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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. |
For the formation of covalent bond, the different in the value of electronegativities should beA. Equal to or less than `1.7`B. More than `1.7`C. `1.7` or moreD. None of these |
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Answer» Correct Answer - a Covalent bond from when electronegativity difference of two atom is equal to `1.7` or less than `1.7`. |
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| 602. |
The geometrical arrangement and shape of `I_3^-` are respectivelyA. Trigonal bipyramidal geometry, linear shapeB. Hexagonal structure, linear shapeC. Triangular planar geometry, triangular shapeD. Tetrahedral geometry, pyramidal shape |
| Answer» Correct Answer - B | |
| 603. |
The shape of `[CH_(3)]^(o+)` is_____ . |
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Answer» Correct Answer - Planar The shape of `[CH_(3)]^(o+)` is triangular planar `CH_(3)^(o+),sp^(2)` carbon In carbocations, carbon atom is `sp^(3)` hybridisation and they have a triangular planar shape . |
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| 604. |
Assertion .`N_(3)^(-)` is a weakar base than ` NH_(2)^(-)` . ltbr. Reason . The lone pair of electrons on N atom in `N_(3)^(-)` is in the `sp^(2)`-orbital while in `NH_(2)^(-)`, it is in an ` sp^(3)`- orbital.A. If both assertion and reason are correct, and reason is the correct explanation of the assertion.B. If both assertion and reason are correct , but reason is not the correct explanation of the assertion.C. If assertion is correct, but reason is incorrect .D. If both assertion and reason are incorrect. |
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Answer» Correct Answer - a Basic strength depends upon the availability of lone pairs of electrons and lone pair of electrons is more easily available in `NH_(2)^(-) ` than in ` N_(3)^(-)` as it is present in `sp^(3)` in `NH_(2)^(-) and sp^(2) ` in `N_(3)^(-)` . Hence , R is the correct explanation of A. |
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| 605. |
The bond angle between two hybrid orbitals is `180^(@)` The percentage s-character of hybrid orbital is between .A. `50` and `55%`B. `9` and `12%`C. `22` and `23%`D. `11` and `12%` |
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Answer» Correct Answer - A The bond angle `180^(@)` means sp hybridisation which has `50%` s-character Alternatively refer to Section 2.22 (point3) `cos0 = ((s%)/(s%-100)) implies cos 180^(@) = ((s%)/(s% -100))` `implies - 1 = (s%)/(s%-100) implies :. S% = 50` . |
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| 606. |
The bond angle between two hybrid orgials is ` 105^(@)` . Calculate the percentege of s-character of the hybrid orbital .Given that cos ` 75^(@) = 0 .2588`. |
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Answer» ` cos alpha = - (1)/(m) ` where m = hybridisation index ` therefore - (1)/(m) cos 105^(@) = cos (180 - 75 ) = - cos ^(@) = - 0.2588 or m = (1)/(0.2588) = 3.86` % s-character ` = (1)/(1+m) xx10- = (1)/(1+3.86) xx100 = 20.58%` Alternatively , s-character decreases as the bond angle decreases. For example , `{:(" Hybrid orbital" " "sp^(3) " " sp^(2) " "sp),("s-character " " "25% " "33.3% " "50%),(" Bond angle "" " 109.5^(@) " " 120^(@) " "180^(@)):}` Thus, when bond angle decreases below ` 109.5^(2)`, the s-character will dcrease accordingly. Although the decrease is not linear , even then approsimate value can be calculated as follows : Decrease in angle = `120 - 109.5^(@) = 10.5^(@)` Decrease in s-character =` 33.3 - 25 = 83 ` Actual decrease in bond angle =` 109.5^(@) - 105^(@) = 4.5^(@)` ` therefore ` Expected decrease in s-character `= (8.3)/(10.5) xx4.5 = 3.56%` Thus, s-character should decrease by about 3.56% Hence, s-character = ` 25 - 3.56 = 21 .44%` |
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| 607. |
