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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 151. |
Which of the following is correct with respect to bond length of the species ?A. `C_(2) gt C_(2)^(2-)`B. `B_(2)^(+) gt B_(2)`C. `Li_(2)^(+) gt Li_(2)`D. `O_(2) gt O_(2)^(-)` |
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Answer» Bond order : `C_(2) = 2 , C_(2)^(2-) = 3 , B_(2) = 1 , B_(2)^(+) = 0.5 ` , `Li_(2) = 1 , Li_(2)^(+) = 0.5 , N_(2)= 3 , N_(2)^(+) = 2.5 , O_(2) = 2 ` `O_(2)^(-) = 1.5` . Greater the bond order , shorter is the bond length |
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| 152. |
The single and multiple bond radii of some elements given in the following table Calculate the bond lengths in (a) `SCI_(2)` (b) `NH_(3)` (c ) `CH_(2)CI_(2)` (d) `HOCI` (e) `HCN` (f )`H_(3)PO_(4)` (g) `CH_(3)NH_(2)` . |
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Answer» Covalent bond radii from Table 2.24 are merely added (a) `(1.04 + 0.99) = 2.03Å` (b) `(0.70 + 0.28) = 0.98 Å` (c ) `C -H = (0.77 + 0.28) = 1.05Å` `C -CI = (0.77 + 0.99) = 1.76Å` (d) `H -O = 0.94 Å` ` O -CI = 1.65 Å` (e) `H -CI = 1.05Å` `(C -= N) = (0.61 + 0.55) = 1.16 Å` (f) `H -O 0.94 Å` `O - P =1.76 Å` (g) `C -H = 1.05 Å` `C -H = 1.74 A N -H = 0.98 Å` . |
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| 153. |
Presict the shapes of the following molecules on the basis of hybridisation. `BCl_(3),CH_(4),CO_(2),NH_(3)` |
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Answer» `BCl_(3)H=1//2[3+3-0+0]=3(sp^(2))` `CH_(4):H=1//2[4+4-0+0]=4(sp^(3))` `CO_(2):H=1//2[4+0-0+0]=2(sp)` `NH_(3):H=1//2[5+3-0+0]=4(sp^(3))` |
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| 154. |
Arrange the following bonds in order of increasing ionic character giving reason N-H, F-H, C-H and O-H |
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Answer» Increasing ionic character of the given bonds is : Electronegativity difference : `{:(N-H " " lt " "N-H" "lt " "O-H " "lt " "F-H),("(2.5-2.1) (3.0-2.1) (3.5- 2.1) (4.0-2.1)"),( = "0 . 4 = 0.9 = 1.4 = 1. 9 "):} ` Greater is the electronegativity differnce between the two bonded atoms, greater is the ionic character. |
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| 155. |
Which of the following has maximum bond angle ? ` H_(2)O, CO_(2), NH_(3), CH_(4)`. |
| Answer» `CO_(2), 180^(@)` (due to linear structure) | |
| 156. |
Arrange the following bonds in the order of increasing ionic character : C-H,F-H,Br, Na-I , K-F and Li-Cl |
| Answer» `CO_(2), 180^(@)` (due to linear structure) | |
| 157. |
In `N_2H_4` (hydrazine) both the nitrogen atoms areA. trigonal and the molecule is planarB. trigonal but the molecule is non planarC. pyramidal and the molecule is non planarD. none of these |
| Answer» Correct Answer - C | |
| 158. |
Out of `CS_(2)` and `OCS` which have higher dipole moment and why? |
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Answer» The structures of both the moleculse are given as follows : `{:(overset(delta-)O=overset(delta+)C=overset(delta-)S," "overset(delta-)S=overset(delta+)C=overset(delta-)S),("(Polar) ", " (Non-polar)"):}` Both are linear molecule but in OCS, the bond polarities do not completely cancel out because it is not of symmetrical nature. However, they get cancelled in `CS_(2)` molecule due to its linear and symmetrical nature. Thus, OCS is polar and has dipole moment with `CS_(2)` is non-polar and has zero dipole moment. |
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| 159. |
What is meant by the term vaerange bond enthalpy? Why is there difference in bond enthalpy of O-H bond in ethanol `(C_(2)H_(5)OH)` and water `(H_(2)O)` ? |
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Answer» Average bond enthalpy is the bond sissociatio enthalpy of the bonds of the same type ina molecule present in the gaseous state of a substamce. In ethanol molecule, bond enthalpy of O-H bond is not average bond enthalpy because O atom is surrounded by H and C atoms on either sides. However, it is average bond enthalpy in `H_(2)O` molecule since the two O-H bonds are similar. `" "H-overset(H)overset(|)underset(H)underset(|)C-underset(H)underset(|)overset(H)overset(|)C-O-H" "H-O-H` |
