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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.
| 551. |
Which of the following compounds does not follow the octet rule the electron distributionA. `PCl_(5)`B. `PCl_(3)`C. `H_(2)O`D. `PH_(3)` |
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Answer» Correct Answer - A `PCl_(5)` does not follows octeet, rule, it has 10 electrons in its valence shell. |
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| 552. |
Find the bond order of ` Be_2` |
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Answer» `Be_2=sigma1s^2,sigma^**1s^2,sigma2s^(2),sigma^**2s^2` `:.` Bond order of ` Be_2 = 1/2 [ 4 - 4 ] =0` `:. Be_2 ` is not possible . |
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| 553. |
Which among the following have bond order 2.5?A. `O_2`B. `N_2^(2-)`C. `N_2^+`D. `O_2^+` |
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Answer» Correct Answer - C,D B.O. of `N_2^(+)=(N_b-N_a)/2 =(9-4)/2=2.5` B.O. of `O_2^(+)=(N_b-N_a)/2 =(9-4)/2.5` (Both are isoelectronic with 16 electrons each) |
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| 554. |
Which among the following species have bond order zero?A. `F_2^(2-)`B. `Ar_2`C. `He_2^+`D. `H_2^+` |
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Answer» Correct Answer - A,B B.O. of `Ar_2=(N_a-N_a)/2 =(18-18)/2=0` B.O. of `F_2^(2-)=(N_b-N_a)/2 =(10-10)/2=0` |
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| 555. |
Which statement is wrong about `H_2O` ?A. It has high specific heat relative to other liquids or solids due to strong intermolecular H-bondingB. `H_2O` molecule has capaity to form ` 4H`-bondsC. `H_2O` has open cage like structure due to intermolecular H-bonding which give rise to low density to ice than liquid `H_2O`D. `H_2O` has maximum density at `4^@ C` since upto `4^@C` the intermolecular H-bonding persists more and thereby decreasing volume and increasing density . |
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Answer» Correct Answer - D Upto `4^@ C,` the intermolecualr H-bonding keeps on breaking and therebu dereasing the volume and increasing the density . |
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| 556. |
Which of the following species is hypervalent?A. `ClO_4^- `B. `BF_3`C. `SO_4^(2-)`D. `CO_3^(2-)` |
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Answer» Correct Answer - A,C Compounds or ions whose Lewis structure demand more than 8 electrons around atleast one of the atoms are called hypervalent |
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| 557. |
Which one of the following is not a hypervalent compound?A. `PF_5`B. `SF_4`C. `SiCl_4`D. `IF_7` |
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Answer» Correct Answer - C When an atom has a share of more than eight electrons, as does P in `PF_5`, we say that it exhibits an expanded valence shell or octet. This is sometimes referred to as hypervalence. In `SF_4`, sulphur (with 10 valence `e^(-)s`) has an expanded valence shell. In `IF_7`, iodine (with 14 valence `e^(-)s`) has an expanded valence shell. Thus, `PR_5`, `SF_4`, and `IF_7` are examples of superoctet molecules. In `SiCl_4`, silicon (with 8 valence `e^(-)s`) has a complete octet. Thus, it is not a hypervalent compound. |
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| 558. |
Hypervalent compound is :A. `A =B gt C gt D`B. `A= B gt B gt D gtC`C. `a =B =C =D`D. ` C gt D gt A gt B` |
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Answer» Correct Answer - A Hypervalent compounds have expanded octet . |
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| 559. |
A solid element has a specific heat of `1 Jg^(-1) K^(-1)`. The atomic mass of the element isA. 9B. 18C. 27D. 36 |
