CBSE Class 11 Chemistry Chemical Bonding And Molecular Structure Worksheet Set 07

Worksheet Questions with Answers

Question 1. Define the octet rule. Write its significance and its limitations.
Answer:
Octet rule: atoms tend to adjust their valence electrons by transfer or sharing so that each has eight electrons (an octet) in its outermost shell, like the nearest noble gas. Hydrogen, lithium and beryllium aim at a duplet.
Significance: it explains why atoms form ionic and covalent bonds and why the usual valencies of the main-group elements are what they are.
Limitations:
(a) Incomplete octet: in LiCl, BeH₂ and BCl₃ the central atom has fewer than 8 electrons.
(b) Odd-electron molecules such as NO and NO₂ cannot give an octet to every atom.
(c) Expanded octet: elements of period 3 and beyond hold more than 8 electrons, as in PF₅ and SF₆.
(d) It is based on the inertness of noble gases, but Xe and Kr form compounds such as XeF₂ and KrF₂.
(e) It cannot predict the shape of a molecule or its relative stability.
In simple words: Atoms want eight valence electrons, but many molecules are exceptions.

Teacher's Note:
1. Write the five limitations as short points with one example each.
2. Hydrogen follows the duplet rule, not the octet rule.

Question 2. Write the Lewis dot symbols of Mg, Na, B, O, N and Br.
Answer:
The Lewis symbol shows only the valence electrons as dots around the symbol, one on each side first and then in pairs.
Na: 1 valence electron, so 1 dot (unpaired).
Mg: 2 valence electrons (2s²), so 2 dots.
B: 3 valence electrons (2s² 2p¹), so 3 dots, one pair and one single dot.
O: 6 valence electrons (2s² 2p⁴), so 6 dots: two pairs and two single dots.
N: 5 valence electrons (2s² 2p³), so 5 dots: one pair and three single dots.
Br: 7 valence electrons (4s² 4p⁵), so 7 dots: three pairs and one single dot.
In simple words: The number of dots is the number of valence electrons of the atom.

Teacher's Note:
1. For main-group elements the number of valence electrons is the group number (1, 2) or group number − 10 (13 to 17).
2. Do not draw the inner-shell electrons.

Question 3. Use Lewis symbols to show the electron transfer between (a) Ca and O, and (b) Al and N.
Answer:
(a) Ca (2, 8, 8, 2) and O (2, 6). Ca has two electrons more than argon and O needs two electrons to complete its octet. Ca transfers its two valence electrons to O: Ca → Ca²⁺ + 2e⁻ and O + 2e⁻ → O²⁻. The ions combine to form CaO (Ca²⁺ O²⁻).
(b) Al (2, 8, 3) and N (2, 5). Al has three electrons more than neon and N needs three electrons to complete its octet. Al transfers three electrons to N: Al → Al³⁺ + 3e⁻ and N + 3e⁻ → N³⁻. The ions combine to form AlN (Al³⁺ N³⁻).
In simple words: The metal gives away its extra electrons and the non-metal takes exactly the number it needs.

Teacher's Note:
1. The number of electrons given must equal the number received.
2. The compound has a 1 : 1 ratio here because the charges are equal and opposite.

Question 4. Write the Lewis structures of H₂S, SiCl₄, BeF₂, H₃O⁺ and HCOOH.
Answer:
H₂S: 2 + 6 = 8 valence electrons. S is the central atom with two S–H bonds and two lone pairs: H–S–H.
SiCl₄: 4 + 28 = 32 electrons. Si is bonded to four Cl atoms by single bonds, and each Cl has three lone pairs.
BeF₂: 2 + 14 = 16 electrons. F–Be–F, with three lone pairs on each F. Be has only 4 electrons around it (incomplete octet).
H₃O⁺: 3 + 6 − 1 = 8 electrons. O is bonded to three H atoms and has one lone pair; one of the three O–H bonds can be shown as a coordinate bond from O to H⁺.
HCOOH: 1 + 4 + 12 + 1 = 18 electrons. H–C(=O)–O–H: C is bonded to H, to =O (double bond) and to –OH; the O atoms carry two and two lone pairs.
In simple words: Count the valence electrons, join the atoms by single bonds and place the rest as lone pairs.

Teacher's Note:
1. The least electronegative atom (not H) is the central atom.
2. For a cation subtract one electron for each positive charge.

