Explore topic-wise InterviewSolutions in Current Affairs.

This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.

1.

Draw the M.O diagram for oxygen molecule and calculate its bond order and show that O_(2) is paramagnetic.

Answer»

SOLUTION :(i) Electronic configuration of O atom is `1S^(2) 2s^(2) 2p^(4)`
(ii) Electronic configuration of `O_(2)` molecule is `sigma 1s^(2) sigma^(**) 1s^(2) sigma^(**)2s^(2) sigma2p_(x)^(2) pi 2p_(y)^(2) pi2p_(z)^(2) pi^(**)2p_(y)^(1) pi^(**) 2p_(z)^(1)`
(iii) Bond order `= (N_(b) - N_(a))/(2) = (10-6)/(2) =2`
(iv) Molecule has two unpaired ELECTRONS, hence it is paramagnetic.
Discussion
2.

Draw the M.O diagram for oxygen molecule calculate its bond order and show that O_(2) is paramagnetic.

Answer»

Solution :(i) ELECTRONIC configuration of O atom is `1s^(2)2s^(2)2p^(4)`
(ii) Electronic configuration of `O_(2)` molecule is
`sigma 1s^(2)" "sigma**1s^(2)" "sigma2s^(2)" "sigma**2s^(2)" "sigma2p_(x)^(2)" "pi2p_(Z)^(2)" "pi**2p_(y)^(1)" " pi**2p_(z)^(1)`
(iii) Bond order `=(N_(B)-N_(a))/(2)=(10 - 6 )/(2)=2`
(iv) Mole cule has two unpaired electrons, hence it is paramagnetic.
Discussion
3.

Draw the MO diagram for acetylide ion C_(2)^(2-) and calculate its bond order.

Answer»

Solution :Acetylide ion, `C_(2)^(2-)` in ACETYLENE: `UNDERSET(1s^(2)" "2s^(2)" "2p^(4))(C)rarrunderset(1s^(2)" "2s^(2)" "2p^(2))(C^(2-)+2e)`
Electronic configuration of `C_(2)^(2-)` ion is : `SIGMA 1s^(2) sigma **1s^(2)sigma1s^(2) pi2p_(x)^(2)pi2p_(y)^(2)`

Bond order `= (N_(B)-N_(a))/(2)=(8-4)/(2)=2`
Discussion
4.

Draw the MO diagram for acetylide ion C_(2)^(2-) and calculate its bond order .

Answer»

Solution :Acetylide ION `C_(2)^(2-)` in acetylide
`CtoC^(2) + 2E^(-)`
`1s^(2) 2s^(2) 2p^(4) "" 1s^(2)2s^(2) 2p^(2)`
ELECTRONIC configuration of `C^(2-)` ion is :
` sigma_(1s)^(2) , sigma_(1s)^(**2) , sigma_(1s)^(2) , pi_(2p_(x))^(2), pi_(2p_(Z))^(2)`

Bond order `(N_(B) - N_(a))/2 = (8 - 4)/ 2 = 2`
Discussion
5.

Drawthe Lewisstructures of the following molecules and ions and ions and tell in which case/casesthe octet rule is violatedCO_(2), SO_(2), BeCl_(2), NH_(3), AlCl_(3), PCl_(5), CO_(3)^(2-)

Answer»

Solution :
The OCTET RULE is violated in CASE of ` BeCl_(2), AlCl_(3) and PCl_(5)`.
Discussion
6.

Draw the Lewis structures of the species : CN^(-), I_(3)^(-), C_(3) O_(2)(carbon suboxide), HN_(3) (hydrazonic acid ).