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| 160. |
Write the singificance of plus and inus signs in representing the opitals. |
| Answer» Plus and minus signs have been given to identify the nature of the electron waves. Plus (+ ve) sign denotes crest whil negative (-ve) sign denotes trough. | |
| 161. |
The correct order of bond strength is :A. `O_(2)^(-) lt O_(2) lt O_(2)^(+) lt O_(2)^(2-)`B. `O_(2)^(2-) lt O_(2)^(-) lt O_(2) lt O_(2)^(+)`C. `O_(2)^(-) lt O_(2)^(2-) lt O_(2) lt O_(2)^(+)`D. `O_(2)^(+) lt O_(2) lt O_(2)^(-) lt O_(2)^(2-)` |
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Answer» Correct Answer - b Bond order for `O_(2), O_(2)^(+), O_(2)^(-), O_(2)^(2-) are 2, 2.5, 1.5 and 1.0` respectively, Higher is bond order more is bond energy |
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| 162. |
The correct order of bond strength isA. `H_(2)S lt NH_(3) lt BF_(3) lt SiH_(4)`B. `NH_(3)S lt H_(2)S lt SiH_(4) lt BF_(3)`C. `H_(2)S lt NH_(3) lt SiH_(4) lt BF_(3)`D. `H_(2)S lt SiH_(4) lt NH_(3) lt BF_(3)` |
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Answer» Correct Answer - c `h_(2)S lt NH_(3) lt SiH_(4) lt BF_(3)` `92.6^(@) 107^(@)28^(@) 120^(@)` |
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| 163. |
Which of the following exhibits the weakest intermolecular forces?A. HeB. HCIC. `NH_(3)`D. `H_(2)O` |
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Answer» Correct Answer - a Gebnerally zero group elelments arwe linked by the van der walls force .Hence these show weakest intermolecular force |
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| 164. |
A molecule of `H_(2)` exist while that of the `He_(2)` does not. Explain. |
| Answer» Hydrogen atom (Z=1) has only one electron present in the s-orbital `(1s^(1)).` The two such half field atomic orbitals compbine to from a molecular orbital which contains both these electrons. But helium (Z=2) has already a filled orbital `(1s^(2)).` Therefore, the atomic orbitals of the helium atoms donot combine. Thus, a molecule of `H_(2)` exists while that of `He_(2)` does not. | |
| 165. |
Why two hydrogen atoms combine to form ` H_(2)` but two helium atoms do not combine to form ` He_(2)` ? |
| Answer» (i) New forces of attraction and repulsion and (ii) orbital concept ). | |
| 166. |
Explain on the basis of molecular orbital theory why `He_(2)` does not exist. |
| Answer» Bond order for `He_(2) = 0 ` . | |
| 167. |
CO is isoelectronic withA. `NO^(+)`B. `N_(2)`C. `SnCl_(2)`D. `NO_(2)^(-).` |
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Answer» Correct Answer - A::B CO is isolectroinc with `NO^(+) and `N_(2).` Both have fourteen electrons each. |
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| 168. |
Which of the following species have the same shape?A. `CO_(2)`B. `C Cl_(4)`C. `O_(3)`D. `NO_(2)^(-)` |
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Answer» Correct Answer - C::D `O_(3)and NO_(2)^(-)` have angular shapes. |
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| 169. |
Anhydrous `AlCl_(3)` is covalent. From the date given below , predict whether it would remain covalent or became ionic in aqueous solution (Ionizatiion energy for `AlCl_(3) = 5137 kJ mol^(-1), Delta H_("hydration")` for `Al^(3+) = - 4665 kJ mol^(-1), DeltaH_("hydration")` for `Cl^(-)= - 381 KJ mol^(-1))`. |
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Answer» `AlCl_(3) (S) + aqto AlCl_(3) (aq)^(3+) (aq) + 3Cl^(-) (aq)` Total energy released on hydration of 1 mole of `Al^(3+)` ions and 3 moles of `Cl^(-)` ions `= 4665+3xx381 KJ = 5808 mol ^(-1)` Energy required for ionization = 5137 kJ `mol^(-1)` As energy released is greater than the energ required the compound will ionize in aqueous solution . |
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| 170. |
