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.

Boran forms a number of hydrides having the general formulae BnH_(n+4) and BnH_(n+6). These are called boranes the simplest hydride of boran is diborane. Borane contains special types of bonds known as multicentric bonds. Borans have high heat of combustion.Which of the following compound is electron dificient compound.

Answer»

`C_(2)H_(6)`
`SiH_(4)`
`B_(2)H_(6)`
`B_(4)H_(10)`

SOLUTION :`B_(2),H_(6),B_(4)B_(10)` (because .B. is electron EFFICIENT ELEMENT).
Discussion
2.

Boran forms a number of hydrides having the general formulae BnH_(n+4) and BnH_(n+6). These are called boranes the simplest hydride of boran is diborane. Borane contains special types of bonds known as multicentric bonds. Borans have high heat of combustion.The type of hybridisation of boron in diborane is

Answer»

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

SOLUTION :`Sp^(3)`
Discussion
3.

Bones contain calcium ions . What do you think would be the anion associated with them ?

Answer»

SOLUTION :`PO_(4)^(3-)` ION .
Discussion
4.

Bone energy is least in the following

Answer»

`HF`
`HCL`
`HBR`
`HI`

ANSWER :D
Discussion
5.

Bonding formation between two atoms is then envisaged as the progressive overlapping of an atomic orbital from each of the participating atoms, the greater the overlap achieved (the overlap integral), the stronger the bond so formed. In inorgunic benzene (B_(3)N_(3)H_(6)) :

Answer»

only six `(sp^(2)-sp^(2)) sigma` BONDS and THREE `p pi - p pi` coordinate BOND
twelve `(sp^(2)-sp^(2)) sigma` bonds and three `pi -pi` coordinate bond
six `(sp^(2)-sp^(2)) sigma` bonds, six `(sp^(2)-sp) sigma` bonds, and three `p pi- p pi`coordinate bond
six `(sp^(2)-sp^(2)) sigma` bonds, six `(sp^(2)-sp^(2))` coordinate bond

ANSWER :C
Discussion
6.

Bonds that are absent in silicone

Answer»

`Si-O-Si`
`O-Si-R`
`O-Si-R`
`R-Si-R`

ANSWER :B
Discussion
7.

Bonding formation between two atoms is then envisaged as the progressive overlapping of an atomic orbital from each of the participating atoms, the greater the overlap achieved (the overlap integral), the stronger the bond so formed. In which of the following pair both have similarity in bond nngle(s) between adjacent chlorine?

Answer»

`PCI_(3), PCI_(4)^(oplus)`
`PCI_(4)^(oplus), PCI_(5)`
`PCI_(5),PCI_(6)^(oplus)`
`PCI_(4)^(oplus), PCI_(6)^(oplus)`

ANSWER :C
Discussion
8.

Bonding formation between two atoms is then envisaged as the progressive overlapping of an atomic orbital from each of the participating atoms, the greater the overlap achieved (the overlap integral), the stronger the bond so formed. For sigma bond formation the relative overlapping power of :

Answer»

s-orbital is GREATER than p-orbitaJ because s-orbital are closer to NUCLEUS
p-orbitals is greater than s-orbital because electrons of p-orbitals are ORIENTED on INTERNUCLEAR axis
s-orbitals is greater than p-orbital because of spherical shape of s-orbital
p-orbital is greater than s-orbital because electrons of p-orbitals are oriiented on internuclear axis

Answer :D
Discussion
9.

Bond polarity is least in

Answer»

N-H
O-H
H-F
C-H

Answer :D
Discussion
10.

Bond order of O_(2), F_(2), N_(2) respectively are …………….

Answer»

SOLUTION :`2, 1, 3`
Discussion
11.

Bond order of a species is 2.5 and the number of electons in its bonding molecular orbital is found to be 8 The no. of electons in its antibonding molecular orbital is

Answer»

three
four equivalent orbitals at `109^(@)` 28' to each other will be formed.
zero
cannot be calculated FORM the given information.

Solution :Bond ORDER `(1)/(2) (N_(B)-n_(a))`
`2.5 = (1)/(2)(8 - n_(a)) rArr 5 = 8 - n_(a)rArr n_(a)= 8 - 5 = 3`
Discussion
12.

Bond order is maximum among the following

Answer»

`N_2`
`He_2`
`H_2`
`O_2`

ANSWER :A
Discussion
13.

Bond order increases with loss of electron in bonding molecular orbital - State True or False and give reason .

Answer»

SOLUTION :False
Bond ORDER increases only when the electron is LAST from the antibonding MOLECULAR orbital .
Discussion
14.

