Chemistry · Chemical bonding and molecular structure · chapter 4 of 9

Class 11 Chemistry Chapter 4: Chemical Bonding and Molecular Structure MCQs with answers

10 multiple-choice questions on Chemical Bonding and Molecular Structure, each with its answer and the reason, written from the NCERT chapter. For school exams on the CBSE pattern, JEE Main and NEET, which are set from the same NCERT chapter. Open a question's solution to check your working; then take the free test to find your next chapter.

Question 1
The total number of valence electrons to be placed in the Lewis structure of the sulphate ion, SO₄²⁻, is
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Answer: (d) 32

S gives 6 and each of the four O atoms gives 6: 6 + 24 = 30; the two negative charges add two electrons, so 32.

Question 2
In which of the following molecules does the central atom have fewer than eight electrons around it?
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Answer: (c) AlCl₃

Al has only three valence electrons; after three shared pairs with Cl it holds six electrons, an incomplete octet. P in PCl₃, S in H₂S and Si in SiF₄ each reach an octet.

Question 3
In which species are there twelve electrons around the central atom?
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Answer: (a) H₂SO₄

In H₂SO₄ sulphur, a third-period element, has an expanded octet of 12 electrons. S in SCl₂, N in NF₃ and C in CCl₄ have eight.

Question 4
The number of σ bonds and π bonds in a molecule of acrylonitrile, CH₂=CH–C≡N, is
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Answer: (c) 6 σ and 3 π

Every bond between two atoms has one σ bond: three C–H, C=C, C–C and C≡N give 6 σ. The double bond adds one π and the triple bond two π: 3 π.

Question 5
Which pair of elements will form an ionic bond most readily?
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Answer: (d) Cs and F

An ionic bond forms most easily between an element of low ionization enthalpy and one of high negative electron gain enthalpy: caesium and fluorine. The other pairs are non-metals that share electrons.

Question 6
Using the covalent radii C = 77 pm and Cl = 99 pm (single bonds), the expected C–Cl bond length in CH₃Cl is
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Answer: (b) 176 pm

Bond length R = rC + rCl = 77 + 99 = 176 pm. 88 pm is half the sum; 154 pm is the C–C length; 198 pm is twice the Cl radius.

Question 7
In crystals of an element X₂, the X–X bond length within a molecule is 210 pm and the shortest distance between X nuclei of neighbouring molecules is 370 pm. The covalent radius and the van der Waals radius of X are respectively
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Answer: (a) 105 pm and 185 pm

Covalent radius = half the bonded distance = 210/2 = 105 pm. van der Waals radius = half the non-bonded distance between atoms of separate molecules = 370/2 = 185 pm.

Question 8
The enthalpies needed to remove the first, second and third hydrogen atoms from NH₃(g) are 449, 385 and 339 kJ mol⁻¹. The average N–H bond enthalpy in NH₃ is
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Answer: (b) 391 kJ mol⁻¹

The three N–H bonds break with different enthalpies because the chemical environment changes after each step, so the mean is used: (449 + 385 + 339)/3 = 1173/3 = 391 kJ mol⁻¹.

Question 9
Which species has the same bond order as F₂?
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Answer: (c) O₂²⁻

F₂ and O₂²⁻ are isoelectronic and have bond order 1. O₂ has bond order 2; N₂ and NO⁺ have bond order 3.

Question 10
Experiments show that the two nitrogen–oxygen bonds in the nitrite ion, NO₂⁻, are equal in length. Given the average lengths N–O = 136 pm and N=O = 122 pm, the length of each bond in the ion is expected to be
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Answer: (b) between 122 pm and 136 pm

The ion is a resonance hybrid of [O=N–O]⁻ and [O–N=O]⁻; each bond is intermediate between a single and a double bond, as the O–O bonds in O₃ are (128 pm, between 121 and 148 pm).

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