Practice Start Test

Chemical Bonding & Molecular Structure

NEET Chemistry · 95 questions · Page 4 of 10 · Click an option or "Show Solution" to reveal answer

Q31
Predict the correct order among the following :
A bond pair - bond pair > long pair - bond pair > lone pair - lone pair
B lone pair - bond pair > bond pair - bond pair > lone pair - lone pair
C lone pair - lone pair > lone pair - bond pair > bond pair - bond pair
D lone pair - lone pair > bond pair - bond pair > lone pair - bond pair
Correct Answer
Option C
Solution

According to VSEPR theory, the repulsive forces between lone pair and lone pair are grater than between lone pair and bond pair which are further greater than bond pair and bond pair.

Q32
Consider the molecules CH 4 , NH 3 and H 2 O. Which of the given statements is false?
A The H −- O −- H bond angle in H 2 O is smaller than the H −- N −- H bond angle in NH 3 .
B The H −- C −- H bond angle in CH 4 is larger than the H −- N −- H bond angle in NH 3 .
C The H −- C −- H bond angle in CH 4 , the H −- N −- H bond angle in NH 3 , and the H −- O −- H bond angle in H 2 O are all greater than 90 o .
D The H −- O −- H bond angle in H 2 O is larger than the H −- C −- H bond angle in CH 4 .
Correct Answer
Option D
Solution

In CH 4 , there is no lone pair of electrons so it is perfectly tetrahedral in shape and has H-C-H bond angle of 109°28′ but in NH 3 there is one lone pair which due to more repulsion with bond pairs of electrons pushes them more closer thus bond angle decreases to 107°.

In H 2 O, there are two lone pair of electrons which suppress the bond angle more to 104.5°.

Q33
Decreasing order of stability of O 2 , O 2 −- , O 2 + and O 2 2−- is
A O 2 2−- > O 2 −- > O 2 > O 2 +
B O 2 > O 2 + > O 2 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
Option D
Solution

According to molecular orbital theory as bond order decreases stability of the molecule decreases. Bond order =

121 \over 2

(N b - N a ) Bond order of

O2+O_2^+

=

121 \over 2

(10 - 5) = 2.5 Bond order of

O2O_2

=

121 \over 2

(10 - 6) = 2 Bond order of

O2−O_2^-

=

121 \over 2

(10 - 7) = 1.5 Bond order of

O22−O_2^{2-}

=

121 \over 2

(10 - 8) = 1.0 Hence the correct order is

O2+O_2^+

>

O2O_2

>

O2−O_2^-

>

O22−O_2^{2-}
Q34
Which of the following options represents the correct bond order ?
A O 2 −- > O 2 < O 2 +
B O 2 −- < O 2 > O 2 +
C O 2 −- > O 2 > O 2 +
D O 2 −- < O 2 < O 2 +
Correct Answer
Option D
Solution
 \,

Bond order

=12= {1 \over 2}

[N<sub>b </sub> −- N<sub>a</sub>] <br><br>N<sub>b</sub> = No of electrons in bonding molecular orbital <br><br>N<sub>a</sub> == No of electrons in anti bonding molecular orbital <br><br>In O atom 8 electrons present, so in O<sub>2</sub>, 8 ×\times 2 = 16 electrons present. <br><br>Then in

O2+O_2^ +

no of electrons = 15 <br><br>in

O2−O_2^ -

no of electrons = 17 <br><br>

∴    \therefore\,\,\,\,

Molecular orbital configuration of O<sub>2</sub> (16 electrons) is <br><br>

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}^ *

<br><br>

∴    \therefore\,\,\,\,

N<sub>a</sub> = 6 <br><br>N<sub>b</sub> = 10 <br><br>

∴    \therefore\,\,\,\,

BO =

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

<br><br>Molecular orbital configuration of O

2+_2^ +

(15 electrons) is <br><br>

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}^ *

<br><br>

∴    \therefore\,\,\,\,

N<sub>b</sub> = 10 <br><br>N<sub>a</sub> = 5 <br><br>

∴    \therefore\,\,\,\,

BO =

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

= 2.5 <br><br>Molecular orbital configuration of

O2−O_2^ -

(17 electrons) is <br><br>

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}^ *

<br><br>

∴    \therefore\,\,\,\,

N<sub>b</sub> = 10 <br><br>N<sub>a</sub> = 7 <br><br>

∴    \therefore\,\,\,\,

BO =

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

= 1.5 <br><br> <table class=tg> <tbody><tr> <th class=tg-uys7></th> <th class=tg-uys7>