Bond order in He_(2) species is

Answer»

0
1
2
3

Answer :A
Discussion
15.

Bond length order among is

Answer»

`- underset(|)OVERSET(|)C -underset(|)overset(|)C- GT " " gt C= C LT " " gt -C -= C-`
`-C -= C- gt gt C = C lt gt - underset(|)overset(|)C- underset(|)overset(|)C-`
`-underset(|)overset(|)C-underset(|)overset(|)C- = gt C = C gt = - C -= C-`
`-underset(|)overset(|)C-underset(|)overset(|)C- gt gt C= C lt = - C -= C -`

ANSWER :A
Discussion
16.

Bond length of ethane (I), ethene (II), acetylene (III) and benzene (IV) follows the order

Answer»

`I GT II gt III gt IV`
`I gt II gt IV gt III`
`I gt IV gt II gt III`
`III gt IV gt II gt I`

Answer :C
Discussion
17.

Bond length of H_2 is 0.074nm, Bond length of Cl_2 is 1.98A^@. Bond length of HCl is

Answer»

`2.72Å`
`136` pm
1.027
`0.136Å`

ANSWER :B
Discussion
18.

Bond length is more in

Answer»

H - H
D - D
T -T
Same in all

Solution :Due to presence of same number of VALENCY ELECTRONS
Discussion
19.

Bond Length: Internuclear distance between two adjacent atoms in an species is known as bond length, -Bond length depends on: i) size of the atom involved in the bond formation ii) size of the orbitals involved in the bond formation iii) Lone pair-long pair repulsion iv) Resonance v) s-character of combining orbitals with the increasing size of the atoms and atomic orbitals bond length increases. Lone pair-lone pair repulsion increases bond length (if atoms are small sized) whereas resonance can increase some bond lengths and decrease some other bond length. With increasing s-character bond length decreases, whereas with increasing multiplicity of bonds, bond length decreases. However, in some cases, bond lengths are also affected by relative position of bonds (between two similar atoms). Usually but not always with increasing bond length, bond strength (and hence bond dissociation energy) decreases. In which of the following all bonds are not equivalent?

Answer»

`N_(2)O`
`CN_(2)^(2-)`
`N_(3)^(-)`
`NO_(2)^(-)`

ANSWER :A
Discussion
20.

Bond Length: Internuclear distance between two adjacent atoms in an species is known as bond length, -Bond length depends on: i) size of the atom involved in the bond formation ii) size of the orbitals involved in the bond formation iii) Lone pair-long pair repulsion iv) Resonance v) s-character of combining orbitals with the increasing size of the atoms and atomic orbitals bond length increases. Lone pair-lone pair repulsion increases bond length (if atoms are small sized) whereas resonance can increase some bond lengths and decrease some other bond length. With increasing s-character bond length decreases, whereas with increasing multiplicity of bonds, bond length decreases. However, in some cases, bond lengths are also affected by relative position of bonds (between two similar atoms). Usually but not always with increasing bond length, bond strength (and hence bond dissociation energy) decreases. The correct order of B-F bond length follows the sequence

Answer»

`BF_(3) LT BF_(2)OH lt BF_(2)NH_(2) lt BF_(4)^(-)`
`BF_(2) NH_(2) lt BF_(2)OH lt BF_(3) lt BF_(4)^(-)`
`BF_(3) lt BF_(4)^(-) lt BF_(2) OH lt BF_(2) NH_(2)`
`BF_(3) lt BF_(2) NH_(2) lt BF_(2) OH lt BF_(4)^(-)`

Answer :A
Discussion
21.

Bond length in H_2 molecule is 'x' nm. The covalent radius of deuterium is

Answer»

.X. NM
2X nm
4X nm
`x//2` nm

Answer :D
Discussion
22.

Bond is order is 2.5 in A. CN, B. N_(2)^(+) C. NO

Answer»

A,B only
B,C only
A,C only
A,B and C

Answer :D
Discussion
23.

………bond is maximum polar in the following

Answer»

`H_(3)C-Cl`
`H_(3)C-NH_(2)`
`H_(3)C-CH_(3)`
`H_(3)C-H`

ANSWER :A
Discussion
24.

Bond enthalpy of C - H bond is 350 kJ/mol. Calculate the bond enthalpy of C-=C bond in C_(2) H_(2). 2C_((s)) + H_(2(g)) to C_(2) H_(2(g)), Delta H= 225 kJ/mol 2X_((s)) to 2C_((g)) ,""DeltaH= 330 kJ/mol H_(2(g)) to 2H_((g)) , ""DeltaH= 330 kJ/mol

Answer»

1165 kJ/mol
837 kJ/mol
865 kJ/mol
815 kJ/mol

Solution :`DELTA H + 1410 + 330 - (350xx 2)- Delta H _(C -= C) = 255`
`therefore Delta H_(C -= C) = + 815` kJ/mol
Discussion
25.

