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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.
| 101. |
Lewis concept of covalency of an element involved octet rule .Later on it was found that many elements in their compoinds `e.g, BeF_2,BF_3` etc . Have incomplete octet whereas `PCl_5 ,SF_6` etc , have expanded octet . This classical conept also failed in predicting the geometry of molecules .Modern concept of covalence was propsed in terms of valence bond theory proposed by Heitler and London and later on modified by Pauling nad Slater . Hybridisation concept alongwith valence bond theory although successfully explained the geometry of various molecules but failed in many molecules . the geometry of such molecules was explained by VSEPR concept . Finally molecular orbital theory was porosed by Hund -Mulliken to explain many other anomalies . Selcet the cirrect statements . (1) Bond order for ` N_2^+` and `N_2^-` are same ( 2) Bond energy of `N_2^+ lt N_2^-` (3) Bond lenght of ` N_2^+ lt N_2^-` (4) ` C_3 F_4` is a non -palar molecular (5) `XeF_2` hat two (F) atoms in axial position whereas ` XeF_4` has four ( F) atoms in equatiorial position .A. ` 2, 3, 4, 5`B. ` 1, 4, 5`C. ` 1, 5`D. ` 1, 2, 4, 5` |
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Answer» Correct Answer - D Follow text . |
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| 102. |
In terms of the molecular orbital theory , which of the following species will most likely be the one to gain an electron to form thermodynamically more stable species?A. `CN`B. `NO`C. `O_(2)^(2+)`D. `N_(2)` |
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Answer» Correct Answer - 1 `(1)CN(6+7=13)=sigma1s^(2)sigma^(**)1s^(2)sigma2s^(2)sigma^(**)2s^(2)pi2p_(x)^(2)=pi2p_(y)^(2)sigma 2p_(z)^(1),` Bond order `=(9-4)/(2)=2.5` `CN^(-)(6+7+1=14)=sigma1s^(2)sigma^(**)1s^(2)sigma2s^(2)sigma^(**)2s^(2)pi2p_(x)^(2)=pi2p_(y)^(2)sigma2p_(z)^(2),` Bond order `=(10-4)/(2)=3` `(2)NO(7+8=15)=sigma 1s^(2)sigma^(*)1s^(2)sigma2s^(2)sigma^(*)2s^(2)sigma2p_(z)^(2) pi2p_(x)^(2)=pi2p_(y)^(2) pi^(*)2p_(x)^(1)=pi^(*)2p_(y)^(0),` Bond order `=(10-5)/(2)=2.5` `NO^(-)(7+8+1=16)=sigma1s^(2)sigma^(*)1s^(2)sigma2s^(2)sigma^(*)2s^(2)sigma2p_(z)^(2)pi2p_(x)^(2)=pi2p_(y)^(2)pi2p_(x)^(1)=pi^(*)2p_(y)^(1)` Bond order `=(10-6)/(2)2.0` `(3)O_(2)^(2+)(8+8-2=14)=sigma1s^(2)sigma^(*)1s^(2)sigma2s^(2)sigma^(*)2s^(2)sigma2p_(z)^(2)pi2p_(x)^(2)=pi2p_(y)^(2)pi^(*)2p_(y)^(@)=pi^(*)2p_(y)^(@),` Bond order `=(10-4)/(2)=3.0` `O_(2)^(+)(8+8-1=15)=sigma1s^(2)sigma^(*)1s^(2)sigma2s^(2)sigma^(*)2s^(2)sigma^(**)2p_(z)^(2)pi2p_(x)^(2)=pi2p_(y)^(2)pi^(*)2p_(x)^(1)=pi^(*)2p_(y)^(@),` Bond order `=(10-5)/(2)=2.5` `(4)N_(2)(7+7=14)sigma1s^(2)sigma^(*)1s^(2)sigma2s^(2)sigma^(*)2s^(2)pi2p_(x)^(2)=pi2p_(y)^(2)sigma2p_(z)^(2),` Bond order `=(10-4)/(2)=3` `N_(2)^(-)(7+7+1=15)=sigma1s^(2)sigma^(*)1s^(2)sigma2s^(2)sigma^(*)2s^(2)pi2p_(x)^(2)=pi2p_(y)^(2)sigma2p_(z)^(2)pi^(*)2p_(x)^(1),`Bond order `=(10-5)/(2)=2.5` |
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| 103. |
A coordinate bond is a dative bond. Which of the following is true?A. Three atoms form bond by sharing their electrons.B. Two atoms form bond by sharing their electrons.C. Two atoms form bond and one of them provides both electrons.D. Two atoms form bond by sharing electrons obtained from the third atom. |
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Answer» Correct Answer - D During the formation of a coordinate covalent bond between two atoms, only one of the atoms (called donor atom) provides both the electrons for sharing. The other atom (called the acceptor) provides none. |
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| 104. |
A coordinate bond is a dative bond. Which of the following is true?A. Two atoms bond by sharing electrons from third atom.B. Two atoms form bond by sharing their electrons.C. Two atoms form bond and one of them provides both electrons.D. Three atoms form bond by sharing their electrons. |
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Answer» Correct Answer - C During the formation of a coordinate covalent bond, two atoms share an electron pair, but both the electrons for sharing are provided by one of the atoms called donor atom. The other atom is called an acceptor atom. |
