Chemical Bonding and Molecular Structure Test 6

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Chemical Bonding and Molecular Structure Test 6

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Weekly Quiz Competition

Question 1
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Direction (Q. Nos. 1-15) This section contains 15 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE option is correct.
Q. For the
I. benzene (C₆H₆) and
II. borazine (B₃N₃H₆₎
Select the correct statement.

A
C—H bond length in I is identical with N—H and B—H bond lengths

B
The nature of double bond is perfectly identical in both

C
Both the molecules are planar

D
I is non-polar and II is polar

Solution

(C—H) bond lengths are different from (B—H) and (N—H) bond lengths. Both are planar.
Question 2
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Which is not the correct structure of ion?

A

B

C

D

Solution

In structure (a), octet of N-atom is not complete. In all the other structures, octets of N and all the O-atoms are complete.
Question 3
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Which of the following angle corresponds to sp₂ hybridisation?

A
90°

B
120°

C
180°

D
109°

Solution

sp² hybridisation gives three sp² hybrid orbitals which are planar triangular forming an angle of 120° with each other.
The electronic configurations of three elements A, B and C are given below.
Answer the questions from 14 to 17 on the basis of these configurations.
A ls²2s²2p⁶
B ls²2s²2p⁶3s²3p³
C ls²2s²2p⁶3s²3p^s
Question 4
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Which of the following species has tetrahedral geometry?

A
NH₂^–

B
BH₄^–

C
CO₃^2–

D
H₃O⁺

Solution

pyramidal.
Question 5
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Which of the following structures is more acceptable for HNO₃?

A

B

C
Both equally

D
None of these

Solution

Structure is decided based on formal charge.

where, v = valence electrons
s = shared electrons
u = unshared electrons

Structure with at least one neutral atom is favoured.
Question 6
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Consider the following pairs

Q. More stable species in each pair is

A
a

B
b

C
c

D
d

Solution

Greater the resonating structures, greater the stability.
I : (Y)with negative charge has greater delocalisation of n-electrons than HNO₃(X).
Question 7
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Azide ion exhibits an (N—N) bond order of 2 and may be represented by resonance structures I, II and III given below :
Select the correct statement(s) about more contributions,

A
II and III

B
I and III

C
I and II

D
Equal of each

Solution

Structure with least formal charge on each atom, makes greater contributions. Also, structures with at least one neutral atom is also favoured II and III identical.
Question 8
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C—Cl bond in (vinyl chloride) is stabilised in the same way as in

A
benzyl chloride

B
benzoyl chloride

C
chlorobenzene

D
allyl chloride

Solution

Due to delocalisation of π-electrons, (C—Cl) bond is stable and it does not show S_N reactions; Cl directly attached (C=C) bond, i.e. vinyl group.
Question 9
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NO₂ can be represented as
Q. Formal charge on each oxygen atom is

A
-2

B
-1

C
0

D
+1

Solution

Formal charge (F)
where, v = valence electrons = 6
s = shared electrons = 4
u = unshared electron = 4
Question 10
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Which representation for the Lewis structure of HNO₃ is correct?

A

B

C

D

Solution

Only structure a is able to show the valence electrons along with charge distributions. So, ith the Lewis structure of HNO₃ .
Question 11
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By S_N 1 reaction allyl chloride changes to allyl alcohol as shown
Select the correct product isotope of carbon)

A

B

C
Both (a) and (b) are correct

D
None of these is correct

Solution

S_N1 means unimolecular (1) nucleophilic (N) substitution (S). Rate is independent of molar concentration of H2_O but is dependend on molar concentration of

(¹⁴C at changed position due to resonance).
Question 12
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Stability of (C—Cl) bond in chlorobenzene is sim ilar to (C—Cl) bond in

A
benzyl chloride

B
vinyl chloride

C
allyl chloride

D
All of these

Solution

(C—Cl) bond has (C=Cl) double bond character, thus it is stable.

Benzyl carbocation is stable due to resonance, hence (C—Cl) bond is cleaved easily.
Question 13
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Consider the following compounds
I. Vinyl chloride
II. B₃N₃H₆
Delocalisation takes place

A
I, II and III

B
I and II

C
II and III

D
I and III

Solution
Question 14
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Hydrogen bonds are formed in many compounds e.g., H₂O, HF, NH₃. The boiling point of such compounds depends to a large extent on the strength of hydrogen bond and the number of hydrogen bonds. The correct decreasing order of the boiling points of above compounds is :

A
HF > H₂O > NH₃

B
H₂O > HF > NH₃

C
NH₃ > HF > H₂O

D
NH₃ > H₂O > HF

Solution

Strength of H-bonding depends on the electronegativity of the atom which follows the order: F > O > N .
Strength of H-bond is in the order:
H……. F > H…….. O > H…….. N
But each H₂O molecule is linked to 4 other H₂O molecules through H-bonds whereas each HF molecule is linked only to two other HF molecules.
Hence, correct decreasing order of the boiling points is H_zO > HF > NH₃.
Question 15
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Out of the following two structures of N₂O, which is more accurate?

A
I

B
II

C
Both equally

D
None of the above

Solution

Plan Oxygen is more electronegative than oxygen, hence the structure with a negative formal charge on oxygen atom is probably lower in energy than the structure that has a negative formal charge on N-atom is more accurate.
Thus (I) is more accurate than (II).