Question 5. Draw the Lewis structure of nitric acid, HNO₃, and calculate the formal charge on every atom.
Answer:
Valence electrons: 1 + 5 + 18 = 24. The skeleton is H–O–N with two more O atoms on N. N has one N=O double bond, one N–O⁻ bond and the O–H bond.
H: 1 − 0 − 2/2 = 0.
O of the O–H group: V = 6, L = 4, S = 4, so 6 − 4 − 2 = 0.
N: V = 5, L = 0, S = 8, so 5 − 0 − 4 = +1.
Double-bonded O: V = 6, L = 4, S = 4, so 6 − 4 − 2 = 0.
Single-bonded O⁻: V = 6, L = 6, S = 2, so 6 − 6 − 1 = −1.
The total is 0 + 0 + 1 + 0 − 1 = 0, the charge on the neutral molecule.
In simple words: Work out V − L − S/2 for each atom; the answers must add up to the overall charge.

Teacher's Note:
1. Write the V, L and S values in a small table.
2. The +1 on N and −1 on O are formal charges, not real charges.

Question 6. Define lattice enthalpy. Which compound in each pair has the larger lattice enthalpy: (a) LiF or CsI, (b) CaO or KF? Give reasons.
Answer:
Lattice enthalpy is the energy required to completely separate one mole of a solid ionic compound into its gaseous ions, e.g. NaCl(s) → Na⁺(g) + Cl⁻(g), ΔH = +788 kJ mol⁻¹. It is a positive quantity.
(a) LiF. Both have charges of 1+ and 1−, but Li⁺ and F⁻ are the smallest ions in the pair. A short distance between the ions gives a strong attraction, while Cs⁺ and I⁻ are the largest ions.
(b) CaO. Ca²⁺ and O²⁻ carry double charges, while K⁺ and F⁻ carry single charges. The attraction between 2+ and 2− ions is much larger, and the effect of charge outweighs the small difference in size.
In simple words: Strong attraction in the lattice needs small ions and large charges.

Teacher's Note:
1. Compare the charges first and the sizes second.
2. Lattice enthalpy is not the same as the enthalpy of formation.

Question 7. Write the resonance structures of the nitrate ion, NO₃⁻, and state the N–O bond order and the charge on each oxygen atom.
Answer:
NO₃⁻ has 5 + 18 + 1 = 24 valence electrons. In each canonical structure N is bonded to three O atoms: one N=O double bond and two N–O⁻ single bonds. N has a formal charge of +1 and each singly bonded O has −1.
There are three equivalent structures, differing only in which O atom carries the double bond. The actual ion is a resonance hybrid of the three.
Total number of N–O bonds in a structure = 4 (one double + two single). Spread over three positions: bond order = 4/3. All three N–O bonds are equal in length.
The total charge of −1 comes from two O atoms with −1 in each structure, so on average each O carries −2/3 of a charge (and N carries +1).
In simple words: The nitrate ion is a blend of three equal structures, so all N–O bonds are the same.

Teacher's Note:
1. Resonance structures differ only in the position of electrons.
2. Compare with CO₃²⁻, which has the same three equivalent structures.

Question 8. Can the structures (HO)₂P(=O)H and P(OH)₃ be taken as canonical forms of the resonance hybrid of H₃PO₃? Give a reason.
Answer:
No. Canonical forms of a resonance hybrid may differ only in the positions of electrons, and the positions of all the nuclei must be the same.
In these two structures a hydrogen atom has moved from P to O (the H is attached to P in one structure and to O in the other). The atoms have changed position, so the two are different compounds in equilibrium (tautomers), not resonance structures.
In simple words: Resonance moves electrons only, never atoms.

Teacher's Note:
1. Compare the position of every atom in the two structures.
2. If any atom has moved, it is not resonance.

Question 9. Although both CO₂ and H₂O are triatomic molecules, H₂O is bent while CO₂ is linear. Explain this on the basis of dipole moment.
The bond dipoles are shown here.
Answer:
CO₂ has a dipole moment of zero. Each C=O bond is polar, so zero is possible only if the two equal bond dipoles are opposite to each other, i.e. the molecule is linear (O=C=O).
H₂O has a dipole moment of 1.85 D, so its two O–H bond dipoles do not cancel. This is possible only if the two bonds are inclined at an angle, so H₂O is bent (104.5°).
In CO₂ the bond dipoles cancel (resultant = 0). In H₂O they add up (resultant = 1.85 D).
In simple words: Zero dipole moment means a symmetrical, linear shape, and a measurable dipole moment means a bent one.

