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.

The bond between carbon atoms 1 and 2 in the compound : N-=overset(1)(C)-overset(2)(C)H=CH_(2) involves the hybridised carbon as :

Answer»

`SP^(2)` and `sp^(2)`
`sp^(3)` and `sp`
`sp` and `sp^(2)`
`sp` and `sp`.

ANSWER :C
Discussion
2.

The bond between carbon atom (1) and carbon atom (2) in compound N-=underset(1)(C)-underset(2)(CH)=CH_(2), involves the hybridization as

Answer»

`sp^(2)` and `sp^(2)`
`sp^(3)` and sp
sp and `sp^(2)`
sp and sp

Solution :`N -= underset(1)(C)-underset(2)(C)H=underset(3)(C)H_(2)`
`'C_(1)'` FORMS TWO `SIGMA`-bonds, hence it is sp-hybridized `C_(2)` forms three `sigma`-bonds, hence, it is `sp^(2)`-hybridized.
Discussion
3.

The bond between carbon atom (1) and carbon atom (2) in compound N-=C-CH=CH_2 involves the hybrid is:

Answer»

`SP^2` and `sp^2`
`sp^3` and `sp`
`sp and sp^2`
`sp` and `sp`

ANSWER :C
Discussion
4.

The bond angles in molecules H_(2)O, NH_(3), CH_(4) and CO_(2) are in the order :

Answer»

`H_(2)O gt NH_(3)gt CH_(4)gt CO_(2)`
`H_(2)O lt NH_(3)lt CH_(4)lt CO_(2)`
`H_(2)O lt NH_(3)lt CO_(2)lt CH_(4)`
`H_(2)O gt NH_(3)gt CO_(2)gt CH_(4)`

Solution :Bond ANGLES are : `H_(2)O(104.5), NH_(3)(107), CH_(4)(109.5), CO_(2)(180)`.
Discussion
5.

The bond between atoms of two elements of atomic number 37 and 53 is :

Answer»

Covalent
Ionic
Coordinate
Metallic

Answer :B
Discussion
6.

The bond angle O-S-O and hybridisation of sulphur in SO_2 are:

Answer»

`119.5^@, sp^3`
`119.5^@, sp^2`
`109^@ 28. , sp^3`
None

Answer :B
Discussion
7.

The bond angle in XeO_3 molecule decreases from 109^@28^1 to 1030^@ due to

Answer»

GREATER repulsions AMONG lone PAIRS
greater repulsions among l.p and b.p
greater repulsions among bodn pairs
steric effect

ANSWER :B
Discussion
8.

The bond angle in XeF4 molecule is

Answer»

`180^@c`
`120^@`
`100^@`
`90^@`

Answer :D
Discussion
9.

The bond angle in TeBr_(2) is

Answer»

`101.5^(@)`
`98^(@)`
`103^(@)`
`90^(@)`

ANSWER :B
Discussion
10.

The bond angle in PH_3 is less than bond angle in PF_3 This is due to

Answer»

Enhanced replulsions due to the presence of double bond in `PF_3`
Increases bond pair - bond pair REPULSIONS due to MULTIPLE bond
Both (a) and (b)
DISPLACEMENT of ELECTRON could in P - F bond towards F in `PF_3`

Solution :
Discussion
11.

Bond angle in "PH"_(4)^(" "+) is higher than that in "PH"_(3). Why ?

Answer»

Solution :In `PH_(4)^(+)`, phosphorous undergo `SP^(3)` hybridization, but in `PH_(3)` phasphorous uses PURE p-oribitals for bonding. Hence BOND angle in `PH_(3)` is nearly `90^(@)`.
Discussion
12.

The bond angle in NF_(3) (102.3^(@)) is similar than NH_(3) (107.2^(@)). This is because of

Answer»

large size of F COMPARED to H
large size of N compared of F
opposite POLARITY of N in the two molecules
small size of H compared to N.

Solution :
Discussion
13.

The bond angle in H_2S is:

Answer»

`109^@ 28`
`104^@ 51.`
`120^@`
`92.5^@`

Answer :D
Discussion
14.

The bond angle in H_2O (for H-S-H) is

Answer»

same as that of Cl-Be-Cl in `BeCl_2`
greater than H-N-H bond angle in `NH_3`
greater than H-Se-H and less than H-O-H
same as Cl-Sn-Cl in `SnCl_2`

SOLUTION :The bond angle in `BeCl_2=180^@` , in `NH_3=107^@`, in `H_2O=104.5^@`, in `SnCl_2=119^@`
WHEREAS that of
is close to `90^@`
Discussion
15.

The bond angle in H_(2)S is

Answer»

`109^(@)28`
`104^(@)51`
`120^(@)`
`92.5^(@)`

Answer :D
Discussion
16.

