Chemical Bonding and Molecular Structure Test 4

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Chemical Bonding and Molecular Structure Test 4

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Weekly Quiz Competition

Question 1
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Direction (Q. Nos. 1-16) This section contains 16 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE option is correct.
Q. Octet rule is not followed in

A
CCI₄, N₂O₄, N₂O₅

B
BF₃, BeCI₃, NO₂

C
NaCI, MgCI₂, MgO

D
PCI₃, NH₃, H₂O

Solution
Question 2
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Select the correct statement about valence bond approach.

A
Each bond is formed by maximum overlap for its maximum stability

B
It represents localised electron model of bonding

C
Most of the electrons retain the same orbital locations as in separated atoms

D
All of the above are correct statements

Solution

(a) Hs maximum overlap - correct.

(c) Based on valence bond approach,-no prediction of change of structure due to (Ip-Ip), (Ip-bp) repulsions- correct.
Question 3
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In which of the following substances will hydrogen bond be strongest?

A
HCl

B
H₂O

C
HI

D
H₂S

Solution

HC1, HI and H₂S do not from H-bonds. Only H₂O forms hydrogen bonds. One H₂O molecule forms four H-bonding.
Question 4
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The common features among the species CN^-, CO and NO⁺ are

A
    Bond order three and isoelectronic

B
    Bond order three and weak field ligands

C
    Bond order two and acceptors

D
    Isoelectronic and weak field ligands

Solution

According to M.O theory,
CN⁻,CO and NO⁺ are isoelectronic (14e⁻) and bond order is 3.
Hence (a) is the correct answer.
Question 5
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The correct order of increasing bond length of
I. C— H
II. C— O
III. C— C
IV. C=C

A
I< II< III< IV

B
I< IV < II< III

C
Ill < IV < II < l

D
II < I < III < IV

Solution

H is smallest in size thus, (C— H)bond length is least.

Hence, bonding pair in (C— O) is nearer to nucleus decreasing bond length as compared to (C— C).
Thus, I < IV < II < III.
Question 6
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Ratio of σ and π-bonds is maximum in

A
tetracyanomethane

B
benzene

C
1-phenyl-prop-1-ene

D
enolic form of acetone

Solution
Question 7
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In which of the following cases, covalent bonds are cleaved?

A
Boiling of H₂O

B
Melting of KCN

C
Boiling of CF₄

D
Melting of SiO₂

Solution

(a) Boiling of H₂O Intermolecular H-bonds are cleaved.
Question 8
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The compound (NH₃, BF₃) can be easily separated into its individual components because

A
BF₃ is highly reactive

B
NH₃ is highly reactive

C
ESF₃ is a Lewis base

D
BF₃ and NH₃ are joined by weak coordinate bond and have independent existence

Solution

Thus, in (NH₃,BF₃), both have independent existence and thus, it can be separated.
Question 9
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In allene (C₃H₄), the type(s) of hybridisation of the carbon atoms is (are)

A
sp andsp²

B
sp and sp³

C
Only sp²

D
sp² and sp³

Solution

In allene, hybridization of C-atoms:
The two C-atoms H²-C=C=C-H² at terminal have sp³ while the middle have sp .
Question 10
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The bond between two identical nonmetal atoms has a pair of electrons:

A
  unequally shared between the two.

B
  transferred fully from one atom to another

C
  with identical spins

D
  equally shared between them

Solution

When bond is formed between identical atoms, it is a perfectly covalent bond. Electrons will be shared equally among them.
Question 11
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Following is the dot structure of XY₄
Q. X and Y can be

A
Cl, F

B
I, Br

C
Xe, F

D
Si, Cl

Solution

There are three lone pairs on Y and four σ - bonds between X and Y.
Thus, octets of X and Y are complete.
X has four electrons in valence shell.
Y has seven electrons in valence shell.
Thus, X = Si (group 14), y = Cl (group 17)
Question 12
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Hyperconjugation involve overlap of the following orbitals

A
σ-σ

B
σ-π

C
p-p

D
π-π

Solution

Hyperconjugation takes place in allylic compounds in which alkyl group (having at least one H) is attached to bond .

σ-bond of (— CH₃) and π -bond of (ene) are involved, π-bond is formed due to overlapping of p-orbitals. Thus, σ-orbital and p-orbital are involved.
Question 13
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Malleability and ductility of metals can be accounted due to

A
the presence of electrostatic force

B
the crystalline structure in metal

C
the capacity of layers of metal ions to slide over the other

D
the interaction of electrons with metal ions in the lattice

Solution

In metallic bonds, the valence shell electrons are delocalised and shared between many atoms. These delocalised electrons allow the metal atoms to slide past one another without being subjected to strong repulsive forces. The malleability and ductility of metals is due to this sliding capacity of the delocalised electrons.
Question 14
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Polarity in a molecule and hence the dipole moment depends primarily on electronegativity of the constituent atoms and shape of a molecule. Which of the following has the highest dipole moment ?

A
CO₂

B
HI

C
H₂O

D
SO₂

Solution

H₂O will have highest dipole moment due to maximum difference in electronegativity of H and O atoms.
Question 15
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Which of the following is nonpolar?

A
CI₃CCH₃

B
PCI₅

C
CH₂CI₂

D
NH₃

Solution

(b) PCI₅ triangular bipyramidal, symmetrical, no lone pair.
Question 16
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The bond between B and C will be

A
Ionic

B
Covalent

C
Hydrogen

D
Coordinate

Solution

Both B and C are non-metals and therefore, bond formed between them will be covalent.
Question 17
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Direction (Q. Nos. 17) Choice the correct combination of elements and column I and coloumn II are given as option (a), (b), (c) and (d), out of which ONE option is correct.
Q.
Match the orbital overlap figures shown in Column I with the description given in Column II and select the correct answer using the code given below the columns.
[JEE Advanced 2014]

A
a

B
b

C
c

D
d

Solution

(i) (d-d) σ-bonding (ii) (p-dπ bonding)
(iii) (p-dπ anti-bonding) (iv) (ddσ anti-bonding)