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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.
| 501. |
In `XeF_2, XeF_4` amnd `XeF_6`, the number of lone pair of electrons on ` Xe` are respectively :A. `2, 3, 1`B. `1, 2, 3`C. `4, 1, 2`D. `3, 2, 1` |
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Answer» Correct Answer - d `Xe` in `XeF. XeF_(4) and ZXeF_(6) show sp^(3)d^(2) and sp^(3)d^(2)` hybridisation respectively. |
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| 502. |
Which molecule has trigonal planar geometry ?A. `BF_(3)`B. `NH_(3)`C. `PCl_(3)`D. `IF_(3).` |
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Answer» Correct Answer - A `BF_(2)` has trigonal planar geometry. `H=1/2[3+3-0]=6/2=3(sp^(2))` |
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| 503. |
Which moleucle is T-shaped ?A. `BeF_(2)`B. `BCI_(3)`C. `NH_(3)`D. `CIF_(3)` |
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Answer» Correct Answer - D Hybridisation of `:overset(..)CIF_(3) implies (1)/(2) (7 +3) =5 = sp^(3)` d with two lp of electrons It has `T` or arrow shape `(rarr)` . |
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| 504. |
The hybridisation of the central atom in `ICI_(2)^(o+)` is .A. `sp^(3)`B. `sp^(2)`C. `sp^(2)`D. `sp` |
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Answer» Correct Answer - A Hybridisation of `ICI_(2)^(o+) implies (1)/(2) (7 +2 -) implies 4 =sp^(3)` . |
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| 505. |
Explain unusual stabillity of chorohybrate thugh a compound with two or more -OH groups present on one carbon atom is usually unstabel . |
| Answer» The unusual stability of chloral hydrate has been attributed to -I effect of chlorine and to the formation of intramolecular hydrogen bonds `(CI_(3) C=-CH_(3) C - CH (OH)_(2))` . | |
| 506. |
The bond length in `O_2^+, O_2 ,O_2^-` and ` O_2^(2-)` follows the order :A. `O_2^(+) lt O_2 lt O_2^(-) lt O_2^(+)`B. `O_2^(2-) lt O_2^(-) lt O_2 lt O_2^(2+)`C. `O_2^(+) lt O_2^(-) lt O_2 lt O_2^(2-)`D. `O_2^(+) lt O_2^(2-) lt O_2 lt O_2^(-)` |
| Answer» Correct Answer - B | |
| 507. |
The bond order of `CO` and `NO` is .A. 3 and 2B. 3 and 2.5C. 3 and 1.3D. 3 and 3.5 |
| Answer» Correct Answer - B | |
| 508. |
Atomic orbitals of bonded atoms combine to form molecular orbitals. The number of molecular orbitals formed is equal to the number of atomic orbitals taking part in the bond formation. When two atomic orbitals combine, two molecular orbitals are formed one of which has lower energy than the combining orbitals and is called bonding Molecular Orbital (MO). Whereas the other having higher energy than the two combining atomic orbitals is called Anti Bonding Molecular orbitals (ABMO) The two combining atomic orbitals must have comparable energies and should be properly oriented to allow considerable overlapping. If the overlapping is end to end along internuclear axis, the molecular orbital is called sigma and if the overlapping is lateral 1.e., sidewise the molecular orbital is called pie. Just like atomic orbitals, the molecular orbitals also have varying energy levels. Filling of electrons in molecular orbitals takes place following the same rules as followed for filing of atomic orbitals. The order of filling may not be same for all the molecules or their ions. Bond order is a useful parameter for comparing the various characteristics of molecules. Which of the following pair is expected to have the same bond order ?A. `O_2,N_2`B. `O_2^(+),N_2^(-)`C. `O_2^(-),N_2^(+)`D. `O_2^(-),N_2^(-)` |
| Answer» Correct Answer - B | |
| 509. |
