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Chemical Bonding & Molecular Structure

JEE Chemistry · 150 questions · Page 3 of 15 · Click an option or "Show Solution" to reveal answer

Q21
Total number of lone pair of electrons in I3−{\rm I}_3^ - ion is
A 3
B 9
C 6
D 12
Correct Answer
Option B
Solution
∴   \therefore\,\,\,

Total number of lone pair = 9

Q22
Hybridisation of underline atom changes in
A Al‾H3\underline {Al} {H_3} changes to AlH4−AlH_4^-
B H2O‾{H_2}\underline O changes to H3O+H_3O^+
C N‾H3\underline N {H_3} changes to NH4+NH_4^+
D in all cases
Correct Answer
Option A
Solution

(a)

   \,\,\,

AlH3 + H−- →\to AlH

4−_4^ -

Steric number of AlH3 is =

12{1 \over 2}

[3 + 3] = 3

∴   \therefore\,\,\,

AlH3 is sp2 hybridized. Steric number of AlH

4−_4^ -

=

12{1 \over 2}

[3 + 4 +1] = 4

∴   \therefore\,\,\,

AlH

4−_4^ -

is sp3 hybridized. (b)

   \,\,\,

H2O + H+ →\to H3O+ Steric no of H2O =

12{1 \over 2}

(6+ 2) = 4

∴    \therefore\,\,\,\,

H2O s sp3 hybridized. Steric no of H3O+ =

12{1 \over 2}

[ 8 + 3 −-1] = 4

∴   \therefore\,\,\,

H3O+ s also sp3 hybridized. (c)

   \,\,\,

NH3 + H+ →\to NH4+ Steric no of NH3 =

12{1 \over 2}

[5 + 3] = 4

∴   \therefore\,\,\,

hybridization of NH3 is sp3 Steric number of NH4+ =

12{1 \over 2}

[5 + 4 −- ] = 4

∴   \therefore\,\,\,

Hybridization of NH4+ is sp3

Q23
Which of the following are arranged in an increasing order of their bond strengths?
A O2−O_2^- < O2 < O2+O_2^+ < O22−O_2^{2-}
B O22−O_2^{2-} < O2−O_2^- < O2O_2 < O2+O_2^{+}
C O2−O_2^- < O22−O_2^{2-} < O2O_2 < O2+O_2^{+}
D O2+O_2^{+} < O2O_2 < O2−O_2^- < O22−O_2^{2-}
Correct Answer
Option B
Solution

Note : (1)

    \,\,\,\,

Bond strength ∝\propto Bond order (2)

    \,\,\,\,

Bond length ∝\propto

1Bond  order{1 \over {Bond\,\,order}}

(3)

 \,

Bond order

=12= {1 \over 2}

[Nb −- Na] Nb = No of electrons in bonding molecular orbital Na == No of electrons in anti bonding molecular orbital (4)

    \,\,\,\,

upto 14 electrons, molecular orbital configuration is Here Na = Anti bonding electron == 4 and Nb = 10 (5)

    \,\,\,\,

After 14 electrons to 20 electrons molecular orbital configuration is - - - Here Na = 10 and Nb = 10 In O atom 8 electrons present, so in O2, 8 ×\times 2 = 16 electrons present.

Then in

O2+O_2^ +

no of electrons = 15 in

O2−O_2^ -

no of electrons = 17 in

O22−O_2^{2 - }

no of electrons = 18

∴    \therefore\,\,\,\,

Molecular orbital configuration of O2 (16 electrons) is

σ1s2  σ1s2∗ {\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * \,
σ2s2  σ2s2∗ {\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,
σ2pz2  π2px2=π2py2  π2px1∗  =π2py1∗{\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}} = {\pi _{2p_y^2}}\,\,\pi _{2p_x^1}^ * \,\, = \pi _{2p_y^1}^ *
∴    \therefore\,\,\,\,

Na = 6 Nb = 10

∴    \therefore\,\,\,\,

BO =

12[10−6]=2{1 \over 2}\left[ {10 - 6} \right] = 2

Molecular orbital configuration of O

2+_2^ +

(15 electrons) is

σ1s2 σ1s2∗ σ2s2 σ2s2∗ σ2pz2 π2px2 = π2py2 π2px1∗ = π2pyo∗{\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\,\,\,\,

Nb = 10 Na = 5

∴    \therefore\,\,\,\,

BO =

12[10−5]{1 \over 2}\left[ {10 - 5} \right]

= 2.5 Molecular orbital configuration of

O2−O_2^ -

(17 electrons) is

σ1s2 σ1s2∗ σ2s2 σ2s2∗ σ2pz2 π2px2 = π2py2 π2px2∗ = π2py1∗{\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\,\,\,\,

Nb = 10 Na = 7

∴    \therefore\,\,\,\,

BO =

12[10−7]{1 \over 2}\left[ {10 - 7} \right]

= 1.5 Molecular orbital configuration of O

22−_2^{2 - }

(18 electrons) is

σ1s2 σ1s2∗ σ2s2 σ2s2∗ σ2pz2 π2px2 = π2py2 π2px2∗ = π2py2∗{\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\,\,\,\,

Nb = 10 Na = 8

∴    \therefore\,\,\,\,

BO =

12{1 \over 2}

[ 10 −- 8] = 1 As Bond strength ∝\propto Bond order so, correct order is

O22−<O2−<O2<O2+O_2^{2 - } < O_2^ - < {O_2} < O_2^ +
Q24
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
Correct Answer
Option C
Solution

(A) Electronegativity of F is 4 and Electronegativity of H is 2.1.

∴\therefore Electronegativity difference between F and H is = 4 - 2.1 = 1.9 Now Electronegativity of Br is 2.8 and Electronegativity of H is 2.1.

