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.

Bond energies of N-=N, H-H and N-H bonds are 945,463 & 391 kJ mol^(-1) respectively, the enthalpy of the following reactions is : N_(2)(g)+3H_(2)(g)rarr2NH_(3)(g)

Answer»


SOLUTION :`DELTAH = 6xxDeltaH_(N-H) - DeltaH_(N-N) - 3DeltaH_(H-H) = 93 KJ`
2.

Bond energies of F_(2) and Cl_(2) are respectively 36.6 and 58.0 kcal per mole. If the heat liberated in the reaction F_(2)+Cl_(2) to 2FCl is 26.6 kcal , calculate the bond energy of F-Cl bond.

Answer»

SOLUTION :Given that,
`F_(2)+Cl_(2) to 2FCl`, `DeltaH=-26.6` kcal
For reactants
Bond energy of 1 mole of `F-F` bonds `=36.6 ` kcal
Bond energy of 1 mole of Cl-Cl bonds `=58.0` kcal
For PRODUCT
Energy of formation of 2 moles of F-Cl bond `=2XX x`
(where x is the bond formation energy of F-Cl bonds in kcal/mole)
Adding all the HEAT changes , we get `DeltaH` of the given reaction i.e.,
`36.6+58+2xx x=-26.6`
`x=-60.6`kcal
Thus, the bond energy of `F-Cl` bonds is `+60.6` kcal per mole.
3.

Bond dissociation enthalpy of the halogens shows the trend

Answer»

`F-F LT Cl - Cl GT Br - Br gt I - I `
`F - F gt Cl - Cl gt Br - Br gt I-I`
`F - F gt Cl - Cl lt Br - Br gt I-I`
`F - F gt Br - Br gt Cl- Cl gt I -I`

SOLUTION :`F-F lt Cl - Cl gt Br - Br gt I - I `
4.

Bond dissociation enthalpy of H_(2), Cl_(2) and HCl are 434, 242 and 431 kJ mol^(-1) respectively. Enthalpy of formation of HCl is

Answer»

`- 93 KJ mol^(-1)`
245 kJ `mol^(-1)`
93 kJ `mol^(-1)`
`-245` kJ `mol^(-1)`

Solution :`(1)/(2)H_(2)+(1)/(2)Cl_(2)rarrHCl`
`DeltaH=(1)/(2)xx434+(1)/(2)xx242-431`
`=217+121-431=-93 kJ//"MOLE"`.
5.

Bond dissociation enthalpy of F_(2) is less than that of Cl_(2). Explain why ?

Answer»

Solution :`F_(2)` is having HIGHER electron-electron repulsion DUE to its smaller size, as COMPARED to `Cl_(2)`.
6.

Bond dissociation enthalpy of E-H (E = element) bonds is given below. Which of the compounds will act as strongest reducing agent?

Answer»

`NH_3`
`PH_3`
`AsH_3`
`SbH_3`

Solution :WEAKER the E-H bond, STRONGER the reducing AGENT. Hence `SbH_3`is the strongest reducing agent.
7.

Bond dissociation enthalpy of E-H(E = element) bonds is given below: {:("Compound",NH_3,PH_3,AsH_3,SbH_3),(Delta_("diss")"(E-H)",,,,),(kjmol^(-1),389,322,297,255):} Which of the following compounds will act as strongest reducing agent ?

Answer»

`NH_3`
`PH_3`
`AsH_3`
`SbH_3`

Solution :WEAKER the E-H bond, STRONGER the reducing AGENT. Hence `SbH_3`is the strongest reducing agent.
8.

Bond dissociation enthalpy of E - H (E = element) bonds is given below. Which of the compounds will act as strongest reducing agent ?

Answer»

`NH_(3)`
`PH_(3)`
`AsH_(3)`
`SbH_(3)`

SOLUTION :`SbH_(3)` will ACT as STRONGEST reducing agent.
9.