Atomic orbitals of bonded atoms combine to form molecular orbitals. The number of molecular orbitals formed is equal to the number of atomic orbitals taking part in the bond formation. When two atomic orbitals combine, two molecular orbitals are formed one of which has lower energy than the combining orbitals and is called bonding Molecular Orbital (MO). Whereas the other having higher energy than the two combining atomic orbitals is called Anti Bonding Molecular orbitals (ABMO) The two combining atomic orbitals must have comparable energies and should be properly oriented to allow considerable overlapping. If the overlapping is end to end along internuclear axis, the molecular orbital is called sigma and if the overlapping is lateral 1.e., sidewise the molecular orbital is called pie. Just like atomic orbitals, the molecular orbitals also have varying energy levels. Filling of electrons in molecular orbitals takes place following the same rules as followed for filing of atomic orbitals. The order of filling may not be same for all the molecules or their ions. Bond order is a useful parameter for comparing the various characteristics of molecules. In the homonuclear diatomic molecule which of the following sets of M.O. orbitals are grade or un-gradeA. `sigma_(2s) , pi_(2p_x)`B. `sigma_(2s)^(**) , pi_(2p_x)^(**)`C. `sigma_(2s)^(**) , pi_(2p_x)`D. `sigma_(2p_x) , pi_(2p_x)^(**)` |
| Answer» Correct Answer - D | |
| 510. |
Atomic orbitals of bonded atoms combine to form molecular orbitals. The number of molecular orbitals formed is equal to the number of atomic orbitals taking part in the bond formation. When two atomic orbitals combine, two molecular orbitals are formed one of which has lower energy than the combining orbitals and is called bonding Molecular Orbital (MO). Whereas the other having higher energy than the two combining atomic orbitals is called Anti Bonding Molecular orbitals (ABMO) The two combining atomic orbitals must have comparable energies and should be properly oriented to allow considerable overlapping. If the overlapping is end to end along internuclear axis, the molecular orbital is called sigma and if the overlapping is lateral 1.e., sidewise the molecular orbital is called pie. Just like atomic orbitals, the molecular orbitals also have varying energy levels. Filling of electrons in molecular orbitals takes place following the same rules as followed for filing of atomic orbitals. The order of filling may not be same for all the molecules or their ions. Bond order is a useful parameter for comparing the various characteristics of molecules. Which of the following combinations is not allowed (assume z-axis as the internuclear axis ) ?A. 2s and 2sB. `2p_x and 2p_x`C. `2p_z and 2p_z`D. `2p_x and 2p_y` |
| Answer» Correct Answer - D | |
| 511. |
Atomic orbitals of bonded atoms combine to form molecular orbitals. The number of molecular orbitals formed is equal to the number of atomic orbitals taking part in the bond formation. When two atomic orbitals combine, two molecular orbitals are formed one of which has lower energy than the combining orbitals and is called bonding Molecular Orbital (MO). Whereas the other having higher energy than the two combining atomic orbitals is called Anti Bonding Molecular orbitals (ABMO) The two combining atomic orbitals must have comparable energies and should be properly oriented to allow considerable overlapping. If the overlapping is end to end along internuclear axis, the molecular orbital is called sigma and if the overlapping is lateral 1.e., sidewise the molecular orbital is called pie. Just like atomic orbitals, the molecular orbitals also have varying energy levels. Filling of electrons in molecular orbitals takes place following the same rules as followed for filing of atomic orbitals. The order of filling may not be same for all the molecules or their ions. Bond order is a useful parameter for comparing the various characteristics of molecules. In the formation of `N_2^+` from `N_2`, the electron is removed fromA. `sigma` orbitalB. `pi` orbitalC. `sigma^(**)` orbitalD. `pi^(**)` orbital |
| Answer» Correct Answer - A | |
| 512. |