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Answer» Correct Answer - C We know Atomic mass= `(6.4)/("sp. Heat (in calories)")` Also 4.184 J = `~~` 1 calorie 1J=0.239 cal `~~` 0.24 cal `g^(-1) k^(-1)` Atomic mass =6.4/0.24 =26.77 `~~`27.0 |
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| 560. |
From among the following triatomic species the least angle around the central atom is inA. `O_3`B. `I_3^-`C. `NO_2^-`D. `H_2S` |
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Answer» Correct Answer - D In `H_2S` the bond lt H-S-H is close to `90^@` |
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| 561. |
A compound possesses 8% sulphur by mass. The least molecular mass isA. 32B. 64C. 128D. 400 |
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Answer» Correct Answer - D Since sulphur is 8% by mass `therefore` 8g sulphur is present in mass of 100 g or 32 g sulphur will be present in mass of =`100/8xx32=400` Thus, if a minimum of one sulphur atom is present in the molecule, its minimum mass should be 400 amu. |
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| 562. |
Which of the following weighs the least?A. 14 g atoms of nitrogenB. 0.6 moles of `S_8`C. 4 g equivalents of sulphate ionsD. `1.2xx10^24` chloride ions. |
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Answer» Correct Answer - D `1.2xx10^24 Cl^-` ions weigh `=(1.2xx10^24)/(6.02xx10^23)xx35.5=71.0 g` |
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| 563. |
Which of the following types of hybridisation leads to two dimensional geometry of bonds around the atoms ?A. spB. `sp^2`C. `sp^3`D. none of the above. |
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Answer» Correct Answer - B `sp^2` hybrid structures have trigonal planar geometry, which is two dimensional. |
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| 564. |
Which of the following does not have a co-ordinate bond ?A. `H_3O^(+)`B. `PCl_5`C. `O_3`D. `HNO_3` |
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Answer» Correct Answer - B In `PCl_5, 5Cl` atoms are bonded to P by covalent bonds and there is no co-ordinate bond. |
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| 565. |
The correct order of bond order values among the following (i) `NO^(-)` (ii) `NO^(+)` (iii) `NO` (iv) `NO^(2+)` (v) `NO^(2-)`A. (i) lt (iv) lt (iii) lt (ii) lt (v)B. (iv) = (ii) lt (i) lt (v) lt (iii)C. (v) lt (i) lt (iv) = (iii) lt (ii)D. (ii) lt (iii) lt (iv) lt (i) lt (v) |
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Answer» Correct Answer - C `NO(7+8=15e^(-)s)`. Its electron configuration is `KK(sigma2s)^2(sigma^**2s)^2(sigma2p_z)^2(pi2p_x^2=pi2p_y^2)(pi^**2p_x^1=pi^**2p_y^0)` `NO: bo=(10-5)/(2)=2.5` In `NO^(-)`, there is one more `e^(-)` which enters `pi^**2p_y` orbital, thus, `NO^(-): bo=(10-6)/(2)=2` In `NO^(2-)`, there are 2 more `e^(-)s` which enter into antibonding orbitals, thus, `NO^(2-): bo=(10-7)/(2)=1.5` In `NO^(+)`, one `e^(-)` is removed from antibonding orbital: thus, `NO^(+): bo =(10-4)/(2)=3` In `NO^(2+)` one `e^(-)` is lost from bonding `MO` while another `e^(-)` is lost from antibonding `MO`. Thus, `NO^(2+): bo=(9-4)/(2)=2.5` Thus, the correct order of bond order is `NO^(2-) lt NO^(-) lt NO = NO^(2+) lt NO^(+)` i.e., `(v) lt (i) lt (iii) = (iv) lt (ii)` |
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| 566. |
Which has regular terrahedral gementory ?A. `SF_4`B. `BF_4^-`C. `XeF_4`D. `[Ni(CN)_4]^(2-)` |
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Answer» Correct Answer - B B has ` sp^2` -hybridised . |
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| 567. |
Which has the highest dipole moment?A. B. C. D. |
