Chemical Bonding & Molecular Structure — Page 14 | JEE Chemistry

Q131

Match List I with List II .tg .tg LIST I (Molecule/Species) LIST II (Property/Shape) A.

\mathrm{SO_2Cl_2}

I. Paramagnetic B.

\mathrm{NO}

II. Diamagnetic C.

\mathrm{NO_2^-}

III. Tetrahedral D.

\mathrm{I_3^-}

IV. Linear Choose the correct answer from the options given below:

A A-II, B-III, C-I, D-IV

B A-III, B-I, C-II, D-IV

C A-IV, B-I, C-III, D-II

D A-III, B-IV, C-II, D-I

Correct Answer

Option B

Solution

\begin{aligned} & \mathrm{A} \rightarrow \text{ Tetrahedral (III) } \\ & \mathrm{B} \rightarrow \text{ Paramagnetic (I) } \\ & \mathrm{C} \rightarrow \text{ Diamagnetic (II) } \\ & \mathrm{D} \rightarrow \text{ Linear (IV) } \end{aligned}

Q132

Match the with , \mathrm{NO}_2$

List - I		List - II
(B)	Molecules with incomplete octet	(II)	$\mathrm{BCl}_3, \mathrm{AlCl}_3$
(C)	Molecules with incomplete octet with odd electron	(III)	$\mathrm{H}_2 \mathrm{SO}_4, \mathrm{PCl}_5$
(D)	Molecules with expanded octet	(IV)	$\mathrm{CCl}_4, \mathrm{CO}_2$

A A-IV, B-I, C-III, D-II

B A-II, B-IV, C-III, D-I

C A-IV, B-II, C-I, D-III

D A-III, B-II, C-I, D-IV

Correct Answer

Option C

Solution

(A) $\mathrm{A} \rightarrow$ IV (B) B $\rightarrow$ II (C) $\mathrm{C} \rightarrow \mathrm{I}$ (D) $\mathrm{D} \rightarrow$ III

Q133

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

Correct Answer

Option B

Solution

Here $\mu$ total = $\mu$ 1 + $\mu$ 2 $\ne$ 0 $\mu$ total = $\mu$ 1 $-$ $\mu$ 2 = 0 $\mu$ total = $\mu$ 1 + $\mu$ 2 $\ne$ 0 $\mu$ 1 + $\mu$ 2 + $\mu$ 3 = $\mu$ 4 $\mu$ total = 0 In NO2 and O3 have permanent dipole moment as both of them are not symmetric molecule.

In CO2 and SiF4 each individual bond c = O and S- $-$ F have dipole moment but because of symmetric structure individual dipole moment get's cancelled.

Q134

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

A OH2

B SH2

C NH3

D SO2

Correct Answer

Option B

Solution

(a) H2O is sp3 hybridized, and oxygen atom has 2 bond pair and 2 lone pair.

So the angle between two O $-$ H bond is 104.5o (b) 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. (c) NH3 has sp3 hybridization and N atom has 3 bond pair and one lone pair so bond angle is 107o (d) SO2 is sp2 hybridized and bond angle is 117o.

Q135

\mathrm{O}-\mathrm{O}

bond length in

\mathrm{H}_{2} \mathrm{O}_{2}

is

\underline{\mathrm{X}}

than the

\mathrm{O}-\mathrm{O}

bond length in

\mathrm{F}_{2} \mathrm{O}_{2}

. The

\mathrm{O}-\mathrm{H}

bond length in

\mathrm{H}_{2} \mathrm{O}_{2}

is

\underline{Y}

than that of the

\mathrm{O}-\mathrm{F}

bond in

\mathrm{F}_{2} \mathrm{O}_{2}

. Choose the correct option for

\underline{X}

and

\underline{Y}

from those given below :

A

\mathrm{X}

- shorter,

\mathrm{Y}

- longer

B

\mathrm{X}

- longer,

\mathrm{Y}

- shorter

C

\mathrm{X}

- shorter,

\mathrm{Y}

- shorter

D

\mathrm{X}

- longer,

\mathrm{Y}

- longer

Correct Answer

Option B

Solution

.tg .tg Bond Bond Length O - O $1.48 \,\mathop A\limits^o\left(\text{ In }{\mathrm{H}_2 \mathrm{O}_2}\right)$ O - O $1.22 \,\mathop A\limits^o\left(\text{ In }{\mathrm{O}_2 \mathrm{F}_2}\right)$ Thus, $\mathrm{O}-\mathrm{O}$ bond length in $\mathrm{H}_2 \mathrm{O}_2$ is longer than $\mathrm{O}-\mathrm{O}$ bond length in $\mathrm{O}_2 \mathrm{~F}_2$ . .tg .tg Bond Bond Length O - F $158 \,\mathrm{pm}\left(\text{ In }{\mathrm{O}_2 \mathrm{F}_2}\right)$ O - H $95 \,\mathrm{pm}\left(\text{ In }{\mathrm{H}_2 \mathrm{O}_2}\right)$ Thus, the $\mathrm{O}-\mathrm{H}$ bond length in $\mathrm{H}_2 \mathrm{O}_2$ is shorter than $\mathrm{O}-\mathrm{F}$ bond length in $\mathrm{O}_2 \mathrm{~F}_2$ .