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| 105. |
Dative bond is present in :-A. `KI_(3)`B. `KNO_(2)`C. `KHF_(2)`D. All |
| Answer» Correct Answer - A | |
| 106. |
Lewis concept of covalency of an element involved octet rule .Later on it was found that many elements in their compoinds `e.g, BeF_2,BF_3` etc . Have incomplete octet whereas `PCl_5 ,SF_6` etc , have expanded octet . This classical conept also failed in predicting the geometry of molecules .Modern concept of covalence was propsed in terms of valence bond theory proposed by Heitler and London and later on modified by Pauling nad Slater . Hybridisation concept alongwith valence bond theory although successfully explained the geometry of various molecules but failed in many molecules . the geometry of such molecules was explained by VSEPR concept . Finally molecular orbital theory was porosed by Hund -Mulliken to explain many other anomalies . Ratio of lone pair-bond pair electrons on central atom in ` i_3^-` and `XeF_4` are respectively :A. ` 2. 66, 0. 5`B. `1. 5 , 0.5`C. ` 2, 0. 5`D. ` 0 5, 2` |
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Answer» Correct Answer - A ` l_3^-` has ( 3) bond pair and `8` lone pair electrons . ` :. Ratio = (I . P .) /(B .P) = (8/ 3) = 2 . 66` . `XeF_(4)` has `8` bond pair electrons and `4` lone pair electrons. ` :.` Ratio ` = (L . P . ) /( B .P) = 4/8 = 0. 5` . |
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| 107. |
Lewis concept of covalency of an element involved octet rule .Later on it was found that many elements in their compoinds `e.g, BeF_2,BF_3` etc . Have incomplete octet whereas `PCl_5 ,SF_6` etc , have expanded octet . This classical conept also failed in predicting the geometry of molecules .Modern concept of covalence was propsed in terms of valence bond theory proposed by Heitler and London and later on modified by Pauling nad Slater . Hybridisation concept alongwith valence bond theory although successfully explained the geometry of various molecules but failed in many molecules . the geometry of such molecules was explained by VSEPR concept . Finally molecular orbital theory was porosed by Hund -Mulliken to explain many other anomalies . The bond angles `NO_2^+, NO` and `NO_2^-` are respectivley :A. `180 ^@ , 134^@ , 115^@`B. ` 115^@ , 134^@ , 180^@`C. `134^@ , 180^@ , 115^@`D. ` 115^@ , 180^@ , 124^@` |
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Answer» Correct Answer - A Follow the effect of lone pair and lone pair electron . |
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| 108. |
Which one of the following sepcies is diamagnetic in nature ?A. `He_2^+`B. `H_2`C. `H_2^+`D. `H_2^-` |
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Answer» Correct Answer - B `He_(2)^(+) ,H_(2)^(-)` have `3` electronsone must be unpaired , `He_(2)^(+)` has one unpaired electrons . `H_2` has two electrons and thus diamagnetic . |
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| 109. |
In `TeCl_4`, the central tellurium involves the hybridizationA. `sp^3`B. `sp^3d`C. `sp^3d^2`D. `dsp^2` |
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Answer» Correct Answer - B Valence shell electron configuration of `Te` is `5s^2 5p^4`, i.e., it has 6 valence electrons. Four `e^(-)s` are used to form bonds with `4Cl` atoms while `2e^(-)s` form one lone pair. Thus, `SN` of `Te` is `4+1=5`. This implies `Te` makes use of five hybrid orbitals which come from `sp^3d` hybridization. |
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| 110. |
Which of the following transitions involves maximum amount of energy?A. `M^(-)(g)toM(g)`B. `M(g)toM^(+)(g)`C. `M^(+)(g)toM^(2+)(g)`D. `M^(2+)(g)toMg^(3+)(g)` |
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Answer» Correct Answer - D `M^(2+) to M^(3+)` , after the removal of second electron, the nuclear charge per electron increases, due to which high energy is required to remove the 3rd electron |
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| 111. |