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Answer» Correct Answer - A Because of tras-configuration, compounds (3) and (4) possess net zero dipole moment. Compounds (1) and (2) possesses net nonzero dipole moment. However, compound (1) has higher dipole moment on account of highly polar carbonyl group. |
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| 568. |
The bond lengths and bond angles in the molecules of methane, ammonia, and water are given below: This variation in bond angle is a result of (i) the increasing repulsion between H atoms as the bond length decreases (ii) the number of nonbonding electron pairs in the molecule (iii) a nonbonding electron pair having a greater repulsive force than a bonding electron pairA. (i), (ii), and (iii) are correctB. (i) and (ii) are correctC. (ii) and (iii) are correctD. only (i) is correct |
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Answer» Correct Answer - C Option (i) is wrong because the smaller the bond length, the closer the H atoms are, the greater the repulsion between them and, hence, the greater the bond angle. C atom has no lone pair `(109.5^@)`. N atom has one lone pair `(107^@)`, and O atom has two lone pairs `(104.5^@)`. The more the number of lone pairs, the greater the repulsion on bond pairs and, thus, the smaller the bond angle is. Hence, options (ii) and (iii) are correct. |
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| 569. |
The pair of species having identical shape of both species :A. `BF_3, PCl_3`B. `PF_5, IF_5`C. `CF_4, SF_4`D. `XeF_2, CO_2` |
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Answer» Correct Answer - D `XeF_2 (sp^3d) ` and ` CO_2` (sp) are linear molecules . |
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| 570. |
The states of hybridisation of borom and oxygen atoms in boric acid `(H_3 BO_3)` are respecitivelty :A. `sp^3 and sp^3`B. `sp^(2)` and `sp^(3)`C. `sp^3 and sp^2`D. `sp^2 and sp^2` |
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Answer» Correct Answer - B B has ` sp^2` and oxige has ` sp^2` -hybridisation . |
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| 571. |
The hybridization of oxygen atom in `H_2O_2` isA. `sp^3d`B. `sp`C. `sp^2`D. `sp^3` |
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Answer» Correct Answer - D Lewis structure of `H_2O` is `H-underset(..)overset(..)O-underset(..)overset(..)O-H` `SN` of each O atom is `2+2=4`. Thus, the hybridization of each O atom in `H_2O` is `sp^3`. |
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| 572. |
Which one of the following pairs consists of only paramagnetic speciesA. `O_2, NO`B. `O_2^(+)`,`O_2^(2-)`C. `CO, NO`D. `NO, NO^(+)` |
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Answer» Correct Answer - A `O_2^(2-)(18e^(-)s)`, `CO(14 e^(-)s)`, and `NO^(+)(14e^(-)s)` are diamagnetic while `O_2(16e^(-)s)`, `NO(15e^(-)s)`, `O_2^(+)(15e^(-)s)`, and `N_2^(-)(13e^(-)s)` are paramagnetic. Note that `O_2` is unusual as it is an even electron molecule but has `2` unpaired electrons. |
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| 573. |
The ion which is not tetrahedral in shape isA. `BF_4^-`B. `NH_4^+`C. `Cu(NH_3)_4^(2+)`D. `NiCl_4^-` |
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Answer» Correct Answer - C `[Cu(NH_3)_4]^(2+)` is square planar in shape with `Cu^(2+) d sp^2`-hybridised |
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| 574. |
Which of the following compound can form hydrogen bondsA. `CH_4`B. NaClC. `CHCl_(3)`D. `H_2O` |
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Answer» Correct Answer - D `H_(2)O` can form hydrogen bonds rest `CH_(4)` and `CHCl_(3)` are organic compound having no oxygen while NaCl has itself intraionic attraction in the molecule. |
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| 575. |
Which of the following hydrogen bonds are strogent in vapour phaseA. HF---HFB. HF---HClC. HCl---HClD. HF---HI |
