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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.
| 651. |
In forming `(i) N_(2) rarrN_(2)^(o+)` and `O_(2)rarrO_(2)^(o+)` the electrons respectively removed from .A. `(pi^(**)2p_(y) "or " pi^(**) 2p_(x)) " and " (pi^(**)2p_(y)"or" pi^(**)2p_(x))`B. `(pi 2p_(y) "or " pi 2p_(y)) " and " (pi 2p_(y) "or " 2p_(x))`C. `(sigma 2p_(z))` and `(pi^(**)2p_(y))` or `pi^(**)2p_(x)`D. `(pi^(**)2p_(y) "or " pi^(**) 2p_(x)) " and " (pi 2p_(y) " or " pi 2p_(x))` |
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Answer» Correct Answer - C (i) `N_2 to (sigma1s)^(2) (sigma^(**)1s)^(2)(sigma2s)^(2)(sigma^(**)2s)^(2)(pi 2p_x)^(2) (pi 2p_(y))^(2)(sigma2p_(z))^(2)` `N_(2)^(+) to (sigma1s)^(2)(sigma^(**)1s)^(2) (sigma2s)^2(sigma^(**)2s)^(2)(sigma2p_(z))^(1)` (ii) `O_(2) to (sigma1s)^(2)(sigma^(**)1s)^(2)(sigma2s)^(2)(sigma^(**)2s)^(2)(sigma2p_(z))^(2)(pi 2p_(x))^(2) (pi 2p_(y))^(2)(pi^(**)2p_(x))^(1)(pi^(**)2p_(y))^(1)` `O_(2)^(+) to (sigma1s)^(2)(sigma^(**)1s)^(2)(sigma2s)^(2)(sigma^(**)2s)^(2)(sigma2p_(z))^(2)(pi 2p_(x))^(2)(pi 2p_(y))^(2)(pi^(**)2p_(x))^1` In `N_(2)^(+)` , electron removes from `sigma2p_(z)` becuase in `N_(2),pi2p_(x)` and `pi2p_(y)` have lower energy in comparision to `sigma2p_(z)`. in `O_(2)^(+)` electron removes either from `pi^(**)2p_(y)` or `pi^(**) 2p_x`. |
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| 652. |
The paramagnetism of `O_(2)^(+)` is due to the presence of an odd electron in the MOA. `sigma^(**)2s`B. `pi 2p_(y)`C. `pi 2p_(z)`D. `pi^(**)2p_(y)` |
| Answer» Correct Answer - D | |
| 653. |
Specify the co-ordination geometry around and hybridisation of `N` and ` B` complex of `NH_3` and `BF_3`,A. N : tetrahedral , `sp^(3)` , B : tetrahedral ,` sp^(3)`B. N : Pyramidal, `sp^(3)` , : Pyramidal , `sp^(3)`C. N : Pyramidal , `sp^(3)` , : Pyramidal , `sp^(3)`D. N : tetrahedral , `sp^3` , B : tetrahedral , `sp^3` |
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Answer» Correct Answer - A `underset(sp^3)(NH_3)+underset(sp^3)(BF_3)rarr [underset(sp^3)(H_3N) rarr underset(sp^3) (BF_3)]` |
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| 654. |
Maximum bond angle is present in case ofA. `BCl_3`B. `B BF_3`C. `BF_(3)`D. Same for all |
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Answer» Correct Answer - D `BCl_(3),B Br_(3)` and `BF_(3)`, all of these have same structure i.e, trigonal planar (`sp^(2)` hybridization) hence bond angle is same is for all of them (i.e. , equal to `120^(@)`) |
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| 655. |
Which of the following species have maximum number of unpaired electrons ?A. `O_(2)`B. `O_(2)^(+)`C. `O_(2)^(-)`D. `O_(2)^(2-)` |
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Answer» Correct Answer - A `O_(2)` has 2 unpaired electron while `O_(2)^(+)` and `O_(2)^(-)` has one each unpaired electron while `O_(2)^(+)` does not have any unpaired electron. |
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| 656. |
Two or more atoms of the same or different elements chemically combine to form a (i) molecule of an element (ii) molecule of a covalent compound (iii) polyatomic ion (iv) network solidA. (i), (ii)B. (i), (ii), (iii)C. (i), (ii), (iv)D. (i), (ii), (iii), (iv) |
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Answer» Correct Answer - D Two atoms of oxygen combine to form a molecule of oxygen `(O_2)`, two atoms of hydrogen and one atom of oxygen combine together to form a molecule of water `(H_2O)`, one C atom combines with one N atom to form cyanide ion `(CN^(-))`, and a number of Si and O atoms combine to form a network solid (silica, `SiO_2`). |
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| 657. |
Which of the following molecualr species has unpaired electrons ?A. `N_2`B. ` F_2`C. `O_2^(-)`D. `O_2^(2-)` |
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Answer» Correct Answer - C `O_(2)^(-)` has one unpaired electron . |