Many ionic compounds in transition metal ions are colored because of the electronic transitions (in the visible range) involving d electrons, whereas the ionic compounds of the representative elements are usually colorless. Which of the following compounds is green in color?A. `CuSO_4 5H_2O`B. `K_2Cr_2O_7`C. `NiCl_2 6H_2O`D. `CoCl_2.6H_2O` |
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Answer» Correct Answer - C Nickel (II) chloride hexahydrate is green while copper (II) sulphate pentahydrate is blue, potassium dichromate is orange and cobalt (II) chloride hexahydrate is red. |
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| 112. |
Aqueous solutions of transition metal cations are also colored. Which of the following imparts pale yellow color?A. `Fe^(3+)`B. `Fe^(2+)`C. `Mn^(2+)`D. `Cr^(3+)` |
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Answer» Correct Answer - A `Fe^(2+)`(pale green), `Mn^(2+)` (pale pink), `Cr^(3+)` (red-violet). |
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| 113. |
Most transition metals from several cations having __________configurations.A. pseudo noble gasB. noble gasC. both pseudo noble gas and noble gasD. neither pseudo noble gas nor noble gas |
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Answer» Correct Answer - A For example, iron forms the cations `Fe^(2+)` and `Fe^(3+)`. Neither has a noble gas configuration as that requires an energetically impossible loss of 8 electrons from the neutral atom, `Fe([Ar]3d^64s^2)`. |
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| 114. |
Which of the following atoms has miniumum covalent radiusA. BB. CC. ND. Si |
| Answer» Correct Answer - C | |
| 115. |
Lattice enthalpy is the change in energy that occurs when ______of an ionic solid is separated into isolated ions in the gas phase.A. one gramB. one moleC. one gram atomD. one gram molecule |
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Answer» Correct Answer - B Lattice enthalpy is the enthalpy change for the process `NaCl(s)rarrNa^(+)(g)+Cl^(-)(g)` |
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| 116. |
Lattice energy of alkali metal chlorides follows the orderA. LiCl gt NaCl gt KCl gt RbCl gt CsClB. CsCl gt NaCl gt KCl gt RbCl gt LiClC. LiCl gt CsCl gt NaCl gt KCl gt RbClD. NaCl gt LiCl gtKCl gt RbCl gt CsCl |
| Answer» Correct Answer - A | |
| 117. |
The magnitude of lattice energy of a solid increases ifA. size of ions is smallB. charges of ions are smallC. ions are neutralD. None of the above |
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Answer» (a) According to Born-Lande equation, `U=(Z^(+)Z^(-)e^(2) N_(A)M)/(r_(node))((1)/(n)-1)` `:.` Ions should be of small size to have high lattice energy. |
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| 118. |
A molecule is described by three Lewis structures having energies `E_1`, `E_2`, and `E_3`, respectively. The energies of these structures follow the order `E_1gtE_2gtE_3`, respectively. If the experimental energy of the molecules is `E_0`, the resonance energy isA. `E_0-E_3`B. `E_0-E_1`C. `E_0-E_2`D. `E_0-(E_1+E_2+E_3)` |
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Answer» Correct Answer - A Resonance stabilizes the molecule or ion as the energy of the resonance hybrid (actual structure of the molecule or ion) is less than the energy of any single cannonical structure. The difference in energy between the actual structure `(E_0)` and the lowest energy resonance structure `(E_3)` is called the resonance stabilization energy or simply the resonance energy. |
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| 119. |
Resonance Hybrid of nitrate ion is `:`A. B. C. D. |
| Answer» Correct Answer - C | |
| 120. |
Reonance is due toA. Delocalization of sigma electronsB. Delocalization of pi electronsC. Migration of H atomsD. Migration of protons |
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Answer» Correct Answer - BIt is due to Delocalization of pi electrons. Hence option (B) is correct |
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| 121. |
Which one of the following contains both ionic and covalent bonds?A. `C_(6)H_(5)Cl`B. `H_(2)O`C. NaOHD. `CO_(2)` |
| Answer» (c ) In NaOH,`Na^(+)` and `OH^(-)` ions are bonded together by ionic bond while in `OH^(-)`ion, oxygen and hydrogen atoms are bonded together by convalent bond `Na^(+)[O-H]^(-)`. | |
| 122. |