Atomic orbitals of bonded atoms combine to form molecular orbitals. The number of molecular orbitals formed is equal to the number of atomic orbitals taking part in the bond formation. When two atomic orbitals combine, two molecular orbitals are formed one of which has lower energy than the combining orbitals and is called bonding Molecular Orbital (MO). Whereas the other having higher energy than the two combining atomic orbitals is called Anti Bonding Molecular orbitals (ABMO) The two combining atomic orbitals must have comparable energies and should be properly oriented to allow considerable overlapping. If the overlapping is end to end along internuclear axis, the molecular orbital is called sigma and if the overlapping is lateral 1.e., sidewise the molecular orbital is called pie. Just like atomic orbitals, the molecular orbitals also have varying energy levels. Filling of electrons in molecular orbitals takes place following the same rules as followed for filing of atomic orbitals. The order of filling may not be same for all the molecules or their ions. Bond order is a useful parameter for comparing the various characteristics of molecules. The bond order (BO) in `B_2` molecule is |
| Answer» Correct Answer - B | |
| 513. |
Atomic orbitals of bonded atoms combine to form molecular orbitals. The number of molecular orbitals formed is equal to the number of atomic orbitals taking part in the bond formation. When two atomic orbitals combine, two molecular orbitals are formed one of which has lower energy than the combining orbitals and is called bonding Molecular Orbital (MO). Whereas the other having higher energy than the two combining atomic orbitals is called Anti Bonding Molecular orbitals (ABMO) The two combining atomic orbitals must have comparable energies and should be properly oriented to allow considerable overlapping. If the overlapping is end to end along internuclear axis, the molecular orbital is called sigma and if the overlapping is lateral 1.e., sidewise the molecular orbital is called pie. Just like atomic orbitals, the molecular orbitals also have varying energy levels. Filling of electrons in molecular orbitals takes place following the same rules as followed for filing of atomic orbitals. The order of filling may not be same for all the molecules or their ions. Bond order is a useful parameter for comparing the various characteristics of molecules. In which of the following ionization processes, the bond order has increased and the magnetic behaviour has changed ?A. `N_2toN_2^(+)`B. `C_2toC_2^(+)`C. `NOtoNO^(+)`D. `O_2toO_2^(-)` |
| Answer» Correct Answer - C | |
| 514. |
What is the bond order between C and O in `CO` ? |
| Answer» Correct Answer - 3 | |
| 515. |
The order of energies of following combination (a)2HHe (b)`H_2+He_2` (c )`He_2+2H` and (d)`H_2+2He` isA. `d lt a ltbltc`B. `d ltbltaltc`C. `cltaltblt d`D. `c lt b ltaltd` |
| Answer» Correct Answer - A | |
| 516. |
When two atoms combine to form a molecule .A. Energy is releasedB. Energy is absorbedC. Energy is neither released nor absorbedD. Energy may either released or absorbed |
| Answer» Correct Answer - A | |
| 517. |
The combination of atoms takes place so thatA. They can gain two electrons in the outermost orbitB. They get eight electrons in the outermost orbitC. They acquire stability by lowering of energyD. They get eighteen electrons in the outermost orbit. |
| Answer» Correct Answer - C | |
| 518. |
The combination of atoms occur because they tendA. To decrease number of electrons in the outermost orbitB. To attain an inert gas configurationC. To remain same number of electrons in the outer most orbitD. To attaiin 18 electrons in the outermost orbit |
| Answer» Correct Answer - B | |
| 519. |
Which of the following is the correct order of dipole moment? A. `I gt II gt III`B. `II gt I gt III`C. `III gt I gt II`D. `III gt II gt I` |