Which of the following is not a correct statement ? .A. Ionic compounds are electrically netural.B. Boilling point of an ionic compound is more than a covalent compound .C. Melting point of a covalent compound is more than an ionic compoundD. Ionic compound are soluble in polar solvent . |
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Answer» Correct Answer - C Due to electrostatic force of attraction the melting points and boiling points of ionic compounds are high . |
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| 608. |
Write the electronic configuration of `O_(2)` molecule in the singlet state. |
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Answer» Ordinary ` O_(2)` moleucle has the electronic configuration `sigma_(1s)^(2) sigma_(1s)^(**2) sigma_(2s)^(2) sigma_(2p_(z))^(**2) .pi_(2P_(x))^(2)pi_(2P_(y))^(2)pi_(2P_(x))^(**1)pi_(2P_(y))^(**2)` Multiplicity = (2S + 1 ) Where S = total electron spin ` = 2 ((1)/(2) + (1)/(2)) + 1 = 3 ` " "` [ S = (1)/(2) + (1)/(2) ` because two electons have parallel spins] Hence , it is in the triplet state if two electrons pair up ` S = (1)/(2) - (1)/(2) = 0 ` (because they will have opposite spin) Spin multiplicity = `2S + 1 = 0 + 1 = 1 ` This is called singlet state . Hence, electronic configuration of `O_(2)` molecule in the single state will be `sigma_(1s)^(2) sigma_(1s)^(**2) 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))^(**0)` |
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| 609. |
The formula of a compound is `A_(2)B_(5).` The number of electrons in the outermst shells of A and B respectively are:A. 6 and 3B. 5 and 6C. 5 and 2D. 2 and 3 |
| Answer» Correct Answer - B | |
| 610. |
An atom of an element A has three electron in its outer shell and `B` has six electron in outermost shell. The formula of the compound formed between these two will beA. `A_(6)B_(6)`B. `A_(2)B_(3)`C. `A_(3)B_(2)`D. `A_(2)B` |
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Answer» Correct Answer - b In this case the valence electron in the atom `A` is three and hence its generally `3`. In the atom `B` the number of valence electron is six. Hence its valency is usually `2` Hence the formula of the molecule formed from `A and B` could br `A_(2)B_(2)`. An example of two such elements are `AI` and `O` and the formula of Aluminium oxide is `AI_(2)O_(3)`. |
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| 611. |
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. `A_(2)B_(3)`B. `A_(2)B_(6)`C. `A_(2)B`D. `A_(3)B_(2)` |
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Answer» Correct Answer - A `[A:]_(2) [:B]_(3)` |
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| 612. |
Four elements A, B, C and D form a series of compounds having the formulae AB, ` B_(2), CB_(3), DB_(2)` and `DB_(3)`. If the jumbled up atomic numbers of A, B, C and D are 13, 19 , 26 and 35 , What are the ordered atomic numbers of A, B C and C ? |
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Answer» ` 13 = Al = 1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p_(x)^(1)` (can lose `1 e ^(-))` ` 19 = K = 1ss^(2) 2s^(2) 2p^(6) 3s^(2) 3p^(6) 4s^(1) (can lose `1e ^(-))` 26 =` Fe = 1s^(2)2s^(2) 2p^(6) 3s^(2) 3p^(6)3d^(6) 4s^(2) ` (can lose ` 2e^(-) or 3 r^(-))` 35 ` = Br = 2s^(2) 2s^(2) 3p^(6) 3s^(2)3p^(6) 3d^(10)4s^(2) 4p_(x)^(2) 4p_(y)^(2) 4p_(z)^(1) ` (can gain or share ` 1 e ^(-)` ) Compunds formed will be ` KBr , Br_(2), AlBr_(3), Fe Br_(2) and FeBr_(3).` Hence ,` A = K = 19 , B= Br = 35 , C = Al = 13 , D = Fe = 26.` |
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| 613. |
The pair of elements which form ionic bond is .A. `C +CI`B. `H +F`C. `Na + Br`D. `O +H` |
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Answer» Correct Answer - C Sodim `IE` is less Br electron affinity is more The type of bond formed is ionic bond . |
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| 614. |