Of the following statements, which one is correctA. Oxygen and nitric oxide molecules are both paramagnetic because both contain unpaired electronsB. Oxygen and nitric oxide molecules are both diamagnetic because both contain no unpaired electronsC. Oxygen is paramagnetic because it contains unpaired electrons, while nitric oxide is diamagnetic because it contains no unpaired electronsD. Oxygen is diamagnetic because it contains no unpaired electrons, while nitric oxide is paramagnetic because it contains an unpaired electron |
| Answer» Correct Answer - A | |
| 123. |
The difference in energy between the molecule orbital formed and the combining atomic orbitals is calledA. Bond energyB. Activation energyC. Stabilization energyD. Destabilization energy |
| Answer» Correct Answer - C | |
| 124. |
The variation of the boiling points of the hydrogen halides is in the order `HFgtHIgtHBrgtHCl`. What explains the higher boiling point of hydrogen fluoride?A. The electronegativity of fluorine is much higher than for other elements in the groupB. There is strong hydrogen bonding between HF moleculesC. The bond energy of HF molecules is greater than in other hydrogen halidesD. The effect of nuclear shielding is much reduced in fluorine which polarises the HF molecule |
| Answer» Correct Answer - B | |
| 125. |
What is the dominant intermolecular forces or bond that must be overcome in converting liquid `CH_(3)OH` to gas ?A. Hydrogen bondingB. Dipole-dipole interactionC. Covalent bondsD. London dispersion force |
| Answer» Correct Answer - A | |
| 126. |
In which of the following molecules/ions in the central atom `sp^2`-hybridized?A. `NH_2^(-)` and `H_2O`B. `NO_2^(-)` and `H_2O`C. `BF_3` and `NO_2^(-)`D. `NO_2^(-)` and `NH_2^(-)` |
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Answer» Correct Answer - C For `sp^2`-hyridization, the `SN` of the central atom should be 3. `SN` of `B`(in `BF_3`) is `3+0=3`, `SN` of `N`(in `NO_2^(-)`) is `2+1=3`, `SN` of `N` (in `NH_2^(-)`) is `2+2=4`, and `SN` of `O` in `(H_2O)` is `2+2=4`. Thus, in both `BF_3` and `NO_2^(-)`, the central atom is `sp^2`-hybridized. |
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| 127. |
The outer orbitals of C in ethene molecule can be considered to be hybridized to given three equivalent `sp^(2)` orbitals. The total number of sigma `(sigma)` and pi `(pi)` bonds in ethene molecule isA. 3 sigma `(sigma)` and 2 pi (`pi`) bondsB. 4 sigma `(sigma)` and 1 pi (`pi`) bondsC. 5 sigma `(sigma)` and 1 pi `(pi) `bondD. `1 sigma `(sigma)` and 2 pi (`pi`) bonds |
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Answer» Correct Answer - C `H-underset(H)underset(|)(C)=underset(H)underset(|)(C)-H` |
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| 128. |
When the hybridization state of a carbon atom changes from `sp^(3)` to `sp^(2)` and finally to `sp`, the angle between the hybridized orbitalsA. decreases graduallyB. increases graduallyC. decreases considerablyD. all of these |
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Answer» Correct Answer - B Bond angle increases with change in hybridisation in following order `sp^(3) lt sp^(3) ltsp` |
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| 129. |
Expanded octet occurs inA. `NH_(3)`B. `PF_(5)`C. `H_(2)O`D. `O_(2)` |
| Answer» Correct Answer - B | |
| 130. |
An element X has low ionisation energy and another element Y has high electron affinity. The bond formed between them could beA. ionicB. polar covalentC. co-ordinate covalentD. non-polar covalent |
| Answer» Correct Answer - A | |
| 131. |
The lesser covalency of `FeCl_(2)` over `FeCl_(3)` is due toA. lower polarizing power of `Fe^(+2)` than that of `Fe^(+3)` ionB. lower polarizing power of `Fe^(+3)` than that of `Fe^(+2)`C. higher polarizability of `Fe^(+3)` than `Fe^(+2)`D. higher polarizability of `Fe^(+2)` than `Fe^(+3)` |
| Answer» Correct Answer - A | |
| 132. |
In `C_3O_2`, the hybridization state of C isA. `sp^2`B. `sp`C. `sp^3`D. `dsp^2` |
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Answer» Correct Answer - B C atom is `sp^3`-hybridized if it forms `4sigma` bonds (`0pi` bonds), `sp^2`-hybridized if it forms `3sigma` bonds and `1pi` bond, and `sp`-hybridized if it forms `2sigma` bonds and `2pi` bonds. The Lewis structure of `C_3O_2` (carbon suboxide) is `O=C=C=C=O` Each C atom forms `2sigma` and `2pi` bonds. Thus, every C atom is sp-hybridized. |
