iCON Education - Chemical Bonding & Molecular Structure

2018 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 JEE Main 2018 (Online) 15th April Morning Slot

In the molecular orbital diagram for the molecular ion, N₂⁺, the number of electrons in the $\sigma $_{2p_z} molecular orbital is :

A.

0

B.

1

C.

2

D.

3

2018 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 JEE Main 2018 (Online) 15th April Morning Slot

The decreasing order of bond angles in BF₃, NH₃, PF₃ and I₃^- is :

A.

I₃^- > NH₃ > PF₃ > BF₃

B.

I₃^- > BF₃ > NH₃ > PF₃

C.

BF₃ > I₃^- > PF₃ > NH₃

D.

BF₃ > NH₃ > PF₃ > I₃^-

2018 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 JEE Main 2018 (Online) 15th April Morning Slot

JEE Main 2018 (Online) 15th April Morning Slot Chemistry - Chemical Bonding & Molecular Structure Question 215 English

In hydrogen azide (above) the bond orders of bond (I) and (II) are :

A.

$\eqalign{ & (\mathrm{I})\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(\mathrm{II}) \cr & < 2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, > 2 \cr} $

B.

$\eqalign{ & (\mathrm{I})\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(\mathrm{II}) \cr & > 2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, < 2 \cr} $

C.

$\eqalign{ & (\mathrm{I})\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(\mathrm{II}) \cr & > 2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, > 2 \cr} $

D.

$\eqalign{ & (\mathrm{I})\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(\mathrm{II}) \cr & < 2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, < 2 \cr} $

2018 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 JEE Main 2018 (Online) 15th April Morning Slot

Identify the pair in which the geometry of the species is $T$-shape and square - pyramidal, respectively :

A.

ClF₃ and $I{O_4}^ - $

B.

ICl₂^- and ICl₅

C.

XeOF₂ and XeOF₄

D.

IO₃^- and IO₂F₂^-

2017 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 JEE Main 2017 (Online) 9th April Morning Slot

The group having triangular planar structures is :

A.

BF₃, NF₃, $CO_3^{2 - }$

B.

$CO_3^{2 - }$, $NO_3^ - $, SO₃

C.

NH₃, SO₃, $CO_3^{2 - }$

D.

NCl₃, BCl₃, SO₃

2017 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 JEE Main 2017 (Online) 8th April Morning Slot

Which of the following is paramagnetic ?

A.

NO⁺

B.

CO

C.

$O_2^{2 - }$

D.

B₂

2017 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 JEE Main 2017 (Online) 8th April Morning Slot

sp³d² hybridization is not displayed by :

A.

BrF₅

B.

SF₆

C.

[CrF₆]^3$-$

D.

PF₅

2017 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 JEE Main 2017 (Offline)

Which of the following species is not paramagnetic?

A.

CO

B.

O₂

C.

B₂

D.

NO

2016 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 JEE Main 2016 (Online) 10th April Morning Slot

The bond angle H - X - H is the greatest in the compound :

A.

CH₄

B.

NH₃

C.

H₂O

D.

PH₃

2016 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 JEE Main 2016 (Online) 9th April Morning Slot

The group of molecules having identical shape is :

A.

SF₄ , XeF₄ , CCl₄

B.

ClF₃ , XeOF₂ , XeF$_3^ + $

C.

BF₃ , PCl₃ , XeO₃

D.

PCl₅ , IF₅ , XeO₂F₂

2016 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 JEE Main 2016 (Offline)

The species in which the $\mathrm{N}$ atom is in a state of $s p$ hybridization is :

A.

$\mathrm{NO}_2^{+}$

B.

$\mathrm{NO}_2^{-}$

C.

$\mathrm{NO}_3^{-}$

D.

$\mathrm{NO}_3^{-}$

2015 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 JEE Main 2015 (Offline)

The intermolecular interaction that is dependent on the inverse cube of distance between the molecule is:

A.

ion-dipole interaction

B.

London force

C.

hydrogen bond

D.

ion-ion interaction

2013 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 JEE Main 2013 (Offline)

In which of the following pairs of molecules/ions, both the species are not likely to exist?

A.

$H_2^+$ , $He_2^{2-}$

B.

$H_2^-$ , $He_2^{2-}$

C.