(a) In a polar solvenbt ,` PCl_(5)` undergoes an ionization reaction as follows : ` 2 PCl_(5) hArrPCl_(4)^(+) + PCl_(6) ^(-)` Wht will be the geometrical shape of each species present In the equalilbrium maxture ? (b) Why does ` PCl_(5)` exist as ` [ PCl_(4)]^(+) [PCl_(6)]^(-)` ? |
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Answer» `X = (1)/(2) [{{:(" NO of valence electrons "),(" of the central atom "):}}+{{:(" NO of monovalent atoms/group "),(" surrounding the central atom "):}}]` `- {{:("Charge on the cation if the given "),(" species is a polyatomic cation "):}}+{{:(" Charge on the anion if the given "),(" species is a polyatomic anion"):}}` For `PCl_(5), X = (1)/(2) [5 + 5] - 0 + 0 = 5 (sp^(3) d ,`bipyramidal) For `PCl_(4_^(+) , X = (1)/(2) [ 5 + 4 - 1 + 0 ] = 4 (sp^(3), `tetrahedra) For ` PCl_(6) ^(-), X = (1)/(2) [ 5 + 6 - 0 + 1] = 6 (sp^(3) d^(2)` , octahedral) (b) ` PCl_(5)` has different axial and equatorial bohnd lengths and therefore , has an unsymmetrical structure Which is not stable . By dissociation, it changes into tetrahedral ` ([PCl_(4)]^(+)` and octahedral `[PCl_(6)]^(-)` Which are stable . |
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| 615. |
In a polar solvent, `PCl_(5)` undergoes an ionisatin reaction as : `" "2PCl_(5)hArrPCl_(4)^(+)+PCl_(6)^(-)` What are the geometrical shapes of the species involved in the equilibrium mixture ? |
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Answer» The geometry can be predicted on the basis of hydrisation (H) which may be calculated as: `H=1/2` [No. of electrons in the valence shell of the atom + No. of monovalent atoms surrounding the atom- charge on cation + charged on anion] `PCl_(5) :" "H=1/2[5+5-0+0]=10/2=5(sp^(3)d)` The geometry of the molecule is trigonal bi pyramidal. `PCl_(4)^(+) :" "H=1/2[5+4-1+0]=8/2=4(sp^(3))` The geometry of the ion is tetrahedral. `PCl_(6)^(-) :" "H=1/2[5+6-0+1]=12/2=6(sp^(3)d^(2))` The geometry of the ion is octahedral. |
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| 616. |
In the electronic structure of `H_2SO_4`, the total number of unshared electrons isA. 20B. 16C. 12D. 8 |
| Answer» Correct Answer - A | |
| 617. |
Silicon carbide is aA. Molecular solidB. Covalent solidC. Ionic solidD. None of the above |
| Answer» Correct Answer - B | |
| 618. |
Assertion `H_(2)O` has maximum density at `4^(@)C` Hence in water ice will sink to the bottom at `4^(@)C` Reasoning Up to `4^(@)C` more and more hydrogen bonds are formed between `H_(2)O` molecules .A. If both `(A)` and `(R )` are correct and `(R )` is the correct explanation of `(A)`B. If both `(A)` and `(R )` are correct and `(R )` is the correct explanation of `(A)`C. If `(A)` is correct but `(R )` is incorrectD. If `(A)` is incorrect but `(R )` is correct |
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Answer» Correct Answer - D Due to H-bonding ice forms cage like structure Thus ice occupies more volume add `4^(@)C` molecules come closer Hence density of water is highest at `4^(@)C` Therefore ice floats over water . |
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| 619. |
`sigma2s, pi^(**) (2p_(x))` and `pi(2p_(x))` are gerade `MO` . |
| Answer» Correct Answer - F | |
| 620. |
The bond order in `NO` is `2 . 5`, while that in`NO^+` is 3 Which statement is true ?A. Bond length is unpredictableB. Bond length in `NO` is greater than in `NO^(+)`C. Bond length in `NO^(+)` is equal to than in `NO`D. Bond length in `NO^(+)` is greater than in `NO` |
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Answer» Correct Answer - b Higher is the bond order, lesser is bond length. |
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| 621. |
Which of the following is insolube in waterA. AgFB. AgIC. `KBr`D. `CaCI_(2)` |
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Answer» Correct Answer - b Solubility depends on relative hydration and lattice energies `AgI` ha slower `Delta H_(hydration)` compared to lattic energy |
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| 622. |