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| 658. |
The total number of electrons that take part in forming the bond in `N_(2)` is .A. 2B. 4C. 6D. 10 |
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Answer» Correct Answer - C Nitrogen form triple bond `N-=N` in which 6 electron take part. |
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| 659. |
Arrange the following ions in the order of decreasing `X-O` bond length where X is the central atom:A. `ClO_(4)^(-),SO_(4)^(2-),PO_(4)^(3-),SiO_(4)^(4-)`B. `SiO_(4)^(4-),PO_(4)^(3-),SO_(4)^(2-),ClO_(4)^(4-)`C. `SiO_(4)^(4-),PO_(4)^(3-),ClO_(4)^(-),SO_(4)^(2-)`D. `SiO_(4)^(4-),SO_(4)^(2-),PO_(4)^(3-),ClO_(4)^(-)` |
| Answer» Correct Answer - B | |
| 660. |
`H_2S` is more acidic than `H_2O`. The reason isA. `O-H` bond is stronger than `S-H` bondB. `O` is more electronegative than SC. `H-S` bond is weaker than `O-H` bondD. `O-H` bond is weaker than `H-S` bond |
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Answer» Correct Answer - A::C To release `H^(+)` (proton), the `H-E` (E stands for element which may be O or S) bond must be broken. On account of relatively large size of S atom, the `H-S` bond is weaker than the `O-H` bond and, hence, easy to break. |
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| 661. |
Maximum bond angle is present in case ofA. `BBr_3`B. `BCl_3`C. `BF_4`D. same in all |
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Answer» Correct Answer - D All these molecules `(BX_3)` possess trigonal planar shape as well as geometry. Therefore, they all possess the same bond angle of `120^@`. |
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| 662. |
Which of the following molecular species has unpaired electrons(s) ? .A. `N_(2)`B. `F_(2)`C. `O_(2)^(-)`D. `O_(2)^(2-)` |
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Answer» Correct Answer - C Only `O_(2)^(-)` contains unpaired electrons in antibonding molecular orbital. `(sigma1s)^(2)(sigma^(**)1s)^(2)(sigma2s)^(2)(sigma^(**)2s)^(2)(sigma2p_(z))^(2)(pi 2p_(x))^(2)(pi 2p_(y))^(2)(pi^(**)2p_(x))^(2)(pi^(**)2p_(y))^(1)` This shows one unpaired electron. |
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| 663. |
When two atoms combine to form a molecule .A. One molecular orbitalB. Two molecular orbitalC. Three molecular orbitalD. Four molecular orbital |
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Answer» Correct Answer - B One bonding M.O. and one anti-bonding M.O. |
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| 664. |
Identify the least stable ion amongst the following:A. `Li^(-)`B. `Be^(-)`C. `B^(-)`D. `C^(-)` |
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Answer» Correct Answer - B `Be^(-)` is attaining `2s^(2)sp^(1)` configuration i.e., losing its fulfilled stability of `2s^(2)` configuration. |
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| 665. |
The least stable in amongst the following is :A. `Li^(-)`B. ` Be^(-)`C. ` B^(-)`D. `C^(-)` |
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Answer» Correct Answer - B `Li^(-) : 1s^2 , 2s^2 (EA_2=-ve),` `Be^(-) : 1s^2 , 2s^2 2p^1 (EA_2 = +ve)`. |
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| 666. |
Identify the least stable ion amongst the following:A. `Be^(-)`B. `Li^(-)`C. `B^(-)`D. `C^(-)` |
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Answer» Correct Answer - A `Li^(-) rArr 1S^(2),2S^(2)(EA_(1)=-ve)` `Be^(-) rArr =1S^(2),2S^(2)2P^(1)(EA_(2)=+ve)` |
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| 667. |
Using `MO` theory predict which of the following sepcies has the shortest bond length ?A. `O_(2)^(2+)`B. `O_(2)^(+)`C. `O_(2)^(-)`D. `O_(2)^(2-)` |
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Answer» Correct Answer - A Bond order `O_(2)^(+)=(10-5)/2=2.5` `O_(2)^(-)=(10-7)/2=1.5` `O_(2)^(2-)=(10-8)/2=1` `O_(2)^(2+)=(10-4)/2=3` Bond order `prop1/("bond length")` So, `O_(2)^(2+)` has the shortest bond length. |