Assertion . H-S-H bond angle in `H_(2)S` is closer to ` 90^(@)` but H-O-H bond angle in `H_(2)O is 104.5^(@)`. Reason . Lp-lp repulsion us stronger ub `H_(2)S ` than in `H_(2)O`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 - b Correct explanation . Bond pair -bond pair repulsion are greater in `H_(2)O` than in `H_(2)S` because O-atom is smaller in size than S-atom . |
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| 171. |
Statement-1. The boiling point of `NH_(3) ` lies between that of `SbH_(3) and BiH_(3)` . Statement -2. `PH_(3)` has much lower boiling than `NH_(3)` but it increases from `PH_(3) ` to ` AsH_(3)` to `SbH_(3)` to BiH_(3)` due to increase in van dar Waals forcesA. Statement-1 is Ture , Statement-2 is Ture , Statement-2 is a correct explanation for Statement-1.B. Statement-1 is Ture , Statement-2 is Ture , Statement-2 is not a correct explanation for Statement-1.C. Statement-1 is True , Statement-2 is False .D. Statement-1 is False, Statement-2 is True. |
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Answer» Correct Answer - d Correct statent -1 .Boiling point of `NH_(3)` is less ltbgt than that of `BiH_(3)` as well as `SbH_(3)` |
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| 172. |
The pair of compounds having identical shapes for their molecules is:A. `CH_(4), SF_(4)`B. `BCl_(3), ClF_(3)`C. `XeF_(2),ZnCl_(2)`D. `SO_(2),CO_(2)` |
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Answer» Correct Answer - C `XeF_(2)and ZnCl_(2)` molecules are both linear in shape. |
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| 173. |
Which has maximum dipole moment?A. B. C. D. |
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Answer» Correct Answer - a Dipole moment `prop (1)/("bond angle")` |
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| 174. |
Which of the following molecules has one unpiared electron in antibonding orbitals ? .A. `CO`B. `O_(2)^(Θ)`C. `O_(2)^(o+)`D. `NO` |
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Answer» Correct Answer - B::C::D Refer to Table `2.20 and For `NO` refer to Table 2.23 . |
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| 175. |
When `N_(2)` goes to `N_(2)^(+)`, the `N-N` bond distance …………, and when `O_(2)` goes to `O_(2)^(+)` the `O-O` bond distance……. |
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Answer» Correct Answer - Increases, decrease Inceases decreases In `N_(2)` to `N_(2)^(o+)` bond order decreases fro m3 to 2.5 thus bond length increases) In case of `O_(2)` to `O_(2)^(o+)` bond order increases from thus bond length decreases ) . |
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| 176. |
When `N_(2)` goes `N_(2)^(o+)` the `N-N` bond distance________and when `O_(2)` goes to `O_(2)^(o+)` the `O-O` bond distance _______. |
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Answer» Correct Answer - Increases, decrease When `N_(2)` goes to `N_(2)^(o+)` the `N-N` bond distance increases and when `O_(2)` goes to `O_(2)^(o+)` the `O-O` bond distance The electrons is removed from a bonding molecular orbital in nitrogen and form an antibonding molecular orbital in oxygen . |
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| 177. |
Bond order of `O_(2), O_(2)^(-) and O_(2)^(2-)` is in orderA. `O_(2)^(-) lt O_(2)^(2) lt O_(2) lt O_(2)^(+)`B. `O_(2)^(2-) lt O_(2)^(-) lt O2 lt O_(2)^(+)`C. `O_(2)^(+) lt O_(2) lt O_(2)^(-) lt O_(2)^(2)`D. `O_(2) lt O_(2)^(+) lt O_(2)^(-) lt O_(2)^(2-)` |
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Answer» Correct Answer - b `{:(O_(2), 16, 2.0),(O_(2)^(+), 15, 2.5), (O_(2)^(-),15, 1.5), (O_(2)^(2-),18,1.0):}` Thus is in `O_(2)^(2) lt ,O_(2)^(-) lt O_(2)^(+)` |
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| 178. |
Among the following the pair in which the two species are not isostuctural isA. `SiF_(4) and SF_(4)`B. `IO_(3)^(-) and XeO_(3)`C. `BH_(4)^(-) and NH_(4)^(+)`D. `PF_(6)^(-)` and `SF_(6)` |
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Answer» Correct Answer - c Both `NH_(4)^(+)` and `[BF_(4)]^(-)` have tetrabedral shape due to `sp^(2)` hybridization |