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Answer» Correct Answer - A A compound having maximum electronegativity element will form strong hydrogen bond. |
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| 576. |
Which one of the following molecules contains no `pi` - bond ?A. `NO_(2)`B. `CO_(2)`C. `H_(2)O`D. `SO_(2)` |
| Answer» Correct Answer - C | |
| 577. |
In which one of the following molecules , the central atom said to adopt `sp^2` hybridisation ?A. `BeF_(2)`B. `BCl_(3)`C. `C_(2)H_(2)`D. `NH_(3)` |
| Answer» Correct Answer - B | |
| 578. |
Assertion : Bond order can assume any value number including zero. Reason :Higher the bond order ,shorter is bond length and greater is bond energy.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 |
| Answer» Correct Answer - B | |
| 579. |
The correct order of increasig `C-O` bond length of `CO, CO_(3)^(2-), CO_(2)` isA. `CO_(3)^(2-)ltCO_(2)ltCO`B. `CO_(2)ltCO_(3)^(2-)ltCO`C. `COltCO_(3)^(2-)ltCO_(2)`D. `COltCO_(2)ltCO_(3)^(2-)` |
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Answer» Correct Answer - d |
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| 580. |
The acid having O - O bond isA. `H_(2)S_(2)O_(3)`B. `H_(2)S_(2)O_(6)`C. `H_(2)S_(2)O_(8)`D. `H_(2)S_(4)O_(6)` |
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Answer» Correct Answer - C `H-O-overset(O)overset(||)underset(O)underset(||)(S)-O-O-overset(O)overset(||)underset(O)underset(||)(S)-O-H`(Marshall acid) |
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| 581. |
According to `MOT` whch of the following statement about magnetic character and bond order is corrent regarding `O_(2)^(o+)` .A. Paramagnetic and bond order `lt O_(2)`B. Paramagnetic and Bond order `gt O_(2)`C. Diamagnetic and Bond order `lt O_(2)`D. Diamagnetic and Bond order `gt O_(2)` |
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Answer» Correct Answer - B `O_(2): sigma1s^(2),sigma^(**)1s^(2),sigma2s^(2),sigma^(**)2s^(2),sigma2p_(x)^(2),{{:(pi 2p_(y)^(2)),(pi 2p_(z)^(2)):}{{:(pi^(**)2p_(y)^(1)),(pi^(**)2p_(z)^(1)):}` Bond order =`(10-6)/2` ltBrgt (Two unpaired electrons in antibonding molecular orbital) `O_(2)^(+): sigma1s^(2),sigma^(**)1s^(2),sigma2s^(2),sigma^(**)2s^(2),sigma2p_(x)^(2),{{:(pi 2p_(y)^(2)),(pi 2p_(z)^(2)):}{{:(pi^(**)2p_(y)^(1)),(pi^(**)2p_(z)^(0)):}` Bond order `=(10-5)/2=2.5` (one unpaired electron in antibonding molecular orbital) hence `O_(2)` as well as `O_(2)^(+)` is paramagnetic , and bond order of `O_(2)^(+)` is greater than that of `O_(2)`. |
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| 582. |
Which one of them is the weakest ?A. ionic bondB. covalent bondC. metallic bondD. van der Waals forces |
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Answer» Correct Answer - D Bond order of `He_2` is zero. |
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| 583. |
The `IE_(1)` of `Li` is `5.4 eV` and `IE_(1)` of `H` is `13.6 eV`. Calculate the charge acting on the outermost electron of `Li` atom. |
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Answer» For `Li`. electronic configureation si ` 1s^2 , 2s^2 ,` so given ionisation energy value is for `(n=2)`. We know that : energy of `n^(th)` level `E_(eff)^(2)xxE_(1)` `[( "Here" E_(1)="Energy of first orbit of" "H-atom"),(Z_(eff)="Effective nuclear cahrge")]` or ` Z_(eff) = n sqrt(D_2/E_1)` (Given: `E_1 =- 13. 6eV ,E_2 =- 5. 39eV ` and ` (n=2)` ` Z_(eff) = 2 xx sqrt((5. 39))/(13.6) = 1..26` Thus effective nuclear charge si ` 1.26eV1` because `2s`. |
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| 584. |
What type of hybridization is involved in `XeF_2`?A. `sp^3d`B. `dsp^3`C. `sp^3d^2`D. `d^3sp^3` |