Bone energy of C,C bond in highest in

Answer»

`H_3C-CH_3`
`H_2 C=CH_2`
`CH-=CH`
`C_2 H_5Cl`

ANSWER :C
Discussion
26.

Bond energy is maximum in

Answer»

`F_2`
`N_2`
`O_2`
`Br_2`

ANSWER :B
Discussion
27.

Bond energy is highest in the overlaping

Answer»

`SP^(2)-s`
`sp^(2)-s`
`sp-s`
`sp^(3)-s`

ANSWER :C
Discussion
28.

Bond energy is highest in the molecule

Answer»

`F_2`
`Br_2`
`I_2`
`Cl_2`

ANSWER :D
Discussion
29.

Bone energy is highest in the molecules

Answer»

`F_2`
`Br_2`
`I_2`
`Cl_2`

ANSWER :D
Discussion
30.

Bond energy : Energy required to break a bond (equal to the energy released when a bond is formed) is named bond energy. Its unit is Kcal/mole. Stability of bond energy 00 % s characteroo(1)/(% "P character")oo "bond order" oo(1)/("bond length") Higher is the number of bonds at a place, higher is the bond energy. Which are true in terms of bond energy ? (i)CH_(3) -Fgt CH_(3) -Clgt CH_(3)- Brgt CH_(3)-1(ii) H_(3) C - O lt H_(3) C - CH_(3) lt CH_(2) = CH -O - CH_(3) (iii) CH_(3) - CH_(2) -O - CH_(3) lt CH_(2) =CH-O -CH_(3) (iv) C- N gt N gt C = N (v) CH_(3) - CH_(2)- NH_(2) lt CH_(2) = CH -NH_(2) lt CH= C -NH_(2)

Answer»

(i), (II), (IV), (v)
(i), (ii), (III), (v)
(i), (iii), (iv), (v)
all are correct

Answer :B
Discussion
31.

Bond energy : Energy required to break a bond (equal to the energy released when a bond is formed) is named bond energy. Its unit is Kcal/mole. Stability of bond energy 00 % s characteroo(1)/(% "P character")oo "bond order" oo(1)/("bond length") Higher is the number of bonds at a place, higher is the bond energy. Write down rhe bond energy order of C_(2) - C_(3) in (A) CH_(3)- CH_(2)-CH_(2)-CH_(3) "" B ) CH_(3) -CH_(2)-CH= CH_(2) C) CH_(2)=CH-CH=CH_(2) "" D) CH_(2)=CH-C= CH E) CH_(3) - CH_(2) -C= CH "" F) CH = C- C= CH

Answer»

` E GT D gt C gt F gt B gt A `
`A gt C gt D gt B gt E gt F `
`A gt C gt D gt E gt F gt B `
`F gt D gt E gt C gt B gt A `

Answer :D
Discussion
32.

Bond energy : Energy required to break a bond (equal to the energy released when a bond is formed) is named bond energy. Its unit is Kcal/mole. Stability of bond energy 00 % s characteroo(1)/(% "P character")oo "bond order" oo(1)/("bond length") Higher is the number of bonds at a place, higher is the bond energy. Write down bond energy order of C- H bond in (A) -C- H, (B) = C-H, (C) = C - H

Answer»

`A GT B gt C `
`A LT B lt C `
`B gt A gt C `
`A gt C gt B `

ANSWER :B
Discussion
33.

Bond energies of H_2, F_2 and HF are respec tively 104.2, 36.6 and 134.6 k cal "mol"^(-1). If the electronegativity value of hydrogen is 2.1, calculate the electro-negativity value of fluorine.

Answer»


ANSWER :3.7
Discussion
34.

Bond energies of H_(2),Cl_(2) and HCl are respectively "104, 58 and 103 k cal mol"^(-1). Calculate Pauling's electronegativity of chlorine.

Answer»

Solution :AVERAGE of bond energies of `H_2` and `Cl_2` is the CALCULATED bond energy of `HCl = (104 + 58)/(2)`
`= 81 K cal mol^(-1)`
EXPERIMENTAL bond energy of `HCl = 100 k cal mol^(-1)`
`DELTA -` Bond (resonance ) stabilisation energy = 100- 81 = 19 k cal `mol^(-1)`
`X_1 - X_2 = 0.208 sqrt(Delta) = 0.208 sqrt(19)`
`= 0.208 xx 4.35 = 0.90`
Since Pauling.s ELECTRONEGATIVITY of hydrogen is 2.1 that of chlorine `=2.1 + 0.9 = 3.0`
Discussion
35.