$H_2^{2+}$ , $He_2$

D.

$H_2^-$ , $He_2^{2+}$

2013 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 JEE Main 2013 (Offline)

Stability of the species Li₂, $Li_2^−$ and $Li_2^+$ increases in the order of:

A.

Li₂ < $Li_2^+$ < $Li_2^-$

B.

$Li_2^+$ < $Li_2^-$ < Li₂

C.

$Li_2^-$ <$Li_2^+$ < Li₂

D.

$Li_2^-$ < Li₂ < $Li_2^+$

2012 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 AIEEE 2012

In which of the following pairs the two species are not isostructural ?

A.

$CO_3^{2-}$ and $NO_3^-$

B.

$PCl_4^{+}$ and $SiCl_4$

C.

PF₅ and BrF₅

D.

$AlF_6^-$ and $SF_6$

2011 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 AIEEE 2011

Among the following the maximum covalent character is shown by the compound :

A.

SnCl₂

B.

AlCl₃

C.

MgCl₂

D.

FeCl₂

2011 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 AIEEE 2011

The hybridization of orbitals of N atom in $NO_3^-$, $NO_2^+$ and $NH_4^+$ are respectively :

A.

sp , sp², sp³

B.

sp², sp , sp³

C.

sp , sp³, sp²

D.

sp², sp³, sp

2011 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 AIEEE 2011

The structure of IF₇ is :

A.

trigonal bipyramid

B.

octahedral

C.

pentagonal bipyramid

D.

square pyramid

2009 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 AIEEE 2009

Using MO theory, predict which of the following species has the shortest bond length?

A.

$O_2^+$

B.

$O_2^-$

C.

$O_2^{2-}$

D.

$O_2^{2+}$

Correct Answer: D

Explanation:

Note :

(1) $\,\,\,\,$ Bond length $ \propto $ ${1 \over {Bond\,\,order}}$

(2) $\,$ Bond order $ = {1 \over 2}$ [N_b $-$ N_a]

N_b = No of electrons in bonding molecular orbital

N_a $=$ No of electrons in anti bonding molecular orbital

(4) $\,\,\,\,$ upto 14 electrons, molecular orbital configuration is

AIEEE 2009 Chemistry - Chemical Bonding & Molecular Structure Question 235 English Explanation 1

Here N_a = Anti bonding electron $=$ 4 and N_b = 10

(5) $\,\,\,\,$ After 14 electrons to 20 electrons molecular orbital configuration is - - -

AIEEE 2009 Chemistry - Chemical Bonding & Molecular Structure Question 235 English Explanation 2

AIEEE 2009 Chemistry - Chemical Bonding & Molecular Structure Question 235 English Explanation 2

Here N_a = 10

and N_b = 10

In O atom 8 electrons present, so in O₂, 8 $ \times $ 2 = 16 electrons present.

Then in $O_2^ + $ no of electrons = 15

in $O_2^ - $ no of electrons = 17

in $O_2^{2 - }$ no of electrons = 18

$\therefore\,\,\,\,$ Molecular orbital configuration of O₂ (16 electrons) is

${\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * \,$ ${\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,$ ${\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}} = {\pi _{2p_y^2}}\,\,\pi _{2p_x^1}^ * \,\, = \pi _{2p_y^1}^ * $

$\therefore\,\,\,\,$N_a = 6

N_b = 10

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {10 - 6} \right] = 2$

Molecular orbital configuration of O$_2^ + $ (15 electrons) is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * \, = \,\pi _{2p_y^o}^ * $

$\therefore\,\,\,\,$ N_b = 10

N_a = 5

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {10 - 5} \right]$ = 2.5

Molecular orbital configuration of O$_2^ {2+ }$ (14 electrons) is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^0}^ * \, = \,\pi _{2p_y^o}^ * $

$\therefore\,\,\,\,$ N_b = 10

N_a = 4

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {10 - 4} \right]$ = 3

Molecular orbital configuration of $O_2^ - $ (17 electrons) is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^2}^ * \, = \,\pi _{2p_y^1}^ * $

$\therefore\,\,\,\,$ N_b = 10

N_a = 7

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {10 - 7} \right]$ = 1.5

Molecular orbital configuration of O $_2^{2 - }$ (18 electrons) is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^2}^ * \, = \,\pi _{2p_y^2}^ * $

$\therefore\,\,\,\,$ N_b = 10

N_a = 8

$\therefore\,\,\,\,$ BO = ${1 \over 2}$ [ 10 $-$ 8] = 1

As Bond length $ \propto $ ${1 \over {Bond\,\,order}}$

So $O_2^{2+}$ has the shortest bond length.