∴\therefore Electronegativity difference between Br and H is = 2.8 - 2.1 = 0.7 As in HF Electronegativity difference is more then it's polar nature is also more.

(B) At equilibrium pure H2O produce very less H+ and OH- ions.

(C) Chemical bond formation takes place when forces of attraction overcome the forces of repulsion.

(D) In covalenrt bond, sharing of electons take place.

Transfer of electrons take place in ionic bond.

Q25
In which of the following pairs the two species are not isostructural ?
A CO32−CO_3^{2-} and NO3−NO_3^-
B PCl4+PCl_4^{+} and SiCl4SiCl_4
C PF5 and BrF5
D AlF6−AlF_6^- and SF6SF_6
Correct Answer
Option C
Solution

PF5 ⇒\Rightarrow sp3d hybridized and shape is trigonal bipyramidal.

BrF5 ⇒\Rightarrow 5 bond pair and 1 lone pair present so sp3d2 hybridized and shape is square pyramidal.

Q26
A square planar complex is formed by hybridisation of which atomic orbitals ?
A s, px, py, dxy
B s, px, py, dx2 - y2
C s, px, py, dz2
D s, px, py, dyz
Correct Answer
Option B
Solution

Hybridization of square planar complex is dsp2 .

In square planar complex all 4 surrounding atoms and central atom are in the same plane and let this plane is x −- y plane.

As it is dsp2 hybridized, so it has 1 d orbital, 1 s orbital and 2 p orbital. s orbital is non-directional, so we just write it as s.

In px the two lobes are along x-axis and in py two lobes are along y-axis One d orbital is dx2 −- y2, it means out of four lobes of d orbital two along x-axis and two along y-axis.

Q27
In which of the following species the interatomic bond angle is 109o28' ?
A NH3, BF4-
B NH4+, BF4-
C NH3, BF3
D NH2-1, BF3
Correct Answer
Option B
Solution

Bond angle 109o 28' means Regular Tetrahedral geometry and hybridization is sp3.

Steric Number (SN) for sp3 hybridization is 4.

In sp3 molecules can have different shapes which is decided by SN number (1)

    \,\,\,\,

4 bond pair in a molecule then angle between bonds 109o 28' (2)

    \,\,\,\,

3 bond pair and 1 lone pair in a molecule then angle between bonds 107o. (3)

    \,\,\,\,

2 bond pair and 2 lone pair in a molecule then angle between bonds 104.5o (4)

    \,\,\,\,

1 bond pair and 3 lone pair in a molecule then bond angle is undefined as there is only one bond. (1)

    \,\,\,\,

In NH3, 3 Bond pair(BP) +1 lone pair (LP) present so angle between bond 107o. (2)

    \,\,\,\,

BF

4−_4^ -

, 4 bond pair present so angle is 109o 28'. (3) In NH

4+_4^ +

, 4 bond pair present so angle between bond is 109o 28' (4)

    \,\,\,\,

BF3 has sp2 hybridization. So bond angle is 120o. (5)

    \,\,\,\,

In NH

2−_2^ -

, 2 bond pair and 2 lone pair present, so bond angle is 104.5o So,

NH+NH{}^ +

and BF

4−_4^ -

has bond angle 109o 28'.

Q28
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+
Correct Answer
Option D
Solution

Molecular orbital configuration of NO (15 electrons) is

σ1s2 σ1s2∗ σ2s2 σ2s2∗ σ2pz2 π2px2 = π2py2 π2px1∗ = π2pyo∗{\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\,\,\,\,

Nb = 10 Na = 5

∴    \therefore\,\,\,\,

BO =

12[10−5]{1 \over 2}\left[ {10 - 5} \right]

= 2.5 Similarly Molecular orbital configuration of NO+ (14 electrons) is

σ1s2 σ1s2∗ σ2s2 σ2s2∗ σ2pz2 π2px2 = π2py2 {\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}}\,
∴    \therefore\,\,\,\,

Nb = 10 Na = 4

∴    \therefore\,\,\,\,

BO =

12[10−4]{1 \over 2}\left[ {10 - 4} \right]

= 3 Note : (2)

    \,\,\,\,

Bond length ∝\propto

1Bond  order{1 \over {Bond\,\,order}}

So, bond length in NO > NO+

Q29
In which of the following pairs of molecules/ions, both the species are not likely to exist?
A H2+H_2^+ , He22−He_2^{2-}
B H2−H_2^- , He22−He_2^{2-}
C H22+H_2^{2+} , He2He_2
D H2−H_2^- , He22+He_2^{2+}
Correct Answer
Option C
Solution
H22+H_2^{2+}

,

He2He_2

Both have zero bond order. Thus, they do not exist.

Q30
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
Correct Answer
Option C
Solution

1.

In H2S molecule, central atom S is a 3rd period element and according to Dragos rule, when a 3rd period or higher period element overlap with a small element whose electronegetivity is low then that over lapping cannot be a effective overlapping, because of higher size of 3rd or higher periods element and smaller size of low electronegative element.

Here in H2S, H atom has very low electronegativity and smaller in size so it create more negative charge around sulphur(S) atom and size of S−-2 ion increases, because of this higher size difference efficiency overlapping is not possible.

So, any hybridization do not happen in between S and H atom.

Lone pair electrons of S atom is present in the pure orbital like s, px, py, pz and it look like this : As we know the angle between and py is 90o, so bond angle is 90o.

2.

In NH3, 3 Bond pair(BP) + 1 lone pair (LP) present so angle between bond 107o.

3.

    \,\,\,\,

BF3 has sp2 hybridization.

So bond angle is 120o.

4.

Here Si is sp3 hybridised and shape is regular tetrahedral and bond angle 109o 28'

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