Bond dissociation enthaply of the first H-S bond inhydrogen sulphide is 376 "Kj"//"mole". The enthalpies of formatin of H_(2)S(g)and S(g) are-20.0and277.0 "kj"//"mole" respectively. The enthalpy of formation of gaseous hydrogen atomis 218 "Kj"//"mole". Using above information, answer following questions : The bond disscociation enthalpy of the free radical HS is :

Answer»

`138kJ//"MOLE"`
`276kJ//"mole"`
`357kJ//"mole"`
`376kJ//"mole"`

ANSWER :C
Discussion
36.

Bond dissociation enthaply of the first H-S bond inhydrogen sulphide is 376 "Kj"//"mole". The enthalpies of formatin of H_(2)S(g)and S(g) are-20.0and277.0 "kj"//"mole" respectively. The enthalpy of formation of gaseous hydrogen atomis 218 "Kj"//"mole". Using above information, answer following questions : The enthalpy of formation of free radical HS is :

Answer»

`138kJ//"MOLE"`
`276kJ//"mole"`
`-10KJ//"mole"`
`357kJ//"mole"`

ANSWER :a
Discussion
37.

The bond dissociation energies for Cl_(2),I_(2) and Icl are 242.3, 151 and 211.3 kJ/mol respectively. The enthalpy of sublimation of iodine is 62.8 kJ/mol. What is the standard enthalpy of formation of ICl_((s)) ?

Answer»

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

ANSWER :A
Discussion
38.

Bond dissociation enthalpy of H_(2), CI_(2) and HCI 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 :The REACTION for formation of HO can be written as
`H_(2) + CI_(2) to 2HCI`
`H- H + CI- CI to 2(H- CI)`
Substituting the given values, we GET enthalpy of formation of
`2HCI = - (862-676) = - 186` kJ
`therefore` Enthalpy of formation of
`HCI = (-186)/(2) "kJ"= -93` kJ
Discussion
39.

Bond dissociation enthalpy of E-H (E=element) bonds is given below. Which of the compounds will act as strongest reducing agent ? Compound {:("Compound",NH_(3),PH_(3),AsH_(3),SbH_(3)),(Delta_("diss")(E-H)//kJmol^(-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)`.
Discussion
40.

Bond dissociation energy ofCH_(3)-Hbond is103Kcalmol^(-1)and0.055cal//grespectively ,If specific heat of iodine is 24cal//g at 200^(@),Calcualte its value at250^(@)C is

Answer»

`3.61Kcalmol^(-1)`
`-3.61Kcalmol^(-1)`
`33.61Kcalmol^(-1)`
`-33.61Kcalmol^(-1)`

Solution :`CH_(4)(G)=CH_(3)(g)+H("ATOM,g")`
`(DeltaH_(f))_(CH_(3),H)CH_(3)-H=103`
`=DeltaH_(f)(CH_(3),g)+DeltaH_(f)(H "atom, g")-DeltaH_(f)(CH_(4),g)`
`(DeltaH_(f))_(CH_(3)g)=103-(103)/(2)+(-17.889)=33.611Kcalmol^(-1)`
`:.` HEAT of formation of methyl radical is
`DELTAH=33.611Kcalmol^(-1)`
Discussion
41.

Bond angles in PB r_(3)(101.5^(@)),(PB l_(3)(100^(@)) and PF_(3)(97^(@)) decrease with increase in electronegativities of the surrounding atoms, however, bond angles in BF_(3),BC l_(3) andB Br_(3) do not change with change in electronegativities of the surrounding atoms. explain with reason.

Answer»

Solution :
(i) `PX_(3)` has a trigonal pyramidal geometry with increase in electronegativity of the surrounding halonge (X) atoms, bond pairs are oriented more towards halogen atoms, resulting in decrease in bond PAIR-bond pair repulsion. hence X-P-X bond angles decrease in the order: `PB r_(3)(101.5^(@)) gt PB l_(3)(100^(@)) gt PF_(3)(97^(@))`
(ii) `BX_(3)` has a trigonal planar geometry, where 3 halogen atoms are located at 3 corners of an equilateral TRIANGLE. with increse in electronegativity of halogen atom, bond pair tend to concentrate more towards the halogen atoms resulting in decrease in bond pair-bond pair repulsion. since all 3 halogen atoms lie on the same plane, forming 3 equivalent B-X BONDS, there is no change in the X-B-X bond ANGLE `(120^(@))`.
Discussion
42.

