain dative bond. |
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| 518. |
The elements fo groups 16 and 17 whose atoms have the largest negative electron gain enthalpies would be expected to form_________by gaining electrons to give___________configurations.A. polyatomic ions, noble gasB. polyatomic ions, pseudo noble gasC. monatomic ions, noble gas, or pseudo noble gasD. monoatomic ions, noble gas |
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Answer» Correct Answer - C An atom of a group 17 element (valence shell configuration `ns^2 np^5`) pics up one electron to give 1-anion `(ns^2 np^6)`, examples are `F^(-)` and `Cl^(-)`. Heavier halide ions (`Br^(-)` onwards) possess psudo-noble-gas configurations. Note that hydrogen also forms compounds of the 1-ion, `H^(-)`. The hydride ion `H^(-)` has a `1s^2` configuration, like the noble gas atom helium. An atom of a group 16 element (valence shell configuration `ns^2np^4`) picks up two electrons to give 2-anion `(ns^2np^6)`, examples are `O^(2-)` and `S^(2-)`. |
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| 519. |
No compounds of representative elements are found with ions having charges greater than the group number becauseA. these elements do not exhibit variable valencyB. once atoms of these elements have lost their valence electrons, they have stable noble gas or pseudo noble gas configurationsC. they are `s-` and p-block elementsD. they have good shielding electrons |
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Answer» Correct Answer - B The `Delta_iH_2` of Na is nearly ten times greater as the electron in this case must be taken from `Na^(+)`, which has a neon configuration. Mg and Al atoms show similar behaviors as the energy needed to take an electron from either of the ions that result (`Mg^(2+)` and `Al^(3+)`) is extremely high. |
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| 520. |
Group 13 elements show less tendency to form ionic compounds than do group 1 and 2 elements, which primarily form ionic compounds becauseA. they are p-block elementsB. they fall after d-block elementsC. the loss of successive electrons from an atom requires increasingly more energyD. none of these |
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Answer» Correct Answer - C Boron forms no compounds with `B^(3+)` ions. The bonding is normally convalent. However, the tendency to form ions becomes greater goind down any column of the periodic table because of decreasing `Delta_iH`. The reamining elements of group 13 do form compounds containing `3+` ions |
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| 521. |
There is a tendency for the elements of groups 13 to 15 of higher periods, particulary period 6, to form compounds with ions having a positive charge of ___________less than the group number.A. oneB. threeC. fourD. two |
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Answer» Correct Answer - D Ions with charge equal to the group number minus two are obtained when the np electrons of an atom are lost but the `ns^2` electrons are retained. This is referred to as inert pair effect. Tl of group 13 prefers to form compounds with `1+` charge. In group 14, a few compounds of `4+` ions are known because the energy required to form ions is so great. However, tin and lead commonly form compounds with `2+` ions (ionic charge is equal to the group number minus two). `SnCl_2` is an ionic compound but `SnCl_4` is a covalent compound. Bismuth, in group 15, a metallic, element forms compounds containing `Bi^(3+)` ions (ionic charges is equal to the group number minus two). |
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| 522. |
Which of the following is not paramagnetic ?A. `N_2^+`B. `CO`C. `O_2^-`D. NO. |
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Answer» Correct Answer - B It has 6+8 = 14 electrons to be filled in bonding orbitals. |
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| 523. |
The transition of an electron from a 4s orbital to ls orbital in hydrogen atom causesA. photoelectric effectB. a Lyman lineC. increase in kinetic energy of electronD. conversion of `H^+` to H atom |
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Answer» Correct Answer - B The transition from higher to first orbit causes appearance of Lyman line |
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| 524. |
Which of the following is planar?A. `XeO_4`B. `XeO_3F`C. `XeO_2F_2`D. `XeF_4` |
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Answer» Correct Answer - D It has `sp^3d^2` hybridisation with 2 bond pairs and 2 lone pairs |
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| 525. |
Among the following the electron deficient compound isA. `C Cl_4`B. `PCl_5`C. `BeCl_2`D. `BCl_3` |
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Answer» Correct Answer - D It has incomplete octet. |
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| 526. |
In which one of the following species , the central atom has the tuype of hybdridiztion which is not the same as that present in other three?A. `PCl_5`B. `SF_4`C. `I_3^-`D. `SbCl_5^(2-)` |
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Answer» Correct Answer - D Central atoms in `PCl_5`, `SF_4`, `I_3^(-)` are `sp^3 d` hybridized but central Sb atom in `SbCl_5^(2-)` is `sp^3d^2` hybridized |
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| 527. |
In which one of the following species , the central atom has the tuype of hybdridiztion which is not the same as that present in other three?A. `PCl_(5)`B. `SF_(4)`C. `I_(3)^(-)`D. `SbCl_(5)^(2-)` |
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Answer» Correct Answer - D In `SbCl_(5)^(2-)` sb is `sp^(3)d^(2)` hybridised, and rest have `sp^(3)d` |
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| 528. |
Which of the following is not correct?A. `Na^(+)+Cl^(-)rarrNaCl`B. `Ca^(+)+2F^(-)rarrCaF_(2)`C. `Na^(+)+F^(-)rarrNaF`D. `Ca^(+)+2Na^(+)rarrCaNa_(2)` |
| Answer» (d) Both Ca and Na are metals and form positive ions, which repel each other. Hence,the given reaction is not feasible. | |
| 529. |
Which of the following is least soluble in water ?A. `BeF_(2)`B. `SrF_(2)`C. `CaF_(2)`D. `MgF_(2)` |
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Answer» Correct Answer - B The solubility order is `BeF_(2) gt MeF_(2) gt CaF_2 gt SrF_(2)` so `SrF_(2)` is least soluble. |
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| 530. |
The molecular formula of chloride of a metal M is `MCl_3`. The formula of its carbonate would beA. `MCO_(3)`B. `M_(2)(CO_(3))_(3)`C. `M_(2)CO_(3)`D. `M(CO_(3))_(2)` |
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Answer» Correct Answer - B As the molecular formula of chloride of a metal M is `MCl_(3)`, it is trivalent so formula of its carbonate will be `M_(2)(CO_(3))_(3)`. |
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| 531. |
The strongest bond isA. IonicB. CovalentC. Hydrogen bondD. Metallic |
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Answer» Correct Answer - A The order of the bond strength (strongest to weakest) is ionic bond gt covalent bond gt coordinate bond gt metallic bond gt hydrogen bond. |
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| 532. |
In the formation of a molecule, only the outer shell electrons take part in chemical combination and are known asA. Valence electronsB. inner electronsC. inert electronsD. reactive electrons |
| Answer» (a) Outer shell is lknown as valence shell and its electrons are called valence electrons. | |
| 533. |
According to VSEPR theory, the most probable shape of the molecule having 4 electrons pairs in the outer shell of the central atom isA. LinearB. TetrahedralC. HexahedralD. Octahedral |
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Answer» Correct Answer - B Central atom having four electron pairs will be of tetrahedral shape. |
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| 534. |
Which of the following statement is correctA. `SF_(4)` is polar and non-reactiveB. `SF_(6)` is non-polar and very reactiveC. `SF_(6)` is a strong fluorinating agentD. `SF_(4)` is prepared by fluorinating `SCl_(2)` with `NaF` |
| Answer» Correct Answer - B | |
| 535. |
Choose the correct statementA. Anion polarisation is more pronounced by highly charged cationB. small cation has minimum capacity to polarise an anionC. small anion has maximum polarizabilityD. none of these |
| Answer» Correct Answer - A | |
| 536. |
The bonds between P atoms and Cl atoms in `PCl_5` are likely to beA. Ionic with no covalent characterB. covalent with some ionic characterC. covalent with non ionic characterD. ionic with some metallic character |
| Answer» Correct Answer - B | |
| 537. |
The correct sequence of increasing covalent character is represent byA. LiCl lt NaCl lt`BeCl_(2)`B. `BeCl_(2)` ltNaCl ltLiClC. NaCl lt LiCl lt`BeCl_2`D. `BeCl_2` ltLiCl ltNaCl |
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Answer» Correct Answer - C Order of polarising power `Be^(++) gt Li^(+) gt Na^(+)` Hence order of covalent character `BeCl_(2) gt LiCl gt NaCl` |
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| 538. |
Covalent compounds are generally.........in waterA. solubleB. insolubleC. dissociatedD. hydrolysed |
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Answer» Correct Answer - B Water is a polar solvent while covalent compounds are non-polar so they are usually insoluble in water. |
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| 539. |
The species having diamagnetic nature and bond order `1. 0` is :sA. ` O_(2)^(2-)`B. `O_(2)^(+)`C. `O_(2)^(2-)`D. `O_2` |
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Answer» Correct Answer - A M.O. configuration of `O_2: sigma1s^2 :sigma 1s_(2)^(**) 1s^2` `sigma 2s^2, sigma^(**) 2s^2 , sigma 2p_x^2 [(pi 2p_y^2),(pi^(**) 2p_y^1)][(pi^(**) 2p_y^1),(pi^(**) 2p_z^1)]` |
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| 540. |
The correct order of increasing covalent character is :A. `BeCl_2 lt NaCl lt LiCl`B. `NaCl lt LiCl lt BeCl_2`C. `BeCl_2 lt LiCl lt NaCl`D. `LiCl lt NaCl lt BeCl_2` |