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| 668. |
Which of the following bonds require the largest amount of bond energy to dissociate the atoms concerned ?A. H-H bond in `H_2`B. C-H bond in `CH_4`C. N = N bond in `N_2`D. O=O bond in `O_2` |
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Answer» Correct Answer - C Out of the given five choices, `N-= N` bond has highest B.E. (943 J `mol^(-1)`). B.E of H-H bond is 433 kJ `mol^(-1)`. |
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| 669. |
`CH_(2)Cl-CHCl_(2)`A. spB. `sp^(2)`C. `sp^(3)`D. `sp^(2)d` |
| Answer» Correct Answer - C | |
| 670. |
Peroxide ion....... (i) Has five completely filled antibonding molecular orbitals (ii) Is diamagnetic (iii) Has bond order one (iv) Is isoelectric with neon Which one of these is correctA. (iv) and (iii)B. (i),(ii) and (iv)C. (i),(ii) and (iii)D. None of these |
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Answer» Correct Answer - D (None of the given options is correct) Peroxide ion is `O_(2)^(2-)` Total electrons=18 E.c.=`sigma1s^(2)sigma^(**)1s^(2)sigma2s^(2)sigma^(**)2s^(2)sigma(2p_(x))^(2)pi (2p_(y))^(2)` `pi^(**)(2p_(x))^(2)pi^(**)(2p_(y))^2` `B.O. =(10-8)/2=1` options (ii) and (iii) are correct. |
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| 671. |
According to molecular orbital theory, the paramagnetism of `O_(2)` molecule is due to the presence ofA. Unpaired electrons in the bonding `sigma` molecular orbitalB. Unpaired electrons in the antibonding `sigma` molecular orbitalC. unpaired electron in the bonding `pi` molecular orbitalD. Unpaired electrons in the antibonding `pi` molecular orbital |
| Answer» Correct Answer - D | |
| 672. |
Why `XeF_(5)^(-)` is planar ? |
| Answer» To avoid strong repulsion of lone pair and bond pair. | |
| 673. |
Which of the following pairs of species are isostructural ?A. `SO_(4)^(2-) and BF_(4)^(-)`B. `NH_(3) and NH_(4)^(+)`C. `CO_(3)^(2-) and CO_(2)`D. `CH_(4) and BF_(3)` |
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Answer» Correct Answer - A `SO_(4)^(2-) and BF_(4)^(-) " have " sp^(3)` hybridisation and are tetrahedral in shape. |
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| 674. |
Given below is the bond angle in various types of hybridisation. Mark the bond angle which is not correctly matched.A. `dsp^(2) - 90^(@)`B. `sp^(3)d^(2) - 90^(@)`C. `sp^(3) d - 90^(@)`D. `sp^(3) - 109.5^(@)` |
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Answer» Correct Answer - C In `sp^(3)` d, the bond angle is `120^(@) and 90^(@)`. |
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| 675. |
The increasing d-character in hybridisation of Xe in `XeF_(2), XeF_(4), XeF_(6)` isA. `sp^(2) , sp^(3) d, sp^(3) d^(2)`B. `sp^(3) d, sp^(3) d^(2), sp^(3) d^(3)`C. `sp^(3) d^(2), sp^(3) d, sp^(3) d^(3)`D. `sp^(2), sp^(3), sp^(3) d` |
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Answer» Correct Answer - B `XeF_(2) - sp^(3) d`, Total no. of valence electrons = 22 ` 22/8 = 2 (Q) + 6(R ) , 6/2 = 3 (Q)` ` X = 2 + 3 = 5` Hybridisation is `sp^(3)` d. `XeF_(4) - sp^(3) d^(2)` Total no. of electrons in outermost shells `= 8 + 28 = 36 36/8 = 4 (Q) + 4(R) , 4/2 = 2 (Q) + 0 (R)` ` X = 4 + 2 + 0 = 6` Hybridisation is `sp^(3) d^(2)`. ` XeF_(6) - sp^(3) d^(3)` Total no. of valence electrons = ` 8 + 42 = 50 50/8 = 6 (Q) + 2 2 (R) , 2/2 = 1 (Q) , X = 6 + 1 = 7` Hybridisation is `sp^(3) d^(3)`. |
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| 676. |
Arrange the following in the increasing order of their bond order:A. `O_2 , O_2^(+), O_2^(-) and O_2^(2-)`B. `O_2^(2-), O_2^(-), O_2, O_2^(+)`C. `O_2^(+),O_2,O_2^(-),O_2^(2-)`D. `O_2, O_2^(+), O_2^(-), O_2^(2-)` |
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Answer» Correct Answer - B Increasing bond order is `O_2^(2-) (1.0) lt O_2^(-) (1.5) lt O_2(2.00) lt O_2^(+) (2.5)` |
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| 677. |
`RbO_2` isA. Peroxide and paramagneticB. Peroxide and magneticC. Super oxide and paramagneticD. Super oxide and diamagnetic. |
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Answer» Correct Answer - C `RbO_2` means `Rb^+` and `O_2^(-)` ion. `O_2^(-)` ion is super oxide ion. It has 17 electrons and is paramagnetic. |
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| 678. |
Sideways overlap of `p-p` orbitals formsA. Sigma bondB. Pi bondC. Coordinate bondD. H-bond. |
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Answer» Correct Answer - B Sidewise overlap of p orbital, results `pi` bonds |
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| 679. |
The molecule , .A. has intermolecular H-bondingsB. has intramolecular H-bondingC. reduces Tollens reagentD. is steam -volatile |
| Answer» Correct Answer - B::D | |
| 680. |
The number of unpaired electrons in a parmamagnetic diatomic molecyle of an element with atomic number ` 16` is :A. 4B. 1C. 2D. 3 |
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Answer» Correct Answer - C `S_2` molecule is paramagnetic like `O_2` having two unpaired electrons . |
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| 681. |
Using `MO` theory predict which of the following sepcies has the shortest bond length ?A. `O_2^(2+)`B. `O_2^(+)`C. `O_2^(-)`D. `O_2^(2-)` |
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Answer» Correct Answer - A The MO configuration for `O_2` is `KK(sigma2s)^2(sigma^**2s)^2(sigma2p_z)^2(pi2p_x^2=pi2p_y^2)(pi^**2p_x^1=pi^**2pi_y^1)` It has `N_b` 10 and `N_a` 6. Thus, `O_2^(2+) bo=(10-4)/(2)=3` `O_2^(+) bo=(10-5)/(2)=2.5` `O_2^(-) bo=(10-7)/(2)=1.5` `O_2^(2-) bo=(10-8)/(2)=1.0` Since bo is inversely related to bond length, `O_2^(2+)` with the highest bo of 3 has the shortest bond length. |
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| 682. |
The bond angle and dipole moment of water respectively are `:`A. `102.5^@`, `1.56D`B. `107.5^@, 1.56D`C. `109.5^@, 1.84D`D. `104.5^@, 1.84D` |
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Answer» Correct Answer - D The steric number of O atom in water is `2+2=4`. Thus, `H_2O` has tetrahedral geometry but angular shape. Due to lone pair-lone pair repulsion, the bond angle is reduced from `109.5^@` to `104.5^@`. Due to contribution of electronic dipoles and bond dipoles, it has high dipole moment. |
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| 683. |
The number of water molecule(s) derectly bonded to the metal centre in `CuSO_4.5H_2O` isA. `2`B. `3`C. `4`D. `5` |
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Answer» Correct Answer - D The actual description of the compound is `[Cu(H_2O)_4]SO_4* * * * * * * H_2O` The fifth water molecule is hydrogen bonded to `SO_4^(2-)` ion. |
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| 684. |
Which of the following statements is (are) correct ?A. `PH_5` and `BiCl_5` does Not existB. ` p pi -` d pi bonds are present in ` SO-2`C. `SeF_4` and `CH_4` has same shapeD. ` I_2^+` has bent geometry |
| Answer» Correct Answer - A::B::D | |
| 685. |
How many sigma and pi bonds are present in the linear chain compound which has the formula `C_5H_4` and contains both double and triple bonds?A. 6 sigma and 6 piB. 8 sigma and 2 piC. 6 sigma and 4 piD. 8 sigma and 4 pi |
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Answer» Correct Answer - D According to the formula `C_5H_4`, the Lewis structure for a linear chain compound containing both double and triple bonds will be `CH-=C-CH=C=CH_2` or `H-C-=C-overset(H)overset(|)C=C=overset(H)overset(|)C-H` Every single bond is a sigma bond and every multiple bond contains one sigma bond. Thus, it has 8 sigma bonds and 4 pi bonds. |
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| 686. |
In `PO_(4)^(3-)` the formal charge on each O-atom and `P-O` bond order respectively are .A. `-0.75,1.25`B. `-0.75,1.0`C. `-0.75,0.6`D. `-3,1.25` |
