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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.
751. |
Given that `""^(r)Ca^(2+) = 114 ` pm and `""^(r)CO_(3)^(2-) = 185` pm , calculate the lattice energy of ` Ca CO_(3)`. |
Answer» Applying Kapustinskii equation `U_(L) = 120250 ((v*|z_(+)|*|z_(-) |)/(d)) (1 - (34.5)/(d))` Where v = No. of ions per formula unit ` Z_(+), Z_(-) = ` charge on the cation and anion respectively `d = r_(+) + r_(-)`, i.e., sum of ionic radii in pm `U_(L)` = lattice energy in kJ `mol^(-1)` Here, ` v = 2 , z_(+) = + 2, z_(-) = - 2, z_(-) = 2 , i.e., |z_(+)| = 2 , |z_(-)| = 2 ` ` d = r_(+) + r_(-) = 144 + 185 = 299 pm` `U_(L) = 120250 ((2xx2xx2)/(299))(1 - (34.5)/(299)) = 2846 " kJ mol "^(-1)`. |
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752. |
Assertion (A) : Though the central atom of both `NH_(3) and H_(2)O ` molecules are `sp^(3)` hybridised , yet H-N-H bond angle is greater than that of H-O-H.A. A and R both are correct, and R is the correct ecplanation of A.B. A and R both are correct, But R is not the correct ecplanation of A .C. A is true but R is false .D. A and R both are false. |
Answer» Correct Answer - A | |
753. |
Both Na and H occur in group 1 of the periodic table ,yet melting point of HCI is `-114^(@)C.` Why ? |
Answer» Through Na and H are present in the group 1, they differ in their electronegativities. The electronegativity of H is 2.1 and that of Na is 0.9 Dut to large electronegativity difference between Na and Cl (3.0-0.9=2.1), NaCl is ionic in nature and exists as a crystalline solid with very high melting point `(800^(@)C).` On the other hand, HCl is polar covale due to small electronegativity difference (3.0-2.1=0.9) and exists in the gaseous state at room temperature. order to sokidify it, a very low temperature `(-114^(@)C)` in needed. | |
754. |
What is an ionic bond ? With two suitable examples, explain the diference between an ionic and a covalent bond ? |
Answer» The two points of difference between ionic and covalent bonds are given : (i) in the ionic bond formation, there is transference of one of more electrons from one atom to the other (e.e. NaCl) while in the covalent bond formation, there is electon sharing `(e.g. C Cl_(4)).` (ii) ionic bond is non-directional while covalent bond has directional nature. |
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755. |
`H_(2)O` is depolar, wheras `BeF_(2)` is not. it becauseA. The electronegativity of `F` is greater than that of OB. `H_(2)O` involves hydrogen bending whereas `BeF_(2)` is a discrete moleculeC. `H_(2)O` is linear and `BeF_(2)` is angularD. `H_(2)O` is angular and `BeF_(2)` is linear |
Answer» Correct Answer - d In a linear symmertical molecule like `BeF_(2)` the bond angle between three atoms is `180^(@)` , hence the polarity due to one bond is cancelled by the equal polairity due to other bond. While it is not so in the case of angular molecules like `H_(2)O`. |
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756. |
Describe the hybridisation in case of `PCl_(2).` Why are the axial bonds longer as compared to equatorial bonds ? |
Answer» The hybridisation in the molecule of `PCl_(5)` because the central atom is the same in both the molecules and the two halogens chlorine (Cl) and fluorine (F) also belong to the same group (group 17). They anexpected to have same hybridisation as well as same geometry. The P atom is `sp^(3)` d hybridised and the molecile has tregonal bipyrramidal geometry. The details of hybridisation of `PF_(5)` have been given in sections 4.14 | |