Bond dissociation enthalpy of E-H (E = element) bonds is given below. Which of the compounds will act as strongest reducing agent? {:("Compound",NH_(3),PH_(3),AsH_(3),SbH_(3)),(Delta_(diss)(E-H)//kJ mol^(-1),389,322,297,255):}

Answer»

`NH_(3)`
`PH_(3)`
`AsH_(3)`
`SbH_(3)`

Solution :Weaker the E-H BOND, STRONGER the REDUCING agent, i.e., `SbH_(3)`.
10.

Bond dissociation energy of F_(2) is less than that of CI_(2). Explain. Or F_(2) has lower bond dissociation enthalpy than CI_(2). Why ?

Answer»

Solution :Due to SMALLER SIZE, the LONE pairs of ELECTRONS on the F-atoms repel the bond pair of the F-F bond. In CONTRAST, because of comparatively large size of CI atoms, the lone pairs on the CI atoms do not repel the bond pair of CI-CI bond. As a rsult, F-F bond energy is lower than that of CI-CI bond energy.
11.

Bond dissociation energies of HF, HCl, HBr follow the order

Answer»

`HClgtHBrgtHF`
`HFgtHBrgtHCl`
`HFgtHClgtHBr`
`HBrgtHClgtHF`

ANSWER :C
12.

Bond dissociation energies, of HF,HCI, HBr follow the order

Answer»

`HCl gt HBR gt HF`
`HRgtHBr gt HCl`
`HF gt HCl gt HBr`
`HBr gt HCl gt HF`

SOLUTION :`HF gt HCl gt HBr gt HI`.
13.

Bond dissociation energies of HF, HCl , HBr follow the order

Answer»

`HCL gt HBR gt HF`
`HF gt HBr gt HCl`
`HF gt HCl gt HBr`
`HBr gt HCl gt HF`

Answer :C
14.

{:("Bond" ,"Bond dissociation energy" ("kJ mole"^(–1))),( C-I ,240 ""to "Element A"),( C-II ,328 ""to "Element B"), (C-III, 276 ""to "Element C"),( C-IV ,485 ""to "Element D"):} Elements A, B, C and D, which element has the smallest atom ?

Answer»

I
III
II
IV

ANSWER :D
15.

Bond angles in SCl_(2) and OF_(2) respectively are

Answer»

`107^(0), 101.5^(0)`
`103^(0), 109.5^(0)`
`101.5^(0), 105^(0)`
`103^(0), 103^(0)`

ANSWER :D
16.

Bond angle is the highest in the molecule

Answer»

`XeO_4`
`XeF_4`
`XeO_3`
`XeF_2`

ANSWER :D
17.

What is the bond angle in SO_2 molecule ?

Answer»

`120^@`
`90^@`
`180^@`
`109^@ 28.`

Answer :A
18.

Bond angle is minimum for

Answer»

`H_2O`
`H_2S`
`H_2Se`
`H_2Te`

ANSWER :D
19.

Bond angle in PH_(4)^(+)is higher than that in PHz. Why?

Answer»

Solution : Both `PH_4^+`and `PH_3`involve `sp^3`hybridisation of P atom. In `PH_4^+`all the four orbitals are BONDED, WHEREAS in PH, there is a lone pair of electrons on P. In `PH_4^+` , the HPH bond angle is TETRAHEDRAL angle of `109.5^@`. But in `PH_3` , lone pair-bond pair repulsion is more than bond pair-bond pair repulsion so that bond angles becomes LESS than normal tetrahedral angle of `109.5^@`. The bond angle in `PH_3`has been found to be about `93.6^@`
20.

Bond angle in PH_4^(+) is higher than that in PH_3. Why?

Answer»

Solution :P in `PH_3` is `sp^(3)`-hybridised. It has three bond pairs and one lone pair around P. Due to stronger lone pair-bond pair repulsions than bond pair-bond pair repulsions, the tetrahedral angle decreases from `109^(@)28. to 93.6^(@)`. As a result, `PH_3` is pyramidal. HOWEVER, when it REACTS with a PROTON, it forms `PH_4^(+)` ion which has four bond pairs and no lone pair. Now, there are no lone pair-bond pair repulsions. Only four identical bond pair-bond pair interactions exist. `PH_4^(+)` therefore assumes tetrahedral geometry with a bond angle of `109^(@)28.`. This explains why the bond angle in `PH_4^(+)` is higher than in `PH_3`
21.

