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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.
| 51. |
KF combines with HF to form `KHF_(2)` . The compound contains the speciesA. ` K_(+).F^(-) and H^(+)`B. `K^(+), F^(-) and HF`C. `K^(+) and [HF_(2)]^(-)`D. `[KHF]^(+) and F_(2)` . |
| Answer» Correct Answer - C | |
| 52. |
Why does type of overlap given in the following figure not result in bnd formation ? |
| Answer» In the first case, the ares under + + ovelap is equal to the area under +- pverlap. The two cancel out and the net overlap is zero, Bond formation does not take place. In the second case, no overlap is possible in `p_(x)and p_(y)` orbitals. Therefore, no bond is formed. | |
| 53. |
Why does type of overlap given in the following figure not result in bond formation? |
| Answer» In the figure(I), area of ++ overlap is equal to +- overlap, so net overlap is zero, while in figure (II), there is no overlap due to different symmetry. | |
| 54. |
It is a common observation that many compounds containing hydrogen attached to highly electronegative elements such as oxygen, nitrogen, or fluorine, often exhibit unexpected properties, such as relatively high melting points, boiling points, viscosity, solubility in water etc Such an unexpected behaviour can be explained on the basis of hydrogen bonding. When a hydrogen atom is attached to a highly electronegative element of small size such as F. N, 0. etc. the electronegative atoms strongly attracts the shared pair of electrons towards it self. As a result, the hydrogen atom becomes slightly positive and the electronegative element becomes slightly negative when two molecules of such a substance say HF come close to each other, the negatively charged fluorine atom of one molecule attracts the positively charged hydrogen atom of the other molecule. A hydrogen atom, thus links to highly electronegative atoms, one by a strong covalent bond and the other by weak electrostatic attraction as shown ahead `overset(delta+)H-overset(delta-)F....overset(delta+)H-overset(delta-)F`, the dotted line represents a hydrogen bond. `NH_3` has much higher boiling point than `PH_3` becauseA. `NH_3` has larger molecular massB. `NH_3` undergoes umbrella massC. `NH_3` forms hydrogen bondD. `NH_3` contains ionic bonds whereas `PH_3` contain covalent bonds |
| Answer» Correct Answer - C | |
| 55. |
It is a common observation that many compounds containing hydrogen attached to highly electronegative elements such as oxygen, nitrogen, or fluorine, often exhibit unexpected properties, such as relatively high melting points, boiling points, viscosity, solubility in water etc Such an unexpected behaviour can be explained on the basis of hydrogen bonding. When a hydrogen atom is attached to a highly electronegative element of small size such as F. N, 0. etc. the electronegative atoms strongly attracts the shared pair of electrons towards it self. As a result, the hydrogen atom becomes slightly positive and the electronegative element becomes slightly negative when two molecules of such a substance say HF come close to each other, the negatively charged fluorine atom of one molecule attracts the positively charged hydrogen atom of the other molecule. A hydrogen atom, thus links to highly electronegative atoms, one by a strong covalent bond and the other by weak electrostatic attraction as shown ahead `overset(delta+)H-overset(delta-)F....overset(delta+)H-overset(delta-)F`, the dotted line represents a hydrogen bond. Hydrgen bonding is not present inA. glycerineB. waterC. hydrogen sulphideD. ammonia |
| Answer» Correct Answer - C | |
| 56. |