O2−O_2^-

</th> <th class=tg-uys7>

O2O_2

</th> <th class=tg-uys7>

O2+O_2^+

</th> </tr> <tr> <td class=tg-uys7><span style=font-weight:bold>B.O. :</span></td> <td class=tg-uys7>1.5</td> <td class=tg-uys7>2.0</td> <td class=tg-uys7>2.5</td> </tr> </tbody></table>

Q35
Which of the following species contains equal number of σ\sigma - and π\pi -bonds ?
A (CN) 2
B CH 2 (CN) 2
C HCO 3 −-
D XeO 4
Correct Answer
Option D
Solution

Number of σ\sigma bonds = 4 Number of π\pi bonds = 4

Q36
Which of the following pairs of ions are isoelectronic and isostructural ?
A SO 3 2−- , NO 3 −-
B ClO 3 −- , SO 3 2−-
C CO 3 2−- , SO 3 2−-
D ClO 3 −- , CO 3 2−- ,
Correct Answer
Option B
Solution

<table class=tg> <tbody><tr> <th class=tg-hgcj>Species</th> <th class=tg-hgcj>Hybridisation</th> <th class=tg-hgcj>Shape</th> <th class=tg-amwm>No. of e<sup>-</sup></th> </tr> <tr> <td class=tg-s6z2>

SO32−SO_3^{2-}

</td> <td class=tg-s6z2>sp<sup>3</sup></td> <td class=tg-s6z2>Pyramidal</td> <td class=tg-baqh>42</td> </tr> <tr> <td class=tg-s6z2>

ClO3−ClO_3^{-}

</td> <td class=tg-s6z2>sp<sup>3</sup></td> <td class=tg-s6z2>Pyramidal</td> <td class=tg-baqh>42</td> </tr> <tr> <td class=tg-s6z2>

CO32−CO_3^{2-}

</td> <td class=tg-s6z2>sp<sup>2</sup></td> <td class=tg-s6z2>Triangular planar</td> <td class=tg-baqh>32</td> </tr> <tr> <td class=tg-s6z2>

NO3−NO_3^{-}

</td> <td class=tg-s6z2>sp<sup>2</sup></td> <td class=tg-s6z2>Triangular planar</td> <td class=tg-baqh>32</td> </tr> </tbody></table>

Q37
Maximum bond angle at nitrogen is present in which of the following ?
A NO 2 +
B NO 3 −-
C NO 2
D NO 2 −-
Correct Answer
Option A
Solution

<table class=tg> <tbody><tr> <th class=tg-hgcj>Species</th> <th class=tg-s6z2>

NO3−NO_3^-

</th> <th class=tg-s6z2>

NO2NO_2

</th> <th class=tg-s6z2>

NO2−NO_2^-

</th> <th class=tg-baqh>

NO2+NO_2^+

</th> </tr> <tr> <td class=tg-hgcj>Hybridisation</td> <td class=tg-s6z2>sp<sup>2</sup></td> <td class=tg-s6z2>sp<sup>2</sup></td> <td class=tg-s6z2>sp<sup>2</sup></td> <td class=tg-baqh>sp(linear)</td> </tr> <tr> <td class=tg-amwm>Bond Angle</td> <td class=tg-baqh>120<sup>o</sup></td> <td class=tg-baqh>134<sup>o</sup></td> <td class=tg-baqh>115<sup>o</sup></td> <td class=tg-baqh>180<sup>o</sup></td> </tr> </tbody></table>

