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Chemical Bonding & Molecular Structure

JEE Chemistry · 150 questions · Page 5 of 15 · Click an option or "Show Solution" to reveal answer

Q41
The structure of IF7 is :
A trigonal bipyramid
B octahedral
C pentagonal bipyramid
D square pyramid
Correct Answer
Option C
Solution

The structure of IF7 pentagonal bipyramidal having sp3d3 hybridisation.

Q42
Stability of the species Li2, Li2−Li_2^− and Li2+Li_2^+ increases in the order of:
A Li2 < Li2+Li_2^+ < Li2−Li_2^-
B Li2+Li_2^+ < Li2−Li_2^- < Li2
C Li2−Li_2^- <Li2+Li_2^+ < Li2
D Li2−Li_2^- < Li2 < Li2+Li_2^+
Correct Answer
Option C
Solution

Li2 =

σ1s2  σ1s2∗ {\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * \,
σ2s2 {\sigma _{2{s^2}}} \,

∴\therefore Bond order =

12(4−2){1 \over 2}\left( {4 - 2} \right)

= 1

Li2+Li_2^+

=

σ1s2  σ1s2∗ {\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * \,
σ2s1 {\sigma _{2{s^1}}} \,

∴\therefore Bond order =

12(3−2){1 \over 2}\left( {3 - 2} \right)

= 0.5

Li2−Li_2^-

=

σ1s2  σ1s2∗ {\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * \,
σ2s2  σ1s1∗ {\sigma _{2{s^2}}} \,\,\sigma _{1{s^1}}^ * \,

∴\therefore Bond order =

12(4−3){1 \over 2}\left( {4 - 3} \right)

= 0.5 The bond order of

Li2+Li_2^+

and

Li2−Li_2^-

is same but

Li2+Li_2^+

is more stable than

Li2−Li_2^-

because

Li2+Li_2^+

is smaller in size and has 2 electrons in antibonding orbitals whereas

Li2−Li_2^-

has 3 electrons in antibonding orbitals. Hence

Li2+Li_2^+

is more stable than

Li2−Li_2^-

.

Q43
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
Correct Answer
Option C
Solution

Hydrogen bond is a type of strong electrostatic dipole- dipole intersection and dependent on the inverse cube of distance between the molecular ion-dipole - interaction

∝1r3.\propto {1 \over {{r^3}}}.
Q44
The bond angle H - X - H is the greatest in the compound :
A CH4
B NH3
C H2O
D PH3
Correct Answer
Option A
Solution

CH4 - sp3 hybridisation - Bond angle (109o28') NH3 - sp3 hybridisation - Bond angle ( 107.8o) H2O - sp3 hybridisation - Bond angle ( 104.5o) PH3 - No hybridisation - Bond angle ( 90o) Note : According to DRAGO RULE, lone pair of electrons are in pure s- orbital of the atom P.

That is why PH3 has no hybridisation.

Q45
The group of molecules having identical shape is :
A SF4 , XeF4 , CCl4
B ClF3 , XeOF2 , XeF3+_3^ +
C BF3 , PCl3 , XeO3
D PCl5 , IF5 , XeO2F2
Correct Answer
Option B
Solution

H =

12{1 \over 2}

[VE + MA – c + a] ClF3 , H =

12{1 \over 2}

(7 + 3 – 0 + 0) = 5 (sp3d) XeOF2 , H =

12{1 \over 2}

(8 + 2 – 0 + 0) = 5 (sp3d) XeF3+ , H =

12{1 \over 2}

(8 + 3 – 1 + 0 ) = 5 (sp3d) All molecules have sp3d hybridization and 3 bond pair + 2 lone pairs.

Hence all have identical stucture (T-shape).

Q46
Match List I with List II .tg .tg LIST I (Compound/Species) LIST II (Shape/Geometry) A. SF4\mathrm{SF_4} I. Tetrahedral B. BrF3\mathrm{BrF_3} II. Pyramidal C. BrO3−\mathrm{BrO_3^-} III. See saw D. NH4+\mathrm{NH_4^+} IV. Bent T-Shape Choose the correct answer from the options given below:
A A-III, B-II, C-IV, D-I
B A-III, B-IV, C-II, D-I
C A-II, B-IV, C-III, D-I
D A-II, B-III, C-I, D-IV
Correct Answer
Option B
Solution
 A → Sea-saw (III)  B → Bent T-shape (IV)  C → Pyramidal (II)  D → Tetrahedral (I) \begin{aligned} & \text{ A } \rightarrow \text{ Sea-saw (III) } \\ & \text{ B } \rightarrow \text{ Bent T-shape (IV) } \\ & \text{ C } \rightarrow \text{ Pyramidal (II) } \\ & \text{ D } \rightarrow \text{ Tetrahedral (I) } \end{aligned}
Q47
The group having triangular planar structures is :
A BF3, NF3, CO32−CO_3^{2 - }
B CO32−CO_3^{2 - }, NO3−NO_3^ - , SO3
C NH3, SO3, CO32−CO_3^{2 - }
D NCl3, BCl3, SO3
Correct Answer
Option B
Solution