757. |
Differentiate between `VB` theory and Lewis concept. |
Answer» The valence bond approach for the covalent bond formation takes into account the energy states of the bonding atomic orbitals. However, the Lewis concept is silent about the energy. It is based on the electron sharing in the bonding atoms. | |
758. |
Suppose an element X has Lewis symbol as `overset(.)(X)` . Identify the element X from the following optionsA. NaB. MgC. AlD. Si |
Answer» Correct Answer - A | |
759. |
In `I_(3)^(-)` , Lewis base isA. `I_(2)`B. `I_(2)^(-)`C. `I_(2)^(+)`D. `I^(-)` |
Answer» Correct Answer - D As `I^(-)` ion donates electron pair into empty `sp^(3)d` orbital, hence`I^(-)` ion acts as Lewis base. |
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760. |
Which of the following species has a linear shape ?A. `NO_(2)^(+)`B. `O_(3)`C. `NO_(2)^(-)`D. `SO_(2)` |
Answer» Correct Answer - A `O=underset(+)N=O` is linear . All others are bent molecules/ions. |
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761. |
Write the significance of a plus and a minus sign shown in representing the orbitals . |
Answer» As orbitals are respresented by weve functions, a plus sign in an orbital represents a +ve weve founction and a minus sign represents a -ve weve founction . |
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762. |
Use molecular orbital theory to explain why the `Be_(2)` molecules do not exist? |
Answer» The electronic configuration of Beryllium is `1s^(2)2s^(2)`. The molecular orbital electronic configuration for `Be_(2)` molecule can be written as : `underset (1s)sigma^(2) underset(1s)sigma^(*2) underset(2s)sigma^(2) underset(2s)sigma^(*2)` Hence, the bond order for `Be_(2)` is `(1)/(2)(N_(b)-N_(a))` Where, `N_(b)`= Number of electrons in bonding orbitals `N_(a)`=Number of electrons in anti-bonding orbitals `therefore` Bond order of `Be_(2)=(1)/(2)(4-4)=0` A negative or zero bond order means that the molecules is unstable. Hence, `Be_(2)` molecule does not exist. |
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763. |
You are given the following species `C_(2)^(+), O_(2)^(2+), Be_(2), C_(2), O_(2)^(2-), C_(2)^(-) ` Arrange them in order of increasing bond strength giving reason. |
Answer» `C_(2)^(+)(11)= sigma_(1s)^(2)sigma_(1s)^(**2)sigma_(2s)^(2)sigma_(2s)^(**2) (pi_(2P_(x))^(2)= pi_(2p_(y))^(1))` Bond order = ` (1)/(2) (N_(b)- N_(a)) = (1)/(2) (7-4) = 1.5` `C_(2)^(2+)(14)= sigma_(1s)^(2)sigma_(1s)^(**2)sigma_(2s)^(2)sigma_(2s)^(**2)sigma_(2p_(z))^(2) (pi_(2P_(x))^(2)= pi_(2p_(y))^(1))` Bond orger ` = (1)/(2) (10 - 4) = 3 ` `Be_(2) (8) = sigma_(1s)^(2)sigma_(1s)^(**2)sigma_(2s)^(2)sigma_(2s)^(**2)` Bond order `= (1)/(2) (4-4) = 0 ` `C_(2)(12)= sigma_(1s)^(2)sigma_(1s)^(**2)sigma_(2s)^(2)sigma_(2s)^(**2)sigma_(2p_(z))^(2) (pi_(2P_(x))^(2)= pi_(2p_(y))^(2))` Bond order ` = (1)/(2) (8-4) = 2 ` `O_(2)^(2-) (18) = 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))^(**2))` Bond order ` (1)/(2) (10 - 8 ) = 1 ` `C_(2)^(-)(13)= sigma_(1s)^(2)sigma_(1s)^(**2)sigma_(2s)^(2)sigma_(2s)^(**2) (pi_(2P_(x))^(2)= pi_(2p_(y))^(2))sigma_(2p_(z))^(1)` Bond order ` = (1)/(2) (9 - 4) = 2.5 ` Greater the bond order, greater is the bond strength . Hence, increasing order of bond strength will be `Be_(2) lt O_(2)^(2-) lt C_(2)^(+) lt C_(2) lt C_(2)^(-) lt O_(2)^(2+)` . |
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764. |
Match the species in Column I with the bond order in Column II `{:(" Column I Column II " ),((i) NO " (a)" 1.5),((ii) CO" (b)" 2.0),((iii)O_(2)^(-)" "(c) 2.5),((iv) O_(2)" (d)" 3.0):}` |