2008 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 AIEEE 2008

Which of the following pair of species have the same bond order?

A.

CN^- and NO⁺

B.

CN^- and CN⁺

C.

$O_2^-$ and CN^-

D.

NO⁺ and CN⁺

2008 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 AIEEE 2008

The bond dissociation energy of B - F in BF₃ is 646 kJ mol^-1 whereas that of C - F in CF₄ is 515 kj mol^-1. The correct reason for higher B - F bond dissociation energy as compared to that of C - F is

A.

stronger $\sigma$ bond between B and F in BF₃ as compared to that between C and F in CF₄

B.

significant $p\pi - p\pi$ interaction between B and F in BF₃ whereas there is no possibility of such interaction between C and F in CF₄

C.

lower degree of $p\pi - p\pi$ interaction between B and F in BF₃ than that between C and F in CF₄

D.

smaller size of B - atom as compared to that of C- atom.

2007 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 AIEEE 2007

Which of the following hydrogen bonds is the strongest?

A.

O−H…….N

B.

F−H…….F

C.

O−H…….O

D.

O−H…….F

2007 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 AIEEE 2007

The charge/size ratio of a cation determines its polarizing power. Which one of the following sequences represents the increasing order of the polarizinig order of the polarizing power of the cationic species, K⁺, Ca²⁺, Mg²⁺, Be²⁺?

A.

Mg²⁺ < Be²⁺ < K⁺ < Ca²⁺

B.

K⁺ < Ca²⁺ < Mg²⁺ < Be²⁺

C.

Be²⁺ < K⁺ < Ca²⁺ < Mg²⁺

D.

Ca²⁺ < Mg²⁺ < Be²⁺ < K⁺

2007 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 AIEEE 2007

Which of the following species exhibits the diamagnetic behaviour?

A.

$O_2^{2−}$

B.

NO

C.

$O_2^+$

D.

O₂

2007 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 AIEEE 2007

In which of the following ionization processes, the bond order has increased and the magnetic behaviour has changed?

A.

$C_2 \to C_2^+$

B.

$N_2 \to N_2^+$

C.

$NO \to NO^+$

D.

$O_2 \to O_2^+$

Correct Answer: C

Explanation:

(A) Moleculer orbital configuration of $C_2$ (12 electrons)

= ${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}$

$\therefore\,\,\,\,$N_a = 4

N_b = 8

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {8 - 4} \right] = 2$

Here no unpaired electron present, so it is diamagnetic.

Moleculer orbital configuration of $C_2^+$ (11 electrons)

= ${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^1}}$

$\therefore\,\,\,\,$N_a = 4

N_b = 7

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {7 - 4} \right] = 1.5$

Here 1 unpaired electron present, so it is paramagnetic.

(B) Moleculer orbital configuration of $N_2$ (14 electrons)

= ${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,{\sigma _{2p_z^2}}$

$\therefore\,\,\,\,$N_a = 4

N_b = 10

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {10 - 4} \right] = 3$

Here no unpaired electron present, so it is diamagnetic.

Moleculer orbital configuration of $N_2^+$ (13 electrons)

= ${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,{\sigma _{2p_z^1}}$

$\therefore\,\,\,\,$N_a = 4

N_b = 9

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {9 - 4} \right] = 2.5$

Here 1 unpaired electron present, so it is paramagnetic.

(C) Moleculer orbital configuration of NO (15 electrons) is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}}\,= \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * = \,\,\pi _{2p_y^0}^ * $

$\therefore\,\,\,\,$N_a = 5

N_b = 10

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {10 - 5} \right] = 2.5$

Here is 1 unpaired electron, So it is paramagnetic.