Answer»

Solution :(i) `CN^(-) [: C:::N:]^(-)`
(ii) `I_(3)^(-) [:UNDERSET(. .)overset(. .)I-underset(. .)overset(. . . .)I-underset(. .)overset(. .)I:]^(-)underset("AcceptorDonor")(("":underset(. .)overset(. . )I-underset(. .)overset(. . )I:+:underset(. .)overset(. .)I:"")^(-)),i.e., [I-IlarrI]^(-)`
(III) `C_(3) O_(2) [:overset(. .)O= C = C = C = overset(. . )O:]`
(iv) `HN_(3) [ H-overset(. . )N=overset(+)N=underset(. .)overset(-)N: harr H-underset(. .)overset(-)overset(. .)N-overset(+)N-=N:]`
Discussion
7.

Draw the Lewis structures for the following species. HNO_(3)

Answer»

SOLUTION :(i) `NO_(3)^(-)`
Discussion
8.

Draw the Lewis structures for the following molecules and ions: H_2S,SiCl_4,BeF_2,CO_3^2, HCOOH

Answer»

SOLUTION :
Discussion
9.

Draw the Lewis structures for the following molecules and ions: H_(2)S, "SiCl"_(4), BeF_(2), CO_(3)^(2-), HCOOH

Answer»

SOLUTION :
Discussion
10.

Draw the lewis structures for the following molecules and ions: H_(2)S, SiCl_(4), BeF_(2), CO_(3)^(2-), HCOOH.

Answer»

SOLUTION :
.
Discussion
11.

Draw the Lewis structures for the following molecules and ions : H_(2) S , SiCl_(4), BeF_(2) , CO_(3)^(2-) , HCOOH

Answer»

SOLUTION :
Discussion
12.

Draw the Lewis structures for the following. (i) SO_(4)^(2-) (ii)O_(3)

Answer»

SOLUTION : (i) `SO_(4)^(2-)`
Discussion
13.

Draw the lewis structures for (i) Nitrous acid (HN0_(2)) (ii) Phosphoric acid (iii) Sulphur troxide (SO_(3))

Answer»

SOLUTION :
Discussion
14.

Draw the Lewis dot structures for sulphurtrioxide.

Answer»

SOLUTION :LEWIS DOT STRUCTURE for `SO_(3)`
Discussion
15.

Draw the Lewis structure of N, C, O and He.

Answer»

Solution :`*underset(.)OVERSET(..)N*`
Lewis Structure of Nitrogen atom
Similarly , Lewis dot structure of carbon, oxygen can be drawn as shown below.
`*underset(.)overset(.)C*""*underset(.)overset(..)O*`
Lewis Structures of C & O atoms
Only EXCEPTION to this is helium which has only two electrons in its VALENCE shell which is REPRESENTED as a pair of DOTS (duet) .
`overset(..)He`
Lewis Structures of He atoms
Discussion
16.

Draw the lewis structure of (i) Nitrogen (ii) Carbon (iii) Oxygen.

Answer»

Solution :`*UNDERSET(*)OVERSET(* *)N*`
Lewis structure of nitrogen ATOM
`* underset(*)overset(*)C* *underset(*)overset(* *)O:`
Lewis structure of C & O atoms
Discussion
17.

Draw the lewis structure of (i) Ammonia (ii) Methane (iii) Dinitrogen pentoxide

Answer»

SOLUTION :
Discussion
18.

Draw the lewis structure of PCl_(5) and SF_(6)

Answer»

SOLUTION :
Discussion
19.

Draw the Lewis dot structure of HCNmolecule.

Answer»

Solution :Solution . STEP .1. Total NUMBER of valence electons in HCN = 1 + 4 + 5 = 10 `(""_(1)H= 1, ""_(6)C= 2 , 4,""_(7)N =2 ,5)`
Step 2. Skeletal structure is HCN(C is least electronegative).
Step 3. Putting ONE shared pair of electons between H and C and between C and N, and the remaining
as LONE pairs, we have

In the structure, duplet of H is complete but octets of CAND N are not complete .Hence, multiple bonding is
required between C and N. Octets of C and N will be complete if three is triple bond between C and N. Thus.
.
Discussion
20.

