Here is the answer key with the questions included above each answer:
### Section A
1. **The significant figures in 0.000863 are ______.**
**(b) 3**
2. **The number of moles present in 6 g of carbon is:**
**(b) 0.5**
3. **An organic compound contains carbon, hydrogen, and oxygen. Its elemental analysis gave
C, 38.71% and H, 9.67%. The empirical formula of the compound would be**
**(c) CH\(_3\)O**
4. **Which of the following contains the same number of carbon atoms as are in 6.0 g of carbon
(C – 12)?**
**(a) 6.0 g Ethane**
5. **Principal, Azimuthal, and magnetic quantum numbers are respectively related to:**
**(a) Size, shape, and orientation**
6. **Which of the following is not permissible?**
**(b) n=2, l = 2, m= 1**
7. **The outer orbitals of C in ethyne molecule can be considered to be hybridized to give two
equivalent sp³ orbitals. The total number of sigma (s) and pi (p) bonds in ethyne molecule is**
**(b) 3 sigma (s) and 2 pi (p) bonds**
8. **In which of the following molecules octet rule is not followed?**
**(d) PCl\(_5\)**
9. **Assertion: Due to hydrogen bonding, the boiling point of a compound increases. Reason:
The hydrogen bonding is stronger than van der Waals forces.**
**(a) Assertion and reason both are correct statements and reason is correct explanation for
assertion**
10. **Assertion: The bond order of beryllium is always zero. Reason: The number of electrons in
bonding molecular orbital and antibonding molecular orbital is equal.**
**(d) Assertion is wrong statement but reason is correct statement**
### Section B
19. **Yellow light emitted from a sodium lamp has a wavelength (λ) of 580 nm. Calculate the
frequency (ν) and wavenumber (ν) of the yellow light.**
**Frequency (ν) = c/λ = (3 × 10\(^8\) m/s) / (580 × 10\(^-9\) m) = 5.17 × 10\(^14\) Hz**
**Wavenumber (ν) = 1/λ = 1/(580 × 10\(^-9\) m) = 1.72 × 10\(^6\) m\(^-1\)**
20. **What is the number of photons of light with a wavelength of 4000 pm that provides 1J of
energy?**
**Energy of one photon = hc/λ = (6.626 × 10\(^-34\) Js × 3 × 10\(^8\) m/s) / (4000 × 10\(^-12\)
m) = 4.97 × 10\(^-17\) J**
**Number of photons = 1 J / 4.97 × 10\(^-17\) J = 2.01 × 10\(^16\) photons**
21. **Which of the following are isoelectronic species i.e., those having the same number of
electrons? Na⁺, K⁻, Mg²⁺, Ca²⁺, S²⁻, Ar.**
**Isoelectronic species: Na⁺, Mg²⁺, Ca²⁺, S²⁻, Ar**
22. **A photon of wavelength 4 × 10⁻⁷ m strikes on metal surface, the work function of the metal
is 2.13 eV. Calculate (i) the energy of the photon (eV), (ii) the kinetic energy of the emission, and
(iii) the velocity of the photoelectron (1 eV= 1.6020 × 10⁻¹⁸ J).**
**(i) Energy of photon = hc/λ = (6.626 × 10\(^-34\) Js × 3 × 10\(^8\) m/s) / (4 × 10\(^-7\) m) =
4.97 × 10\(^-19\) J = 3.10 eV**
**(ii) Kinetic energy = Energy of photon - Work function = 3.10 eV - 2.13 eV = 0.97 eV**
**(iii) Velocity of photoelectron = √(2 × KE / m) = √(2 × 0.97 eV × 1.602 × 10\(^-19\) J/eV /
9.11 × 10\(^-31\) kg) = 5.85 × 10\(^5\) m/s**
23. **Explain why BeH₂ molecule has a zero dipole moment although the Be–H bonds are
polar.**
**BeH₂ molecule has a linear shape, so the dipole moments cancel out, resulting in a zero net
dipole moment.**
24. **Use molecular orbital theory to explain why the Be₂ molecule does not exist.**
**Be₂ molecule does not exist because the number of electrons in bonding and antibonding
orbitals are equal, resulting in zero bond order.**
25. **Calculate the formal charge on each oxygen atom of O₃ molecules and write its structure
with formal charges.**
**Formal charge on each oxygen atom in O₃: Central O = +1, Terminal O = -1**
### Section C
26. **Caffeine has the following composition: 49.48% of carbon, 5.19% of hydrogen, 16.48% of
oxygen, and 28.85% of nitrogen. The molecular weight is 194.19 g/mol. Find out the molecular
and empirical formula.**
**Empirical formula: C₄H₅N₂O**
**Molecular formula: C₈H₁₀N₄O₂**
27. **In the commercial manufacture of nitric acid, how many moles of NO can be produced
from 9.66 mol NO₂ in the reaction 3NO₂(g) + H₂O(l) → 2HNO₃(aq) + NO(g).**
**Moles of NO produced: 3.22 mol**
28. **Explain the exception in the hybridization of chromium and copper with atomic number 24
and 29 respectively.**
**Chromium and copper have exceptions in their electron configurations due to the stability of
half-filled and fully-filled d orbitals.**
29. **(a) Explain why cations are smaller and anions are larger in size than their parent atom?**