In which of the following molecules , the type of hybridization changes whenA. `NH_(3)` cambines with `H^(+)`B. `AIH_(3)` cambines `H^(-)`C. in both casesD. in none of the above |
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Answer» Correct Answer - b In `NH_(3)`, the hybridisation of `N is sp^(3)`. In `NH_(4)^(+)` , the hybridisation remains `sp^(3)`. In `AIH_(3)`, the hybridisation of `AI is sp^(2)`. And in `AIH_(4)^(-)`, it becomes `sp^(2)`. |
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| 623. |
The number and type of bonds between two carbon atoms in `CaC_(2)` are:A. One sigma and one pi-bondB. One sigma and one pi-bondsC. One sigma and one -half pi-bondD. One sigma -bond |
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Answer» Correct Answer - B `Ca^(2+) [C-=C]^(2-)` There is a triple bond between two carbon atoms which menans that there is one `sigma` bond and there are two pi-bonds . |
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| 624. |
`KHF_(2)` exists while `KHCl_(2)` does not. Explain. |
| Answer» HF molecules are hydrogen bonded as H-F…H-F…Therefore, they can associate to from `HF_(2)^(-)` anion which can exist in combination with `K^(+)` ion as `KHF_(2),` But no hydrogen bonding is present in H-Cl molecules. Therefore the anion `HCl_(2)^(-)` and `KHCl_(2)` donot exist. | |
| 625. |
The payramidal geometry is associated with :A. `CH_(4)`B. `NH_(3)`C. `H_(2)O`D. `CO_(2)` |
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Answer» Correct Answer - B `NH_(3)` has pyramidal geometry (N is `sp^(2)` hybridised). |
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| 626. |
Which of the following statements are correct about `CO_(3)^(2-)`?A. The hybridisation of central atom is `sp^(3).`B. Its rresonance structure has one C-O single bond and two C=O double bonds.C. The average formal charge on each oxygen atom is 0.67 units.D. All C-O bond lengths are equal. |
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Answer» Correct Answer - C::D Both these statements are correct. |
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| 627. |
The skeletal structure of `CH_(3)COOH` as shown below is correct but some of the bords are wrongly shown. Write the correct Lewis structures of acetic acid. `H-underset(H)underset(|)overset(H)overset(|)C-underset(* *)overset(":"O":")overset(|)C-underset(* *)O-H` |
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Answer» Only the skeletal arrangement of the atoms in the above structure is correct. But it is not according to the lewis concept as well as tetravalent nature of carbon. The correct structure of acetic acid is `H-underset(H)underset(|)overset(H)overset(|)C-overset(":"O":")overset(|)C-underset(* *)overset(* *)O-H` |
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| 628. |
In which one of the following species the central atom has the type of hybridisation which is not the same as that present in the other three ?A. `PCl_(5)`B. `SF_(4)`C. `I_(3)^(-)`D. `SbCl_(5)^(2-)` |
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Answer» Correct Answer - D `PCl_(5) = sp^(3)d , SF_(4) = sp^(3) d, I_(3)^(-) = sp^(3) d`. `SbCl_(5)^(2-) = sp^(3) d^(2)` Thus option (d) is correct. |
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| 629. |
The skeletal structure of `CH_(3)COOH` as shown below is correct, but some of the bonds are shown incorrectly. Write the correct Lewis structure for acetic acid. |
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Answer» The correct Lewis structure for acetic acid is as follows: `H=underset(H)underset(|)overset(H)overset(|)C-overset( :O: )overset(||)C-underset(..)overset(..)O-H` |
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| 630. |
The skeletal structure of `CH_(3) COOH` as shown below is correct , but some of the bonds are shown incorrectly . Write the correct Lewis structure for a acetic acid. `H= overset(H)overset(|)underset(H) underset(|)C- overset(":O:")overset(|) underset(cdot cdot)C- underset(cdot cdot)O - H` |