OCIO bond angle in CIO_(2)^(-) is

Answer»

`109^(@)28^(1)`
`111^(@)`
`118^(@)`
`115^(@)`

ANSWER :B
Discussion
17.

The bond angle in Cl_2O molecule is:

Answer»

`180^@`
`105^@`
`90^@`
`111^@`

Answer :B
Discussion
18.

What is the bond angle in the molecule of ammonia ?

Answer»

`109^@28'`
`90^@`
`107^@`
`105^@`

Answer :C
Discussion
19.

What is the bond angle in the molecules of ammonia ?

Answer»

`109^@28`.
`104^@31`.
`120^@`
`106^@ 51`

Answer :D
Discussion
20.

The bond angle C- OH in methanol is ………………….. .

Answer»

SOLUTION :`108.09^@`
Discussion
21.

The bond angle and % of d- character is SF_(6) are

Answer»

`120^(@), 20%`
`90^(@), 33%`
`109^(@),25%`
`90^(@), 25%`

Solution :Sulphur hexafluoride involves `sp^(3)d^(2)` hybridisation and posses octahedral geometry.
% of d-character `=((2"no. of d-orbital"))/(6("total hybridised ORBITALS"))xx100=33%`
BOND ANGLE `=90^(@)`
Discussion
22.

The bolume of nitrogen gas v_(m) (measured at S.T.P.) required to cover a sample of silica gel with a mono-molecular layer is 129 cm ^(3) g^(-1) of gel. Calculate the surface area per gram of the gel if each nitrogen molecule occupies 16.2xx10^(-20)m^(2).

Answer»

Solution :`22400cm^(3)` or `N_(2)` at S.T.P. CONTAIN `=6.022xx10^(23)` molecules
`THEREFORE 129 cm ^(3)` of `N_(2)` at S.T.P. will contain `=(6.022xx10^(23)xx129)/(22400)=3.468xx10^(21)` molecules
Area occupied by a single molecule `=16.2xx10^(-20)m^(2)`
`therefore` Area occupied by `6.468 xx10^(21)` molecules of `N_(2)=(16.2 xx10^(-20))XX(3.468xx10^(21))m^(2)=561.8m^(2)`
`therefore` Surface area per gram of gel `=561.8m^(2).`
Discussion
23.

The bond angle and O-O bond length in O_3 are respectively ________.

Answer»

`116.8^(@) and 1.28Å`
`90^(@) and 1.38 Å`
`120^(@) and 1.1 Å`
`60^(@) and 1.5 Å`

ANSWER :A
Discussion
24.

The boliling point of 5% solution (W/W) of non-volatile solute in water is 100.45^(@)C. The boiling point of pure water is 100^(@)C.Calculate th molar mas of th solute (K_(b)"for water=0.52 K kg mol"^(-1)).

Answer»


SOLUTION :`DeltaT_(b)=100.45^(@)C-100^(@)C=0.45 K`
`W_(B)=5G, W_(A)=100-5=95g, 0.095 KG, K_(b)=0.52" K kg mol"^(-1)`
`M_(B)=(K_(b)xxW_(B))/(DeltaT_(b)xxW_(A))=((0.52K" kg mol"^(-1))XX(5g))/((0.45K)xx(0.095 kg))=60.8" g mol"^(-1)`.
Discussion
25.

The boling point of benzone is 353.23K. When 1.80 g of a non-valatioe solute is mixed in 90 g of benzene, the boling point id raised to 354.11 Calculate the molar mass of the solute. Given that K_(b) benzene is 2.53 K kg mol^(-1)

Answer»

Solution :`M_(B)=(K_(b)xxW_(B))/(DeltaT_(b)xxW_(A))`
`W_(B)=1.80 KG,k_(b)=2.53K Kg mol^(-1)`
`DeltaT_(b)=(354.11-353.23)=0.88 K`
`M_(B)=((1.80)XX(2.53 kg mol^(-1)))/((0.88)xx(0.090 kg))=9.5 g mol^(-1)`
Discussion
26.

The boiling point of an azeotropic mixture of water and ethyl alcohol is less than that of the theoretical value of water and alcohol mixture. Hence the mixture shows

Answer»

that SOLUTION is HIGHLY saturated
positive DEVIATION from RAOULT's law
negative deviation from Raoult's law
nothing can be said

Answer :B
Discussion
27.

The boiling points of thio-ethers are____than those of ether:

Answer»

LESSER
Equal
higher
None

Answer :C
Discussion
28.

The boiling points of the alcohols are higher than the alkanes of corresponding molecular weight because .

Answer»

Alcohols can form H-bond(molecular ASSOCIATION)
ALKANES are non-polar
Alcohols are polar
Alcohols have low densities

Answer :A
Discussion
29.

The boiling points of organohalogen compounds are comparatively ___________ than corresponding hydrocarbons.