It is a common observation that many compounds containing hydrogen attached to highly electronegative elements such as oxygen, nitrogen, or fluorine, often exhibit unexpected properties, such as relatively high melting points, boiling points, viscosity, solubility in water etc Such an unexpected behaviour can be explained on the basis of hydrogen bonding. When a hydrogen atom is attached to a highly electronegative element of small size such as F. N, 0. etc. the electronegative atoms strongly attracts the shared pair of electrons towards it self. As a result, the hydrogen atom becomes slightly positive and the electronegative element becomes slightly negative when two molecules of such a substance say HF come close to each other, the negatively charged fluorine atom of one molecule attracts the positively charged hydrogen atom of the other molecule. A hydrogen atom, thus links to highly electronegative atoms, one by a strong covalent bond and the other by weak electrostatic attraction as shown ahead `overset(delta+)H-overset(delta-)F....overset(delta+)H-overset(delta-)F`, the dotted line represents a hydrogen bond. Hydrogen bonding is maximum inA. ethyl chlorideB. triethylamineC. ethanolD. diethyl ether. |
| Answer» Correct Answer - C | |
| 57. |
It is a common observation that many compounds containing hydrogen attached to highly electronegative elements such as oxygen, nitrogen, or fluorine, often exhibit unexpected properties, such as relatively high melting points, boiling points, viscosity, solubility in water etc Such an unexpected behaviour can be explained on the basis of hydrogen bonding. When a hydrogen atom is attached to a highly electronegative element of small size such as F. N, 0. etc. the electronegative atoms strongly attracts the shared pair of electrons towards it self. As a result, the hydrogen atom becomes slightly positive and the electronegative element becomes slightly negative when two molecules of such a substance say HF come close to each other, the negatively charged fluorine atom of one molecule attracts the positively charged hydrogen atom of the other molecule. A hydrogen atom, thus links to highly electronegative atoms, one by a strong covalent bond and the other by weak electrostatic attraction as shown ahead `overset(delta+)H-overset(delta-)F....overset(delta+)H-overset(delta-)F`, the dotted line represents a hydrogen bond. Hydrogen bond is strongest inA. F-H…OB. F-H…FC. O-H…SD. O-H…N |
| Answer» Correct Answer - B | |
| 58. |
It is a common observation that many compounds containing hydrogen attached to highly electronegative elements such as oxygen, nitrogen, or fluorine, often exhibit unexpected properties, such as relatively high melting points, boiling points, viscosity, solubility in water etc Such an unexpected behaviour can be explained on the basis of hydrogen bonding. When a hydrogen atom is attached to a highly electronegative element of small size such as F. N, 0. etc. the electronegative atoms strongly attracts the shared pair of electrons towards it self. As a result, the hydrogen atom becomes slightly positive and the electronegative element becomes slightly negative when two molecules of such a substance say HF come close to each other, the negatively charged fluorine atom of one molecule attracts the positively charged hydrogen atom of the other molecule. A hydrogen atom, thus links to highly electronegative atoms, one by a strong covalent bond and the other by weak electrostatic attraction as shown ahead `overset(delta+)H-overset(delta-)F....overset(delta+)H-overset(delta-)F`, the dotted line represents a hydrogen bond. The molecular containing hydrogen bond isA. HIB. `CuSO_4 . 5H_2O`C. HFD. All of these |
| Answer» Correct Answer - C | |
| 59. |
Assertion(A) - Silicon tetrafluoride. `SiF_4` is non polar even though fluorine is much more electronegative than silicon. Reason(R)-The four bond dipoles cancel one another in `SiF_4`.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 - A | |