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Answer» Correct Answer - a The group moment of `-NO_(2), -OH, - CI and -Me and 3.98 D. 1.6 D. 1.5 Dand 0.4 D` respectively |
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| 520. |
In a compound `A,B` if the element `B` attracts electrons more than element `A` it will tend to be_____charged . |
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Answer» Correct Answer - Negatively Negatively |
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| 521. |
Assertion(A) - `PCl_5` exists but `BiCl_5` does not. Reason(R ) : Bi does not contain d orbitals to expand its octet.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 | |
| 522. |
Increasing order of dipole moment isA. `CF_(4) lt NH_(3) lt NF_(3) lt H_(2)O`B. `CF_(4) lt NH_(3) lt H_(2)O lt NF_(3)`C. `CF_(4) lt NF_(3) lt H_(2)O lt NH_(3)`D. `CF_(4) lt NF_(3) lt NH_(3) lt H_(2)O` |
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Answer» Correct Answer - d `CF_(4)` is symmetrical and non poler `mu = 0`. In `NF_(3)`.the resultant of three `N - F` dipole oppose the effect of lone pairs of electrons of `N` -atom , whereas in `NH_(3)`, the effect of lone pairs of electrons of `N` atom and resultant of three `N - H`dipoles are in the same direction. Therefore`mu_(NH_(3)) (=1.46 D)gt mu_(NH_(3)` `mu_(H_(2)O) = 1.85 D` |
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| 523. |
Acidic strength order `CI_(2)O_(7) gtSO_(3) gtP_(4) O_(10)` . |
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Answer» Correct Answer - T Higher the oxidation state of the central atom more acidic is that compound `overset(+7)(CI_(2)O_(7))gtoverset(+6)(SO_(3))gtoverset(+5)(P_(4)) O_(10)` . |
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| 524. |
Due to hybridisation________hybrid orbitals are formed . |
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Answer» Correct Answer - Equivalent Edquivalent |
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| 525. |
The bond multiplicity leads to______in bond length . |
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Answer» Correct Answer - Decreases Decreases |
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| 526. |
Describe the change in hydridization (if any) of the Al atom in the following reaction : `AlCl_(3) P Cl^(-) to AlCl_(4)^(-)` |
| Answer» E.C. of `._(13)Al = 1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p_(x)^(1)` (Ground state ) or = `1s^(2) 2s^(2) 2p^(6) 3s^(1) 3p_(x)^(1) p_(y)^(1)` (Excited state ). | |
| 527. |
Which of the following is not a correct statement ?A. Every `AB_(2)` molecule does in fact has square pyramid structure.B. Maltiple bonds are always shorter then cotreponding single bonds.C. The electron deficient molecule can act as Lewis acids.D. The canonical structure have no real existence. |
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Answer» Correct Answer - a `IF_(3)` is squre pyramid `(sp^(3) d^(2)- "hybridization in" 1)` in `PCI_(3)` is trigonaal bipyramid (`sp^(3)` d-hybridazition in P). |
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| 528. |
Describe the change in hybridisation (if any) of the `AI` atom in the following: `AlCl_(3)+Cl^(ө)rarr AlCl_(4)^(ө)` |
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Answer» In `AlCl_(3),` the central Al atom is `sp^(2)` hybridised while in the ion `[AlCl_(4)]^(-),` it has `sp^(3)` hybridisation. `underset(H=1//2[3+3-0+0]=3(sp^(2)))(AlCl_(3))" "underset(H=1//2[3+4-0+1]=4(sp^(2)))([AlCl_(4)]^(-))` |
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| 529. |
In which of the following pairs first partner is more soluble in water than second partner ?A. B. NaCl and AgClC. `Be(OH)_(2)` and `Ba(OH)_(2)`D. |
| Answer» Correct Answer - B::D | |
| 530. |
Which of the following is a correct statement ?A. `CO_(2)` is a monomer while `SiO_(2)` is a three dimensional giant moleculeB. Graphite is a nonconductor of electricityC. In `(CH_(3))_(3) , C-N-C` bond angle is approximately `107^(@)` whereas in `(SiH_(3))_(3)` N , `Si-N-Si` bond angle is approximately `120^(@)`D. In `CO_(3)^(2-)` ion all C-O bonds are equal , while in `H_(2)CO_(3)` all C-O bonds are not equal |
| Answer» Correct Answer - A::C::D | |
| 531. |
Which of the following is a correct statement ?A. All molecules will polar bond have dipole momentB. `SnCl_(2)` is non-linear moleculeC. Dipole moment of `CH_(3)Cl` is greater than `CH_(3)F`D. `I_(3)^(-)` has linear shape |
| Answer» Correct Answer - B::C::D | |
| 532. |
In solid `PCl_(5)` exist as `[PCl_(4)]^(+) [PCl_(6)]^(-)`. The hybridisation is P is /areA. `sp^(3)`B. `sp^(3)d`C. `sp^(3)d^(2)`D. `sp^(3) d^(3)` |
| Answer» Correct Answer - A::C | |
| 533. |
Choose the correct statement regarding `PCl_(5)`A. All P-Cl bond lengths are equalB. `P-Cl_("axial ")` bond lengths is more than `P-Cl_("equitorial")`C. `P-Cl_("equitorial")` bond lengths are more than `P-Cl_("axial")`D. `P-Cl_("equitorial")` bond length may be equal or larger than `P-Cl_("axial")` |
| Answer» Correct Answer - B | |
| 534. |
The bond length the species `O_(2), O_(2)^(+) and O_(2)^(-)` are in the order ofA. `O_(2) gt O_(2) gtO_(2)^(-)`B. `O_(2)^(+) gt O_(2)^(-) gt O_(2)`C. `O_(2) gt O_(2)^(+) gtO_(2)^(-)`D. `O_(2)^(-) gt O_(2) gtO_(2)^(+)` |
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Answer» Correct Answer - a As bond order increases bond length decreases, the bond order of species are `= (Number of bonding electron - Number of antibonding electron)/(2)` For `O_(2) = (10-6)/(2) = 2 , O_(2)^(+) = (10-5)/(2) = 2.5` `O_(2)^(-) = (10-7)/(2) = 1.5` So bond order `O_(2)^(+) gt O_(2) gt O_(2)^(-)` and length are `O_(2)^(+) gt O_(2) gt O_(2)^(-)` |
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| 535. |
In which of the following substance will hydrogen bond be strongest ?A. HClB. `H_(2) O `C. HID. `H_(2) S` |
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Answer» Correct Answer - B HCl, HI and ` H_(2)S` do not form H-form H-bonds. Only `H_(2)O` forms H-bonds . |
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| 536. |
Which of the following comnpounds is paramagnetic?A. `CO`B. `NO`C. `O_(2)^(2-)`D. `O_(3)` |
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Answer» Correct Answer - b `NO` has an odd electron bond. Hence it will be paramagenetic. |
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| 537. |
The compound which contains both covalent and coordinate bond isA. `C_(2)H_(5)NC`B. `C_(2)H_(5)CN`C. `HCN`D. None of these |
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Answer» Correct Answer - a Though all compounds have covlent have covalent bonds but there is a coordination bond also between `N and C` in `C_(2)H_(3)NC`. |
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| 538. |
Which one of the following contains ionic , covalent and coordinate bonds ?A. `NaOH`B. NaClC. NaCND. NaNC |
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Answer» Correct Answer - D `Na^(+) N underset(to)=C^(-)` contains ionic , covalent and coodinate bonds. |
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| 539. |
Which of the following species is not paramagnetic ?A. `O_(2)`B. `B_(2)`C. `NO`D. `CO` |
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Answer» Correct Answer - D `O_(2) = 16` electrons . M.O. configuration is ` KK 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))^(1) ` (paramagnetic) `B_(2) = 10 ` electrons. M. O. configuration is `KK sigma (2s)^(2) sigma^(**)(2s)^(2) pi (2p_(x))^(1) pi(2p_(y))^(1)` (paramagnetic) NO = 15 electrons .M.O. configuration is `KK sigma (2s)^(2) sigma^(**)(2s)^(2) sigma (2p_(z))^(2) pi (2p_(x))^(2) pi (2p_(y))^(2)` `pi^(**)(2p_(x))^(1) ` CO = 14 electrons . M.O. configuration is `KK sigma (2s)^(2) pi(2p_(x))^(2)pi(2p_(y))^(2) sigma (2p_(z))^(2) sigma ^(**) (2s)^(2)` (diamagnetic) |
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| 540. |
The bond order of `NO_(3)^(-)` ion is ………… and its shape is ………. |
| Answer» 1.33 planer triangular | |
| 541. |
Which is the weakest among the following types of bondsA. Ionic bondB. Covalent bondC. Metallic bondD. Hydrogen bond |
| Answer» Correct Answer - D | |
| 542. |
Which is (are) correct among the following ? .A. The radius of `CI^(Θ)` ion is `1.56 Å` while that of `Na^(o+)` ion is `0.95Å` .B. The radius of `CI` atom is `0.99` while that of Na atom is `1.54`C. The radius of `CI` atom is `0.99` while that of `CI^( Θ)` ion is `0.81`D. The radius of Na atom is 0.95 while that of `Na^(o+)` ion is `1.54` . |
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Answer» Correct Answer - A::B `r_(CI)ltr_(Na)` [along the period `(rarr)` size decreases] but `r_(CI^(Θ)) gt r_(Na)^(o+)` as in `CI-`atom an an `e^(-)` has been added and an `e^(-)` has been removed from Na, leading to removal of 2s shell which is still there in CI-atom . |
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| 543. |
The energy released when gaseous hydrogen atoms combine together to form one of hydrogen molecules is …………….. `KJ mol ^(-1)` . |
| Answer» Correct Answer - 435.8 | |
| 544. |
`SnCI_(2)` is a non-linear molecule . |
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Answer» Correct Answer - T True `SnCI_(2)` is a non-linear molecules It is V-shaped . |
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| 545. |
The carbon-carbon bond order in benzene isA. OneB. TwoC. One and halfD. One and two alternately |
| Answer» Correct Answer - C | |
| 546. |
The bond order of ` N_(2), N_(2)^(+) , N_(2) ^(-) and N_(2)^(2-)` respectively are ……….., ………….. ,………….. and …………… . |
| Answer» Correct Answer - 3, 2.5 , 2.5 , 2 | |
| 547. |
Which of the following is the most stable resonance structure . |
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Answer» Structure `(c ) ` is the most stabel resonance structure (i) Number of covalent bonds in (a) and (b) `=3` (ii) Number of covalent bonds in (c ) and (d) `=14` (iii) In ( c) Positive charge is no`N` whereas in (d) it is than `O` atom Since `N` uis less `EN` (electronegative) than `O` (c) is more stable (iv) Order of stability `c gt d gt a gt b` . |
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| 548. |
Among the following , which one is a wrong statement ?A. `PH_(5) and BiCl_(5)` do not existB. ` p pi-dpi` bonds are present in `SO_(2)`C. `SeF_(4) and CH_(4)` have same shapeD. `I_(3)^(-)` has bent geometry . |
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Answer» Correct Answer - C `(a) PH_(5)` does not exist whereas `PCl_(5) and PF_(5)` exist This is because Cl and F have p-orbitals to overlap with d-orbitals of P but H does not have p-orbitals . Again `BiCl_(3)` exists but `BiCl_(3)` does not due to inert pair effect (b) ` p pi - d pi` bonds are present in `SO_(2)` (c) `SeF_(4) = sp^(3) d, lp = 1 ` shape = see-saw ` CH_(4) = sp^(3) , lp = 0` , shape = tetrahedral (d) ` I_(3)^(+) = sp^(3), lp = 2 `shape = bent/angular. |
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| 549. |
Which of the following is most stableA. B. C. D. |
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Answer» Correct Answer - a In ( e) and (d) there is `lp-bp` repulsions. The size of methyl group than `F`. Hence (b) experince greater repulsion than (a) |
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| 550. |
The states of hybridisation of boron and oxygen atoms in boric acid `(H_3 BO_3)` are respecitivelty :A. `sp^(3) and sp^(3)`B. `sp^(2) and sp^(3)`C. `sp^(3) and sp^(2)`D. `sp^(2) and sp^(2)` |
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Answer» Correct Answer - b `B` has `sp^(2)` and hydrogen has `sp^(3)` hybridisation |
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