Answer»


ANSWER :HIGHER
Discussion
30.

The boiling points of hydrides of group 16 elements are in the order :

Answer»

`H_(2)O GT H_(2)S gt H_(2)TE gt H_(2)Se`
`H_(2)Te GE H_(2)Se gt H_(2)S gt H_(2)O`
`H_(2)O gt H_(2)Te gt H_(2)Se gt H_(2)S`
`H_(2)Te gt H_(2)O`

ANSWER :C
Discussion
31.

The boiling points of HF, HCI, HBr and HI follow the order

Answer»

`HF gt HCI gt HBr gt HI`
`HF gt HI gt HBr gt HCl`
`HI gt HBr gt HCl gt HF`
`HCl gt HF gt HBr gt HI`

Solution :Boiling POINT depends on the following factors
(1) Stronge H-bonding that increase boiling point.
(2) VAN der Waal.s force increase boiling point. Van der WAALS. force increases with increase in atomic radius.
(3) Increase in the molecular weight increases boiling point.
Factor (2) and (3) give us the following order:
`HI > HBr > HCl`
Factor (1) makes HF highest boiling point containing.
Discussion
32.

The boiling points of H_2O, H_2S , H_2Se and H_2Te are in the following order

Answer»

`H_2O GT H_2S gt H_2Se gt H_2Te`
`H_2Ogt H_2Te gt H_2Se gt H_2S`
`H_2Te gt H_2Se gt H_2S gt H_2O`
`H_2S gt H_2O gt H_2Te gt H_2Se`

ANSWER :B
Discussion
33.

The boiling points of halogens increase with increase in molecular weight, it is because:

Answer»

As the size increases molecules UNDERGO ASSOCIATION leading to HIGHER stability
Bond strength increases due to increase in electronegativity
Van DER Waals. forces increase with increase in number of electrons per mole
None

Answer :C
Discussion
34.

The boiling points of amines are____than those of alcohols of comparable molecular mass due to_______

Answer»

SOLUTION :LOWER, WEAKER H-bonding.
Discussion
35.

The boiling points for aqueous solutions of sucrose and urea are same at constant temperature. If 3 gm of urea is dissolved in its 1 litre solution, what is the weight of sucrose dissolved in its 1 litre solution ?[Urea - 60 gm / mole, sucrose = 342 gm / mole]

Answer»

3.0 GRAM
17.1 gram
6.0 gram
34.2 gram

ANSWER :B
Discussion
36.

The boiling points of acetic anhydride (I), propionic anhydride (II) and butyric anhydride (III) follow the order:

Answer»

`I LT II lt III`
`III lt II lt I`
`I lt III lt II`
` II lt III lt I`

ANSWER :B
Discussion
37.

The boiling points of aldehydes and ketones are lower than that of the corresponding alcohols and acids. Why?

Answer»

SOLUTION :ALDEHYDES and ketones have lower boiling point than alcohols and carboxylic ACIDS because they are not associasted with intermolecular H-bonding WHEREAS alcohols and carboxylic acids are ASSOCIATED with intermolecular H-bonding.
Discussion
38.

The boiling points of aldehydes and ketones lie in between alkanes and alcohols of comparable masses because :

Answer»

Alkanes are polar
Aldehydes and KETONES are non-polar
alkanes are nonpolar and aldehydes and KETONE contain polar gtC=O group and lower alcohols have H-bonding.
Alkanes are held TOGETHER by weak Vander Waals forces (being non - polar) , aldehydes and ketones contain polar. gtC=O group and held together by STRONG dipole-dipole attraction and lower alcohol have H-bonding, which is stronger than dipole dipole attraction.

Answer :D
Discussion
39.

The boiling point of water is exceptionally high because

Answer»

there is COVALENT BOND between H and O
water molecules ASSOCIATE due to hydrogen bonding
water MOLECULE is not linear
water molecule is linear.

Answer :B
Discussion
40.

The boiling point of water is 100^(@)C and it becomes 100.52^(@)C if 3 g of a non-voltile is dissolved in 20 mL of it. Calculatethe molecular weight of the solute. (K_(b) for water is 0.52 K kg mol^(-1), density of water= 1 g mol^(-1)).

Answer»


Solution :`DeltaT_(b)=100.52-100=0.52^(@)C=0.52 K`
`W_(B)= 3 g, W_(A)=20 g [20" mL of water= 20 g"], 0.02 kg, k_(b)=0.52" K kg mol"^(-1)`
`M_(B)= ?`
`M_(B)=(K_(b)xxW_(b))/(DeltaT_(b)xxW_(A))=((0.52" K kg mol"^(-1))xx(3G))/((0.52K)xx(0.02 kg))=150" g mol"^(-1)`.
Discussion
41.