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| 179. |
In an octahedral structure , the pair of d orbitals involved in `d^(2)sp^(2)` hybridization isA. `d_(x^(2)),d_(xz)`B. `d_(xy),d_(yz)`C. `d_(x^(2) - y^(2)),d_(z^(2))`D. `d_(xz),d_(x^(2) - y^(2))` |
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Answer» Correct Answer - c In the formation of `d^(2)sp^(2)` hybrid orbital two `(n - 1)d` orbitals of e.g. set (i.e. `(n - 1)dz^(2) and (n - 1)dx^(2) - y^(2)` orbitals ) one ns three np [np,np, and np] orbitals combine togather and form six `d^(2)sp^(2)` hydral orbitals . |
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| 180. |
Which d-orbital is involved in ` sp^(3) d` hydridisation and why ? |
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Answer» ` d_(z^(2))` . The three planar triangular hybrid orbitals may be assmed to be formed form one s and two p- orbitals `(sp^(2))` and the remaining p-orbital may combine with ` d_(z^(2))` to form two axial hybrid orbitals. |
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| 181. |
The geometry and the type of hybrid orbitals present about the central atom in ` BF_3` is :A. Linear,spB. Trigonal planar, `sp^(2)`C. Tetrahedral,` sp^(3)`D. Pyramidal `sp^(3)` |
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Answer» Correct Answer - B `BF_(3)` has triangular planar molecule due to `sp^(2)` hybridisation . |
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| 182. |
The hydridisation of S in `SO_(4)^(2-)` is same asA. l in `lCl_(4)^(-)`B. S in `SO_(3)`C. P in `PO_(4)^(3-)`D. N in `NO_(3)^(-)` |
| Answer» Correct Answer - C | |
| 183. |
Assertion Carbon tetrachloride dose not form a percipitate of `AgCI` with `AgNO_(3)` solution Reasoning Carbon tetrachloride is a liquid.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 - B `C CI_(4)` is covalent and has no ions therefore no precipitation occurs and is not related to its state . |
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| 184. |
The pair having similar geometry isA. `PCl_(3), NH_(4)^(+)`B. `BeCl_(2), H_(2)O`C. `CH_(4), CCl_(4)`D. `IF_(5), PF_(5)` |
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Answer» Correct Answer - C In both `CH_(4) and CCl_(4)` molecules , C is `sp^(3)` hybridized and thus both have tetrahedral geometry. |
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| 185. |
How do you account for the difference in melting points between (a) and (b) between (c ) and (d) and between these two differences ? . |
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Answer» The crystal forces in (a) and (d) are largely van der Walls. Compounds (a) and (c ) containing the polar `OH` groups from H-bonding (b) in (a) there is intramolecular H-bonding whereas in (c ) it is intermolecular H-bonding which leads to increases in meliting point of (c ) as compared to (a) and (d) (c) In the absence of strong intermolecular H-bonding the difference in m.pt as compared with the reference substance (b) should be small This is due to differences in the crystal structure or due to the van der Waals forces, which would be slightly larger for (b) than for (a) on account of the slightly increased molecular mass . |
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| 186. |
In `MO` energy diagram for heteronuclear diatomic molecule is similar However, the energies of the `AO` s of the atom having higher atomic number being lower, the diagram will be unsymmetrical, but that will not make a difference in the electron count The bond order is half the difference in the number of electrns of the bonding `(sigma and pi)` and anti-bonding `(sigma and pi)` `MO s` For a bond to have been formed the bond order the shorter is the bond distance and the greater is the bond dissociation energey But if the bond order is smae in the above two cases, then the bond distance will be greater and the bond dissocation energy smaller in the case which has more populated anti-bonding orbitals The presence of unpaired electron(s) in a molecualr orbital will make the system paramagnetic Which of the following species is not expected to exist ? .A. `He_(2)^(o+)`B. `H_(2)^(o+)`C. `Be_(2)`D. `Be_(2)^(o+)` |