Bond angles H-O-H and H-O-O- in water and H_(2)O_(2) respectively are

Answer»

`104.5^(@), 104.5^(@)`
`94.8^(@),94.8^(@)`
`104.8^(@),94.8^(@)`
`94.8^(@),104.5^(@)`

SOLUTION :`H-O-H (104.5^(@)) and H-O-O- (94.8) ` in `H_(2)O_(2)`
Discussion
43.

Bond angle of water is reduced from 109.28^(@) ot 104.5^(@).Explain.

Answer»

Solution :(i) Water molecule undergoes `sp^3` hybridisation.
The O- ATOM in `H_2O` molecule contains two bonding pairs and two lone pairs of electrons.
(iii) The overall arrangement for four electron pairs is tetrahedral but the lp-lp repulsions being greater than Ip-lb repulsions in `H_2O`
(iv) The HOH ANGLE is REDUCED to `104.5^(@)` than `109^(@)`28'
(V) The molecule has a bent shape.
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Discussion
44.

Bond angle of chlorine is 198 pm then calculate atomic radius .

Answer»

Solution :CHLORINE is non-metal its atomic RADIUS or covalent radius becomes HALF than bond length
`" Atomic radius " = " Bond length"/2`
` = (198 "PM")/2 = 99 "pm"`
Discussion
45.

Bond angle in PH_(4)^(+) is highre than that in PH_(2). Why ?

Answer»

<P>

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 tetraheldral 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)^(+)` ionn 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)`.
Discussion
46.

Bond angle in PH_(4^(+) is higher than in PH_3 Why?

Answer»

Solution :Phosphorous in both `PH_3` and `PH_4^(+)` is `sp^3` hybridised . Due to the absence of lone, pair -bond pair repulsion and presence of FOUR identical bond pair- bond pair interactions, `PH_4^+` assumes tetrahedral geometry with a bond angle of `109^@ 28.`
But `PH_3` has three bond pairs and one lone pair around P. Due to greater lone pair-bond pair repulsion than bond pair-bond pair repulsion, the tetrahedral angle decreases from `109^@ 28.` to `93.6^@` . As a result `PH_3` is PYRAMIDAL.
`PH_3` -Pyramidal with bond angle of `93.6^@`
`PH_4^(+4)` - Tetrahedral with bond angle of `109^@ 28.`.
Discussion
47.

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

Answer»

Solution :In `PH_(3), P" is "sp^(3)` hybridised . Three ORBITALS are involved in BONDING with three hydrogenatoms and the fourth one contains a lonepair.
As LONE PAIR - bond pair repulsion is stronger than bond pair - bond pair repulsion , the tetrahedralshape associated with `sp^(3)` bonding is CHANGED to pyramid.
`PH_(3)` combines with a proton to form `PH_(4)^(+)`, in which the lone pair is absent.
Due to the absence of lone pair in `PH_(4)^(+)` is higher than the bond angle in`PH_(3)`.
Discussion
48.

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

Answer»

Solution :Phosphorous in both `PH_(3)" and "PH_(4)^(+)` is `SP_(3)`hybridised.
Due to the ABSENCES of lone pair - bond pair repulsion and PRESENTS of four identical bond pair - bond pair interactions. `PH_(4)^(+)` assumes tetrahedral geometry with a bond ANGLE of `109^(@) 28`'
But `PH_(3)` has three bond pairs and one lone pair around P. Due to greater lone pair - bond pair repulsion than bond pair bond pair repulsion the tetra hedral angle decreases from `109^(@) 28' " to " 93.6^(@)` . As a result `PH_(3)` is pyramidal.
Discussion
49.

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

Answer»

Solution :Phosphorous in both `PH_(3) and PH_(4)^(+) is SP^(3)` hybridised. Due to the absence of LONE pair-bond pair repulsion and presence of four identical bond pair - bond pair interactions. `PH_(4)^(+)` assume TETRAHEDRAL geometry with a bond angle of `109^(@)28.`
But `PH_(3)` has three bond pairs and one lone pair around P. Due to greater lone pair-bond pair repulsion than bond pair-bond pair repulsion. the tetrahedral angle decreases from `109^(@)28. to 93.6^(@)`.As a result PH, is pyramidal.
`PH_(3)` - yramidal with bond angle of `93.^(@)`
`PH_(4)^(+)` - Tetrahedral with bond angle of `109^(@)28..`
Discussion
50.

Bond angle in benzene is ………………..

Answer»

SOLUTION :`120^(@)`
Discussion