Moleculer orbital configuration of NO⁺ (14 electrons) is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}}\,= \,{\pi _{2p_y^2}}\,\pi _{2p_x^0}^ * = \,\,\pi _{2p_y^0}^ * $

$\therefore\,\,\,\,$N_a = 4

N_b = 10

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {10 - 4} \right] = 3$

Here is no unpaired electron, So it is diamagnetic.

(D) Molecular orbital configuration of O₂ (16 electrons) is

${\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * \,$ ${\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,$ ${\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}} = {\pi _{2p_y^2}}\,\,\pi _{2p_x^1}^ * \,\, = \pi _{2p_y^1}^ * $

$\therefore\,\,\,\,$N_a = 6

N_b = 10

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {10 - 6} \right] = 2$

Here 2 unpaired electrons present, so it is paramagnetic.

Molecular orbital configuration of O$_2^ + $ (15 electrons) is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * \, = \,\pi _{2p_y^o}^ * $

$\therefore\,\,\,\,$ N_b = 10

N_a = 5

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {10 - 5} \right]$ = 2.5

Here 1 unpaired electrons present, so it is also paramagnetic.

2006 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 AIEEE 2006

In which of the following molecules/ions are all the bonds not equal?

A.

XeF₄

B.

$BF_4^−$

C.

SF₄

D.

SiF₄

2006 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 AIEEE 2006

Which of the following molecules/ions does not contain unpaired electrons?

A.

$O_2^{2−} $

B.

B₂

C.

$N_2^+$

D.

O₂

2006 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 AIEEE 2006

The decreasing values of bond angles from NH₃ (106^o) to SbH₃ (91^o ) down group-15 of the periodic table is due to

A.

increasing bp-bp repulsion

B.

increasing p-orbital character in sp³

C.

decreasing lp-bp repulsion

D.

decreasing electronegativity

2005 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 AIEEE 2005

Lattice energy of an ionic compounds depends upon

A.

Charge on the ion only

B.

Size of the ion only

C.

Packing of ions only

D.

Charge on the ion and size of the ion

2005 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 AIEEE 2005

Which one of the following species is diamagnetic in nature?

A.

$He_2^+$

B.

H₂

C.

$H_2^+$

D.

$H_2^-$

2004 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 AIEEE 2004

The maximum number of 90° angles between bond pair of electrons is observed in

A.

dsp³ hybridization

B.

sp³d² hybridization

C.

dsp² hybridization

D.

sp³d hybridization

2004 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 AIEEE 2004

The states of hybridization of boron and oxygen atoms in boric acid (H₃BO₃) are respectively

A.

sp² and sp²

B.

sp³ and sp³

C.

sp³ and sp²

D.

sp² and sp³

2004 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 AIEEE 2004

The correct order of bond angles (smallest first) in H₂S, NH₃, BF₃ and SiH₄ is

A.

H₂S < SiH₄ < NH₃ < BF₃

B.

H₂S < NH₃ < BF₃ < SiH₄

C.

H₂S < NH₃ < SiH₄ < BF₃

D.

NH₃ < H₂S < SiH₄ < BF₃

2004 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 AIEEE 2004

The bond order in NO is 2.5 while that in NO⁺ is 3. Which of the following statements is true for these two species?

A.

Bond length in NO⁺ is greater than in NO

B.

Bond length is unpredictable

C.

Bond length in NO⁺ in equal to that in NO

D.

Bond length in NO is greater than in NO⁺

2004 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 AIEEE 2004

Which one of the following has the regular tetrahedral structure?
(Atomic nos : B = 5, S = 16, Ni = 28, Xe = 54)

A.

XeF₄

B.

[Ni(CN)₄]^2-

C.

$BF_4^-$

D.

SF₄

2003 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 AIEEE 2003

Which one of the following pairs of molecules will have permanent dipole moments for both members

A.

NO₂ and CO₂

B.

NO₂ and O₃

C.

SiF₄ abd CO₂

D.

SiF₄ abd NO₂

2003 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 AIEEE 2003

The pair of species having identical shapes for molecules of both species is

A.

XeF₂, CO₂

B.

BF₃, PCl₃

C.

PF₅, IF₅

D.

CF₄, SF₄

2003 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 AIEEE 2003

Which one of the following compounds has the smallest bond angle in its molecule?

A.

OH₂

B.

SH₂

C.

NH₃

D.

SO₂

2002 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 AIEEE 2002

In which of the following species is the underlined carbon having sp³ hybridisation?