| 60. |
Assertion : Fluorine`(Fe_(2))` is gas white iodine `(I_(2))` is solid at room temperature. Reason :A large molecule or heavy atom is more polarizable and has larger dispersion forces because it has many electrons some of which are less hightly hekld and are farther from the nucleus.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 and the reason is the correct explanation of the assertion.C. If assertion is true but reason is false.D. if assertion is false but reason is true. |
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Answer» Correct Answer - a London dispersion forms exits among the non-polar molecules like `F_(2),O_(2),I_(2), CI_(2)` ext in sold liquid states .Evan in atoms in molecules which have no permanent dipole, instancules dipole will as a result of momentary inbalances in electron distribution.London force are exermely short range in action and teh weakest of all attractive forces .The london forces increase molecule volume and teh number of polarizable electrons. |
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| 61. |
It is a common observation that many compounds containing hydrogen attached to highly electronegative elements such as oxygen, nitrogen, or fluorine, often exhibit unexpected properties, such as relatively high melting points, boiling points, viscosity, solubility in water etc Such an unexpected behaviour can be explained on the basis of hydrogen bonding. When a hydrogen atom is attached to a highly electronegative element of small size such as F. N, 0. etc. the electronegative atoms strongly attracts the shared pair of electrons towards it self. As a result, the hydrogen atom becomes slightly positive and the electronegative element becomes slightly negative when two molecules of such a substance say HF come close to each other, the negatively charged fluorine atom of one molecule attracts the positively charged hydrogen atom of the other molecule. A hydrogen atom, thus links to highly electronegative atoms, one by a strong covalent bond and the other by weak electrostatic attraction as shown ahead `overset(delta+)H-overset(delta-)F....overset(delta+)H-overset(delta-)F`, the dotted line represents a hydrogen bond. High density of water as compared to ice is due toA. dipole-dipole interactionB. hydrogen bondingC. dipole induced dipole interactionD. None of the above. |
| Answer» Correct Answer - B | |
| 62. |
It is a common observation that many compounds containing hydrogen attached to highly electronegative elements such as oxygen, nitrogen, or fluorine, often exhibit unexpected properties, such as relatively high melting points, boiling points, viscosity, solubility in water etc Such an unexpected behaviour can be explained on the basis of hydrogen bonding. When a hydrogen atom is attached to a highly electronegative element of small size such as F. N, 0. etc. the electronegative atoms strongly attracts the shared pair of electrons towards it self. As a result, the hydrogen atom becomes slightly positive and the electronegative element becomes slightly negative when two molecules of such a substance say HF come close to each other, the negatively charged fluorine atom of one molecule attracts the positively charged hydrogen atom of the other molecule. A hydrogen atom, thus links to highly electronegative atoms, one by a strong covalent bond and the other by weak electrostatic attraction as shown ahead `overset(delta+)H-overset(delta-)F....overset(delta+)H-overset(delta-)F`, the dotted line represents a hydrogen bond. Which is most viscous ?A. `CH_3OH`B. `C_2H_5OH`C. `undersetunderset(CH_2OH)(|)(C )H_2OH`D. None |
| Answer» Correct Answer - C | |
| 63. |