Teacher's Note:
1. Draw the bond dipole arrows to see whether they cancel.
2. Dipole moment is a vector quantity.

Question 10. The dipole moment of HF is 1.78 D and its bond length is 91 pm. Calculate the partial charge on each atom as a fraction of the charge of an electron. (1 D = 3.336 × 10⁻³⁰ C m; e = 1.602 × 10⁻¹⁹ C)
Answer:
μ = 1.78 × 3.336 × 10⁻³⁰ = 5.938 × 10⁻³⁰ C m.
Q = μ ÷ r = 5.938 × 10⁻³⁰ ÷ 91 × 10⁻¹² = 6.53 × 10⁻²⁰ C.
Fraction of an electronic charge = 6.53 × 10⁻²⁰ ÷ 1.602 × 10⁻¹⁹ = 0.41.
So H carries about +0.41e and F about −0.41e. If HF were completely ionic the dipole moment would be 1.602 × 10⁻¹⁹ × 91 × 10⁻¹² ÷ 3.336 × 10⁻³⁰ ≈ 4.4 D.
In simple words: HF has about 41% of a full charge separation, more than HCl, because F is more electronegative.

Teacher's Note:
1. Convert pm to metres and D to C m before dividing.
2. Compare with HCl (0.17e): the larger electronegativity difference gives the larger charge.

Question 11. What is the significance of dipole moment? How does it help to predict the shape of a molecule?
Answer:
Dipole moment (μ = Q × r) measures the polarity of a bond or a molecule.
Uses: (i) to distinguish polar molecules (μ ≠ 0) from non-polar molecules (μ = 0); (ii) to find the shape of a molecule, e.g. CO₂ (μ = 0) is linear, while H₂O (μ = 1.85 D) is bent and BF₃ (μ = 0) is planar; (iii) to estimate the extent of charge separation (ionic character) of a bond.
Symmetrical molecules such as CO₂, BF₃, CH₄ and CCl₄ have polar bonds but μ = 0 because the bond dipoles cancel.
In simple words: A non-zero dipole moment means the molecule is not symmetrical.

Teacher's Note:
1. Give one example of a polar and a non-polar molecule.
2. Dipole moment is expressed in debye (D).

Question 12. Define bond length, bond angle, bond enthalpy and bond order. How does bond order affect bond length and bond enthalpy?
Answer:
Bond length: the equilibrium distance between the nuclei of two bonded atoms.
Bond angle: the angle between the orbitals containing bonding electron pairs around the central atom.
Bond enthalpy: the energy needed to break one mole of bonds of a particular type in the gaseous state (kJ mol⁻¹).
Bond order: the number of bonds between two atoms in a Lewis structure.
As bond order increases, the atoms come closer, so bond length decreases, and the bond is harder to break, so bond enthalpy increases. Example: C–C 154 pm, C=C 133 pm, C≡C 120 pm; N≡N has a bond enthalpy of 941.4 kJ mol⁻¹ against 498 for O=O and 435.8 for H–H.
In simple words: More bonds between two atoms pull them closer and make the bond stronger.

Teacher's Note:
1. Quote the units: pm for length and kJ mol⁻¹ for enthalpy.
2. Bond length is inversely related to bond order.

Question 13. Explain with the help of the electronic configurations how MgCl₂ is formed from magnesium and chlorine.
Answer:
Mg (2, 8, 2) has two valence electrons and Cl (2, 8, 7) needs one electron to complete its octet.
Mg loses two electrons: Mg → Mg²⁺ + 2e⁻ (2, 8, like neon). Each of two Cl atoms gains one electron: Cl + e⁻ → Cl⁻ (2, 8, 8, like argon).
The Mg²⁺ ion and two Cl⁻ ions attract each other by electrostatic force to form the ionic compound MgCl₂, i.e. Mg²⁺ (Cl⁻)₂.
In simple words: One magnesium atom gives one electron to each of two chlorine atoms.

Teacher's Note:
1. The electrovalency of Mg is 2 and that of Cl is 1, which gives the formula MgCl₂.
2. An ionic bond is the electrostatic attraction between the ions.