Draw the Lewis dot structure ofCO_(3)^(2-) ion .

Answer»

Solution :Solution . Step .1 . Total number of valence electons of ` CO_(3)= 4 + 3 xx6 = 22 (""_(6)C= 2 ,4,""_(8)O= 2, 6)`
Step 2 . Total number of electrons to be distributed in ` CO_(3)""^(2-) = 22 + 2 ` (for two units - ve charge) = 24
Step 3. The skeletal sturcture of ` CO_(2)` is
`O OVERSET (O) C O`
Step 4. Putting one SHARED pair of electrons between each C and O and completing the OCTETS of OXYGEN, we have

In this structure, octet of C is not complete. Hence, multiple bonding is required between C and one of the
O- atoms, Drawing a double BOND between C and one O-atom serves the purpose
`(PR_CHE_01_XI_C04_SLV_004_S02)`
Discussion
21.

Draw the Lewis dot structure for the following . (i) SO_(3)""(ii) NH_(3)""(iii) CH_(4)""(iv) N_(2)O_(5)""(v) HNO_(3)

Answer»

SOLUTION :
Discussion
22.

Draw the Lewis dot structure of CO molecule.

Answer»

Solution :Solution. Stop . TOTAL number of valence electrons in ` CO = 4 + 6 = 10 (""_(6) C = 2, 4, ""_(6)C = 2, 4, ""_(8)O= 2, 6)`
Step 2. Skeletal structure is CO.
Step 3. PUTTING asingle bond between C and O, i.e., one shared pair of electrons between C and O, and the
remaining 8 electrons as 3 long pairs on O to COMPLETE its octet and 1 lone pair on C, we have

Step 4. As in this structure , octet of C is not complete, MULTIPLE bonding is required between C and O . Toltbgt Complele the octet of C, triple bond is required between C and O . We should , therefore, shift two lone pairs on O
as shared pairsbetween C and O so that sctet of both C and O remains comlete. thus , the structure should be
Discussion
23.

Draw the lewis structure for Sulphur troxide (SO_(3))

Answer»

SOLUTION :SULPHUR troxide `(SO_(3))`
Discussion
24.

Draw the hyperconjugation resonance structure of (a) CH_(3)CH= CH_(2) (b) CH_(3) overset(+)(C )H_(2) (c ) CH_(3)CH_(2) overset(+)(C )H_(2)

Answer»

Solution :`H- UNDERSET(underset(H)(|))overset(overset(H)(|))(C )-underset(underset(H)(|))overset(overset(H)(|))(C )-underset(underset(H)(|))overset(+)(C )-H HARR H-underset(underset(H)(|))overset(overset(H)(|))(C )-underset(underset(H)(|))overset(H^(+))(C )= underset(underset(H)(|))(C )- H harr H- underset(underset(H)(|))overset(overset(H)(|))(C )- underset(H^(+))overset(overset(H)(|))(C )= underset(underset(H)(|))(C )- H`
Discussion
25.

Draw the lewis structure for Phosphoric acid

Answer»

SOLUTION :PHOSPHORIC acid `(H_(3)PO_(4))` :
Discussion
26.

Draw the lewis structure for Nitrous acid (HNO_(2))

Answer»

SOLUTION :Nitrous ACID `(HNO_(2))` :
Lwis DOT structure
Discussion
27.

Draw the geometrical isomers of above compound.

Answer»

SOLUTION :
Discussion
28.

Draw the formulae of the following compounds and write their IUPAC names :

Answer»

SOLUTION :
Discussion
29.

Draw the first six members of the carboxylic acid homologous series.

Answer»

SOLUTION :(i) HCOOH (ii) `CH_(3)COOH` (iii) `CH_(3)-CH_(2)COOH`
(iv) `CH_(3)-CH_(2)-CH_(2)COOH` (v) `CH_(3)-CH_(2)-CH_(2)-CH_(2)COOH`
(vi) `CH_(3)-CH_(2)-CH_(2)-CH_(2)-CH_(2)-COOH`
Discussion
30.

