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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.
| 601. |
Carbon dioxide is isostructural with which of the following ?A. `HgCl_2`B. `H_2O`C. `SnCl_2`D. `NO_2^(-)` |
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Answer» Correct Answer - A `CO_2` is isostructural with `HgCl_2` |
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| 602. |
The calculated bond order in `H_2^(-)` ion is |
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Answer» Correct Answer - B The bond order in `H_2^-` ion is 1/2. |
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| 603. |
Which of the following has the least dipole moment?A. `C Cl_(4)`B. `CHCl_(3)`C. `CH_(3)CH_(2)OH`D. `CH_(3)COCH_(3)` |
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Answer» Correct Answer - A Carbon tetrachloride has a zero dipole moment because of its regular tetrahedral structure. |
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| 604. |
Which of the following fluorides of Xe has zero dipole moment ?A. `XeF_(2)`B. `XeF_(3)`C. `XeF_(4)`D. `XeF_(6)` |
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Answer» Correct Answer - C In `XeF_(4)`, the four fluorine atoms are present at the corners of a square plane and two lone pair are present at axial positions. All the individual bond moments in such an arragement cancel out, as a result, the molecule has zero dipole moment. |
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| 605. |
There are …………. `pi` bonds in a nitrogen molecule |
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Answer» Correct Answer - 2 |
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| 606. |
Many transition elements also form `3+` ions by losing one `(n-1)d` electron in addition to the two ns electrons exceptA. `Sc`B. `Cr`C. `Fe`D. `Ni` |
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Answer» Correct Answer - D `Ni` is predominantly divalent and ionic. `Sc(3d^14s^2)rarrSc^(3+)([Ar])` `Cr(3d^54s^1)rarrCr^(3+)(3d^3)` `Fe(3d^64s^2)rarrFe^(3+)(3d^5)` |
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| 607. |
In forming ions, the atoms of transition metals generally lose the ns electrons first, then they may lose one or more `(n-i)d` electrons. The `2+` ions are common for the transition elements and are obtained by the loss of the highest energy electrons from the atom exceptA. `Ni`B. `Fe`C. `Cu`D. `Co` |
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Answer» Correct Answer - C `Cu(3d^(10)4s^1)rarrCu^(2+)(3d^9)+2e^(-)` |
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| 608. |
The nature of interparticle forces in benzene isA. Dipole-dipole interactionB. Dispersion forceC. Ion-dipole interactionD. H-bonding |
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Answer» Correct Answer - B Benzene being a non polar molecule, the interparticle forces are van der Waal forces. |
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| 609. |
Inter molecular forces in solid hydrogen areA. Covalent forcesB. van der Waal forces or London dispersion forcesC. Hydrogen bondsD. All of these. |
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Answer» Correct Answer - B `H_2` molecule is non-polar. Thus, in solid state of hydrogen, the interparticle forces are van der Waals forces. |
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| 610. |
The correct order of decreasing polarisability of ion isA. `Cl^(-), Br^(-), I^(-), F^(-)`B. `F^(-),I^(-),Br^(-),Cl^(-)`C. `F^(-),Cl^(-), Br^(-), I^(-) `D. `I^(-),Br^(-),Cl^(-), F^(-)` |
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Answer» Correct Answer - D Larger the size of the anion, greater will be its polarisability. |
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| 611. |
Statement : `SF_4` has lone pair of electron at equatorial position in preference to axial position in the overall trigonal bipyramidal geometry . Explanation : If lone pair is at equatorial position , then only repulsion is minimum .A. `S` is correct by ` E` is wrong .B. `S` is wrong but `E` is wrong .C. Both `S` and `E` are correct and `E` is correct explanation os ` S` .D. Both `S` and ` E` are conrrect and `E` is correct explanation os `S`. |
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Answer» Correct Answer - C …………….do ……………… |
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| 612. |
The pair having similar geometry isA. `PCl_(3),NH_(3)`B. `BeCl_(2),H_(2)O`C. `CH_(4),C Cl_(4)`D. `IF_(5),PF_(5)` |
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Answer» Correct Answer - A::C `PCl_(3) , NH_(3) to ` pyramidal `CH_(4), C Cl_(4) to `Tetrahedral |
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| 613. |
In which pair of species both species do have the similar geometry ? .A. `CO_(2),SO_(2)`B. `NH_(3),BH_(3)`C. `CO_(3)^(2-),SO_(3)^(2-)`D. `SO_(4)^2-,ClO_(4)^-` |
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Answer» Correct Answer - D Species having same hybridization show similar geometry `SO_(4)^(2-)` : Hybridization of ` S to sp^(3)` `ClO_(4)^(-)` : Hybridization of `C l to sp^(3)` |
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| 614. |
The percentage s-character of the hybrid orbitals in methane , ethene are respectivelyA. 25,33,50B. 25,50,75C. 50,75,100D. 10,20,40 |
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Answer» Correct Answer - A % of s-character in `underset((sp^(3)))(CH_(4))=100/4-25, underset((sp^(2)))(C_(2)H_(4))=100/3 =33` `underset((sp))(C_(2)H_(2))=100/2=50` |
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| 615. |
The electronegativity difference between two atoms A and B is `2`, then percentage of covalent character in the molecule isA. `54%`B. `46%`C. `23%`D. `72%` |
| Answer» (a) If the electronegativity difference between two atoms A and B is 2.0, then the percentage of convalent character in the molecule is 54%. | |
| 616. |
Element A has three electrons in the outermost orbit and `B` has six electrons in the outermost orbit The formula of the compound will be .A. `X_(2)Y_(6)`B. `XY_(2)`C. `X_(2)Y_(3)`D. `X_(3)Y_(2)` |
| Answer» (c ) If an atom X has three valence electrons and atom Y has six valence electrons, the compound formed between them will have the formula `X_(2)Y_(3)` like `Al_(2)O_(3)`. | |
| 617. |
During change of `O_(2)` to `O_(2)^(2-)` ion, the electrons add on which of the following orbitals ?A. `pi^(**)` orbitalsB. `pi` orbitalsC. `sigma^(**)` orbitalsD. `sigma` orbitals |
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Answer» Correct Answer - A `O_(2) overset(+1e^(-))toO_(2)^(-)` Molecular orbital configuration of `O_(2)` `sigma1s^(2)sigma^(**)1s^(2)sigma2s^(2)sigm^(**)2s^(2)sigma2p_(z)^(2)pi 2p_x^(2)=pi2p_(y)^(2)` `pi^(**)2p_x^(1)=pi^(**)2p_(y)^(1)` therefore incoming `e^(-)` would enter into `pi^(**)2p_(x)` molecular orbital. |
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| 618. |
The pair of species with the same bond order is :A. `O_(2)^(2-),B_2`B. `O_(2)^(+),NO^(+)`C. `NO,CO`D. `N_(2),O_(2)` |
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Answer» Correct Answer - A Both `O_(2)^(2-)` and `B_(2)` had bond order equal to 1. |
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| 619. |
Which of the following is a planar molecule ?A. `NH_(3)`B. `H_(3)O^(+)`C. `BCl_(3)`D. `PCl_(3)` |
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Answer» Correct Answer - C `BCl_(3)` molecule show `sp^(2)`-hybridization and planar structure. |
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| 620. |
Decreasing order of stability of `O_(2), O_(2)^(-), O_(2)^(+)` and `O_(2)^(2-)` isA. `O_(2)^(+) gt O_(2) gt O_(2)^(-) gt O_(2)^(2-)`B. `O_(2)^(2-) gt O_(2)^(-) gt O_(2) gt O_(2)^(+)`C. `O_(2) gt O_(2)^(+) gt O_(2)^(2-) gt O_(2)^(-)`D. `O_(2)^(-) gt O_(2)^(2-) gt O_(2)^(+) gt O_(2)` |
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Answer» Correct Answer - A `B.O. underset(2.5)(O_(2)^(+)) gt underset(2)(O_(2)) gt underset(1.5)(O_(2)^(-))gt underset(1.0)(O_(2)^(2-))` |
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| 621. |
Which molecules does not show zero dipole momentA. `BF_(3)`B. `NH_(3)`C. `C Cl_(4)`D. `CH_(4)` |
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Answer» Correct Answer - B Ammonium have some dipole moment |
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| 622. |
Which of the following molecules will show dipole moment?A. methaneB. Carbon tetrachlorideC. chloroformD. carbon dioxide |
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Answer» Correct Answer - C Chloroform has 3 chlorine atom and one hydrogen atom attached to the carbon so it is polarised and it will show dipole moment. |
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| 623. |
Which of the following molecules has the maximum dipole moment ?A. `NH_(3)`B. `NF_(3)`C. `CO_(2)`D. `CH_(4)` |
| Answer» Correct Answer - A | |
| 624. |
Which of the following is a correct set with respect to molecule, hybridization, and shape?A. `BeCl_(2),sp^(2)` linearB. `BeCl_(2),sp^(2)` triangular planarC. `BCl_(3),sp^(2)`, triangular planarD. `BCl_(3),Sp^(3)` , tetrahedral |
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Answer» Correct Answer - C `BCl_(3)` boron trichloride molecule show `sp^(2)`-hybridization and trigonal planar structure. |
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| 625. |
Bond length is maximum inA. `O_(2)`B. `O_(2)^(-1)`C. `O_(2)^(+1)`D. `O_(2)^(-2)` |
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Answer» Correct Answer - C B.O. of `O_(2)` is 2, B.O. of `O_(2)^(-)` is 1.5 , B.O. of `O_(2)^(+1)` is 2.5 and of `O_(2)^(2-)` is 1 . |
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| 626. |
Which of the following is not paramagnetic ?A. `S^(-2)`B. `N_(2)^(-)`C. `O_(2)^(-)`D. `NO` |
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Answer» Correct Answer - A `S^(-2)` have all paired electrons so it is diamagnetic. |
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| 627. |
Which of the following is a correct set with respect to molecule, hybridization, and shape?A. `BeCl_2, sp^2`, linearB. `BeCl_2, sp^2`, triangularC. `BCl_3, sp^3`, triangular planarD. `BCl_3, sp^3`, tetrahedral |
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Answer» Correct Answer - C The steric number of `Be` in `BeCl_2` is `2(2+0)`. Therefore, it is linear with `sp` hybridization. The steric number of `B` in `BCl_3` is `3(3+0)`. Therefore, it is trigonal planar with `sp^2` hybridization. |
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| 628. |
The common features anong the species ` CN^- , CO ` and ` NO^+` are :A. bond order 3 and isoelectronicsB. bond order (3) and weak field ligandsC. bond order 2 pi-acceptorD. isoelectric and weak field ligands |
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Answer» Correct Answer - A Each pssesses ` 14` electrons with bond order 3 . |
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| 629. |
Which one of the following is not correct with respect to bond length of the species?A. `C_2 gt C_2^-`B. `B_2^+ gtB_2`C. `Li_2^+ gt Li`D. `O_2 gt O^-` |
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Answer» Correct Answer - D Bond length `prop 1/(BO)` `C_2=2, C_2^(2-)=3 , B_2=1, B_2^(+)=0.5 , Li_2 =1 , Li_2^(-)=1.5` `N_2=3, N_2^(+)=2.5 , O_2=2 , O O_2^(-)=1.5` |
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| 630. |
Which of the following options with respect to increasing bond order is correct ?A. `NO lt C_(2) lt O_(2)^(-) lt B_(2)`B. `C_(2) lt NO lt B_(2) lt O_(2)^(-)`C. `B_(2) lt O_(2)^(-) lt NO lt C_(2)`D. `B_(2) lt O_(2)^(-) lt C_(2) lt NO` |
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Answer» Correct Answer - D `B_(2)` bond order `=(6-4)/(2)=1,O_(2)^(-)` bond order `=(10-7)/(2)=1.5` `C_(2)` bond order `=(8-4)/(2)=2,NO` bond order `=(10-5)/(2)=2.5` |
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| 631. |
Which of the following is diamagnetic?A. superoxide ionB. carbon moleculeC. unipositive ion of nitrogen moleculeD. oxygen molecule |
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Answer» Correct Answer - B Carbon molecule `(C_2)` with `12e^(-)s` is diamagnetic. Though `O_2` has `16e^(-)s`, but it is paramagnetic due to the presence of two unpaired `e^(-)s` in the antibonding `pi MOs`. Superoxide ion `(O_2^(-))` is also paramagnetic as it has one unpaired `e^(-)`. Nitrogen molecule `(N_2)` is diamagnetic but `N_2^(+)` is paramangnetic due to one unpaired electron. |
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| 632. |
Which of the following is isoelectronic with carbon?A. `Na^(+)` and `Ne`B. `K^(+)` and OC. Ne and OD. `Na^(+)` and `K^(+)` |
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Answer» Correct Answer - A `Na^(+)` and Ne are isoelectronic each contain 10 electrons |
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| 633. |
Which one of the following is not correct with respect to bond length of the species?A. `C_(2) gt C_(2)^(2-)`B. `B_(2)^(+) gt B_(2)`C. `Li^(2^(+)) gt Li_(2)`D. `O_(2) gt O_(2)^(-)` |
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Answer» Correct Answer - D Bond order of `{:(C_(2)=2,C_(2)^(2-)=3,B_(2)^(+)=0.5),(B_(2)=1,Li_(2)^(+)=0.5,Li_(2)=1),(N_(2)^(+)=2.5,N_(2)=3,O_(2)=2.0),(O_(2)^(-)=1.5,,):}` Bond order is inversely proportional to the bond length . Greater the bond order, shorter is the bond length therefore, `O_(2)^(-) gt O_(2)` |
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| 634. |
Calculate the theoretical value of bond length in `H -F` , if `f_H` and `r_F` are ` -0.37Å` respectively. Electronegativites of `F` and `H` are `4.0` and `2.1` respectively . |
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Answer» Given, `r-H =0.37 Å ,r-F =0.72 Å , X-F =4.0 X-H =2.1` `d_(A-B) =r-H =r-H + r_F -0.09 (X-F -X_H)` `= 0. 37 + 0. 72 - 0.09 (4.0- 2.1) =0.92 Å`. |
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| 635. |
The following species is repelled by a magnetic fieldA. `O_(2)`B. `H_(2)O`C. `CrO_(2)`D. `Fe_(3)O_(4)` |
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Answer» Correct Answer - B Substance which are weakly repelled by external magnetic field are called diamagnetic substance, e.g. `H_(2)O` |
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| 636. |
Isoelectronic species areA. `K^(+),Cl^(-)`B. `Na^(+),Cl^(-)`C. Na,ArD. `Na^(+),Ar` |
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Answer» Correct Answer - A `K^(+)=1s^(2)2p^(6)3s^(2)3p^(6)` `Cl^(-)=1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)` |
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| 637. |