Q136

Which of the following molecule(s) show/s paramagnetic behavior? A.

\mathrm{O}_2

B.

\mathrm{N}_2

C.

\mathrm{F}_2

D.

\mathrm{S}_2

E.

\mathrm{Cl}_2

Choose the correct answer from the options given below:

A A & C only

B A & E Only

C A & D Only

D B Only

Correct Answer

Option C

Solution

No. of unpaired $\mathrm{e}^{-}$ .tg .tg (A) $\mathrm{O}_2$ 2 (B) $\mathrm{N}_2$ 0 (C) $\mathrm{F}_2$ 0 (D) $\mathrm{S}_2$ 2 (E) $\mathrm{Cl}_2$ 0 If species contain unpaired electron than it is paramagnetic.

So A & D are paramagnetic.

Q137

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

Correct Answer

Option C

Solution

Regular Tetrahedral structure is possible in sp3 hybridization where central atom has 4 bond pair and no lone pair. (a) XeF4 is sp3d2 hybridised and structure is square planar. (b) [Ni(CN)4] $-$ 2 is coordinate compound and oxidation number of Ni is +2.

Electronic configuration of Ni+2 is $=$ [Ar]3d8 But because of CN $-$ ion which is a strong field ligand , it can perform pairing of electron.

And the structure of dsp2 hybridization is square planar.

(C)

\,\,\,\,

BF

_4^ -

, 4 bond pair present so angle is 109o 28' and sp3 hybridised.

So structure is regular tetrahedral. (d) SF4 is sp3d hybridised and structure is see-saw.

Q138

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

A XeF4

B

BF_4^−

C SF4

D SiF4

Correct Answer

Option C

Solution

(a) XeF4 is sp3d2 hybridised with 4 bond pairs and 1 lone pair and structure is square planar.

Here all the bond lengths are equal. (b)

\,\,\,\,

BF

_4^ -

, 4 bond pair present so angle is 109o 28' and sp3 hybridised.

So structure is regular tetrahedral.

Here all the bond lengths are equal. (c) SF4 is sp3d hybridised with 4 bond pairs and 1 lone pair and its expected trigonal bipyramidal geometry gets distorted due to presence of a lone pair of electrons and it becomes distorted tetrahedral or see-saw with the bond angles equal to < 120o and 179o instead of the expected angles of 120o and 180o respectively.

Here axial and equitorial both bonds are presents.

And we know axial bonds are longer and weaker. (d) SiF4 is sp3 hybridisation and regular tetrahedral geometry.

Here all the bond lengths are equal.

Q139

Which of the following species exhibits the diamagnetic behaviour?

A

O_2^{2−}

B NO

C

O_2^+

D O2

Correct Answer

Option A

Solution

Those species which have unpaired electrons are called paramagnetic species.

And those species which have no unpaired electrons are called diamagnetic species. (a)

O_2^{2−}

has 18 electrons. Moleculer orbital configuration of

O_2^{2−}

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}^ *

Here is no unpaired electron so it is diamagnetic. (b) NO has 15 electrons. Moleculer orbital configuration of NO 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}^ *

Here is 1 unpaired electron, So it is Paramagnetic. (c)

O_2

has 16 electrons. Moleculer orbital configuration of

O_2

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}^ *

Here 2 unpaired electron present, so it is paramagnetic. (d)

O_2^{+}

has 15 electrons. Moleculer orbital configuration of

O_2^{+}

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}^ *

Here 1 unpaired electron present, so it is paramagnetic.

Q140

Which of the following species is not paramagnetic?

A CO

B O2

C B2

D NO

Correct Answer

Option A

Solution

Those species which have unpaired electrons are called paramagnetic species. (a) CO has 14 electrons.

Moleculer orbital configuration of CO is

{\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}}

Here is no unpaired electron so it is diamagnetic. (b)

O_2

has 16 electrons. Moleculer orbital configuration of

O_2

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}^ *

Here 2 unpaired electron present, so it is paramagnetic. (c) B2 has 10 electrons.

Molecular orbital configuration of B2 is

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

Here two unpaired electrons present.

So it is paramagnetic. (d) NO has 15 electrons.

Moleculer orbital configuration of NO 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}^ *

Here is 1 unpaired electron, So it is Paramagnetic.