| Answer» `H- overset(H)overset(|)underset(H) underset(|)C- overset(":O:")overset(||) C- underset(cdot cdot)overset(cdot cdot)O - H` | |
| 631. |
The incorrectly metched pair among the following is :A. `{:(" Molecule Shape "),(BrF_(5)" Trigonal bipyramidal "):}`B. `{:(" Molecule Shape "),(SF_(4)" See saw "):}`C. `{:(" Molecule Shape "),(CIF_(3)" T-shape"):}`D. `{:(" Molecule Shape "),(NH_(4)^(+) " Tetrahedral "):}` |
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Answer» Correct Answer - A `BrF_(5)` has square pyramidal shape. |
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| 632. |
In `PO_(4)^(3-)` ion, the formal charge on the oxygen atom of P-O bond isA. `+1`B. -1C. `-0.75`D. `+ 0.75` |
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Answer» Correct Answer - C Formal charge on each O-atoms = `("Total charge ")/("No. of O-atom")=(-3)/(4) = - 0.75`. |
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| 633. |
In `PO_(4)^(3-)` ion the formal charge on the oxygen atom of P-O bond isA. `+1`B. `-1`C. `-0.75`D. `+0.75` |
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Answer» Correct Answer - C c) In `PO_(4)^(3-)` ion, formed charge on each O-atom of P-O bond `=("total charge")/("Noumber of O-atom") = -3/4=-0.75` |
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| 634. |
Assertion LiF is parctically insoluble in water Reasoning LiF has very high lattice energy .A. If both `(A)` and `(R )` are correct and `(R )` is the correct explanation of `(A)`B. If both `(A)` and `(R )` are correct and `(R )` is the correct explanation of `(A)`C. If `(A)` is correct but `(R )` is incorrectD. If `(A)` is incorrect but `(R )` is correct |
| Answer» Correct Answer - A | |
| 635. |
Match the items given in Column I with examples given in Column II. `{:(" Column I Column II " ),((i) "Hydrogen bond (a)C" ),((ii)"Resonance (b) LiF "),((iii)"Ionic solid (c)"H_(2)),((iv)"Covalnet solid (d)HF"),(" "(e)O_(3)):}` |
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Answer» Correct Answer - (i)-(d),(ii)-(e),(iii)-(b),(iv)-(a) Hydrogen bond `to` HF, Resonance `to O_(3)`, Ionic bond `to` LiF, Covalent solid `to` C |
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| 636. |
Match the items given in column i with example given in Column II |
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Answer» (A-4), (B-5), (C-2), (D-1) A) Hydrogen bond `to` HF B) Resonance `to O_(3)` C) Ionic bond `to`Lif D) Covalent solid `to` C |
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| 637. |
For a stable molecule the value of bond order should beA. negativeB. positiveC. zeroD. no relationship of stability and bond order. |
| Answer» Correct Answer - B | |
| 638. |
The difference of energy between the actual structure of molecule and that of the most stable contributing structure is called…………… |
| Answer» Correct Answer - resonance energy | |
| 639. |
Statement-1 : `AlF_(3)` is a high melting point solid whereas `SiF_(4)` is a gas and Statement-2 : Both `AlF_(3)` and `SiF_(4)` are covalent molecules .A. Statement-1 is True , Statement-2 is True , Statement-2 is a correct explanation for Statement-3B. Statement-1 is True , Statement-2 is True , Statement-2 is NOT a correct explanation for Statement-3C. Statement-1 is True , Statement-2 is FalseD. Statement-1 is False , Statement-2 is True |
| Answer» Correct Answer - C | |
| 640. |
Why `AlF_(3)` is a high melting solid wheareas `SIF_(4)` is a gas ? |
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Answer» `AIF_(3)` is an ionic solid due to large differnce in electronegtivies of Al and F whereas `SiF_(4)` is a covalent compound and hence there are only weak van der Walls forces among their molecules . |
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| 641. |
The dipole moment of `H_(2) O ` molecule is ………. Whereas that of ` SF_(4)` is ………… |
| Answer» regular octahedral, distorted tetrahedron | |
| 642. |