Which of the following is an electron-deficient compound?A. `BeCl_2`B. `BCl_3`C. `AlCl_3`D. All of these |
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Answer» Correct Answer - D The Lewis structures of the compounds are `Cl-Be-Cl` `Cl-underset(Cl)underset|B-Cl` `Cl-underset(Cl)underset|Al-Cl` Compounds such as there in which the central atom capable of accommodating `8e^(-)` shares fewer than `8e^(-)` are sometimes referred to as electron-deficient compounds. |
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| 638. |
Which of the following has the shortest bond lengthA. C-HB. C-NC. C-OD. C-C |
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Answer» Correct Answer - A `{:("Bond",,"Bond length (Å)"),(C-H,,1.08),(C-N,,1.47),(C-O,,1.43),(C-C,,1.54),(C-F,,1.35):}` |
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| 639. |
Which of the following is paramagneticA. `N_(2)`B. `C_(2)`C. `N_(2)^(+)`D. `O_(2)^(2-)` |
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Answer» Correct Answer - C `N_2: K K sigma(2s)^(2)sigma^(**)(2s)^(2)pi (2p_9x))^(2)pi (2p_(y))^(2)sigma(2p_(z))^(2)` (diamagnetic) `N_2^(+): K K sigma(2s)^(2)sigma^(**)(2s)^(2)pi(2p_(x))^(2)pi (2p_(y))^(2) ` (diamagnetic) `O_(2)^(2-): K K sigma (2s)^(2) sigma^(**)(2s)^(2)sigma(2p_(z))^(2) pi (2p_(x))^(2)pi (2p_(y))^(2) pi^(**) (2p_(x))^(2) pi^(**)(2p_(y))^(2)` (diamagnetic) |
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| 640. |
Atomic radius and ionic radius of `F_(g)` and `F_(g)^-` are `72` and `136` pm prespectivley. Calculate the ratio and percentage increase in terms of volume during formation of `F_(g)^(-)` form `F_(g)` . |
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Answer» Volume of ` F_((g)) = 4/3 pi xx (136)^3 = 10.54 xx10^6` Voleume of `F_((g))^(-) = (4)/(3) pi xx (136)^(3) =10.54 xx10^6` ` :. V_(F)^(-) =V_(F) xx [(136)/(72)]^3 =V-F xx 6.745` `:.` Increase in volume ` = [10.54 -1.56 ] xx 10^6 =8.98 xx10^6` ` :. %` increase = `(8,98 xx10^6) /(1 .56 xx10^6) xx 100=5. 75 xx10^2`. |
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| 641. |
Bond polarity of diatomic molecule is because of _________.A. Difference in electron affinities of the two atomsB. difference in electronegativities of the two atomsC. Difference in ionisation potentialD. All of these |
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Answer» Correct Answer - B Polarity character is due to the difference in electronegativity of two atoms or molecule |
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| 642. |
Which one of the following is the correct statementA. `O_(2)` molecule has bond order 2 and is diamagneticB. `N_(2)` molecule has bond order 3 and is paramagneticC. `H_(2)` molecule has bond order zero and is diamagneticD. `C_(2)` molecules has bond order 2 and is diamagnetic |
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Answer» Correct Answer - D `O_(2): K K sigma(2s)^(2)sigma^(**)(2s)^(2)sigma(2p_(z))^(2)pi (2p_(x))^(2)pi (2p_(y))^(2) pi^(**)(2p_(x))^(1)pi^(**)(2p_(y))^(1)` Bond order =2, paramagnetic `N_(2): K K sigma(2s)^(2)sigma^(**)(2s)^(2)pi (2p_(x))^(2)pi(2p_(y))^(2)sigma(2p_(x))^(2)` Bond order =3, diamagnetic `H_(2): sigma(1s)^(2)` Bond order=1,diamagnetic `C_(2): K K sigma(2s)^(2)sigma^(**)(2s)^(2))pi (2p_(x))^(2)pi(2p_(y))^(2)` Bond order =2, diamagnetic `He_(2)^(+): sigma (1s)^(2) sigma^(**)(1s)^(1)` Bond order =1//2, paramagnetic |
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| 643. |
Lewis formulas are not normally written for compounds containing___________elements. (i) s-block (ii) p-block (iii) d-block (iv) f-blockA. (ii), (iii), (iv)B. (iii), (iv)C. (ii), (iii)D. (i), (ii), (iii) |