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Answer» Correct Answer - A The ground state valence shell electron configuration of `Xe` is `5s^2 5p^6`. Out of eight valence electrons, two are used to bond two univalent F atoms and the rest 6 valence `e^(-)s` form 3 lone pairs: `F-underset(..)X overset(*)(e)-F` Thus, the steric number of the central `Xe` atom is `2+3=5`. This means five hybrid orbitals are required for bonding in `Xe`. This is possible only through `sp^3d` hybridization. `dsp^3` is excluded because `Xe` is not a d-block element. |
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| 585. |
`XeF_2` is isostructural withA. `ICl_2^(-)`B. `SbCl_3`C. `BaCl_2`D. `TeF_2` |
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Answer» Correct Answer - A The steric number `(SN)` of `XeF_2` is `2+3=5`, thus, it is linear. `SN` of `ICI_2^(-)` is `2+3=5`, thus, it is linear. SN of `SbCl_3` is `3+1=4`, thus, it is trigonal pyramidal. `SN` of `TeF_2` is `2+2=4`, thus, it is angular or bent-shaped. Conclusion is that `XeF_2` (linear) is isostructural with `ICl_2^(-)` (linear). |
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| 586. |
Which of the following compounds possesses zero diple moment ?A. `H_(2) O`B. `C_(6) H_(6)`C. `C Cl_(6)`D. `BF_(3)` |
| Answer» Correct Answer - B::C::D | |
| 587. |
Ionization energy is influenced by :A. size of atomB. charge on the nucleusC. electrons present in inner shellsD. None of these |
| Answer» Correct Answer - B::C::D | |
| 588. |
Which of the following is paramagnetic?A. `O_2^(-)`B. `CN^(-)`C. `NO^(+)`D. `CO` |
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Answer» Correct Answer - A We work out the total number of electrons. If it is an odd number, then the molecule/ion is paramagnetic as all the electrons will not be paired. `O_2^(-)(8+8+1=17)`, `CN^(-)(6+8=14)`, `NO^(+)(7+8-1=14)`, and `CO(6+8=14)`. Therefore, `O_2^(-)` is paramagnetic due to the presence of one unpaired electron, It is also justified by its MO configuration: `KK(sigma1s)^2(sigma^**1s)(sigma2s)^2(sigma^**2s)^2(sigma2p_z)^2(pi2p_x^2=pi2p_y^2)(pi^**2p_x^2=pi^**2p_y^1)`. It has one unpaired `e^(-)` in antibonding `MO(pi^**2p_y)`. |
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| 589. |
Which one of the following molecules contains no `pi` - bond ?A. `H_2O`B. `SO_2`C. `NO_2`D. `CO_2` |
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Answer» Correct Answer - A Only in `H_2O` molecule, the atoms are held together by means of single covalent bonds. Thus, there is no `pi` bond: `H-O-H` `O=S=O` `O=N-O` `O=C=O` |
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| 590. |
The type of bond `s` present in ammonium chloride is (are) :A. `ionic `B. covalentC. co-ordinateD. None of these |
| Answer» Correct Answer - A::B::C | |
| 591. |
Which statement `s` is `are` true ?A. ` PF_3` has higher bond angle than ` PCl_3`B. Dipole moment of `NH_3` is more than `NF_3`C. `I^+` is smaller than `I^-` ionD. `I^-` is smaller than `I^+` ion |
| Answer» Correct Answer - A::B::C | |
| 592. |
Which of the following is a polar moleule ?A. `SF_4`B. `SiF_4`C. `XeF_4`D. `BF_3` |
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Answer» Correct Answer - A Only `SF_4` has an unsymmetrical shape-seesaw. Therefore, it has a net dipole moment. While the rest of the molecules have symmetrical shapes and, hence, zero net dipole moment: `SiF_4` (tetrahedral), `XeF_4` (square planar), and `BF_3` (trigonal planar). |
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| 593. |
The dercreasing valuers of bond angles from `NH_3 (106^@)` to ` SbH_3 ( 101^@)` down the group `15` of the peridic table is due to :A. increase in bp-bp repulsionB. increase in p-orbital character in `sp^3`C. decrease in-pp repulsionD. decrease in electrongativity |
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Answer» Correct Answer - D It is a reason for given fact . |
