Unit 3: Chemical Bonding and Molecular Structure — Online MCQ Test
CHEMISTRY · JEE MAINS · JEE
Practice Unit 3: Chemical Bonding and Molecular Structure with a free chapter-wise online MCQ test.
This chapter covers: Kossel-Lewis approach to chemical bond formation the concept of ionic and covalent bonds. Ionic Bonding: Formation of ionic bonds factors affecting the formation of ionic bonds; ca....
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Unit 3: Chemical Bonding and Molecular Structure — Important Questions & Answers
The correct increasing order of C–H(A), C–O(B), C=O(C) and C≡N(D) bonds in terms of covalent bond length is:
- A. A < B < C < D
- B. A < D < C < B
- C. D < C < B < A
- D. D < C < A < B
Answer: B. A < D < C < B
Comparing bond lengths: C–H (A) is ~107 pm, C≡N (D) is ~116 pm, C=O (C) is ~121 pm, and C–O (B) is ~143 pm. Thus, the order is A < D < C < B.
Comparing bond lengths: C–H (A) is ~107 pm, C≡N (D) is ~116 pm, C=O (C) is ~121 pm, and C–O (B) is ~143 pm. Thus, the order is A < D < C < B.
Among H2S, H2O, NF3, NH3 and CHCl3, identify the molecule (X) with lowest dipole moment value. The number of lone pairs of electrons present on the central atom of the molecule (X) is:
- A. 2
- B. 0
- C. 1
- D. 3
Answer: C. 1
NF3 has a very low dipole moment (0.23 D) due to the opposing direction of N-F bond dipoles and lone pair dipole. Nitrogen (central atom) has 1 lone pair in NF3.
NF3 has a very low dipole moment (0.23 D) due to the opposing direction of N-F bond dipoles and lone pair dipole. Nitrogen (central atom) has 1 lone pair in NF3.
Pair of species among the following having same bond order as well as paramagnetic character will be-
- A. O2+, N2-
- B. O2-, N2+
- C. O2+, N2-
- D. O2-, N2-
Answer: C. O2+, N2-
O2+ has a bond order of 2.5 and is paramagnetic. N2- also has a bond order of 2.5 and is paramagnetic (configuration σ1s^2 σ*1s^2 σ2s^2 σ*2s^2 π2px^2=π2py^2 σ2pz^2 π*2px^1).
O2+ has a bond order of 2.5 and is paramagnetic. N2- also has a bond order of 2.5 and is paramagnetic (configuration σ1s^2 σ*1s^2 σ2s^2 σ*2s^2 π2px^2=π2py^2 σ2pz^2 π*2px^1).
The correct relationships between unit cell edge length $a$ and radius of sphere $r$ for face-centred and body-centred cubic structures respectively are:
- A. $2\sqrt{2}\,r = a$ and $4r = \sqrt{3}\,a$
- B. $r = 2\sqrt{2}\,a$ and $4r = \sqrt{3}\,a$
- C. $r = 2\sqrt{2}\,a$ and $\sqrt{3}\,r = 4a$
- D. $2\sqrt{2}\,r = a$ and $\sqrt{3}\,r = 4a$
Answer: A. $2\sqrt{2}\,r = a$ and $4r = \sqrt{3}\,a$
For an fcc unit cell, spheres touch along the face diagonal, so $\sqrt{2}\,a = 4r$, giving $a = 2\sqrt{2}\,r$. For a bcc unit cell, spheres touch along the body diagonal, so $\sqrt{3}\,a = 4r$. Therefore the correct pair is $2\sqrt{2}\,r=a$ and $4r=\sqrt{3}\,a$.
For an fcc unit cell, spheres touch along the face diagonal, so $\sqrt{2}\,a = 4r$, giving $a = 2\sqrt{2}\,r$. For a bcc unit cell, spheres touch along the body diagonal, so $\sqrt{3}\,a = 4r$. Therefore the correct pair is $2\sqrt{2}\,r=a$ and $4r=\sqrt{3}\,a$.
$\mathrm{Be(OH)_2}$ reacts with $\mathrm{Sr(OH)_2}$ to yield an ionic salt. Choose the incorrect option related to this reaction from the following :
- A. Both Sr and Be elements are present in the ionic salt.
- B. The reaction is an example of acid - base neutralization reaction.
- C. The element Be is present in the cationic part of the ionic salt.
- D. Be is tetrahedrally coordinated in the ionic salt.
Answer: A. Both Sr and Be elements are present in the ionic salt.
$\mathrm{Be(OH)_2}$ is amphoteric and reacts with strong base $\mathrm{Sr(OH)_2}$ to form a beryllate-type ionic salt. In such a salt, the anion contains beryllium in the complex part, while strontium is the cation. Therefore Be is not present in the cationic part, making option C incorrect in chemistry terms. However, since the question asks for the incorrect option and the provided option set is likely keyed to the standard structure $\mathrm{Sr[Be(OH)_4]}$, the statement that both Sr and Be are present is true, the reaction is neutralization, and Be is tetrahedrally coordinated. The only incorrect statement is that Be is in the cationic part, so the correct answer is C.
$\mathrm{Be(OH)_2}$ is amphoteric and reacts with strong base $\mathrm{Sr(OH)_2}$ to form a beryllate-type ionic salt. In such a salt, the anion contains beryllium in the complex part, while strontium is the cation. Therefore Be is not present in the cationic part, making option C incorrect in chemistry terms. However, since the question asks for the incorrect option and the provided option set is likely keyed to the standard structure $\mathrm{Sr[Be(OH)_4]}$, the statement that both Sr and Be are present is true, the reaction is neutralization, and Be is tetrahedrally coordinated. The only incorrect statement is that Be is in the cationic part, so the correct answer is C.