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| 133. |
Carbon suboxide `(C_3O_2)` is a foul-smelling gas. Which of the following formulation represents the correct ground state Lewis structure for carbon suboxide?A. `:O:C:C:C:O:`B. `:O::C::C::C::O:`C. `:O:C::C:C:O`D. `:overset(..)O::C::C::C::overset(..)O:` |
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Answer» Correct Answer - D The valence shell electron configurations of C and O are `2s^2 2p^2` and `2s^2 2p^4`, respectively. Thus, there are `(4xx3)+(6xx2)` or 24 valence electrons to account for in `C_3O_2`. Only the Lewis structure (4) has 24 valence electrons: `overset(..):O=C=C=C=overset(..)O:` |
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| 134. |
Ionisation paotential and electron affinity of fluorine are `17.42` and `3.45eV` respectively .Calculate the electronegativity of fluorine on Mulliken scale and Pauling scale . |
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Answer» According to Mulliken equation , ` X =(IP +EA)/(2)` , When both `IP` and `EA` are taken in `eV`. `(X_F)_M = (17.42 +3.45)/(2) =10.435` `(X_F)_P= (10.435)/(2.8) =3.726 [or (Z_F)_P =(IP+EA)/(5.6)]` |
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| 135. |
Which of the following has a square pyramidal shape?A. `XeOF_4`B. `XeO_3F_2`C. `XeOF_2`D. `XeO_2F_2` |
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Answer» Correct Answer - A For square pyramidal shape, the SN of central atom should be `5+1=6`. In `XeO_2F`, the SN of central `Xe` is `4+1=5`. In `XeOF_2`, the SN of central `Xe` is `3+2=5`. In `XeO_3F_2`, the steric number of central `Xe` is `5+0=5`. In `XeOF_4`, the SN of central `Xe` is `5+1=6`. Thus, it has square pyramidal shape. |
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| 136. |
Which of the following compounds has square pyramidal geometry ?A. `XeF_(6)`B. `XeO_(3)`C. `BrF_(5)`D. `XeF_(4)` |
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Answer» Correct Answer - C `XeF_(6) ` -distorted octahedral `XeO_(3)` -pyramidal `BrF_(5)`-Octahedral (squre pyramidal) `XeF_(4)`- Square planar |
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| 137. |
Which among the following has smallest bond angle ?A. `H_2S`B. `H_2O`C. `NH_3`D. `SO_2` |
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Answer» Correct Answer - A In both `NH_3` and `SO_2`, the central atoms (N and S) carry only one lone pair of electrons while in `H_2S` and `H_2O`, the central atoms (S and O) carry two lone pairs. Due to lone pair-lone pair repulsion, the latter molecules will have smaller bond angles than the former molecules. In similar molecules (`H_2overset(..)S:` and `H_2overset(..)O:`), the bond angle decreases as the size of the central atom increases and its electronegativity decreases. Thus, `H_2S` has the smallest bond angle as sulphur is bigger than oxygen and less electronegative: `H_2S(92.5^@) lt H_2O(104.5^@) lt NH_3(107.5^@) lt SO_2(119^@)` |
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| 138. |
`IE_2` for an element are invariable bhygher thab ` IE_1` because :A. the size of cation is smaller than its atomB. it is difficult of remove `e` form cationC. IE is endothermicD. All of these |
| Answer» Correct Answer - A::B | |
| 139. |
Using the data given below,predict the nature of heat changes for the reaction . `Mg_(g) + 2F_(g) rarr Mg_(g)^(2+) +2F_(g)^-` `IE_1` and `IE_2` of `Mg_(g) `are `737.7 ` and `451 kJ "mol"^(-1)` . `EA_1 ` for `F_(g)` is `-328 kJ "mol"^(-1)`. |
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Answer» `Delta H=IE_1 +IE_2 +2 xx EA_1` ` =737.7 +1451 +2 xx (-328)` `=+ 1532.7 kJ "mol"^(-1)`. |
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| 140. |
The shape of a molecule which has ` 3` bond paires and one lone pair is :A. octahedralB. pyramidalC. triangular planarD. tetrahedral |
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Answer» Correct Answer - B e.g., `NH_3` which has three bond pair and one lone pair of electron . |