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Answer» Correct Answer - A Bond order =`("Total number of bonds between atoms")/("Total number of resonating structure")` `=5/4 =1.25` |
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| 687. |
Point out incorrect statement about resonanceA. Resonance structure should have equan energyB. In resonance structure, the constituents atoms should be in the same positionC. In resonance structure, there should not be the same number of electron pairs.D. Resonance structure should differ only in the location of electrons around the constituent atoms |
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Answer» Correct Answer - C In resonance structure three should be the same number of electron pairs. |
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| 688. |
Which of the following is incorrect regarding resonance?A. The canonical forms have no real existence.B. The molecule exists for a certain fraction of time in one canonical form and for other fractions of time in other canonical forms.C. There is no such equilibrium between the canonical forms as we have between tautomeric forms (keto and enol) is tautomerism.D. The molecule as such has a single structure which is the resonance hybrid of the canonical forms and which cannot as such be depicted by a Lewis structure. |
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Answer» Correct Answer - B A molecule or polyatomic ion for which two or more Lewis formulas with the same arrangement of atoms can be drawn to describe the bonding is said to exhibit resonance. The different Lewis structures drawn are resonance structures of the molecule or ion. The relationship among them is indicated by the double-headed arrow `iff`. This symbol does not mean that the molecule or ion flips back and forth among different Lewis structures. The true structure can be described as an average, or hybrid, of the three. |
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| 689. |
How many covalent bonds are present in a molecule of carbon dioxide?A. TwoB. FourC. SixD. Three |
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Answer» Correct Answer - B The Lewis formula for `overset(ddot)underset(ddot)( :Cl*)+overset(ddot)underset(ddot)( *Cl: )rarr overset(ddot)underset(ddot)( :Cl: )overset(ddot)underset(ddot)(Cl: )` In a Lewis formula, we represent a covalent bond by writing each shared electron pair either as a pair of dots between the two atom symbols or as a dash connecting them. |
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| 690. |
Which statement`s` is `(are)` correct ?A. A pi- bond is weaker than sigma-nondB. A sigma-bond is weaker than pi-bondC. A (double ) bond stronger than single bondD. A covalent bond is stronger than H-bond |
| Answer» Correct Answer - A::C::D | |
| 691. |
The molecule of sulphuric acid containsA. ions, covalent, and coordinate bondsB. ionic and covalent bondsC. covalent and coordinate bondsD. only covalent bonds |
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Answer» Correct Answer - D The most likely Lewis structure of `H_2SO_4` is `H-overset(..)underset(..)O-overset( :O: )overset(||)underset( :O: )underset(||)S-overset(..)underset(..)O-H` It consists of only covalent bonds. Some textbooks describe the less plausible Lewis structure `H-overset(..)underset(..)O-overset( :O: )overset(uarr)underset( :O: )underset(darr)S-overset(..)underset(..)O-H` which contains both covalent and coordinate bonds. |
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| 692. |
Which of the following statement (s) is (are) correct ?A. The peroxide ion has a bond order of `1` while the oxygen molecule has a bond order of `2`B. the peroxide ion has a longer and weaker bond than theoxygen moleculeC. The peroxid ion as the oxygen molecule are paramagneticD. The bond length of peroxide ion is greater than that of the oxygen molecule |
| Answer» Correct Answer - A::B | |
| 693. |