Answer» Correct Answer - (i)-(c),(ii)-(d),(iii)-(a),(iv)-(b) Bond order are : `NO= 2.5, CO = 3, O_(2)^(-) = 1.5, O_(2) = 2.0`. |
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765. |
The bond length the species `O_(2), O_(2)^(+) and O_(2)^(-)` are in the order ofA. `O_(2)^(+) gt O_(2) gt O_(2)^(-)`B. `O_(2)^(+) gt O_(2)^(-) gt O_(2)`C. `O_(2) gt O_(2)^(+) gt O_(2)^(-)`D. `O_(2)^(-) gt O_(2) gt O_(2)^(+)` |
Answer» Correct Answer - D `O_(2)^(-) gt O_(2) gt O_(2)^(+)` Lesser the bond order, more will be the bond length. `B.O. : O_(2)^(-) (1.5),O_(2)(2.0),O_(2)^(+) (2.5).` |
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766. |
`NH_(3)and BF_(3)`from adduct readily because they fromA. Ionic bondB. Covalent bondC. Co-ordinate bondD. Hydrogen bond |
Answer» Correct Answer - C Co-ordianate bond is formed `NH_(3)` and `BF_(3)` because N atom has lone pair while B is electron deficient in nature `H_(3)N to BF_(3)` |
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767. |
The angle between the covalent bonds is maximum in :A. `CH_(4)`B. `BF_(3)`C. `PF_(3)`D. `NH_(3)` |
Answer» Correct Answer - B `BF_(3)` is planar and bond angle `(120^(@))` is the maximum. |
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768. |
Among the following species, identify the isostuctural pairs `NF_(3). NO_(3)^(-), BF_(3), H_(3)O, HN_(3)`A. `[NF_(3),NO_(3)^(-)]` and `[BF_(3)H_(3)O^(+)]`B. `[NF_(3),HN_(3)]` and `[NO_(3)^(-),BF_(3)]`C. `[NF_(3),N_(3)O^(+)]` and `[NO_(3)^(-),BF_(3)]`D. `[NF_(3),H_(3)O^(+)]` and `[HN_(3)^(-),BF_(3)]` |
Answer» Correct Answer - c `NF_(3)` and `H_(2)O` have `sp^(2)` hybridisation `NO_(3)^(-)` and `BF_(3)` have `sp^(2)` hybridisation |
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769. |
The correct increasing bomnd angle among `BF_(3),PF_(3)` and `CIF_(3)` follow the orderA. `BF_(3) lt PF_(3), CIF_(3)`B. `PF_(3) lt BF_(3), CIF_(3)`C. `CIF_(3) lt PF_(3), BF_(3)`D. All have equal bonfd angle |
Answer» Correct Answer - c In `BF_(1)B` is `sp^(2)` hybridised with bond angle `120^(@) C ` in `PF_(2)` P is sp hybridised but bond angle is less than `109^(@)28` due to lone pair bond pair repulsion in `CIF_(2) CI ` is `sp^(3)d hybridised having T-shope with bond angle `90^(@)C` |
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770. |
Predict whether the following molecules are isostructural or not Justify your answer `N(Me)_(3)` and `N(SiH_(3))_(3)` . |
Answer» They are not isostructural `N(Me)_(3)` is pyramidal in shape due to `sp^(3)` hybridisation `N(SiH_(3))_(3)` is trigonal planar Nitrogen atom shows `sp^(2)` hybridisation The structure is stabilised by the donation of lone pair of electrons from nitrogne into vacant d-orbitals of silicon forming `(p pi-d pi)` back bonding . |
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771. |
Which one of the following constitutes a group of the isoelectronic species ?A. `C_(2)^(2-), O_(2)^(-), CO, NO`B. `NO^(+), C_(2)^(2), CN^(-), N_(2)`C. `CN^(-), N_(2)O_(2)^(2-), C_(2)^(2-)`D. `N_(2), O_(2)^(-), NO^(+), CO` |
Answer» Correct Answer - b All the species have the same number of electrons, i.e., `14`. |
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772. |
Assetion Both `N_(2)` and `NO^(o+)` are diamagnetic Reasoning `NO^(o+)` is isoelectronic with `N_(2)` .A. Both A and R true and R is the correct explanation of AB. Both A and R true and R is not a correct explanation of AC. A is true but R is falseD. A is false but R is true |
Answer» Correct Answer - B | |
773. |