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Answer» Correct Answer - C Bond order in `Be_(2)` is zero . Hence it does not exist . |
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| 187. |
The critical temperature of water is higher than that of `O_(2)` because the `H_(2)O` molecule has .A. Fewer electrons than `O_(2)`B. Three covalent bondsC. Two covalent bondsD. Dipole moment |
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Answer» Correct Answer - C::D Critical temperature of water is higher than that of `O_(2)` because `H_(2)O` molecule has dipole moment Its two covalent bonds are at an angle of `104.5^(@)` to each other and the dipole moment vectors do not cancel each other . |
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| 188. |
Which of the following statements regarding the concept of resonance is not correct?A. The different resonting structures of a molecule have fixed arrangement of atomic nuclei .B. The differnet resonating structures differ in the arrangement of electrons .C. None of the individual resonating structures explains the verious characteristics of the molecule .D. The hybrid structures have equal contribution from all the resonating structures . |
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Answer» Correct Answer - D Statement (d) is wrong because resonating strucures have different stabilities and therefore their contributions to the stabilities and therefore their contributions to the hybrid structures are different . |
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| 189. |
Which of the following species does not have tetraahedral geometry ?A. `BH_(4)^(-)`B. `NH_(2)^(-)`C. `CO_(3)^(2-)`D. `H_(3)O^(+)` |
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Answer» Correct Answer - C `BH_(4)^(-):H=^(1//2)[5+4-1+0]=4(sp^(3))` `NH_(2)^(-):H=^(1//2)[5+2-0+1]=4(sp^(3))` `CO_(3)^(2-):H=^(1//2)[4+0-0+2]=3(sp^(2))` `OH_(3)^(+):H=^(1//2)[6+3-1+0]=4(sp^(3))` `CO_(3)^(2-)` ion has trigonal geometty. |
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| 190. |
A plant virus was found to consist of uniform cylindrical particles `100A` in diameter and `4000 A` long The virus has a specific volume `0.314 cm^(3) g^(-1)` If the virus particle is considered to be one molecule, what is its molecular weight ? . |
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Answer» `V = pi r^(2) h = (3.14) ((100Å)/(2))^(2) ( 400Å)` ` = 3.14 xx 10^(7) (Å)^(3) ((10^(-8)cm)/(Å))^(3)` `= 3.14 xx 10^(-17) cm^(3)` Therefore, the specific volume is `0.314 cm^(3) g^(-1)` of molecule. If the specific volume is `3.14 xx 10^(-17)cm^(3)` The weight per molecule is `=((1g)/(0.314cm^(3)))(3.14 xx 10^(-7) cm^(3))` ` = 10^(-16) "molecule"^(-1)` Molecular weight of virus `=(10^(-6) "g molecule"^(-1))` `((6.02 xx 10^(23 "molecules"))/(mo1))` ` =6.02 xx 10^(7) g mo1^(-1)` . |
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| 191. |
Which of the following pairs do not constitute resonance structures ? |
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Answer» In resonance the position of atoms or nuclei remains fixed, so pairs in (b) constiute resonance. In other pairs, the positions of atoms are changing, therefore (a),(c ), and (d) pairs do not constitute resonance structures (a) They constitute functional isomerism(nitro to nitrite) (b) They constitute resonance structures (c ) They constitute tautomerism (keto and enol) and also show functional isomerism (d) They constitute positional isomerism (position of double bond has changed) . |
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| 192. |
Strongest hydrogen bonding is present inA. HFB. HClC. HBrD. HI |
| Answer» Correct Answer - A | |
| 193. |
Resonance structure of a molecule cannot haveA. Identical arrangement of atomsB. Nearly same energy contentC. The same number of paired electronsD. Identical bonding |