(A) BF3 : sp2 Hybridization : Triangular Planar NF3 : sp3 Hybridization : Tetrahedral

CO32−CO_3^{2 - }

: sp2 Hybridization : Triangular Planar (B)

CO32−CO_3^{2 - }

: sp2 Hybridization : Triangular Planar

NO3−NO_3^ -

: sp2 Hybridization : Triangular Planar SO3 : sp2 Hybridization : Triangular Planar (C) NH3 : sp3 Hybridization : Tetrahedral SO3 : sp2 Hybridization : Triangular Planar

CO32−CO_3^{2 - }

: sp2 Hybridization : Triangular Planar (D) NCl3 : sp3 Hybridization : Tetrahedral BCl3 : sp2 Hybridization : Triangular Planar SO3 : sp2 Hybridization : Triangular Planar

Q48
sp3d2 hybridization is not displayed by :
A BrF5
B SF6
C [CrF6]3−-
D PF5
Correct Answer
Option D
Solution

Hybridization (X) =

12{1 \over 2}

[ VE + MA – c + a ] where, VE = No. of valence electrons of central atom MA = No. of monovalent atoms/groups surrounding the central atom, c = Charge on the cation, a = Charge on the anion (A) [BrF5] : X =

12{1 \over 2}

[ 7 + 5 - 0 + 0] = 6 = sp3d2 hybridized (B) [SF6] : X =

12{1 \over 2}

[ 6 + 6 - 0 + 0] = 6 = sp3d2 hybridized (C) [CrF6]3−- : Cr+3 = [Ar]3d3 (D) [PF5] : X =

12{1 \over 2}

[ 5 + 5 - 0 + 0] = 5 = sp3d hybridized Note : [CrF6]3−- shows d2sp3 hybridization not sp3d2.

So option (C) should also be the correct answer.

Q49
According to molecular orbital theory, which of the following will not be a viable molecule?
A He22+{\rm H}e_2^{2 + }
B He2+{\rm H}e_2^{ + }
C H2−{\rm H}_2^{- }
D H22−{\rm H}_2^{2 - }
Correct Answer
Option D
Solution

Note : According to molecules orbital theory, when a molecule have bond order = 0 then that molecule does not exist. (a)

   \,\,\,

Configuration of

He22+He_2^{2 + }

(2 electrons) is =

σ1s2{\sigma _{1{s^2}}}
∴   \therefore\,\,\,

Bond order =

12{1 \over 2}

(2 −-0) = 1 (b)

   \,\,\,

Configuration of

He2+He_2^ +

(3 electrons) is =

σ1s2{\sigma _{1{s^2}}}
σ1s1∗\sigma _{1{s^1}}^ *
∴   \therefore\,\,\,

Bond order =

12{1 \over 2}

(2 −-1) = 0.5 (c)

   \,\,\,

Configuration of

H2−H_2^ -

(3 electrons) is =

σ1s2{\sigma _{1{s^2}}}
σ1s1∗\sigma _{1{s^1}}^ *

∴\therefore Bond order =

12{1 \over 2}

(2 −- 1) = 0.5 (d)

∴   \therefore\,\,\,

Configuration of

He22−He_2^{2 - }

is =

σ1s2  σ1s2∗{\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ *
∴   \therefore\,\,\,

Bond order =

=12= {1 \over 2}

(2 −- 2) = 0

∴   \therefore\,\,\,
H22−H_2^{2 - }

is not a viable molecule.

Q50
The incorrect geometry is represented by :
A BF3 - trigonal planar
B H2O - bent
C NF3 - trigonal planar
D AsF5 - trigonal bipyramidal
Correct Answer
Option C
Solution

NF3 has trigonal pyramidal geometry.

Here central atom nitrogen has sp3 hybridization and one lone pair of electron.

Because of lone pair there exist repulsion force between lone pair and bond pair of electrons, that is why bond angles are lower than tetrahedral bond angle.

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