Out of `CH_(3)^(o+),H_(3)O^(o+),NH_(3),CH_(3)^(Theta)` the species which is not isoelectronic is________. |
Answer» Correct Answer - `CH_(3)^(o+)` `(CH_(3)^(o+) = 9 e^(-), H_(3) O^(o+) = 10 e^(-) ,NH_(3) = 10 e^(-) ,CH_(3)^(Theta) = 10e^(-))` . |
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774. |
Assertion : Carbon has unique ability to form `p pi-p pi` multiple bonds with itself and with other atomic of small size and high electronegativety. Reason : Heaviur elements of group `14th` do not form `p pi -p pi` bonds because their atomic orbital are too large and diffuse to have effective sideways overapping.A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertion.C. If assertion is true but reason is false.D. If assertion is false but reason is true. |
Answer» Correct Answer - b Both Assertion and Reason are true but Reason is not true explantion (directly not but indirectly true) of Assertion. Atomic size of carbon is smaller and thus is has effective overiapping with itself and with other atom of small size and high electromatively .Few example of multiple bonding are C= C, C = C,C= O, C = S and C = N heaver element do not form `p pi-pi m` bonds because their atomic orbitals are too large and diffuse to have effective overlapping. |
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775. |
Which bond angle `theta` would result in the maximum dipole moment for the triatomic `YXY`?A. `theta = 90^(@)`B. `theta = 120^(@)`C. `theta = 150^(@)`D. `theta = 180^(@)` |
Answer» Correct Answer - a The dipole moment of two dipole inclined at an angle `theta` is given by the eqquation `mu = sqrt(X^(2) + Y^(2) + 2XY cos theta)` `cos 90^(@) = 0`. Since the angle increases more nad more`-ve` and hence resulant decreasd. Thus dipole moment is maximum when `theta = 90^(@)` |
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776. |
Assertion(A) - In `N_2` molecule, the N atoms are bonded by one sigma and two `pi`-bonds. Reason(R) :N atoms assume `sp^2` hybrid state to constitute `N_2` molecule.A. Both A and R true and R is the correct explanation of AB. Both A and R true and R is not a correct explanation of AC. A is true but R is falseD. A is false but R is true |
Answer» Correct Answer - C | |
777. |
If the electron pair forming a bond between two atoms and B is not in the center then the bond is ?A. Single bondB. Polar bondC. None- polar bondD. `pi-`bond |
Answer» Correct Answer - b When electronegativity difference is more between two is joined atoms then covlent bond because polar and electron pair forming a bond do not remain in the centre. |
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778. |
The total number of electrons that take part in forming the bond in `N_(2)` is .A. `2`B. `4`C. `6`D. `100` |
Answer» Correct Answer - C Six because a triple covalent bond is formed between two nitrogen atoms . |
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779. |
The order of decreasing boiling point is `Xe gtCH_(4) gtNe gtHe gtH_(2)` . |
Answer» Correct Answer - T | |
780. |
The boiling points of methanol, water and dimethyl ether are respectively `65^@ C, 100^@ C` and `34.5^@ C`. Which of the following best explains thesewide variations in b.p. ?A. The molecular mass increases from water (18) to methanol (32) to diethyl ether (74)B. The extent of H-bonding decreases from water to methanol while it is absent in etherC. The extent of intramolecular H-bonding decreases from ether to methanol to waterD. The density of water is 1.00 g `ml^(-1)` methanol `0.7914 g ml^(-1)`- and that of diethyl ether is `0.7137 g ml^(-1)` |