The boiling point of water ,ethyl alcohol and diethyl ether are 100°C , 78.5°C and 34.6°C respectively. Intermolecular forces will be in order of

Answer»

`WATER gt Ethyl ALCOHOL gt Diethyl ether`
`Ethyl alcohol gt Water gt Diethyl ether`
`Diethyl ether gt Ethyl alcohol gt Water`
`Diethyl ether gt Water gt Ethyl alcohol`

ANSWER :A
Discussion
42.

The boiling point of water becomes 100.52^(@)C if 1.5 g of a non-volalite solute is dissolved in 100 mL of it. Calculate the molecular weight of the solute. (K_(b) for water=0.6 K kg mol^(-1)). (density of waer-1 g moL^(-1)).

Answer»


Solution :`DeltaT_(b)=100.52-100=0.52^(@)C=0.52 K`
`W_(B)=1.5g, W_(A)=100g=0.1 kg ""[1" mL of water"~~1G]`
`K_(b)=0.6K//m, M_(B)=?`
`M_(B)=(K_(b)xxW_(B))/(DeltaT_(b)xxW_(A))=((0.6" K kg MOL"^(-1))xx(1.5g))/((0.52 K)xx(0.1 kg))=17.31" g mol"^(-1)`
Discussion
43.

The boiling point of water at high altitude is low, because

Answer»

the TEMPERATURE is low
the ATMOSPHERICPRESSURE is low
the temperature is high
the ATMOSPHERIC pressure is high

Answer :A::B::C
Discussion
44.

The boiling point of water at high altitude is low, because-

Answer»

the TEMPERATURE is LOW.
the ATMOSPHERIC PRESSURE is low.
the temperature is HIGH.
the atmospheric pressure is high.

Answer :B
Discussion
45.

The boiling point of water (100^@ C) becomes, 100.52^@ C if 3 g of non-volatile solute is dissolved is 200 ml of it: Calculate the molecular weight of the solid. (K_b = 0.6 K/m)

Answer»

Solution :`DELTA T_b = 100.52 -100=0.52 ^@ C K`
` W_B =3g`
`W_A =200 ml`
` M_B=?`
` K_b = 0.6 K//m`
` DeltaT_b =K_b xx(W_b xx 1000 )/( M_B W_A)`
` 0.52 =( 0.6 xx 3xx 1000)/( 200xx 0.52 )`
`=( 90 )/( 0.52)`
`=173.07//Mol .`
Discussion
46.

The boiling point of water at 760 mm pressure is 373 K. The vapour pressure of water at 298 K is 23 mm. If the vaporisation enthalpy is 40.656 KJ/mole, then what will be boiling point of water at 23 mm. pressure ?

Answer»

250 K
298 K
51.6 K
12.5 K

SOLUTION :Boiling POINT is the temperature at which the vapour PRESSURE will be EQUAL to the atmospheric pressure.
Discussion
47.

The boiling point of solution containing 68.4 g of sucrose ("molar mass" = 342 g mol^(-1)) in 100 g of water is (K_(b) "for water" = 0.512 K. kg mol^(-1))

Answer»

`100.02^(@)C`
`98.98^(@)C`
`101.02^(@)C`
`100.512^(@)C`

Solution :`DELTA T_(b) = K_(b) XX m = K_(b) xx W_(2)/(M_(2) xx W_(1) (kg))`
`=0.512 xx 68.4/(342 xx 0.1) = 1.024 ^(@) C `
` :. T_(b) = T_(b)^(o) + Delta T_(b) =100 ^(@) C + 1.024^(@)C=101.024^(@)C`
Discussion
48.

The boiling point of p-nitrophenol is hgiher than that of o-nitrophenol because

Answer»

`NO_(2)` group at p-position behaves in a DIFFERENT way from that at o- position
INTRAMOLECULAR hydrogen bonding exists inp-nitrophenol
there is INTERMOLECULAR hydrogen bonding in p-nitrophenol
p-nitrophenol has a HIGHER molecular weight than o-nitrophenol.

Solution :The b.pt of p-nitrophenol is higher than that of o-nitrophenol because in p-nitrophenol there is intermolecular H-bonding in o-nitrophenol it is intramolecular H=bonding.
Discussion
49.

The boiling point of methanol is greater than that of Methyl thiol because

Answer»

there is intramolecular hydrogen BONDINGIN methanol and intermolecular hydrogen bondingin methyl thiol
there is intermolecularhydrogen BONDING in methanol and no hydrogen bonding in methyl thiol
THEREIS no hydrogen bonding inmethanoland intermolecular hydrogen bonding in methly thiol
there is intramolecularhydrogen bonding in methanol and no hydrogen bonding in methyl thiol

Answer :B
Discussion
50.

The boiling point of n-alkanes is …. Than the branched chain alkanes of the same molecule wieght

Answer»

More
Equal
Less
None of these

Answer :A
Discussion