Which of the following is not the correct representation of resonance? A. Only IB. Only IIC. Both I and IID. None of these |
| Answer» (d) Both representation of resonating structure in molecules of `CO_(2)` and `CO_(3)^(2-)` are correct. | |
| 694. |
The relationship between the dissociation energy of `N_2` and `N_2^+` isA. dissociation energy of `N_2^+` gt dissociation energy of `N_2`B. dissociation energy of `N_2` = dissociation energy of `N_2^+`C. dissociation energy of `N_2` gt dissociation energy of `N_2^+`D. dissociation energy of `N_2`can either be lower or higher than the dissociation energy of `N_2^+` |
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Answer» Correct Answer - C The dissociation energy will be more when the bond order will be greater and bond order `prop` dissociation energy Molecular orbital configuration of `N_2(14)=sigma1s^2 , overset***sigma1s^2, sigma2s^2 , overset***sigma2s^2, sigma2p_y^2~~ pi2p_z^2,pi2p_x^2` So, bond order of `N_2=(N_b-N_a)/2=(10-4)/2=3` and bond order of `N_2^(+)=(9-4)/2=2.5` As the bond order of `N_2` is greater than `N_2^+` so , the dissociation energy of `N_2` will be greater than `N_2^+` |
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| 695. |
Mark the incorrect statement in the following .A. The bond order in the species `O_2, O_2^+` and `O_2^-` decreases as `O_2^(+) gt O_2 gt O_2^-`B. The bond energy in a diatomic molecule always increases when an electron is lostC. Electrons in antibonding MO contribute to repulsion between two atomsD. With increase in bond order, bond length decreases and bond strength increases. |
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Answer» Correct Answer - B When a diatomic molecule lost electron , then its bond order may increase or decrease, so its bond energy may decrease or increase. |
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| 696. |
Which of the following statement is not correct ?A. Double bond is shorter than a single bondB. Sigma bond is weaker than a `pi`-bondC. Double bond is stronger than a single bondD. Covalent bond is stronger than hydrogen bond |
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Answer» Correct Answer - B Sigma bond is always stronger than `pi`-bond because the extent of overlapping is maximum in sigma bond formation. |
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| 697. |
By hybridization, we mean the hybridization ofA. electronsB. atomic orbitalsC. atomsD. protons |
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Answer» Correct Answer - B Hybridization is the mixing of atomic orbitals in an atom (usually a central atom) to generate a set of new atomic orbitals, called hybrid orbitals in an atom (usually a central atom) to generate a set of new atomic orbitals, called hybrid orbitals. |
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| 698. |
Structrue of `ICO_2^-` is :A. trigonalB. distorted trigonal bipyramidC. octahedralD. square planar |
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Answer» Correct Answer - B ` ICl_2^(-)` has `sp^3` d-hybridaised state ( i.e., trigonal bipyramid stage but distroted due to lone paire of electrons on I atom ) |
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| 699. |
The shape of `ClO_3^(-)` isA. triangular pyramidalB. tetrahedralC. triangular planarD. triangular bipyramidal |
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Answer» Correct Answer - A The steric number of `Cl` in `ClO_3^(-)` is four (3 atoms bonded +1 lone pair). Therefore, it has tetrahedral geometry but triangular pyramidal shape. |
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| 700. |
Which one in the following is not the resonance structure of `CO_(2)`A. O=C=OB. `.^(-)O-C-=O^(+)`C. `.^(+)O-=C-O^(-)`D. `O-=C=O` |
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Answer» Correct Answer - D Choice (a),(b),© are the resonance structure of `CO_(2)` |
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