Answer» Correct Answer - B | |
781. |
Which of the following two are isostructural ?A. `XeF_(2),IF_(2)^(-)`B. `NH_(3),BF_(3)`C. `CO_(3)^(2-),SO_(3)^(2-)`D. `PCl_(5), ICl_(5)` |
Answer» Correct Answer - a Both `XeF_(2)` and `IF_(2)^(-)` are linear and therefore isostructural. |
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782. |
High boiling point of water is due to :A. Its high specific heatB. Hydrogen bonding between the moleculesC. Weak dissociation of water moleculesD. Its high dielectric constant |
Answer» Correct Answer - B | |
783. |
The number of valence electrons in carbon atom isA. ZeroB. 2C. 6D. 4 |
Answer» Correct Answer - D | |
784. |
Elements X,Y and Z have 4,5 and 7 valence electrons respectively, (i) Write the molecular formula of the compounds formed by these elements individually with hydrogen (ii) which of these compounds will have the highest dipolw moment ? |
Answer» `Hunderset(H)overset(H)(underset(* * )overset(* * )(":X"":"))H" "Hunderset(H" "H)overset(H)(underset(* *)overset(* *)(":"Y":"))H" "underset(* *)overset(* *)(":"Z":")H` The compound Z- H has maximum dipole moment value since it is linear. |
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785. |
Which of the following pair contains iso-structural species ?A. `CH_(3)^(-)and CH_(3)^(+)`B. `NH_(4)^(+)and NH_(3)`C. `SO_(4)^(2-)and BF_(4)^(-)`D. `NH_(2)^(-)and BeF_(2)` |
Answer» Correct Answer - C Both `SO_(4)^(2-)and B F_(4)^(-)` have tetrahedral geometries. |
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786. |
The molecule which has dipole moment isA. `CH_(2) CI_(2)`B. `BF_(3)`C. `NF_(3)`D. `CIO_(2)` |
Answer» Correct Answer - b `BF_(3)` is trigonal planar and hence has zero dipole moment |
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787. |
Which of the following has greater bond length ? .A. `P-O`B. `S-O`C. `CI-O`D. `O=O` |
Answer» Correct Answer - A | |
788. |
Which of the following oxyacids of phosphorouse are monoprotic (mono basic) ? .A. `H_(3)PO_(3)`B. `H_(3)PO_(3)`C. `H_(3)PO_(2)`D. `H_(4)P_(2)O_(7)` |
Answer» Correct Answer - C::D | |
789. |
`AICI_(3)` is covalent while `AIF_(3)` is ionic This can be justified on the basic of .A. The valence bond theoryB. Fajans rulesC. The molecuar orbital theoryD. Hydration energy |
Answer» Correct Answer - B | |
790. |
The `C -C` bond length is `1.54 Å C =C` bond length is `1.33 Å` What is the circumference of benzene ring ? Bond length between single and double bonds `= 1.4 Å` .A. `(3 xx 1.54 + 3 xx 1.33) Å`B. `(4 xx 1.54 + 2 xx 1.33) Å`C. `(6 xx 1.4)Å`D. `(4 xx 1.33 +2 xx 1.54)Å` |
Answer» Correct Answer - C Due to resonance the `C -C` bond length in benzyne is in between single and double bonds `1.4Å` Hence circumference ` = 6 xx 1.4 Å` . |
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791. |
The correct bond order in the following species is:A. `O_(2)^(+)ltO_(2)^(-)ltO_(2)^(2+)`B. `O_(2)^(-)ltO_(2)^(+)ltO_(2)^(2+)`C. `O_(2)^(2+)ltO_(2)^(+)ltO_(2)^(2-)`D. `O_(2)^(2+)ltO_(2)^(-)ltO_(2)^(2+)` |
Answer» Correct Answer - B `O_(2)^(-)ltO_(2)^(+)ltO_(2)^(2+)` BO `1.5 lt 2.5 lt3.0` |
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792. |
Calculate the percent ionic character of HCL . Given that the observed dipole moment is 1 . 0 3 D and bond length of HCl is 1 .275 Å`. |
Answer» If HCl were 100% ionic and would carry charge equal to one unit, viz, ` 4.8 xx10^(-10)` esu. As bond length of HCl is ` 1.275 Å` , its dipole moment for 100% ionic character would be `mu_("ionic") = qxxd = 4.8xx10^(10) esu xx 1.275xx10^(-8) cm = 6 . 12 xx10^(18)` "esu cm "= 6.12 D `mu_("observed")= 1 . 0 3 D ("Given")` ` therefore % ` ionic character = `(mu_("observed"))/(mu_("ionic"))xx 100 = (1.03)/(6.12) xx100 = 16.83 %` |