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

Answer»

Solution :P in `PH_(3)` is `SP^(3)`-hybridized. It has three bond pairs and one lone PAIR around P. Due to stronger lone pair-bond pair repulsions than bond pair-bond pair repulsions, the tetrahedral angle decreases from `109^(@)`- 28' to `93.6^(@)`. As a result, `PH_(3)` is pyramidal. However, when it REACTS with a proton, it forms `PH_(4)^(+)` ion which has four bond pairs and no lone pair. Due to the absence of lone pair-bond pair repulsions and presence of four identical bond pair-bond pair interactions, `PH_(4)^(+)` assumes tetrahedral geometry with a bond angle of `109^(@)`-28'. This explains why the bond angle in `PH_(4)^(+)` is higher than in `PH_(3)`.
22.

Bond angle in PH_3 si closer to 90^(@) while that in NH_3 is 104.5^(@).Which of the following best explains this structural feature?

Answer»

Due to larger size of the LONE pair electron cloud, there is larger lone pair - bond pair REPULSION in `PH_3` compared to `NH_3`
Higher electronegativity of nitrogen concentrates the bond pair electron cloud near the central ATOM which increases the bond pair - bond pair repulsion which in turn decreases the bond angle in `NH_3`
Energy difference between 3s an,d 3P orbitals is quite high and hence the lone pair on phosphorous prefers to occupy unhybridized s- orbital rather than hybridized `sp^3` hydridized orbital which causes its s-orbital energy to increase.
Phosphorous FORMS `ppi-dpi` bonds while nitrogen does not.

Answer :C
23.

Bond angle in PH_(3) is :

Answer»

1. GREATER than in `PF_(3)`
2. smaller than in `PCl_(3)`
3. larger than in `BF_(3)`
4. same as in `NH_(3)`

Answer :A
24.

Bond angle in PH4+ is higher that in PH_(3). Why?

Answer»

Solution : In both `PH_(4)^(+) and PH_(3)` , atom phosphorus is in `SP^(3)`.hybridised state. `PH_(4)^(+)`is TETRAHEDRAL in shape and bond angle is `109^(@) 28^(1)`In `PH_(3)`due to the presence of one lone pair of electrons bond angle deceases from tetrahedral angle.
25.

Bond angle in O_3 molecule is :

Answer»

`108. 29^0`
`108.28^0`
`116.90^0`
`120^0`

ANSWER :C
26.

Bond angle in is PH_4^+ higher than that inPH_3

Answer»

<P>

Solution :The HYBRIDISATION of P in both `PH_3` and `PH_4^+` is `sp^3.PH_3`CONTAINS THREE bond pairs and one lone pair of electrons. DUE to lone pair-bond pair repulsion, the bond angle is less than the tetrahedral angle in `PH_3`. But in `PH_4^+`there are four bond pairs and no lone pair. So its bond ,angle is almost the same as that of tetrahedral angle.
27.

Bond angle in (CH_(3))_(3) N is little more than the bond angle in NH_(3) . Explain.

Answer»

SOLUTION :In ammonia bond angle is `107^(@)` . In `(CH_(3))_(3)` N it is nearly `108^(@)` it is due to presence ofthree bulky -`CH_(3)` GROUPS .
28.

Bond angle in alkynes is :

Answer»

`109^@ 28'`
`180^@`
`120^@`
`360^@`

Answer :B
29.

Bond angle in alkenes is equal to :

Answer»

`120^@`
`109^@28' `
`180^@`
`60^@`

ANSWER :A
30.