Draw the fisher projection formula for tartaric acid.

Answer»

SOLUTION :
Discussion
31.

Draw the dash line structure, condensed structure and bond line structure of 1,3-dimethyl cyclopentane.

Answer»

SOLUTION :
Discussion
32.

Draw the conformations of ethane usingnewman projectionformula method

Answer»

SOLUTION :
Discussion
33.

Draw the complete structures of bromomethane, bromoethane, 2 -bromopropane and tertbutyl bromide. Arrange them in order of increasing steric hindrance.

Answer»

Solution :(i) Bromomethane, `CH_3 Br: H- UNDERSET(underset(H)|)overset(overset(H)|)C - Br`
(ii) Bromoethane, `CH_3 CH_2 Br:H - underset(underset(H)|)overset(overset(H)|)C - underset(underset(H)|)overset(overset(H)|)C - Br`
(iii) 2- Bromopropane, `CH_3 CH Br CH_3 : H - underset(underset(H)|)overset(overset(H)|)C- underset(underset(Br)|)overset(overset(H)|)C - underset(underset(H)|)overset(overset(H)|)C- H`
(IV) tert-Butyl bromide, `(CH_3)_3 Br: H- underset(underset(H)|)overset(overset(H)|)C - underset(underset(H-underset(underset(H)|)C-H)|)overset(overset(H-overset(overset(H)|)C-H)|)C - Br`
Increasing order of STERIC hindrance :
Bromomethane < Bromoethane < 2-Bromopropane < tert-Butyl bromide.
Discussion
34.

Draw the complete structural formula, condensed structure and bond line structure of (i) n-propanol (ii) 1, 3-butadiene.

Answer»

SOLUTION :
Discussion
35.

Draw the cis, trans isomeric structures of 1,3-butadiene.

Answer»

SOLUTION :
Discussion
36.

Draw the cis and trans structures of hex-2-ene. Which isomer will have higher b.p. and why?

Answer»

SOLUTION :
cis-Hex-2-ene has a higher dipole moment than trans-Hex2-ene. DUE to higher dipole moment, it will have a stronger dipole-dipole interaction as COMPARED to trans-Hex-2-ene. Consequently, the cis-Hex-2-ene will have a higher BOILING POINT.
Discussion
37.

Draw the cis and trans structures of hex-2-ene. Which isomer will have higher b.p. and why ?

Answer»

Solution :`CH_(3)CH=CHCH_(2)CH_(2)CH_(3)` is hex-2-ene. Their geometrical isomers are as FOLLOWS :

Boiling of cis-isomer is more.

`{:("Cis-Hex-2-ene",|,"Dipole moment is more."),("Trans-Hex-2-ene","Dipole moment is more.",),(THEREFORE " Boiling point is more",therefore " Boiling point is less",):}`
Where , boiling point `prop` dipole moment `(mu)`
`therefore` cis-isomer boiling point `GT` boiling point of trans.
Boiling point, ions in liquid `prop` boiling point
Negative meltingpoint `prop` INTERNUCLEAR force
`therefore` Tras MELTING point is more.
Discussion
38.

Draw the cis-and trans-structures for hex-2-ene. Which isomer will have higher b.p. and why ?

Answer»

Solution :The STRUCTURES of cis-and trans-isomer of hex-2-ene are :

The boiling point of a molecule DEPENDS upon dipole-dipole interactions. Since cis-isomer has HIGHER dipole moments , THEREFORE, it has higher boiling point.
Discussion
39.

Draw structures of all the acyclic isomeric ethers corresponding to the molecular formula C_(5)H_(12)O.

Answer»

Solution :D.B.E. = 1/2 [5(4-2) + 12 (1-2) + 1 (2-2)] + 1 = 0.
Since D.B.E. = 0. therefore, `C_(5)H_(12)O` REPRESENTS only SATURATED ETHERS. The following six isomers are possible.
Discussion
40.