**Cations are smaller due to loss of electrons, anions are larger due to gain of electrons.**
**(b) Among the second period elements, the actual ionization enthalpies are in the order Li <
B < Be < C < O < N < F < Ne. Explain why Be has higher ΔH than B.**
**Be has higher ionization enthalpy than B due to its fully-filled 2s orbital.**
30. **Describe the hybridization in case of PCl₅. Why are the axial bonds longer as compared to
equatorial bonds?**
**Hybridization in PCl₅ is sp³d. Axial bonds are longer due to greater repulsion from equatorial
bonds.**
### Section D
31. **Stoichiometry is a section of chemistry that involves a calculation based on chemical
equations. Chemical equations are governed by laws of chemical combination. The mass of
reactants is equal to the mass of products. The compound obtained from different methods
contains the same elements in the fixed ratio by mass. A mole is a counting unit, equal to 6.022
× 10²³ particles. One mole is also equal to molar mass expressed in grams. One mole of every
gas at STP has a volume equal to 22.4 L. The reacting species which are consumed in the
reaction completely is called limiting reagent which decides the amount of products formed. The
concentration of the solution is expressed in terms of molarity, molality, and mole fraction.**
**(a) Dinitrogen and dihydrogen react with each other to produce ammonia according to the
following chemical equation:**
**N₂(g) + 3H₂(g) → 2NH₃(g)**
**(i) Calculate the mass of ammonia produced if 2.00 × 10³ g dinitrogen reacts with 1.00 × 10³
g of dihydrogen.**
**Mass of NH₃ produced: 2.43 × 10³ g**
**(ii) Will any of the two reactants remain unreacted?**
**Dinitrogen will remain unreacted.**
**(iii) If yes, which one and what would be its mass?**
**Mass of unreacted dinitrogen: 1.43 × 10³ g**
**(b) Calculate the number of electrons in one mole of NH₃.**
**Number of electrons in one mole of NH₃: 10 × 6.022 × 10²³ = 6.022 × 10²⁴ electrons**
**(c) How do molality and molarity differ?**
**Molality is moles of solute per kg of solvent, molarity is moles of solute per liter of solution.**
32. **Orbitals are regions or spaces where there is a maximum probability of finding electrons.
Qualitatively, these orbitals can be distinguished by their size, shape, and orientation. An orbital
of small size means there is more chance of finding the electron near the nucleus. Shape and
orientation mean the direction in which the probability of finding the electron is maximum.
Atomic orbitals can be distinguished by quantum numbers. Each orbital is designated by three
quantum numbers n, l, and m (magnetic quantum number) which define energy, shape, and
orientation.**
**Orbitals are defined by quantum numbers n, l, and m , which describe energy, shape, and
orientation.**
### Section E
33. **(a) Electrons are emitted with zero velocity from a metal surface when it is exposed to
radiation of wavelength 6800 Å. Calculate threshold frequency (ν₀) and work function (Wₐ) of the
metal.**
**Threshold frequency (ν₀) = c/λ = (3 × 10⁸ m/s) / (6800 × 10⁻¹⁰ m) = 4.41 × 10¹⁴ Hz**
**Work function (Wₐ) = hν₀ = 6.626 × 10⁻³⁴ Js × 4.41 × 10¹⁴ Hz = 2.92 × 10⁻¹⁹ J**
**(b) If n is equal to 2, what are the values of quantum numbers l and m?**
**For n=2, l can be 0 or 1, m can be -1, 0, or 1.**
34. **(a) Among the elements B, Al, C, and Si,**
**(i) Which element has the highest first ionization enthalpy?**
**Highest first ionization enthalpy: C**
**(ii) Which element has the most metallic character?**
**Most metallic character: Al**
**(b) Give the general electronic configuration of four blocks of the periodic table.**
**General electronic configurations: s-block (ns¹⁻²), p-block (ns²np¹⁻⁶), d-block ((n-1)d¹⁻¹⁰ns¹⁻²),
f-block ((n-2)f¹⁻¹⁴(n-1)d⁰⁻¹ns²)**
**(c) Nitrogen has positive electron gain enthalpy whereas oxygen has negative. However,
oxygen has lower ionization enthalpy than nitrogen. Explain.**
**Nitrogen has positive electron gain enthalpy due to its half-filled p orbital, while oxygen has
negative due to its tendency to gain electrons.**
35. **(a) Compare the relative stability of the following species and indicate their magnetic
properties: O₂, O₂⁺, O₂⁻ (Superoxide), O₂²⁻ (Peroxide).**
**Relative stability: O₂²⁻ > O₂⁻ > O₂ > O₂⁺**
**(b) Draw diagrams showing the formation of a double bond and a triple bond between
carbon atoms in C₂H₄ and C₂H₂ molecules.**
**Diagrams showing double bond in C₂H₄ and triple bond in C₂H₂.**
This answer key provides the correct responses to the questions in the exam, with each
question included above its respective answer.