Which is more polar `-CO_(2) or N_(2) O ` ? Give reason . |
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Answer» `N_(2) O ` is more polar than `CI_(2)` . This is because `CO_(2)` is linear and symmetrical . Its net dipole moment is zero `(Ooverset(larr)=Coverset(rarr)(=O)) . N_(2) O` is linear but unsymmetrical . It is considered as a reasonance hybrid of the following two structrure : `:overset(-)overset(. .)N= overset(+)N= overset(. .)O: harr :N-=overset(+)N-overset(-)overset(. .)underset(. .)O:` It has a net dipole moment of 0.116 D. |
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| 643. |
Name the shapes of the following molecules : `CH_(4) , C_(2)H_(2)m CO_(2)` |
| Answer» `CH_(4) ` = tetrahedral, ` C_(2) H_(2)` = cylindrical, `CO_(2) = linear . | |
| 644. |
Among ` KO_2 , AlO_(2)^(-) BaO_2` and `NO_2^+` unpaired electron is present in :A. `NO_(2)^(+)` and `BaO_(2)`B. `KO_(2)` and `AlO_(2)^(+)`C. `KO_(2)` onlyD. `BaO_(2)` only |
| Answer» Correct Answer - C | |
| 645. |
Which of the following is paramagentic?A. `O_(2)^(-)`B. `CN`C. `CO`D. `NO^(+)` |
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Answer» Correct Answer - a `CN^(-), CO and NO^(+)` are isolectronic with `14` electron each and there is no unpaired electrons in the `MO` configuration of these species. So these are diamagnetic , `O_(2)^(+)` is paramagnetic due to the presence of one unpaired electron. `(CN^(-), CO, CO^(+))14 = sigma 1s^(2), sigma ** 1s^(2), sigma 2s^(2), sigma** 2s^(2), sigma 2p_(x)^(2)` `pi 2p_(z)^(2) = pi 2p_(z)^(2)` (No unpaired electron diamagnetic) `O_(2^(+)) = sigma 1s^(2), sigma ** 1s^(2), sigma 2s^(2), sigma** 2s^(2), sigma 2p_(x)^(2)` ` pi 2p_(x)^(2) = pi 2p_(z)^(2), pi ** 2p_(y)^(2) = pi ** 2p_(z)^(1)` (One unpaired electron , paramagnetic) |
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| 646. |
Among ` KO_2 , AlO_(2)^(-) BaO_2` and `NO_2^+` unpaired electron is present in :A. `NO_(2)^(+) and BaO_(2)`B. `KO_(2) and AIO_(2)^(-)`C. `KO_(2)` onlyD. `BaO_(2)` only |
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Answer» Correct Answer - c `KO_(4)` is the only supperoxide among the given oxides therefore it will be paramagnetic due to the presence of an odd electron. |
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| 647. |
Among `KO_(2), AlO_(2)^(-), BeO_(2) and NO_(2)^(+)`, unpaired electron is present inA. `NO_(2)^(+) and BaO_(2)`B. `KO_(2) and AlO_(2)^(-)`C. `KO_(2)` onlyD. `BaO_(2)` only |
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Answer» Correct Answer - C Electrons present : `NO_(2)^(+) = 7 + 16 - 1 = 22 , BaO_(2) = 56 + 16 = 72 ` `AlO_(2)^(-) = 13 + 16 + 1 = 30 ` and `KO_(2) = 19 + 16 = 35` Thus , olny `KO_(2)` is odd electron species. |
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| 648. |
Basic strength order `NH_(3) gtPH_(3) gtAsH_(3) gtBiH_(3)` . |
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Answer» Correct Answer - T Basic strength decreases down the group 15 hydrides . |
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| 649. |
Arrange the following types of interation in order of decreasing stability (a) Covalent bond (b) van ber Waals force (c ) H-bonding (d) Dipole interation (e) Ionic bond . |
| Answer» Ionic bond gt Covalent gt H-bonding gt Dipole gt van der Waals `(e gt a gt c gt d gt b)` . | |
| 650. |
Which of the following has been arrange in order of detereasing dipole moment?A. `CH_(3)CI gt CH_(3)F gt CH_(3)Br gt CH_(3)I`B. `CH_(3)F gt CH_(3)CI gt CH_(3)Br gt CH_(3)I`C. `CH_(3)CI gt CH_(3)Br gt CH_(3)I gt CH_(3)F`D. `CH_(3)F gt CH_(3)CI gt CH_(3)I gt CH_(3)Br` |
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Answer» Correct Answer - a Dipole moment depends on the electronegativity of the elements as it is the product charge on one of the ions and the distance between them. Dipole moment of `CH_(2)CI` is greater than `CH_(2)F` due to more charge separation (a fact) |
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