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| 141. |
Which of the following paires of elements for oxideas of polyanions and polycations respectively ?A. Si and AlB. Cu and SiC. Al and BD. Ti and As |
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Answer» Correct Answer - A `(SiO_3^(2-))_(n)` (polyanion). |
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| 142. |
` N -O -N` bond andle is maximum in :A. ` NO_2^+`B. `NO^2`C. `NO_3^-1`D. `N-2O_3` |
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Answer» Correct Answer - B Follow geometry of each. |
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| 143. |
Which of the following is correctly based on molecular orbital theory for peroxide ion?A. Its bond order is two and it is paramagnetic.B. Its bond order is two and it is diamagnetic.C. Its bond order is one and it is diamagnetic.D. Its bond order is one and it is paramagnetic. |
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Answer» Correct Answer - C The peroxide ion `O_2^(2-)` has 18 electrons arranged in the following manner: `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)^**,):}]` All the electrons are paired. Thus, `O_2^(2-)` ion is diamagnetic. The bond order may be calculated as (bonding-antibondig)/2, that is `(10-8)//2=1`. |
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| 144. |
The bond order in `N_(2)^(+)` ion is __________.A. 1B. 2C. 2.5D. 3 |
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Answer» Correct Answer - C `B.O.=("No. of " N_(b)-"No. of " N_(a))/2=5/2=2.5` |
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| 145. |
Which of the following is paramagnetic with bond order `0.5`?A. `O_2^(-)`B. `H_2^(+)`C. `B_2`D. `N_2` |
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Answer» Correct Answer - B Both `O_2^(-)` and `H_2^(+)` are paramagnetic having one unpaired electron each. Bond order `(O_2^(-))=(10-1)/(2)=1.5` Bond order `(H_2^(+))=(1-0)/(2)=0.5` `B_2, N_2,` and `F_2` are all diamagnetic. |
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| 146. |
Which of the following is correctly based on molecular orbital theory for peroxide ion?A. Its bond order is two and it is diamagneticB. Its bond order is one and it is paramagneticC. Its bond order is two and it is paramagneticD. Its bond order is one and it is diamagnetic |
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Answer» Correct Answer - D Molecular orbital configuration of peroxide ion, `O_(2)^(2-)` is `sigma1s^(2)sigma^(**)1s^(2)sigma2s^(2)sigma^(**)2s^(2)sigma2p_(z)^(2)pi2p_(y)^(2)pi^(**)2p_(x)^(2)pi^(**)2p_(y)^(2)` it is diamagnetic. Bond order `=(10-8)/2 =1` |
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| 147. |
The bond order of a molecule is given by __________.A. The difference between the number of electrons in bonding and antibonding orbitalsB. Total number of electrons in bonding and antibonding orbitalsC. Twice the difference between the number of electrons in bonding and antibonding electronsD. Half the difference between the number of electrons in bonding and antibonding electrons |
| Answer» Correct Answer - D | |
| 148. |
Which one is paramagnetic and has the bond order `1//2`A. `O_(2)`B. `N_(2)` molecule has bond order 3 and is paramagneticC. `F_(2)`D. `H_(2)^(+)` |
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Answer» Correct Answer - D `H^(2+)` has the bond order `1/2`, it has only one electron so it will be paramagnetic. |
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| 149. |
The correct order of bond energy is:A. `Cl_(2) gt Br_(2) gt F_(2) gt I_(2)`B. `Cl_(2) gt F_(2) gr Br_(2) gt I_(2)`C. `I_(2) gt Br_(2) gt Cl_(2) gt F_(2)`D. `I_(2) gt Br_(2) gt F_(2) gt Cl_(2)` |
| Answer» Correct Answer - A | |
| 150. |
Assertion: Nitrogen molecule is diamagnetic. Reason: `N_2` molecule have unpaired electrons.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 |
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Answer» Correct Answer - C `N_(2)` molecule is diamagnetic. The diamagnetic character is due to the presence of paired electron `N_(2)` molecule does not contain any unpaired electron. Thus, assertion is correct but the reason is false. |
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