Draw structure of diboren.

Answer»

SOLUTION :
Discussion
41.

Draw structure of Al_2Cl_6 and give uses of AlCl_3.

Answer»

Solution :
USE : Being a Lewis acid, it ACT as CATALYST in Friedal craft ALKYLATION and acylation and also in electrophilic aromatic substitution REACTION.
Discussion
42.

Draw orbital diagrams for atoms with the following electronic configurations. 1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p^(6) 4s^(2) 3d^(7)

Answer»

SOLUTION :
Discussion
43.

Draw orbital diagrams for atoms with the following electronic configurations. 1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p^(5)

Answer»

SOLUTION :
Discussion
44.

Draw orbital diagrams for atoms with the following electronic configurations. 1s^(2) 2s^(2) 2p^(5)

Answer»

SOLUTION :
Discussion
45.

Draw Newman projection formulae staggered forms of butane and compare their stability.

Answer»

SOLUTION :Newman.s PROJECTIONS of BUTANE are :
Discussion
46.

Draw Lewis structures of following molecules and ions. H_(2)S, SiCl_(4), BeF_(2), CO_(3)^(2-), HCOOH

Answer»

SOLUTION :
Discussion
47.

Draw MO diagram of CO and calculate its bond order .

Answer»

Solution :(i) ELECTRONIC configuration of C ATOM: `1s^(2) 2s^(2)2P^(2)`
Electronic configuration of O atom: `1s^(2)2s^(2)2p^(4)`
(ii) Electronic configuration of CO molecule is: `sigma 1s^(2)" " sigma ** 1s^(2)" "sigma2s^(2)" "sigma**2s^(2)" "pi 2p_(y)^(2)" "pi2p_(z)^(2)" "sigma2p_(x)^(2)`
(iii) Bond order `= (N_(b)- N_(a))/(2)(10-4)/(2)=3`
(iv) Molecule has no unpaired ELECTRON, hence it is DIAMAGNETIC.
Discussion
48.

Draw Lewis Structures for the following molecules. (a) H_(2)S "(b) " SiCl_(4) (c) BeF_(2) " and" (d) HCOOH

Answer»


ANSWER :4
Discussion
49.

Draw graph of volume (V) against Temperature (T) n^(+) constant pressure (P) and explain.

Answer»

Solution :GRAPH is straight LINE & positive slop.
`THEREFORE V prop T ""`……(constant p and v)
Graph indicated that ..At constant PRESSURE VOLUME of fixed amount of any gas is directly proportional to its Absolute temperature...
Discussion
50.

Draw formulas for the first five members of each homologous series beginning with the following compounds. (a) HCOOH (b) CH_(3)COCH_(3) (c ) H- CH = CH_(2)

Answer»

Solution :(a) (i) HCOOH (ii) `CH_(3)COOH` (iii) `CH_(3)-CH_(2)-COOH` (iv) `CH_(3)-CH_(2)-CH_(2)-COOH` (V) `CH_(3)-CH_(2)-CH_(2)-CH_(2)-COOH`
(b) (i) `CH_(3)COCH_(3)` (ii) `CH_(3)COCH_(2)CH_(3)` (iii) `CH_(3)COCH_(2)CH_(2)CH_(3)` (iv) `CH_(3)COCH_(2)CH_(2)CH_(2)CH_(3)`
(v) `CH_(3)COCH_(2)CH_(2)CH_(2)CH_(2)CH_(3)`
(c ) (i) `H-CH= CH_(2)`
(ii) `CH_(3)-CH= CH_(2)`
(iii) `CH_(3)-CH_(2)-CH= CH_(2)`
(iv) `CH_(3)-CH_(2)-CH_(2)-CH= CH_(2)`
(v) `CH_(3)- CH_(2) - CH_(2) - CH_(2) - CH= CH_(2)`
Discussion