Bond angle, bond length and hybridisation in SO_(3) molecule respectively are

Answer»

`119.5^(0), 143 nm, SP^(2)`
`119.5^(0), 143 pm, sp^(2)`
`119.5^(0), 143 pm, sp^(3)`
`119.5, 143 A^(0), sp^(2)`

ANSWER :B
31.

Bombardment of aluminum by alpha-particle leads to its artificial disintegration in two ways. (I) and (II) as shown. Products X, Y and Z respectively are, {:(._(13)^(27)Aloverset((ii))rarr ._(15)^(30)P + Y),((i) darr "" darr),(._(14)^(30)Si+ X"" ._(14)^(30)Si + Z):}

Answer»

Proton , neutron, positron
Neutron, positron, proton
Proton, positron, neutron
Positron, proton, neutron

Solution :`{:(""._(13)^(27)AL overset((ii) rarr 2 He^(4))rarr ._(15)^(30)P + underset((y)"neutron")(._(0)^(1)N)),((i) rarr ._(2)He^(4) darr"" darr),(""._(14)^(30)SI + underset((x)"proton")(._(1)^(1)p)"" ._(14)^(30)Si + underset((z)"positron")(._(1)^(0)e)):}`
32.

Bombardment of aluminium of alpha- particle leads to its artificial disintegration in two ways (i) and (ii) as shown below. Product X,Y, and Z, respectively, are ._(14)Si^(30)+Xoverset((i))larr._(13)Al^(27)overset((ii))rarr._(15)P^(30)+Y rarr ._(14)Si^(30)+Z

Answer»

Proton, neutron, positron
Neutron, positron, proton
Proton, positron, neutron
Positron, proton, neutron

Solution :`underset(._(2)He^(4)darr(i))(._(13)Al^(27)overset(._(2)He^(4)))rarr underset(darr(II))(._(15)P^(30)+._(0)N^(1)(y)`
`._(14)SI^(30)+._(1)p(x)^(1)``._(14)Si^(30)+._(1)e^(+0)(Z)`
`:. x:` Protons, `y:` neutron, `z:` positron
33.

Bombardment of aluminium by alpha-particles lead to its artificial disintegration in two ways, (and (it) as shown. Products X, Y and Z respectively are:

Answer»

proton, neutron, positron
neutron, positron, proton
proton, positron, neutron
positron, proton, neutron

Solution :`""_(13)^(27)Al+""_(2)^(4)alphato""_(14)^(30)Si+""_(1)^(1)p[X]`
`""_(13)^(27)Al+""_(2)^(4)alphato""_(15)^(30)P+""_(0)^(1)n[Y]`
`""_(15)^(30)Pto""_(14)^(30)Si+""_(+1)^(0)beta[Z]`
34.

Bombardment of aluminium by alpha- particles leads to its artificial disintegration in two ways , (i) and (ii) as shows . Product X , Y and Z respectively are .

Answer»

proton , neutron , positron
neutron , positron , proton
proton , positron , neutron
positron , proton , neutron

Solution :`""_(13) Al^(27) + overset( (""_(2) He^(4)))(to) ""_(14) SI^(30) + ""_(+1) X^(1) (""_(+) X^(1) = p) , ""_(13) Al^(27) + overset((""_(2) He^(4)))(to) ""_(15) p^(30) + ""_(0) y^(1) (""_(0) y^(1) = ""_(0) n^(1))`
`""_(15) P^(30) to ""_(14) Si^(1) + ""_(+1) Z^(0) (""_(+1) Z^(0)= ""_(+) e^(0))`
35.

Bombardment of aluminiumby alpha - particle leads to its artificial disintegration in two way (i) and (ii) as shown.Products X, Y and Z respectively are,

Answer»

proton, neutron, positron
neutron, positron, proton
proton, positron, neutron
positron, proton, neutron

Solution :`._(2)He^(4) +._(13)^(27)Al rarr ._(14)^(30)Si +._(Z)^(A)X`
Applying nuclear charge & mass NUMBER balance `z=1,A=1`
`:. X=rArr ._(1)H^(1)` or `._(1)P^(1)`
`._(15)^(27)Al+._(2)He^(4) rarr ._(15)^(30)P+._(Z)^(A)Y`
Applying nuclear charge and mass number balance Z=0, A=1
`:. Y rArr ._(0)n^(1)`
`._(15)^(30)P rarr ._(14)^(30)Si +Z`
Applying nuclear charge and mass number balance
`Z rArr ._(+1)e^(0)`
36.