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793. |
In `BaC_(2)`_______sigma and________pi bonds are present between two C-atoms . |
Answer» Correct Answer - One,two One two `[overset(Theta)C -= overset(Theta)C] Ba^(+2)` (one sigma and 2pi) . |
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794. |
In a pole molecule , the ionic is `4.8 xx 10^(-10)` esu. If the inter distance is `1 Å` unit, then the dipole moment isA. `41.8` debyeB. `4.18` debyeC. `4.8` debyeD. `0.48` debye |
Answer» Correct Answer - c Given ionic charge `= 4.8 xx 10^(-10)` esu and ionic distance ` = 1 Å = 10^(-8)` we know that dipole moment = ionic distance `= 4.8 xx 10^(-8)` esu per cm = 4.8 debye. |
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795. |
Which are the species in which central atom undergoes `sp^(3)` hybridisation ? .A. `SF_(4)`B. `SCI_(2)`C. `SO_(4)^(2-)`D. `H_(2)O` |
Answer» Correct Answer - B::C::D Hybridisation of `:overset(..)SF_(4) =(1)/(2) (6 +4) = 5 = sp^(3) d` (b) Hybridisation of `:overset(..)SCI_(2) = (1)/(2) (6 +2) = 4 = sp^(3)` (c ) Hybridisation of `SO_(4)^(2-) = (1)/(2) (6 + 0 +2) = 4 = sp^(3)` (d) Hybridisation of `H_(2) overset(..)O: = (1)/(2) (6 + 2) = 4 = sp^(3)` . |
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796. |
Ortho-Nitrophenol is less soluble in water than p- and m- Nitrophenols becauseA. Melting point of o-Nitrophenol is lower than those of m- and p-isomersB. o-Nitrophenol is more volatile is steam than m-and p-isomers .C. o-Nitrophenol shown Intermolecular H- bondingD. o-Nitrophenol shows Intermolecular H-bonding |
Answer» Correct Answer - C o-Nitrophenol is less soluble in water than p- and m-nitrphenols because o-nitrphenol shows intramolecular hydrogen bonding . |
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797. |
The cyanide ion `CN and N_(2)` are isoelectronic, but in contrast to `CN^(-), N_(2)` is chemically inert, because ofA. Low and energyB. Absence of bond polarityC. Usymmetrical electron distributionD. Presence of more number of electrons in bonding orbitals |
Answer» Correct Answer - B Similar atoms are linked to each other and thus there is no polarity . |
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798. |
Ortho -nitrophenol is less soluble in water than `p`-and `m-`nitrophenols becauseA. ortho nitrophenol shows intramolecular H- bondingB. Ortho Nitrophenol shows intramolecular H- bondingC. melting point of o-nitrophenol is lower than those of m-and p-isomersD. o-nitro phenol is more volatile in steam than those of m- and p-isomers. |
Answer» Correct Answer - A | |
799. |
The `CI - C - CI` angle in `1, 1, 2, 2`, tetrachloroethone and tetrachloromethane respectively will be about:A. `109.5^(@)` and `90^(@)`B. `120^(@)` and `109.5^(@)`C. `90^(@)` and `109.5^(@)`D. `109.5^(@)` and `120^(@)` |
Answer» Correct Answer - B | |
800. |
The `CI - C - CI` angle in `1, 1, 2, 2`, tetrachloroethone and tetrachloromethane respectively will be about:A. `120^(@)` and `109.5^(@)`B. `90^(@)` and `109.5^(@)`C. `109.5^(@)` and `90^(@)`D. `109.5^(@)` and `120^(@)` |
Answer» Correct Answer - a In `1, 1, 2, 2` tratachloroethene, each carbon is `sp^(2)`- hybridized thus having bond angle `120^(@)` and in tetrachloromethane carbon is `sp^(3)` - hybridized, hence bond angle is `1095.5^(@)`. |
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