| Answer» Correct Answer - D | |
| 194. |
Which of the following pair constitute resonance structure?A. B. C. D. `H_(3)C-CH=CH-CH_(3)` and `H_(3)C-CH_(2)-CH=CH_(2)` |
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Answer» Correct Answer - b In resonance , the positive of atom remains fixed so pairs in (b) constitute resonance. In other pairs , the positive of atoms keeps changing therefore (a), ( c) and (d) do not constitute resonance |
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| 195. |
Which of the following pairs do not constitute resonance structures ? A. B. C. D. |
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Answer» Correct Answer - B In resonance the position of atoms or nuclei remains fixed, so pairs in (b) constiute resonance. In other pairs, the positions of atoms are changing, therefore (a),(c ), and (d) pairs do not constitute resonance structures (a) They constitute functional isomerism (nitro to nitrite) (b) They constitute resonance (c ) They constitute tautomerism (keto and enol) and also show functional isomerism (b) They constitute positional isomerism (position of double bond has changed) . |
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| 196. |
Which of the following cannot exist on the basic of `MO` theory ? .A. `H_(2)^(o+)`B. `He_(2)^(o+)`C. `He_(2)`D. `O_(2)` |
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Answer» Correct Answer - C Bond order of `He_(2)` is zero . |
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| 197. |
According to `MO` ThoryA. `O_(2)^(+)` is paramagnetic and bond order is greater than `O_(2)`B. `O_(2)^(+)` is paramagnetic and bond order is less than `O_(2)`C. `O_(2)^(+)` is diamagnetic and bond order is less than `O_(2)`D. `O_(2)^(+)` is diamagnetic and bond order is more than `O_(2)` |
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Answer» Correct Answer - a In `O_(2)` molecule , the total number of electron `= 16` Electronic distrubution in molencular orbital of `O_(2) = sigma 1s^(2), sigma 2s^(2), sigma **2s^(2), sigma p1_(s)^(2), pi 2 p _(y), pi 2p_(z), pi **2p_(y)^(1), pi**2p_(z)^(1)` Bond order in `O_(2) = (1)/(2) [N_(b) - N_(a)] = (1)/(2) [10-6] = 2.0` In `O_(2)^(+) = sigma 1s^(2), sigma **1s^(2), sigma 2s^(2), sigma **2_(s)^(2), sigma p _(s)^(2), pi 2s_(y)^(2), pi 2p_(z)^(2),pi **2p_(y)^(1)` It is paramagnetic due to presence of unpaired orbital. Bond order in `O_(2)^(+) = (1)/(2) [10-5] = 2.5` The bond order of `O_(2)^(+)` is `2.5` which is greater than bond order of `O_(2) (2.0)`. Hence, `O_(2)^(+)` is paramagenetic and order greater than `O_(2)` |
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| 198. |
According to `MO` theory,A. `O_(2)^(+)` is paramagnetic and b.o. is ggreater than that of `O_(2)`B. `O_(2)^(+)` is paramagnetic and b.o. is less than that of `O_(2)`C. `O_(2)^(+)` is diamagnetic and b.o. is less than that of `O_(2)`D. `O_(2)^(+)` is diamagnetic and b.o. is more than that of `O_(2).` |
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Answer» Correct Answer - A is the correct answer. |
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| 199. |
According to `MO` theory,A. `O_(2)^(o+)` is paramagnetic and bond order is greater than `O_(2)`B. `O_(2)^(o+)` is paramagnetic and bond order is greater than `O_(2)`C. `O_(2)^(o+)` is paramagnetic and bond order is greater than `O_(2)`D. `O_(2)^(o+)` is paramagnetic and bond order is more than `O_(2)` |
| Answer» Correct Answer - A | |
| 200. |
A simplified applified of `MO` theory to the hypotheritical molecule `OF` would gives its bond order as:A. 2B. 1.5C. `1.0`D. `0.5` |
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Answer» Correct Answer - b `OF` is derivavtive of `O_(2)` and isoelectronic with `O_(2)^(-)` So `(sigma 1*s)^(2) (sigma 2s)^(2) (sigma*2s)^(2) (sigma 2p_(z)^(2) = pi 2p_(y)^(2)) (pi*2p_(s)^(2) = pi *2p_(y)^(1))` The bond order of `1//2 (10-7) = 1.5` |
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