Boiling points of the alkyl halides decrease in the order :

Answer»

`RI GT RBR gt RVl gt RF`
`RF gt RCl gt RBr gt RI`
`RI gt RBr gt RF gt RCl `
`RI gt RVl gt RBr gt RF `

ANSWER :a
37.

Boilingpointsof the followingcompoundsfollowthe order

Answer»

`CH_(3) CH_(3) lt CH_(3) NH_(2)lt CH_(3) lt HCOOH`
`CH3NH_(3) lt CH_(3)OHlt CH_(3) CH_(3) lt HCOOh`
`CH_(3) OH lt CH_(3) CH_(3) lt CH_(3) NH_(2) lt HCOOH`
`HCOCH lt CH_(3) NH_(2) lt CH_(2) OH lt CH_(3) CH_(2) CH_(3)`

Solution :Boilingpoint OFCOMPOUND`prop` molecularweight
38.

Boiling points of nitroalkanes are much higher than those of hydrocarbons of comparable mass - give reasons.

Answer»

Solution :NITROALKANES are POLAR in nature `(mu = 3 - 4D)` and thus have greater dipolar attraction than hydrocarbons. This results in HIGHER values of b.p. `(mu =" DIPOLE moment")`
39.

Boiling points of isomeric amines follow the order :

Answer»

PRIMARY `GT` SECONDARY `gt` Tertiary
Secondary `gt` Primary `gt` Tertiary
Tertiary `gt` Secondary `gt` Primary
Tertiary `gt` Primary `gt` Secondary

Answer :A
40.

Boiling points of carboxylic acids are:

Answer»

LOWER than corresponding ALCOHOLS
Higher than corresponding alcohols
Equal to that of corresponding alcohols
None of the above STATEMENT is correct

Answer :B
41.

Why B.P. of aldehydes and ketones are lower than corresponding alcohols ?

Answer»

Solution :The ALDEHYDES and KETONES are polar compounds having sufficient intermolecular dipole-dipole Interactions between the opposite ends of `C=O` dipoles . But these dipole-dipole interactions are weaker than the intermolecular hydrogen bonding in alcohols and carboxylic ACID. SO, their B.P. are lower than CORRESPONDING alcohols and acids.
42.

Boiling points of aldehydes and ketones are higher than that of ethers of comparable molar masses due to

Answer»

PRESENCE of intermolecular H-bonding
Presence of intermolecular H-bonding
Strong dipole - dipole INTERACTIONS
LONDON dispersion forces

Solution :Carbonyl COMPOUNDS are more than polar than ethers
43.

Boiling points of alcohols are generally high. This is due to

Answer»

hydrogen-bonding INTERMOLECULAR ATTRACTIONS
dipole-dipole attractions
PATH of the above
NONE of the above

Answer :C
44.

Boiling point of water is defined as the temperature at which:

Answer»

VAPOUR PRESSURE of water is equal to that on ONE atmospheric pressure
Bubbles are formed
Steam comes out
None

Answer :A
45.

Boiling point of water at 750 mm Hg is 99.63^(@)C. How sucrose is to be added to 500 g of water such that it boils at 100^(@)C. Molal elevation constant for water is 0.52 K kg mol^(-1).