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| 594. |
The correct order of N-O bond lengths in NO, `NO_2^- , NO_3^-` and `N_2O_4` isA. `N_2O_4 gt NO_2^(-) gt NO_3^(-) gt NO`B. `NO gt NO_3^(-) gt N_2O_4 gt NO_2^-`C. `NO_3^(-) gt NO_2^(-) gt N_2O_4 gt NO`D. `NO gt N_2O_4 gt NO_2^(-) gt NO_3^(-)` |
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Answer» Correct Answer - C As the bond order increases, bond length decreases and bond order is highest for NO, i.e. 2.5 and least for `NO_3^-` , i.e. 1.33. So, the order of bond length is `underset"1.33"(NO_3^-) gt underset"1.5"(NO_2^-)gt underset"1.5"(N_2O_4)gt underset"2.5"(NO)` |
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| 595. |
Heterolytic bond fission in `C_2H_6` gives carbonium and carbonion ions . The hybridisation of carbon atoms in these ions is :A. `sp^3`B. `sp^2`C. `sp`D. `sp^3 , sp^2` |
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Answer» Correct Answer - D `C_(2)H_(6) underset("Bond fission")overset("Heterolyte")(rarr) underset(sp^(2))(CH_(3)^(+))+underset(sp^(3))(CH_(3)^(-))` |
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| 596. |
The limiting line Balmer series will have a frequency ofA. `3.29xx10^(15) s^(-1)`B. `-3.65xx10^(14) s^(-1)`C. `8.22xx10^(14) s^(-1)`D. `-8.22xx10^(14) s^(-1)` |
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Answer» Correct Answer - C The frequency of a line is given by equation `v=3.29xx10^15[1/n_1^2-1/n_2^2]` For limiting line, `n_1=2 and n_2=oo` for Balmer series. |
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| 597. |
Although the electron gain enthalpy of `N(2s^2 2p63)` is positive, the `N^(3-)` ion `(2s^2 2p^6)` is stable in the presence of certain positive ions such asA. `Na`B. `Li`C. `K`D. `Rb` |
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Answer» Correct Answer - B `Li^(+)`, on account of high charge density leads to the formation of lithium nitide, `Li_3N`. Allkaline earth metals also form nitrides on account of high lattice enthalpy. |
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| 598. |
An equimolar mixture of Nitrogen gas and water vapours is taken in a 2 litre flask at `27^(@)`C and `1.23 xx 10^(-2)` atm. pressure. What is the mass of the gas at `-27^(@)`C?A. `2.8xx10^(-4)g`B. `5.2xx10^(-3) g`C. `1.4xx10^(-2) g`D. `0.07 g` |
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Answer» Correct Answer - C The number of moles of gaseous mixture, `n=(PV)/(RT)=(1.23xx10^(-2) X^2)/(0.082xx300)` `=1 xx10^(-3)` moles As the mixture is equimolar. `therefore .^nN_2=0.5xx10^(-3)` and `.^nN_2=0.5xx10^(-3)` moles At `-27^@C` water vapours changes to water solid `therefore` mass of gas shall be only due to `N_2` gas `=28xx5xx10^(-4)=1.4xx10^(-2) g` |
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| 599. |
The structure of orthophosphoric acid isA. `H-O-overset(O)overset(uarr)underset(H)underset(|)underset(O)underset(|)(P)-O-H`B. `Olarroverset(H)overset(uarr)underset(H)underset(|)underset(O)underset(|)(P)-O-H`C. `Olarr overset(H)overset(|)underset(H)underset(|)(P)-O-H`D. `H-O-overset(O)overset(uarr)(P)=O` |
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Answer» Correct Answer - A `H_(3)PO_(4)` is orthophosphoric acid. `H-O-overset(O)overset(uarr)underset(H)underset(|)underset(O)underset(|)(P)-O-H` |
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| 600. |
Which is the weakest among the following types of bondsA. ionicB. covalentC. metallicD. H-bond |
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Answer» Correct Answer - D H-bond is weakest bond because its bond dissociation energy is very low as compared to other given bonds `(10 kJ mol^(-1))`. |
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