Q38
The correct bond order in the following species is
A O 2 + < O 2 −- < O 2 2+
B O 2 −- < O 2 + < O 2 2+
C O 2 2+ < O 2 + < O 2 −-
D O 2 2+ < O 2 −- < O 2 +
Correct Answer
Option B
Solution
 \,

Bond order

=12= {1 \over 2}

[N<sub>b </sub> −- N<sub>a</sub>] <br><br>N<sub>b</sub> = No of electrons in bonding molecular orbital <br><br>N<sub>a</sub> == No of electrons in anti bonding molecular orbital <br><br>In O atom 8 electrons present, so in O<sub>2</sub>, 8 ×\times 2 = 16 electrons present. <br><br>Then in

O2+O_2^ +

no of electrons = 15 <br><br>in

O2−O_2^ -

no of electrons = 17 <br><br>in

O22+O_2^{2 + }

no of electrons = 14 <br><br>

∴    \therefore\,\,\,\,

Molecular orbital configuration of O<sub>2</sub> (16 electrons) is <br><br>

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}^ *

<br><br>

∴    \therefore\,\,\,\,

N<sub>a</sub> = 6 <br><br>N<sub>b</sub> = 10 <br><br>

∴    \therefore\,\,\,\,

BO =

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

<br><br>Molecular orbital configuration of O

2+_2^ +

(15 electrons) is <br><br>

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}^ *

<br><br>

∴    \therefore\,\,\,\,

N<sub>b</sub> = 10 <br><br>N<sub>a</sub> = 5 <br><br>

∴    \therefore\,\,\,\,

BO =

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

= 2.5 <br><br>Molecular orbital configuration of

O2−O_2^ -

(17 electrons) is <br><br>

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}^ *

<br><br>

∴    \therefore\,\,\,\,

N<sub>b</sub> = 10 <br><br>N<sub>a</sub> = 7 <br><br>

∴    \therefore\,\,\,\,

BO =

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

= 1.5 <br><br>Molecular orbital configuration of O

22+_2^{2 + }

(14 electrons) is <br><br>

σ1s2 σ1s2∗ σ2s2 σ2s2∗ σ2pz2 π2px2 = π2py2 π2px0∗ = π2py0∗{\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}^ *

<br><br>

∴    \therefore\,\,\,\,

N<sub>b</sub> = 10 <br><br>N<sub>a</sub> = 4 <br><br>

∴    \therefore\,\,\,\,

BO =

12{1 \over 2}

[ 10 −- 4] = 3 <br><br> <table class=tg> <tbody><tr> <th class=tg-s6z2></th> <th class=tg-s6z2>

O2−O_2^-

</th> <th class=tg-s6z2>

O2O_2

</th> <th class=tg-s6z2>

O2+O_2^+

</th> <th class=tg-baqh>

O22+O_2^{2+}

</th> </tr> <tr> <td class=tg-s6z2><span style=font-weight:bold>B.O. :</span></td> <td class=tg-s6z2>1.5</td> <td class=tg-s6z2>2.0</td> <td class=tg-s6z2>2.5</td> <td class=tg-baqh>3.0</td> </tr> </tbody></table>

Q39
Which one of the following species has plane triangular shape ?
A N 3
B NO 3 −-
C NO 2 −-
D CO 2
Correct Answer
Option B
Solution

Hybridisation of N 3 is sp, so it is linear in shape.

Hybridisation of NO 3 – is sp 2 , so it is plane triangular in shape.

Hybridisation of NO 2 – is sp, so it is linear in shape.

Hybridisation of CO 2 is sp, so it is linear in shape.

Q40
Which of the following molecules has the maximum dipole moment ?
A CO 2
B CH 4
C NH 3
D NF 3
Correct Answer
Option C
Solution

Even though for CO 2 and CH 4 the C - O and C - H bonds are polar but due to their symmetrical structure they have zero dipole moment.

In NH 3 , H is less electronegative than N and hence the dipole moment of each N-H bond is towards N which is in the similar direction of lone pair but for NF 3 the dipole moment of N-F bond is in opposite direction of lone pair which reduce the value of net dipole moment.

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