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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.
| 401. |
Which of the following is paramagnetic ?A. `O_2`B. `N_2`C. `O_2^(-2)`D. `H_2` |
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Answer» Correct Answer - A `O_2` has two unpaired electrons in `pi^**` molecular orbitals. |
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| 402. |
During change of `NO^(+)to NO`, the electron is added toA. `sigma`-orbitalB. `pi`-orbitalC. `sigma^(**)` orbitalsD. `pi^(**)` orbital |
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Answer» Correct Answer - D M.O. configuration of `NO^(+)` is `(sigma1s)^(2)(sigma^(**)1s)^(2)(sigma2s)^(2)(sigma 2p_(z))^(2)(pi2p_(x))^(2)(pi2p_(y))^(2)` and M.O. configuration of NO is `(sigma1s)^(2)(sigma^(**)1s)^(2)(sigma2s)^(2)(sigma^(**)2s)^(2)(sigma2p_(z))^(2)(pi2p_(x))^(2)(pi2p_(y))^(2)(pi^(**)2p_(y))^(1).` |
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| 403. |
The bond order in `O_2^-`ion isA. 2B. 1C. 2.5D. 1.5 |
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Answer» Correct Answer - D The electronic configuration of `O_(2)^(-)` `O_2^(-) to (sigma1s)^(2) (sigma^(**)1s)^(2)(sigma2s)^(2)(sigma^(**)2s)^(2)(sigma2p_(z))^2` `(pi2p_(x))^(2)(pi 2p_(y))^(2)(pi^(**)2p_(x))^(2)(pi^(**)2p_(y))^1` Bond order =1/2 `[N_(b)-N_a]` =1/2 [10-7]=3/2=1.5 |
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| 404. |
Which of the following is not correct `:`A. Carbon `-` carbon bond length in `CaC_(2)` will be more than that in `CH_(2)C CH_(2)`B. `O-O` bond length in `Na_(2)O_(2)` will be more than that in `KO_(2)`.C. `O-O` bond length in `O_(2)[PtF_(5)]` will be less than in `KO_(2)`D. `N-O` bond length in `NO` gaseous molecule will be smaller than that bond length in `NOCl` gaseous molecule. |
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Answer» Correct Answer - 1 In `CaC_(2)` there is `C=C` , while in `CH_(2)C CH_(2)`, there is only `C=C`. `KO_(2)=K^(+)+O_(2)^(-)` `" "`Bond order `=1.5` `Na_(2)O_(2)=2Na^(+)+O_(2)^(2-)` `" "`Bond order `=1.0` `O_(2)(Pt F_(6))O_(2)^(+)+[Pt F_(6)]^(-)` `" "`Bond order `=2.5` N Bond order `=2.5` while in `NOCl`, bodn order `=2` |
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| 405. |
Find the bond order of ` B_2` |
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Answer» ` B_2=sigma1s^2,sigma^**1s^2,sigma2s^(2),sigma^**2s^2, [(pi2p_y^1),(pi2p_z^1)]` `:.` Bond order of `B_2 1/2 xx [ 6 - 4 ] =1 `. |
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| 406. |
Bond order and magnetic moment of `CO^+` isA. 2.5 and paramagnetic momentB. 3.5 and diamagnetic momentC. 3.5 and paramagnetic momentD. 2.5 and diamagnetic moment |
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Answer» Correct Answer - C HOMO of CO is a `sigma`-non-bonding orbital with a fair degree of antibonding characters. Loss of one electrons lowers the anti-bonding electron number. |
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| 407. |
The paramagnetic behaviour of `B_2` is due to the presence ofA. 2 unpaired electrons in `pi_(n) MO`B. `2 unpaired electrons in `pi^(**) MO`C. 2 unpaired electrons in `sigma^(**)MO`D. 2 unpaired electrons in `sigma_(b)MO` |
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Answer» Correct Answer - A `B_(2)`(electrons=10) `sigma(1s)^(2),sigma^(**)(1s)^(2),sigma(2s)^(2),sigma^(**)(2s)^(2),pi(2p_(x))^(1),pi(2p_(y))^(1)` Thus paramagnetic behaviour of `B_(2)` is due to the presence of two unpaired electrons in `pi`-bonding molecular orbital. |
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| 408. |
Which one among the following does not have the hydrogen bond?A. PhenolB. Liquid `NH_(3)`C. WaterD. `HCl` |
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Answer» Correct Answer - d |
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| 409. |
According to molecular orbital theory, which of the following statements is (are) correct?A. `C_(2)^(2-)` is expected to be diamagneticB. `O_(2)^(2+)` is expected to have a longer bond length than `O_(2)`C. `N_(2)^(+)` and `N_(2)^(-)` have the same bond orderD. `He_(2)^(+)` has the same energy as two isolated He atoms |
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Answer» Correct Answer - a,c |
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| 410. |
In the process, `O_2^+ to O_2^(+2)` +e the electron lost is fromA. Bonding `pi `- orbitalB. Antiboonding `pi`-orbitalC. `2p_(z)` orbitalD. `2p_(x)` orbital |
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Answer» Correct Answer - B Electron lost form antibonding `pi` orbital |
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| 411. |
Consider the following molecules `underset(I)(O_(2)),O_(2)underset(II)(("AsF"_(6))),underset(III)(KO_(2))` Choose the correct answer.A. The correct decreasing bond order is `II gt I gt III`.B. The correct decreasing order of bond length is `III gt II gt I`.C. The bond strength of I is less than that of III.D. Bond dissociation energy is highest in case of III. |
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Answer» Correct Answer - A In `O_(2)(AsF_(6)), O_(2)^(+)` has bond order 2.5 . `O_(2)` has bond order 2.0 . In `KO_(2), O_(2)^(-)` has bond order 1.5 Higher the bond order, smaller is the bond length. Hence, the order is `O_(2)(AsF_(6)) gt O_(2) gt KO_(2)`. |
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| 412. |
Match list (Molecules) with list II (Boiling points) and select the correct answer `{:("List I","List II"),((a)NH_3,"(1)290 K"),((b)PH_3,"(2)211 K"),((c )AsH_3,"(3)186 K"),((d)SbH_3,"(4)264 K"),((e)BiH_3,"(5)240 K"):}`A. a-3, b-2,c-5, d -4,e-1B. a -5, b -3, c-2, d-4, e - 1C. a - 1,b-4,c -5, d-2, e-3D. a- 1, b-2, c-3, d-4, e-5 |
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Answer» Correct Answer - B `NH_3` due to hydrogen bonding has higher boiling point than `PH_3`. From `PH_3` to `BiH_3`, boiling point increases as the size of the central atom increases and so van der Waals forces increase. The increase is so high that they have boiling points higher than `NH_3`. Hence the order of boiling points is `PH_3`lt `AsH_3`lt `NH_3` lt `SbH_3` lt`BiH_3`. b-3,c-2, 2-5, d-4, e-1 |
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| 413. |
The species having bond order different from that in `CO` isA. `NO^(+)`B. `NO^(-)`C. `N_2`D. `CN^(-)` |
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Answer» Correct Answer - B Isoelectronic species usually have the same bond order. `NO^(+), CN^(-)`, and `N_2` are isoelectronic with `CO`(14 electrons each). Hence, all of them possess the same bond order which is 3. On the other hand, `NO^(-)` has 16 electrons. Its configuration is identical to that of `O_2`, hence, its bond order is 2. |
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| 414. |
Which of the following species exhibits the diamangetic behaviour ?A. ` O_2^(2-)`B. ` O_2^+`C. `O_2`D. `NO` |
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Answer» Correct Answer - A This species has `18e^(-)` which are filled in such a way that all moecular orbitals are fully filled, so diamagnetic. `sigma1 s^2 sigma^(**) 1 s^2 , sigma 2s^2 , sigma^(**) 2 s^2 , sigma 2p_(z)^(2), pi2p_(x)^(2)=pi2p_(y)^(2), pi^(**) 2p_(x)^(2)=pi^(**)2p_(y)^(2)` |
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| 415. |
Which hass //have magnetic moment ?A. `[Fe(H_2O)_6]^(2+)`B. `[Ni(CN)_4]^(2-)`C. `[Fe(CN)_6]^(3-)`D. `O_2` |
| Answer» Correct Answer - A::C::D | |
| 416. |
Among the following , the paramagnetic compound is :A. `N_2O`B. `Na_2O_2`C. `O_3`D. `KO_2` |
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Answer» Correct Answer - D `N_2O(7+7+8=22)` and `O_3(3xx8=24)` contain even number of electrons which are all paired. `(Na^(+))_2(O_2^(2-))` contains peroxide ion which is diamagnetic. `K^(+)(O_2^(-))` contains superoxide ion which is paramagnetic `O_2^(-)impliessigma1s^2, sigma^**1s^2, sigma2s^2sigma^**2s^2sigma2p{:(pi2p_x^2,pi^**2p_x^2,,,),(pi2p_y^2,pi^**2p_y^1,,,):}` |
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| 417. |
Which of the following is not a correct statement?A. Every `AB_5` molecule has square pyramidal structure.B. Multiple bonds are always shorter than the corresponding single bonds.C. The electron-deficient molecules can act as Lewis acids.D. The cannonical structure has no real existence. |
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Answer» Correct Answer - A `AB_5` molecule having one lone pair on the central atom (e.g., `BrF_5`) has square pyramidal shape but `AB_5` molecule having zero lone pair on the central atom (e.g., `PCl_5`) has trigonal bipyramidal shape. |
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| 418. |
Which of the following statements is `(are)` correct ?A. `CH_3^(+)` shows `sp^(2)` -hybridization whereas `CH_3^-` shows `Sp^2` -hybridiszationB. `NH_(4)^(+)` has a regular tetrahedral geometyC. `sp^(2)` -hybridiszed orbitals have eqwual `s` and p- characterD. Hybridized orbitals always form sigma-bonds |
| Answer» Correct Answer - A::B::D | |
| 419. |
In which of the following ionixation processes , the bond order has increased and the magnetic behaviour has changed ?A. `C_2 rarr C_2^+`B. No `rarr NO^+`C. ` O_2 rarr O_2^+`D. `N_2rarr N_2^+` |
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Answer» Correct Answer - B In `NO rarr NO^+` , electron is removed from antibonding molecular orbitial so bond order increases and nature changes from paramagnetic to diamgnetic . |
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| 420. |
In which of the following ionixation processes , the bond order has increased and the magnetic behaviour has changed ?A. `O_2rarr O_2^(+)`B. `NO rarr NO^(+)`C. `N_2 rarrN_2^(+)`D. `C_2 rarr C_2^(+)` |
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Answer» Correct Answer - B `NO(15e^(-)s)implies KKsigma2s^2, sigma^**2s^2, sigma2p_z^2{:(pi2p_x^2,,),(pi2p_y^2,,):}]{:(pi^**2p_x^1,,),(pi^**2p_y^0,,):}` It is paramangnetic with bond order `=(10-5)/(2)=2.5` `NO^(+)` loses the unpaired `e^(-)` and becomes diamagnetic. Its bond order `=(10-4)/(2)=3` `O_2` (paramagnetic, `bo=2`)`rarrO_2^(+)` (paramagnetic, `bo=2.5`) `N_2` (diamagnetic, `bo=3`)`rarrN_2^(+)` (paramagnetic, `bo 2.5`) `C_2` (diamagnetic, `bo=2`)`rarrC_2^(+)` (paramagnetic, `bo 1.5`) |
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| 421. |
How many `sigma and pi` bonds are present in tetra cyanoethylene ?A. Nine `sigma` and nine `pi`B. Five `pi` and nine `sigma`C. Nine `sigma` and seven`pi`D. Eight `sigma` and eight `pi` |
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Answer» Correct Answer - A There are `9sigma` and `9pi` bonds in `(N-=C)_2 C=C (C-= N)_2` |
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| 422. |
Assertion :- Bond Dissociation energy is `F_(2) gt Cl_(2)` Reason:- `Cl_(2)` have more electronic repulsion than `F_(2)`.A. If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion.B. If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion.C. If Assertion is True, but the Reason is False.D. If both Assertion & Reason are false |
| Answer» Correct Answer - D | |
| 423. |
Which of the following statement (s) is trueA. Higher the bond order lesser the bond lengthB. Higher the bond order greater the bond lengthC. Higher the bond order lesser the bond energyD. Higher the bond order lesser the number of bonds |
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Answer» Correct Answer - A Bond order ` prop` bond energy `prop 1/("Bond length")` |
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| 424. |
Which of the following is incorrect regarding the MO theory?A. The number of molecular orbitals formed is always equal to the number of atomic orbitals combined.B. The more stable the bonding molecular orbitals, the less stable the corresponding antibonding molecular orbital.C. In a stable molecule, the number of electrons in bonding molecular orbitals is always equal to that in antibonding molecular orbitals.D. Like an atomic orbital, each molecular orbital can accommodate up to two electrons with opposite spins in accordance with the Pauli exclusion principle. |
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Answer» Correct Answer - C In a stable molecule, the number of bonding electrons is always greater than the number of antibonding electrons. |
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| 425. |
Which of the following statements is not regarding bonding molecular orbitalsA. Bonding molecular orbitals possess less energy than the atomic orbitals from which they are formedB. Bonding molecular orbitals have low electron density between the two nucleiC. Electron in bonding molecular orbital contributes to the attraction between atomsD. They are formed when the lobes of the combining atomic orbitals have the same sign |
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Answer» Correct Answer - B Bonding M.O. has maximum electron density between two nuclei. |
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| 426. |
The most suitable method of separation of a mixture of ortho and para nitrophernol in the ratio `1:1` is :A. distillationB. crystallisationC. vaporisationD. colour spectrum |
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Answer» Correct Answer - A Ortho nitrophenol has lower biling point (intramolecular H-bonding ) than para nitrophenlo (intermolecular H-bonding ). |
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| 427. |
In `C_2H_6` (ethane), the `C-C` sigma `(sigma)` bond is formed by _______overlap.A. `p-p`B. `sp-s`C. `s-s`D. `sp^3-sp^3` |
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Answer» Correct Answer - D The Lewis structure of `C_2H_6` is `H-overset(H)overset(|)underset(H)underset(|)C-overset(H)overset(|)underset(H)underset(|)C-H` Every C atom is `sp^3`-hybridized as it forms `4sigma` bonds. Thus, the `C-C` bond is `sp^3-sp^3` sigma bond while the `C-H` bonds are `sp^3-s` bonds. |
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| 428. |
Molecular orbitals energy level diagram consists ofA. bonding molecular orbitalsB. antibonding molecular orbitalsC. nonbonding molecular orbitalsD. molecular orbitals in order of increasing energy |
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Answer» Correct Answer - D In order to write the electron configuration of any molecule, we must first arrange the molecular orbitals that are formed in the molecule in order of increasing energy. Once the order, i.e., MO energy level diagram is known, we can use the rules (used in atomic structure) to guide us in filling these molecular orbitals with electrons. |
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| 429. |
In the formation of `N_2^+` the electron is lost form a :A. ` sigma`-orbitalB. `pi-`orbitalC. `sigma^x` -orbitalD. `pi^x-`orbital |
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Answer» Correct Answer - A M.O. configuration of `N_2` is `sigma1s^2 , sigma^(**) 1s^2 , sigma2s^2 , sigma^(**) 2s^2 ,pi 2p_y^2 pi 2p_z^2,sigma2p_x^2` M.O. configuration of `N_(2)^(+)` is `sigma1s^2 , sigma^(**) 1s^2 , sigma2s^2 , sigma^(**) 2s^2 ,pi 2p_y^2 pi 2p_z^2,sigma2p_x^1` . |
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| 430. |
`N_2` accept electron and convert into `N_2^-`, where this electron goesA. Antibonding `pi` molecules orbitalsB. Bonding `pi` molecular orbitalC. `sigma` bonding molecular orbitalsD. `sigma`-anti-bonding molecular orbital |
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Answer» Correct Answer - A `N_2` molecule has 14 electrons. The molecular orbitals electronic configuration of the molecule is as `N_2: K K (simga2s)^(2),(sigma^(**)2s)^(2),(pi2p_(x))^(2),(pi 2p_(y))^(2)~~(sigma2p_(z))^(2)` `N_(2)^(-)` ions is formed when `N_(2)` accept an electron hence, it has 15 electrons. the molecular orbital electronic cofiguration of the molecule is as : `N_(2)^(-) : K K (sigma2s)^(2),(sigma^(**)2s)^(2)~~(pi 2p_(x))^(2),(pi 2p_(y))^(2),(sigma2p_(z))^(2),(pi^(**)2p_(x))^(1)` Hence, this electrons goes to antibonding `pi` molecular orbital. |
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| 431. |
In `O_2^-, O_2 ` and `O_2^(-2)` molecular species, the total number of antibonding electrons respectively areA. 7,6,8B. 1,0,2C. 6,6,6D. 8,6,8 |
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Answer» Correct Answer - A Molecular orbitals electronic configuration of these species are : `O_(2)^(-)(17e^(-))=sigma1s^(2)sigma^(**)1s^(2),sigma2s^(2)sigma^(**)2s^(2),sigma2p_(x)^(2),pi2p_(y)^(2),pi2p_(z)^(2),pi^(**)2p_(y)^(2)pi^(**)2p_(z)^(1)` `O_(2)(16e)=sigma1s^(2)sigma^(**)1s^(2),sigma2s^(2)sigma^(**)2s^(2),sigma 2p_(x)^(2),pi2p_(y)^(2),pi 2p_(z)^(2)pi^(**)2p_(y)^(1)pi^(**)2p_(z)^(1)` `O_(2)^(2-)(18e) =sigma1s^(2)sigma^(**)1s^(2),sigma2s^(2)sigma^(**)2s^(2),sigma2p_(x)^(2),pi2p_(y)^(2),pi2p_(z)^(2)pi^(**)2p_(y)^(2)pi^(**)2p_(z)^(2)` Hence, number of antibonding electrons are 7,6 and 8 respectively. |
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| 432. |
The `ICl` molecule isA. purely electrovalentB. purely covalentC. polar with negative end on iodineD. polar with negative end on chlorine |
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Answer» Correct Answer - D Due to the electronegativity difference. |
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| 433. |
Which of the following is the most polar bondA. O-HB. S-HC. N-HD. C-H |
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Answer» Correct Answer - A More the difference in electronegativity of atoms, bond between them will be more polar |
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| 434. |
The ionic radii of `N^(3-) O^(2-), F^-, Na^+` follows the orderA. `N^(3-) gt F^(-) gt O^(2-)`B. `O^(2-) gt N^(3-) gt F^(-)`C. `O^(2-) gt F^(-) gt N^(3-)`D. `N^(3-) gt O^(2-) gt F^(-)` |
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Answer» Correct Answer - D Greater is the negative charge and size of anion, higher is it polarising power. |
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| 435. |
Ionic radii of :A. `Ti^(4+) lt Mn^(7+)`B. `.^(35)Cl^(-) lt ^(37)Cl^(-)`C. `K^(+) lt Cl^(-)`D. `P^(3+) gt p^(5+)` |
| Answer» Correct Answer - D | |
| 436. |
The species having bond order different from that in `CO` isA. `NO^(-)`B. `NO^(+)`C. `CN^(-)`D. `N_(2)` |
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Answer» Correct Answer - a |
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| 437. |
Which is the most covalentA. C-OB. C-BrC. C-SD. C-F |
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Answer» Correct Answer - C C-S will be most covalent. Covalent character depends on the size of atoms |
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| 438. |
Strongest bond isA. C-CB. C-HC. C-ND. C-O |
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Answer» Correct Answer - C Multiple bonds have more bond energy so `C-=N` will be the strongest. |
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| 439. |
Which type of bond is present in `H_(2)S` moleculeA. Ionic bondB. covalent bondC. co-ordinateD. All of three |
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Answer» Correct Answer - B Electronic configuration of `._(16)S^(32) =1s^(2),2s^(2),2p^(6),3s^(2),3^(4)`. In the last orbit it has only 6 electron. So it require 2 electron to complete its octet, therefore it share 2 electron with two hydrogen atom and forms 2 covalent bond with it. |
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| 440. |
The major binding force of diamond, silicon and quartz isA. Electrostatic forceB. Electrical attractionC. Covalent bond forceD. Non-covalent bond force |
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Answer» Correct Answer - C Diamond, silicon and quartz molecule bounded by covalent bond. |
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| 441. |
The correct order of dipole moment is :A. `CH_(4) lt NF_(3) lt NH_(3) lt H_(2)O`B. `NF_(3) lt CH_(4) lt NH_(3) lt H_(2)O`C. `NH_(3) lt NF_(3) lt CH_(4) lt H_(2)O`D. `H_(2)O lt NH_(3) lt NF_(3) lt CH_(4)` |
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Answer» Correct Answer - A The dipole moment of of `CH_(4)=0D, NF_(3)=0.2 D,NH_(3)=1.47D` and `H_(2)O=1.85D`. Therefore, the correct order of the dipole moment is `CH_(4) lt NHF_(3)lt NH_(3) lt H_(2O` |
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| 442. |
The actual geometry of `NO_(2)^(-)` isA. PlanarB. LinearC. V shapeD. Tetrahedral |
| Answer» Correct Answer - C | |
| 443. |
The geometry of `H_(2)S` and its dipole moment are :A. Angular and non-zeroB. Angular and zeroC. Linear and non-zeroD. Linear and zero |
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Answer» Correct Answer - A In `H_(2)S,S` is `sp^(2)` hybridized, hence, `H_2S` will have angular geometry and non-zero value of dipole moment. |
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| 444. |
In the compound `CH_(2)=CH-CH_(2)-CH_(2)-C-=CH` the `C_(2)-C_(3)` bond is ofA. `sp-sp^(2)`B. `sp^(3)-sp^(3)`C. `sp-sp^(3)`D. `sp^(2)-sp^(3)` |
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Answer» Correct Answer - D `CH_(3)=underset(sp^(2))(CH)-underset("hybridised")underset(sp^(2))(CH_(2))-CH_(2)-C-=CH` |
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| 445. |
The geometry of ` H_2 S` and its dipole moment are :A. angular and non-zeroB. angular and zeroC. linear and non-zer0D. linear and zero |
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Answer» Correct Answer - a |
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| 446. |
The order of dipole moment of the following molecules isA. `CHCl_(3) gt CH_(2)Cl_(2) gt CH_(3)Cl gt C Cl_(4)`B. `CH_(2)Cl_(2) gt CH_(3)Cl gt CHCl_(3) gt C Cl_(4)`C. `CH_(3)Cl gt CH_(2)Cl_(2) gt CHCl_(3) gt C Cl_(4)`D. `CH_(2)Cl_(2) gt CHCl_(3) gt CH_(3)Cl gt C Cl_(4)` |
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Answer» Correct Answer - D `C Cl_(4)` has zero dipole moment because of symmetric tetrahedral structure. `CH_(3)Cl` has slightly higher dipole moment which is equal to 1.86 D . Now `CH_(3)Cl` has less electronegativity then `CH_(2)Cl_(2)`. But `CH_(2)Cl_(2)` has greater dipole moment than `CHCl_(3)`. |
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| 447. |
In the compound `CH_(2)=CH-CH_(2)-CH_(2)-C-=CH` the `C_(2)-C_(3)` bond is ofA. `sp-sp^(2)`B. `sp^(3),sp^(3)`C. `sp-sp^(3)`D. `sp^(2)-sp^(3)` |
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Answer» Correct Answer - d |
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| 448. |
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)` respectivelyD. `sp^(2),sp^(3)` and `sp` respectively |
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Answer» Correct Answer - b |
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| 449. |
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)` respectivelyD. `sp^(2),sp^(3)` and `sp ` respectively |
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Answer» Correct Answer - B For `NO_(2)^(+) : H=1/2(5+0+0-2) =2 :. Sp` hybridization For `NO_(3)^(-) : H=1/2[5+0+1-0] ==3 :. Sp^(2)`-hybridization For `NH_(4)^(+) : H=1/2 [5+4+0-1]=4 , :. Sp^(3)` hybridisation |
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| 450. |
The covalency of oxygen in hydronium ion is equal to number ofA. covalent bondsB. co-ordinate bondsC. covalent bonds and co-ordinate bondsD. valence electrons |
| Answer» In hydronium ion both co-ordinate and covalent bonds are present. Hence, the covalency of oxygen in hydronium ion is equal to number of covalent bonds and co-ordinate bonds. | |