A.

AIEEE 2002 Chemistry - Chemical Bonding & Molecular Structure Question 225 English Option 1

B.

AIEEE 2002 Chemistry - Chemical Bonding & Molecular Structure Question 225 English Option 2

C.

AIEEE 2002 Chemistry - Chemical Bonding & Molecular Structure Question 225 English Option 3

D.

AIEEE 2002 Chemistry - Chemical Bonding & Molecular Structure Question 225 English Option 4

2002 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 AIEEE 2002

Which of the following statements is true?

A.

HF is less polar than HBr

B.

absolutely pure water does not contain any ions

C.

chemical bond formation takes place when forces of attraction overcome the forces of repulsion

D.

in covalency transference of electron takes place

2002 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 AIEEE 2002

A square planar complex is formed by hybridisation of which atomic orbitals ?

A.

s, p_x, p_y, d_xy

B.

s, p_x, p_y, d_{x² - y²}

C.

s, p_x, p_y, d_z²

D.

s, p_x, p_y, d_yz

2002 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 AIEEE 2002

Hybridisation of underline atom changes in

A.

$\underline {Al} {H_3}$ changes to $AlH_4^-$

B.

${H_2}\underline O $ changes to $H_3O^+$

C.

$\underline N {H_3}$ changes to $NH_4^+$

D.

in all cases

2002 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 AIEEE 2002

Which of the following are arranged in an increasing order of their bond strengths?

A.

$O_2^-$ < O₂ < $O_2^+$ < $O_2^{2-}$

B.

$O_2^{2-}$ < $O_2^-$ < $O_2$ < $O_2^{+}$

C.

$O_2^-$ < $O_2^{2-}$ < $O_2$ < $O_2^{+}$

D.

$O_2^{+}$ < $O_2$ < $O_2^-$ < $O_2^{2-}$

Correct Answer: B

Explanation:

Note :

(1) $\,\,\,\,$ Bond strength $ \propto $ Bond order

(2) $\,\,\,\,$ Bond length $ \propto $ ${1 \over {Bond\,\,order}}$

(3) $\,$ Bond order $ = {1 \over 2}$ [N_b $-$ N_a]

N_b = No of electrons in bonding molecular orbital

N_a $=$ No of electrons in anti bonding molecular orbital

(4) $\,\,\,\,$ upto 14 electrons, molecular orbital configuration is

AIEEE 2002 Chemistry - Chemical Bonding & Molecular Structure Question 255 English Explanation 1

Here N_a = Anti bonding electron $=$ 4 and N_b = 10

(5) $\,\,\,\,$ After 14 electrons to 20 electrons molecular orbital configuration is - - -

AIEEE 2002 Chemistry - Chemical Bonding & Molecular Structure Question 255 English Explanation 2

AIEEE 2002 Chemistry - Chemical Bonding & Molecular Structure Question 255 English Explanation 2

Here N_a = 10

and N_b = 10

In O atom 8 electrons present, so in O₂, 8 $ \times $ 2 = 16 electrons present.

Then in $O_2^ + $ no of electrons = 15

in $O_2^ - $ no of electrons = 17

in $O_2^{2 - }$ no of electrons = 18

$\therefore\,\,\,\,$ Molecular orbital configuration of O₂ (16 electrons) is

${\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * \,$ ${\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,$ ${\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}} = {\pi _{2p_y^2}}\,\,\pi _{2p_x^1}^ * \,\, = \pi _{2p_y^1}^ * $

$\therefore\,\,\,\,$N_a = 6

N_b = 10

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {10 - 6} \right] = 2$

Molecular orbital configuration of O$_2^ + $ (15 electrons) is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * \, = \,\pi _{2p_y^o}^ * $

$\therefore\,\,\,\,$ N_b = 10

N_a = 5

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {10 - 5} \right]$ = 2.5

Molecular orbital configuration of $O_2^ - $ (17 electrons) is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^2}^ * \, = \,\pi _{2p_y^1}^ * $

$\therefore\,\,\,\,$ N_b = 10

N_a = 7

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {10 - 7} \right]$ = 1.5

Molecular orbital configuration of O $_2^{2 - }$ (18 electrons) is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^2}^ * \, = \,\pi _{2p_y^2}^ * $

$\therefore\,\,\,\,$ N_b = 10

N_a = 8

$\therefore\,\,\,\,$ BO = ${1 \over 2}$ [ 10 $-$ 8] = 1

As Bond strength $ \propto $ Bond order so, correct order is

$O_2^{2 - } < O_2^ - < {O_2} < O_2^ + $

2002 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 AIEEE 2002

In which of the following species the interatomic bond angle is 109^o28' ?