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Answer» Correct Answer - B For a given negative ion `(Cl^(-))` the covalent character is directly related to the polarizing power of cation which increases in the order `Na^(+) lt Li^(+) lt Be^(2+)` Thus, covalent character increases in the order `NaCl lt LiCl lt BeCl_2` |
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| 541. |
Which pair is isostructivral and possesses same number of bone pair of electron on central atom ?A. ` IF_5 ` and `XeOF_4`B. `NH_3 ` and `ClO_3^-`C. ` SnCl_4` and `Cl_4^-`D. `AlCl_3` and `SO_2` |
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Answer» Correct Answer - A I and `Xe` have `sp^3 d^2-`hybridisation , each having one line pair of electron. |
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| 542. |
With which of the given pairs, `CO_(2)` resemblesA. `HgCl_(2),C_(2)H_(2)`B. `HgCl_(2),SnCl_(4)`C. `C_(2)H_(2),NO_(2)`D. `N_(2)O` and `NO_(2)` |
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Answer» Correct Answer - A All have linear structure. O=C=O,Cl-Hg-Cl,`HC-=CH` |
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| 543. |
Which of the following is covalent?A. `H_2`B. `KCl`C. `Na_2S`D. `CaO` |
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Answer» Correct Answer - A Hydrogen gas is a covalent substance. It is made up of `H_2` molecules, in which two H atom share an electron pair: `H:H` or `H-H` Rest of the substances are ionic compounds: `K^(+)Cl^(-), (Na^(+))_2S^(2-)`, and `Ca^(2+)O^(2-)` |
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| 544. |
Which of the following statements is correct for covalent bondsA. Electrons are shared between two atomsB. it may be polar or non-polarC. Direction is non-polarD. Valency electrons are attracted |
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Answer» Correct Answer - A::B NaH and `CaH_(2)` are ionic hydride an `B_(2)H_(6)` and `NH_(3)` covalent hydride. |
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| 545. |
Which of the following statements is correct for `CsBr_3` ?A. It is a covalent compoundB. It contains `Cs^(3 +)` and `Br^-` ionsC. It contains ` Cs^+` and `Br_3^-` ionsD. It contains ` Cs^+ , Br^-` and lattice `Br_2` molecule |
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Answer» Correct Answer - C `CsBr_3underset(larr)(rarr)Cs^+ +Br_3^-` |
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| 546. |
Which of the following compounds does not follow the octet rule?A. `SF_2`B. `CIF_3`C. `NCl_3`D. `C Cl_4` |
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Answer» Correct Answer - B The electronic basis of the octet rule is that the one s and three p orbitals in the valence shell of an atom can accommodate a maximum of eight electrons. The valence shell of Cl has `n=3`, so it also has vacant `3d` orbitals available that can be involved in bonding. It is for this reason that Cl (and many other representative elements of period 3 and beyond) can exhibit an expanded valence shell. By contrast, elements in the second row of the periodic table can never exceed eight electrons in their valence shells, because each atom has only one s and three p orbitals in that shell. Thus, we understand why `PF_5` can exist but `NF_5` cannot. |
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| 547. |
The octet rule is not followed inA. `F_(2)`B. NaFC. `CaF_(2)`D. `BF_(3)` |
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Answer» Correct Answer - D `BF_(3)` does not have octet, it has only six electrons so it is electron deficient compound. |
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| 548. |
Why is `H_(2)S` more acidic than water ?A. O is more electronegativity than SB. O-H bond is stronger than S-H bondC. O-H bond is weaker than S-H bondD. None of these |
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Answer» Correct Answer - B The acidity of hydrides of VI group elements increases from top to bottom as the bond strength X -H decreases from top to bottom `H_(2)O lt H_(2)S lt H_(2)Se lt H_(2)Te` |
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| 549. |
Which of the following molecules is formed without following the octet rule?A. `SF_6`B. `IF_7`C. `BeCl_2`D. All of these |
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Answer» Correct Answer - D In `SF_6` and `IF_7`, the central atoms S and I, respectively, have expanded valence shell while in `BeCl_2`, the central Be atom has an incomplete octet. Thus, all these molecules violate the octet rule. |
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| 550. |
In which of the following molecules does the central atom not follow the octet rule?A. `CO_2`B. `BF_3`C. `H_2O`D. `PCl_3` |
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Answer» Correct Answer - B In `BF_3`, the central `B` atom has only six valence electrons, i.e., its octet is not complete. |
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