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Answer» Correct Answer - B The Lewis structures are not normally written for compounds containing d-block and f-block elements because these elements utilize d or f orbitals (or both) in bonding as well as s and p orbitals. Thus, they can accommodate more than eight valence electrons. |
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| 644. |
Which one of the following is paramagneticA. `H_(2)O`B. `NO_(2)`C. `SO_(2)`D. `CO_(2)` |
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Answer» Correct Answer - B `NO_(2)` has unpaired electrons so it would be paramagnetic. |
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| 645. |
The bond order of `O_(2)` and `O_(2)^(+)` are _________ respectively.A. 2B. 1.5C. 3D. 3.5 |
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Answer» Correct Answer - A Electronic configuration of `O_(2)` is `O_(2)=(sigma1s)^(2)(sigma^(**)1s)^(2)(sigma2s)^(2)(sigma^(**)2s)^(2)(sigma^(**)2s)^(2)(sigma2p_z)^(2)` `(pi2p_(x)^(2)-=pi2p_(y)^(2))(pi^(**)2p_(x)^(1)-=pi^(**)2p_(y)^(1))` Hence bond order `=1/2[N_(b)-N_(a)]=1/2[10-6]=2` |
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| 646. |
Covalent radius of `._(82)Pb` is `1.53Å`. Calculate its electroneagitity at Allred -Rochow sale . |
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Answer» We have `(X_(Pb) _(AR)) = 0.744 + (0.359 Z_(eff))/(r^(2))` `._(82)PB =1s^2 , 2s^2 . 2p^2 . 3s^3 3p^6, 3p^6 underset(60) (3d^(10)) . 4s^2 4p^6 4d^(10) 4f^(14)` `5s^2 underset(18)(5p^6) 5d^(10) , 6s^2 underset(4) ( 6p^2)` ` sigma = [ 0. 35 xx 3] + [0. 85 xx 18+_ 1.0 xx 60 ] = 7.6 . 35` and `Z_(eff) = 82.00 - 76. 35 = 5.65, r =153 Å` `:. (X_(Pb)) _(AR) =0.44 + ( 0. 359xx 5.65) /((1.530^2)) = 1.61` . |
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| 647. |
Which one does not exhibit paramagnetismA. `ClO_(2)`B. `ClO_(2)^(-)`C. `NO_(2)`D. `NO` |
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Answer» Correct Answer - B `ClO_(2)^(-)` has all paired electrons hence it does not show paramagnetic. |
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| 648. |
Which of the following molecular orbitals has two nodal planesA. `sigma 2s`B. `pi 2p_(y)`C. `pi^(**) 2p_(y)`D. `sigma^(**)2p_(x)` |
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Answer» Correct Answer - C `overset(**)(pi)2p_(y)` has two nodal planes. |
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| 649. |
Which of the following statement is true ?A. HF is less polar than HBr`B. Absolutely pure water does not contain any ionsC. Chemical bond formation taken place when forces of attraction overcome the forces of repulsionD. In covalence transfer of electrons takes place |
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Answer» Correct Answer - C It is a fact . |
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| 650. |
`BF_3` and `NH_3` combine readily because of the formation ofA. a dative bondB. an ionic bondC. a hydrogen bondD. a covalent bond |
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Answer» Correct Answer - A Although boron trifluoride `(BF_3)` is stable, it has a tendency to pick up an unshared electron pair from an atom in another compound, as shown by its reaction with ammonia: `F-underset(F)underset(|)overset(F)overset(|)B+:underset(H)underset(|)overset(H)overset(|)NrarrF-underset(F)underset(|)overset(F)overset(|)(B^(-))-^(+)N-H` The formation of this compound satisfies the octet rule for B, N, and F atoms. The `B-N` bond in the above compound is a dative bond because both electrons for sharing are contributed by the N atom. |
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