A molecule of fluorine is formed by :A. the axial p-p orbital overlapB. the sidewise p-p orbital overlapC. the axial s-s orbital overlap.D. the axial s-p orbital overlap. |
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Answer» Correct Answer - A `F_(2)` moecule is formed by axial p-p overlap. |
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| 64. |
Statement 1: There are ten valence electrons on the sulphur atom in `SF_(4)` molecule. Statement 2: The structure of `SF_(4)` molecule is distorted trigonal bipyramidal.A. Statemetn-1 is true, Statrment-2 is also true, Statement -2 is the correct explanation of statement-13B. Statement -1 is true , Statement 2 is also true, Statement-2 is not the correct ezplanation of Statement-13C. Statement-1 is true, Statement-2 is false.D. Statement-1 is false, Statement-2 is true. |
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Answer» Correct Answer - b In `SF_(4)` molecule, the S atom has four shared electron pairs and one lone pair (8+2=10 electrons). |
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| 65. |
The shape of `SF_(3)Cl_(3)` molecule is:A. trgonal bi-pyramidalB. cubicC. octahedralD. thetrahedral. |
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Answer» Correct Answer - C is the correct answer. |
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| 66. |
An ether is more volatile then alcohol having same molecular fromula . This is due to :A. Intermolecular H-bonding in ethersB. Intermolecular H-bonding in alcobolsC. Dipolar character of ethersD. Resomance character in alchole |
| Answer» Correct Answer - b | |
| 67. |
Assertion Among two cations of similar size the polarising power of cation with pseudo noble gas configuration is larger than cation with noble gas configuration Reasoning Polarising power of `Ag^(o+)` is more than `K^(o+)` .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 - A Cations with pseudo gas configuration has high `Z_(eff)` than cation with nodal gas conguration due to the presence of d-orbitals hence `Ag^(o+)` ion having more `Z_(eff)` than `K^(o+)` and is smaller than `K^(o+)` ion . |
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| 68. |
Assertion: Boron always froms covalent bond. Reason: The small size of `B^(3+)` favours formation of covalent bond.A. Statemetn-1 is true, Statrment-2 is also true, Statement -2 is the correct explanation of statement-11B. Statement -1 is true , Statement 2 is also true, Statement-2 is not the correct ezplanation of Statement-11C. Statement-1 is true, Statement-2 is false.D. Statement-1 is false, Statement-2 is true. |
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Answer» Correct Answer - a Statement 2 is is correct explanation foe Statement 1. |
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| 69. |
In which of the following pairs of molecules centreal atom is `sp^(2)` hybrisised ?A. `NO_(2)^(-)and NH_(3)`B. `BF_(3)and NH_(2)^(-)`C. `HN_(2)^(-)and H_(2)O`D. `BF_(3)and NH_(2)^(-)` |
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Answer» Correct Answer - B For `BF_(3),H=^(1//2)[3+3-0+0]=3(sp^(2))` For `NO_(2)^(-)ion,H=^(1//2)[5+0-0+1]=3(sp^(2))` Option A. |
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| 70. |
Which of the following is a polar molecule ?A. `BfF_(3)`B. `SF_(4)`C. `SiF_(4)`D. `XeF_(4)` |
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Answer» Correct Answer - B `SF_(4)` has a distorted geometry due to presence of a lone pair of electons and hence is polar . All other molecules are symmertical and have zero net dipole moment . |
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| 71. |
The maximum number of ` 90^@` angles hetween bond pair -bond pair of electron is observed in :A. `sp^(2)d^(2)` hybridisationB. `sp^(2)d` hybridisationC. `dsp^(2)` hybridisationD. `dsp^(3)` hybridisation |
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Answer» Correct Answer - a `sp^(3)D^(2)` -hybricisation gives octahedral geometry having maximum no of `90^(@)` angle (six) |
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| 72. |
The charge/size ratio of a cation determines its. polarizing power. Which one of the following sequences respresents the increasing order of the polarizing power of the cationic species ,`K^(+), Ca^(2+), Mg^(2+), Ba^(2+)`?A. `Ca^(2+) lt Mg^(2+) lt Be^(2+) lt K^(+)`B. `Mg^(2+) lt Be^(2+) lt K^(+) lt Ca^(2+)`C. `Be^(2+) lt K^(+) lt Ca^(2+) lt Mg^(2+)`D. `K^(+) lt Ca^(2+) lt Mg ^(2+) lt Be^(2+)` |
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Answer» Correct Answer - D The polarizing power increases when charge/size ratio of cation increases which in turn increases as the charge increases and size decreases , i.e., `K^(+) lt Ca^(2+) lt Mg^(2+) lt Be^(2+)` . |
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| 73. |
Assertion . The `HF_(2)^(-)` ion exists in the solid state and also in the liquid state but not in aqueous solution . Reason . The magnitue of hydrogen bonds inbetween HF -HF moelcules is weaker than that inbetween HF and `H_(2) O` molecules.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 R is the correct explanation of A . |
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| 74. |
Assertion Both o-hydroxy benzaldehyde and p-hydroxy benzaldehyde haveA. 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 - b Correct explanation , o-hydroxybenzaldehyde forms intramolecular hydrogen bond whereas p- hydroxybenzaldehyde forms inermolecular hydrogen bonds. |
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| 75. |
If the electron configuration of an element is `1s^(2), 2s^(2), 2p^(6), 3s^(2), 3p^(2), 3d^(2), 4s^(2)`, the four electrons involved in chemical bond formation will beA. `3p^(6)`B. `3p^(6), 4s^(2)`C. `3p^(6),3d^(2)`D. `3d^(2), 4s^(2)` |
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Answer» Correct Answer - D d) The given electronic configuration shows that an element is vanadium (Z=22). It belongs to d-block of the periodic table. In transition table. In transition elements i.e., of d-block elements, electrons of ns and (n-1)d subshell take part in bond formation. |
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| 76. |
How many H-bonds are formed by each `H_(2) O ` molecule and how many water moleculas are attached to each water molecule and in what diraction ? |
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Answer» Each `H_(2) O ` molecula forms four H-bonds, two O-atom and two with H-atoms. Futher , each ` H_(1)O` molecule in linked to four ` H_(2)O` molecules through H-bonds tetrahedrally |
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| 77. |
`HCN` and `HNC` moleculas have equal number ofA. lone pair and `sigma` bondsB. `sigma` bonds and `pi` bondsC. `pi` bonds and lone pairD. lone pairs, `sigma` bonds and `pi` bonds |
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Answer» Correct Answer - d `H - C = N`: has`1` lone `2 sigma` bonds and two `pi` bonds `H - N overset (+) (rarr) = C overset (-) (rarr)` : has `1` lone `2 sigma` bonds and two `pi` bonds |
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| 78. |
Which of the following orderof energies of molecular orbitals of `N_(2)` is correct?A. `(pi2p_(y) lt (pi2p_(z))lt(pi2p_(x))=(pi2p_(y))`B. `(pi2p_(y))gt(pi2p_(z))gt(pi2p_(x))=(pi2p_(y))`C. `(pi2p_(y))lt(pi2p_(z)) lt(pi2p_(x))=(pi2p_(y))`D. `(pi2p_(y))gt(pi2p_(z))lt(pi2p_(x))=(pi2p_(y))` |
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Answer» Correct Answer - A a) The correct increasing order of energies of molecular orbitals of `N_(2)` is given below `pi` is `ltpi^(star)1sltpi2sltpi^(star)2slt(pi2p_(x) = pi2p_(y))lt pi2p_(z)lt(pi^(star)p_(x)=pi^(star)2p_(y))ltpi^(star)2p_(z)` |
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| 79. |
Sodium metal vaporises on heating and the vapour will have diatomic molecular of sodium `(Na_(2))`. What type of bonding is presetn in these moleculas ? Justify your answer . |
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Answer» Covalent bonds are present in `Na_(2)` . This can be explained on the basis of molecular orbital theory . Electronic configuration of `""_(11)Na = 1s^(2) 2s^(2) 2p^(6) 3s^(1)` . The 3s atomic orbitals of the two sodium atoms combine to form `sigma _(3s) and sigma _(3s)^(**)` molecular orbitals . The two electrons entre into `sigma_(3s)` . Hence, bond order = 1 . Thus , the two sodiium atoms are linked by a single covalent bond . |
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| 80. |
The pair of species with the same bond order is :A. NO,COB. `N_(2),O_(2)`C. `O_(2)^(2-),B_(2)`D. `O_(2)^(+),NO^(+)` |
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Answer» Correct Answer - c (c ) Bond order are `sigma_(2s)^(2)sigma**_(2s)^(2)sigma_(2p_(z))^2pi_(2p_(x))^2pi_(2p_(y))^2pi**_(2p_(x))^1pi**_(2p_(y))^1` and For `O_2^(-):K K 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))^1` |
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| 81. |
Of the two compounds shown below , the vapour pressure of B at a particular temperature is A. higher than that of AB. lower than that of AC. same as that of AD. depends on the amout and size of verssel |
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Answer» Correct Answer - b Than is - o -nitro phenol will only have an intermolecular H-bonding and therefore would exist as a moment .ON the other hand (A) that is p-nitrophenol would have an in termolecular molecule .As a result (B) will have a highest vapour presure |
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| 82. |
Diamagnetic species are those which contain no unapired electrons. Which among the following are diagmagnetic?A. `N_(2)`B. `N_(2)^(2-)`C. `F_(2)^(+)`D. `O_(2)^(-)` |
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Answer» Correct Answer - A::D (a,d) a) Electronic configuration of `N_(2) = sigma1s^(2), sigma^(star)1s^(2), sigma2s^(2), pi2p_(x)^(2)=pi2p_(y)^(2), sigma2p_(z)^(2)`. It has unpaired electron indicates diamagnetic species. b) Electronic configuration of `N_(2)^(2-)` ion =`sigma1s^(2), sigma^(star)s^(2), sigma2s^(2), sigma^(star)2s^(2), pi2p_(x)^(2)=pi2_(y)^(2), sigma2p_(z)^(2), pi^(star)p_(x)^(1)=pi^(star)2p_(y)^(2)` It has two unpaired electrons, paramagnetic in nature. c) Electronic configuration of `O_(2)` = `sigma1s^(2), sigma^(star)2s^(2), pi2p_(s)^(2), sigma^(star)2s^(2), sigma2p_(z)^(2), pi2p_(x)^(2)=pi2p_(y)^(2), pi^(star)2p_(x)^(1)=pi^(star)2p_(y)^(2)` The presence of two unpaired electrons shows its paramagnetic nature. d) Electronic configuration of `O_(2)^(2-) = sigma1s^(2), sigma^(star)1s^(2), sigma2s^(2), sigma2p_(z)^(2), pi2p_(x)^(2)=pi2p_(y)^(2), pi^(star)2p_(x)^(2)=pi^(star)2p_(y)^(2)` It contains no unpaired electron, therefore it is diamagnetic in nature. |
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| 83. |
`CO_(2)` has same geometry as .A. `Hg_(2)Cl_(2)`B. `NO_(2)`C. `C_(2)H_(2)`D. `C_(3)H_(6)` |
| Answer» Correct Answer - A::C | |
| 84. |
Which molecular out of the following does not contain unpaired electrons ?A. `N_(2)^(+)`B. `O_(2)`C. `O_(2)^(2-)`D. `B_(2)` |
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Answer» Correct Answer - C `N_(2)^(+) = KK sigma_(2s)^(2) sigma_(2s)^(**2) pi_(2_(p_(x)))^(2) pi_(2_(p_(y)))^(2) sigma_(2_(p_(z)))^(1)` `O_(2) = KK sigma_(2s)^(2) sigma_(2s)^(**2) sigma_(2p_(z))^(2) pi_(2_(p_(x)))^(2) pi_(2_(p_(y)))^(2) pi_(2_(p_(x)))^(**1) pi_(2_(p_(y)))^(**1) ` `O_(2)^(2-) = KK sigma_(2s)^(2) sigma_(2s)^(**2) sigma_(2p_(z))^(2) pi_(2_(p_(x)))^(2) pi_(2_(p_(y)))^(2) pi_(2_(p_(x)))^(**2) pi_(2_(p_(y)))^(**2) ` `B_(2) = KK sigma_(2s)^(2) sigma_(2s)^(**2)pi_(2_(p_(x)))^(1) pi_(2_(p_(y)))^(1) ` thus , `O_(2)^(2-)` has no unpaired electrons. |
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| 85. |
Explain the sturcture of `CO_(2)` molecule in terms of reasonance, |
| Answer» Correct Answer - `3.38 : 1` | |
| 86. |
The pair of species with the same bond order is :A. `O_(2)^(2-), B_(2)`B. `O_(2)^(+) , NO^(+)`C. `NO,CO`D. `N_(2) , O_(2)` |
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Answer» Correct Answer - A Bond order are `O_(2)^(2-) = 1.0, B_(2) = 1.0, O_(2)^(+) = 2.5 ` ` NO^(+) = 3 NO = 2.5 , CO = 3, N_(2) = 3, O_(2) = 2 ` Thus , `O_(2)^(2-) and B_(2)` have tha same bond order viz 1. |
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| 87. |
Write vitriol is not isomorphous with .A. `K_(2)SO_(4)`B. `MgSO_(4)`C. `CaSO_(4)`D. `H_(2)SO_(4)` |
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Answer» Correct Answer - A::B::C::D White vitriol is `Zn SO_(4) . 5H_(2)O` . |
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| 88. |
Consider the following species `CN^(-),CN^(-),NO and `CN`. Which one of these will hqave the highest bond order ?A. `NO`B. `CN^(-)`C. `CN^(+)`D. `CN` |
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Answer» Correct Answer - b `{:("Ion species order" ,"Total electron", "Bond"),(NO,15,2.5),(CN^(-),14,3),(CN^(+),12,2),(CN,13,2.5):}` |
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| 89. |
The hybridization of atomic orbitals of nitrogen is `NO_(2)^(+), NO_(3)^(-)`, and `NH_(4)^(+)` respectively areA. `sp,sp^(3),and sp^(2)` respectivelyB. `sp,sp^(2),and sp^(3)` respectivelyC. `sp^(2),sp,and sp^(3)` rspectivelyD. `sp^(2),sp^(3),and sp` respectively |
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Answer» Correct Answer - B `NO_(2)^(+)=1//2[5+0-1+0]=4//2=sp` `NO_(3)^(-)=1//2[5+0-0+1]=6//2=sp^(2)` `NH_(4)^(+)=1//2[5+4-1+0]=8//4=sp^(3)` |
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| 90. |
Identify the least stable among the followingA. `Li^(-)`B. `Be^(-)`C. `B^(-)`D. `C^(-)` |
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Answer» Correct Answer - B `Li^(-)=1s^(2)2s^(2)` `Be^(-)=1s^(2)2s^(2)2p_(y)^(1)` `Be^(-)=1s^(2)2s^(2)2p_(x)^(1)2p_(y)^(1)` `C^(-)=1s^(2)2s^(2)2p_(x)^(1)2p_(z)^(1)` |
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| 91. |
Which of the following statements are correct about `CO_(3)^(2-)` ?A. The hybridisation of central atom is `sp^(3)`B. Its resonance structrue has cone C-O single bond and two C= O double bondsC. The average formal charge on each oxygen atom in 0 . 67 unitsD. All C-O bond length are equal . |
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Answer» Correct Answer - C::D Hybridisation involved is `sp^(2)` . Hence ,(a) is wrong . Resonance structure has both C-O bond of equal length . Hence , (b) is wrong. formal charge on each O-atom = `(" Total charge ")/(" No, of O-atoms " ) = (-2)/(3) = -0.67 ` units All C-O bond lengths are equal as mentioned above . |
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| 92. |
Diamagnetic species are those which contain no unpaired electrons. Which among the following are diamagnetic ?A. `N_(2)`B. `N_(2)^(2-)`C. `O_(2)`D. `O_(2)^(2-)` |
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Answer» Correct Answer - A::D `N_(2) and O_(2)^(2-)` have no unpaired electrons |
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| 93. |
Which one of the following molecules is planar?A. `NF_(3)`B. `NCI_(3)`C. `PH_(3)`D. `BF_(3)` |
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Answer» Correct Answer - d Molecule Hybridized state (a) `NF_(3) H = (1)/(2) (7+1) = 4 implies N is sp^(3)` hybridized (b) `NCI_(3) H = (1)/(2) (5+3) = 4 implies N is sp^(3)` hybridized (c) `PH_(3) H = (1)/(2) (5+3) = 4 implies P is sp^(3)` hybridized (d) `BF_(3) H = (1)/(2) (3+5) = 3 implies B is sp^(2)` hybridized and is planar. |
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| 94. |
Blue vitriol hasA. Ionic bondB. Coordinate bondC. Hydrogen bondD. All the above |
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Answer» Correct Answer - d Blue vitriol is `CuSO_(4). SH_(2)O` has all types of bonds. |
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| 95. |
The common features among the species ` CN^- , CO ` and ` NO^+` are :A. bond order three and isoelectronicB. bond order three and weak field ligandsC. bond order tow and `pi`-acceptorsD. isoelectronic and weak field ligands. |
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Answer» Correct Answer - A is the correct answer. |
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| 96. |
Consider the following species : `CN^(+), CN^(-), NO and CN` Which one of these will have the highest bond order ?A. `CN^(+)`B. `CN^(-)`C. NOD. CN |
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Answer» Correct Answer - B `NO (15e^(-)) : sigma (1s)^(2) , 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))^(1) = pi^(**) (2p_(y))^(0)` B.O. ` = (10 - 5)/(2) = 2.5 ` `CN^(-) (14 e^(-)) : sigma(1s)^(2), sigma^(**) (1s)^(2), sigma (2s)^(2)` , ` sigma^(**) (2s)^(2) , pi (2p_(x))^(2) = pi (2p_(y))^(2) ` `sigma (2p_(z))^(2)` B.O. ` = (10- 4)/(2) = 3 ` ` CN (13 e^(-)) : sigma(1s)^(2), sigma^(**) (1s)^(2), sigma (2s)^(2)` , ` sigma^(**) (2s)^(2) , pi (2p_(x))^(2) = pi (2p_(y))^(2) ` `sigma (2p_(z))^(1)` B.O. `= (9-4)/(2) = 2.5 ` `CN (12 e^(-)) : sigma(1s)^(2), sigma^(**) (1s)^(2), sigma (2s)^(2)` , `sigma^(**) (2s)^(2) , pi (2p_(x))^(2) = pi (2p_(y))^(2) ` B.O. ` = (8 - 4)/(2) = 2 ` Hence , `CN^(-)` has the highest bond order . |
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| 97. |
The common features among the species ` CN^- , CO ` and ` NO^+` are :A. Bond order three and isoelectronicB. Bond order three and weak field ligandsC. Bond order two and pi acceptorsD. Isoelectronic and weak field ligands |
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Answer» Correct Answer - A For bond order refer Number of electrons in `CN^(Theta), CO` and `NO^(o+)` are `CN^(Theta) rArr 6 + 7 +1 =14` `CO rArr 6 +8 =14` `NO^(o+) hArr 7 +8 -1 = 14` Hence are isoelectronic Bond order of `CN^(Theta), CO` and `NO^(o+) =3` . |
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| 98. |
The common featrues among the species `CN^(-), CO and CO^(+)` areA. Bond order three and isoelectronicB. Bond order three and weak-field ligandsC. Bond order two and `pi-`acceptorD. Isoelectronic and weak-field ligands |
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Answer» Correct Answer - A Each of them has 14 electrons and bond order = 3 |
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| 99. |
The common features among the species ` CN^- , CO ` and ` NO^+` are :A. bond order three and isoelectronicB. bond order three and weak ligandsC. bond order two and `pi` acceptorsD. isoelectronic and weak fieldligands |
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Answer» Correct Answer - a Bond order three and isoelectronic. |
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| 100. |
In forming `C_2 to C_2^(+)` and (ii) `O_2 to O_2^(+)`, the electrons respectively are removed fromA. `(pi_(2pz)^** or pi_(2px)^**) and (pi_(2py)^** or pi_(2px)^**)`B. `(pi_(2py)^** or pi_(2px)^**) and (pi_(2py)^** or pi_(2px))`C. `(pi_(2py) or pi_(2px)) and (pi_(2py)^** or pi_(2px)^**)`D. `(pi_(2py)^** or pi_(2px)^**) and (pi_(2py) or pi_(2px))` |
| Answer» Correct Answer - C | |