Answer»

Solution :Here, elevation of boiling POINT
`DELTA T_(b)=(100+273)-(99.63+273)=0.37 K`
Mass of water, `w_(1)=500 g`
Molar mass of sucrose `(C_(12)H_(22)O_(11))`,
`M_(2)=11xx12+22xx1+11xx16=34 g mol^(-1)`
Molal elevation constant,
`K_(b)=0.52 " K KG mol"^(-1)`
We know that :
`Delta T_(f)=(K_(b)xx1000xx w_(2))/(M_(2)xx w_(1))`
`w_(2)=(Delta T_(b)xx M_(2)xx w_(1))/(K_(b)xx1000)=(0.37xx342xx500)/(0.52xx1000)`
= 121.7 (approximately)
HENCE, 121.67 g of sucrose is to be added.
46.

Boiling point of water at 750mm Hg is 99.63^@C How much sucrose is to be added to 500g of water such that it boils at 100^@C K_b(water)= 0.52K kg mol^(-1)

Answer»

Solution :ELEVATION of boiling point,`Delta T_b =100-99.63 = 0.37^@`
`K_b` of water = 0.5 K kg`MOL^(-1)`
Molar MASS of sucrose `C_12H_22O_11= 342g mol^(-1)`
`Delta T_b = (1000K_bW_2)/(M_2W_1)`
`thereforeW_2` (mass of SOLUTE)=`(Delta T_b M_2 W_1)/(1000K_b)= (0.37xx342xx500)/(1000K_b)`=121.67g
47.

Boiling point of water at 750 mm is 99.63^@ C. How much of sucrose is to be added to 500 g of water so that it boils at 100^@ C. (K_b =0.052 )

Answer»

Solution :`DELTA T_b = 100-99.63 =0.37K`
`K_b= 0.52 `
` M_B ` for `(C_(12) H_(22) O_(11)) =342 `
`W_A=500 g`
` W_B=(Delta T_b xxM_b xx W_A )/(1000 xx K_b)`
`=(0.37 xx342 xx 500 )/(1000 xx 0.52 )`
`=( 0.37 xx 171 )/( 0.52 )`
`= 121.7`g
48.

Boiling point of water at 750 mm Hg is 99.63^@C . How much sucrose is to be added to 500 gof water such that it boils at 100^@C ? [Molal elevation constant of water is 0.52 K kg "mol"^(-1) ]

Answer»

Solution : Elevation in BOILING point required `(Delta T_b) = 100 - 96.63^@ = 3.37^@ `
Mass of solvent (water), `w_1`= 500 g
MOLAR mass of solvent, `M_1 = 18 g "mol"^(-1)`
Molar mass of solute, `C_12H_22O_11 = 342 g "mol"^(-1)`
APPLYING the formula, `M_2= (1000 K_b w_2)/(w_1 Delta T_b) " or " w_2 = (M_2 xx w_1 xx Delta T_b)/(1000 xx K_b)`
Substituting the values, we get
`w_2 = (342 g"mol"^(-1) xx 500 g xx 0.37 K)/(1000 g kg^(-1) xx 0.52 K kg "mol"^(-1)) = 121.67 g `
49.

Boiling point of water at 750 mm Hg is 99.63^(@)C. How much sucrose is to be added to 500 g of water such that it boils at 100^(@)C? Molal elevation constant for water is "0.52 K kg mol"^(-1).

Answer»

Solution :Elevation in boiling point required `(DeltaT_(b))=100-99.63^(@)=0.37^(@)`
Mass of solvent (water), `w_(1)=500g`
`"Molar mass of solvent, "M_(1)="18 g mol"^(-1),"Molar mass of solute, "C_(12)H_(22)O_(11)="342 g mol"^(-1)`
`"Applying the formula, "M_(2)=(1000K_(b)w_(2))/(w_(1)DeltaT_(b))`
`"or"w_(2)=(M_(2)xxw_(1)xxDeltaT_(b))/(1000xxK_(b))=("342 g mol"^(-1)xx500 g xx0.37K)/("1000 g kg"^(-1)xx0.52"K kg mol"^(-1))=121.7g.`
50.

Boiling point of phosphine is higher than that of ammonia.

Answer»

Solution :BOILING POINT of PHOSPHINE is LOWER than that of ammonia