A.

NH₃, BF₄^-

B.

NH₄⁺, BF₄^-

C.

NH₃, BF₃

D.

NH₂^-1, BF₃

2025 JEE Mains Numerical

iCON Education HYD, 79930 92826, 73309 72826 JEE Main 2025 (Online) 29th January Evening Shift

Total number of non-bonded electrons present in NO₂^$-$ ion based on Lewis theory is ______.

2025 JEE Mains Numerical

iCON Education HYD, 79930 92826, 73309 72826 JEE Main 2025 (Online) 22nd January Morning Shift

The number of molecules/ions that show linear geometry among the following is __________.

$\mathrm{SO}_2, \mathrm{BeCl}_2, \mathrm{CO}_2, \mathrm{~N}_3^{-}, \mathrm{NO}_2, \mathrm{~F}_2 \mathrm{O}, \mathrm{XeF}_2, \mathrm{NO}_2^{+}, \mathrm{I}_3^{-}, \mathrm{O}_3$

2024 JEE Mains Numerical

iCON Education HYD, 79930 92826, 73309 72826 JEE Main 2024 (Online) 9th April Evening Shift

Total number of electrons present in $\left(\pi^*\right)$ molecular orbitals of $\mathrm{O}_2, \mathrm{O}_2^{+}$ and $\mathrm{O}_2^{-}$ is ________.

2024 JEE Mains Numerical

iCON Education HYD, 79930 92826, 73309 72826 JEE Main 2024 (Online) 9th April Morning Shift

The total number of species from the following in which one unpaired electron is present, is _______.

$\mathrm{N}_2, \mathrm{O}_2, \mathrm{C}_2^{-}, \mathrm{O}_2^{-}, \mathrm{O}_2^{2-}, \mathrm{H}_2^{+}, \mathrm{CN}^{-}, \mathrm{He}_2^{+}$

Correct Answer: 4

Explanation:

Species	Unpaired e
$ \mathrm{N}_2 $	0
$ \mathrm{O}_2 $	2
$ \mathrm{C}_2^{-} $	1
$ \mathrm{O}_2^{-} $	1
$ \mathrm{O}_2^{2-} $	0
$ \mathrm{H}_2^{+} $	1
$ \mathrm{CN}^{-} $	0
$ \mathrm{He}_2^{+} $	1

2024 JEE Mains Numerical

iCON Education HYD, 79930 92826, 73309 72826 JEE Main 2024 (Online) 8th April Evening Shift

Number of molecules having bond order 2 from the following molecules is _________.

$\mathrm{C}_2, \mathrm{O}_2, \mathrm{Be}_2, \mathrm{Li}_2, \mathrm{Ne}_2, \mathrm{~N}_2, \mathrm{He}_2$

Correct Answer: 2

Explanation:

To determine the number of molecules with a bond order of 2 from the given molecules, we need to first calculate the bond order for each molecule. The bond order can be determined using Molecular Orbital Theory (MOT). The bond order is given by the formula:

$\text{Bond Order} = \frac{\text{Number of bonding electrons} - \text{Number of anti-bonding electrons}}{2}$

Let's evaluate each molecule individually:

1. $\mathrm{C}_2$:

For $\mathrm{C}_2$, the total number of electrons is 12. The molecular orbital configuration will be $\left( \sigma_{1s} \right)^2 \left( \sigma_{1s}^* \right)^2 \left( \sigma_{2s} \right)^2 \left( \sigma_{2s}^* \right)^2 \left( \pi_{2p_x} \right)^2 \left( \pi_{2p_y} \right)^2$. There are 8 bonding electrons and 4 anti-bonding electrons:

$\text{Bond Order} = \frac{8 - 4}{2} = 2$

2. $\mathrm{O}_2$:

For $\mathrm{O}_2$, the total number of electrons is 16. The molecular orbital configuration will be $\left( \sigma_{1s} \right)^2 \left( \sigma_{1s}^* \right)^2 \left( \sigma_{2s} \right)^2 \left( \sigma_{2s}^* \right)^2 \left( \sigma_{2p_z} \right)^2 \left( \pi_{2p_x} \right)^2 \left( \pi_{2p_y} \right)^2 \left( \pi_{2p_x}^* \right)^1 \left( \pi_{2p_y}^* \right)^1$. There are 10 bonding electrons and 6 anti-bonding electrons:

$\text{Bond Order} = \frac{10 - 6}{2} = 2$

3. $\mathrm{Be}_2$:

For $\mathrm{Be}_2$, the total number of electrons is 8. The molecular orbital configuration will be $\left( \sigma_{1s} \right)^2 \left( \sigma_{1s}^* \right)^2 \left( \sigma_{2s} \right)^2 \left( \sigma_{2s}^* \right)^2$. There are 4 bonding electrons and 4 anti-bonding electrons:

$\text{Bond Order} = \frac{4 - 4}{2} = 0$

4. $\mathrm{Li}_2$:

For $\mathrm{Li}_2$, the total number of electrons is 6. The molecular orbital configuration will be $\left( \sigma_{1s} \right)^2 \left( \sigma_{1s}^* \right)^2 \left( \sigma_{2s} \right)^2$. There are 4 bonding electrons and 2 anti-bonding electrons:

$\text{Bond Order} = \frac{4 - 2}{2} = 1$

5. $\mathrm{Ne}_2$:

For $\mathrm{Ne}_2$, the total number of electrons is 20. The molecular orbital configuration will be $\left( \sigma_{1s} \right)^2 \left( \sigma_{1s}^* \right)^2 \left( \sigma_{2s} \right)^2 \left( \sigma_{2s}^* \right)^2 \left( \sigma_{2p_z} \right)^2 \left( \pi_{2p_x} \right)^2 \left( \pi_{2p_y} \right)^2 \left( \pi_{2p_x}^* \right)^2 \left( \pi_{2p_y}^* \right)^2 \left( \sigma_{2p_z}^* \right)^2$. There are 10 bonding electrons and 10 anti-bonding electrons:

$\text{Bond Order} = \frac{10 - 10}{2} = 0$

6. $\mathrm{N}_2$:

For $\mathrm{N}_2$, the total number of electrons is 14. The molecular orbital configuration will be $\left( \sigma_{1s} \right)^2 \left( \sigma_{1s}^* \right)^2 \left( \sigma_{2s} \right)^2 \left( \sigma_{2s}^* \right)^2 \left( \sigma_{2p_z} \right)^2 \left( \pi_{2p_x} \right)^2 \left( \pi_{2p_y} \right)^2$. There are 10 bonding electrons and 4 anti-bonding electrons:

$\text{Bond Order} = \frac{10 - 4}{2} = 3$

7. $\mathrm{He}_2$:

For $\mathrm{He}_2$, the total number of electrons is 4. The molecular orbital configuration will be $\left( \sigma_{1s} \right)^2 \left( \sigma_{1s}^* \right)^2$. There are 2 bonding electrons and 2 anti-bonding electrons:

$\text{Bond Order} = \frac{2 - 2}{2} = 0$

Thus, the molecules with a bond order of 2 from the given list are $\mathrm{C}_2$ and $\mathrm{O}_2$. Therefore, the number of molecules having bond order 2 is 2.

2024 JEE Mains Numerical

iCON Education HYD, 79930 92826, 73309 72826 JEE Main 2024 (Online) 8th April Morning Shift

Number of molecules from the following which are exceptions to octet rule is _________.

$\mathrm{CO}_2, \mathrm{NO}_2, \mathrm{H}_2 \mathrm{SO}_4, \mathrm{BF}_3, \mathrm{CH}_4, \mathrm{SiF}_4, \mathrm{ClO}_2, \mathrm{PCl}_5, \mathrm{BeF}_2, \mathrm{C}_2 \mathrm{H}_6, \mathrm{CHCl}_3, \mathrm{CBr}_4$

Chemistry Chapters

Chemical Bonding & Molecular Structure

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation: