Here are the answers to the Class 11 CBSE Chemistry ques on paper shown in the images, with
detailed explana ons for key sec ons:
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### **SECTION A [1x8=8]**
1. **Write all the four quantum numbers for the last electron of potassium atom.**
- Answer: n = 4, l = 0, m = 0, s = +1/2
Explana on: Potassium (K, Z=19) has the electron configura on [Ar] 4s¹. The last electron occupies
the 4s orbital.
2. **Calculate the number of moles of carbon and hydrogen atoms in 3 moles of methane (CH₄).**
- Answer: C = 3 moles, H = 12 moles
Explana on: CH₄ contains 1 C atom and 4 H atoms. So, 3 moles of CH₄ will have 3 moles of C and 4
x 3 = 12 moles of H.
3. **Draw the shape of the following molecules according to VSEPR theory: SiF₄, NH₃, PCl₅.**
- Answer:
- SiF₄: Tetrahedral
- NH₃: Trigonal pyramidal
- PCl₅: Trigonal bipyramidal
Explana on: These shapes are based on the steric number and lone pairs of electrons around the
central atom.
4. **Write the IUPAC name and symbol for the element with atomic number 120.**
- Answer: IUPAC Name: Unbinilium (Ubn)
Explana on: According to IUPAC, element 120 is temporarily named Unbinilium.
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### **SECTION B [2x10=20]**
5. **Calculate the number of carbon atoms in 3.42 g of C₁₂H₂₂O₁₁.**
- Answer: \( 7.15 \ mes 10^{22} \) carbon atoms
Explana on: The molar mass of C₁₂H₂₂O₁₁ (sucrose) is 342 g/mol. For 3.42 g, we calculate the
moles and then mul ply by Avogadro's number to get the number of carbon atoms.
\( \frac{3.42}{342} = 0.01 \, \text{mol} \).
Number of atoms = \( 0.01 \ mes 6.022 \ mes 10^{23} \ mes 12 = 7.15 \ mes 10^{22} \).
6. **How many significant figures are there in the following:**
(a) 0.00025
(b) 6.022 × 10¹³
- Answer:
- (a) 2 significant figures (because leading zeros do not count).
- (b) 4 significant figures (since all digits in scien fic nota on are significant).
7. **What would be the IUPAC name and symbol for the element with atomic number 127?**
- Answer: Unbisep um (Ubs)
Explana on: Following IUPAC rules for naming elements, "Un" stands for 1, "bi" for 2, and "sept"
for 7.
8. **Which of the following has the largest bond angle in the pairs?**
(a) NH₃, PH₃
(b) BeCl₂, BBr₃
- Answer:
- (a) NH₃ (because of higher electronega vity of nitrogen leading to more lone pair repulsion)
- (b) BBr₃ (trigonal planar shape leads to 120° bond angle, while BeCl₂ is linear with 180° bond
angle).
9. **A microscope using suitable photons is employed to locate an electron with an uncertainty in
posi on of 0.1Å. What is the uncertainty involved in the measurement of its velocity?**
- Answer:
Use Heisenberg's uncertainty principle:
\( \Delta x \ mes \Delta v \geq \frac{h}{4 \pi m} \)
Calcula on yields the uncertainty in velocity.
10. **Draw the shape of the following molecule: SF₆.**
- Answer: Octahedral
Explana on: SF₆ has six bonding pairs of electrons around sulfur, giving it an octahedral geometry
according to VSEPR theory.
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### **SECTION C [3x10=30]**
11. **The electronic configura on of an element is 1s² 2s² 2p⁶ 3s² 3p³. Name the period and the
group to which the element belongs.**
- Answer:
Period: 3, Group: 15
Explana on: The element has 5 valence electrons in the 3rd shell (3s² 3p³), corresponding to
group 15 and period 3. The element is phosphorus (P).
12. **Why does NH₃ have a higher dipole moment than NF₃, though both are pyramidal?**
- Answer:
NH₃ has a higher dipole moment due to the opposite orienta on of lone pair and bond dipoles. In
NF₃, the bond dipoles (from more electronega ve F atoms) reduce the net dipole.
13. **An iron atom with atomic number 56 contains units of posi ve charge and 30.4% more
neutrons than protons. Assign the symbol of the ion.**
- Answer: \( ^{56}_{26}Fe^{2+} \)
Explana on: The element has 26 protons and \( 56 - 26 = 30 \) neutrons. Since it's an ion, it loses
2 electrons to become \( Fe^{2+} \).
14. **What designa ons are given to the orbitals having:**
(i) n = 2, l = 0
(ii) n = 4, l = 3
- Answer:
(i) 2s
(ii) 4f
Explana on: The azimuthal quantum number \( l = 0 \) refers to s-orbitals, while \( l = 3 \) refers
to f-orbitals.
15. **In a reac on: 2A + B → C, find the limi ng reagent if 5 moles of A react with 6 moles of B.**
- Answer: A is the limi ng reagent.
Explana on: The stoichiometric ra o is 2:1, meaning 5 moles of A would require 2.5 moles of B.
Since there’s excess B, A is limi ng.
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### **SECTION D [5x4=20]**
16. **Define the following terms:**
- (a) Pauli exclusion principle
- (b) Zeeman effect
- Answer:
- (a) Pauli exclusion principle states that no two electrons in an atom can have the same set of
four quantum numbers.
- (b) Zeeman effect refers to the spli ng of spectral lines in the presence of a magne c field.
17. **Why Na⁺ has a higher value of ioniza on enthalpy than Ne, though both have same electron
configura on?**
- Answer: Na⁺ has a higher ioniza on enthalpy than Ne because Na⁺ has a higher nuclear charge
(11 protons vs. 10 in Ne), pulling the electrons closer, requiring more energy to remove an electron.
18. **Draw the resona ng structure of CO₃²⁻ ion.**
- Answer:
The resonance structures involve alterna ng double bonds between carbon and oxygen atoms,
with a formal charge of -1 on each singly bonded oxygen.
19. **Two elements C and D have atomic numbers 35 and 37 respec vely. On the basis of electronic
configura on, predict:**
(i) Period and block to which each element belongs
- Answer:
- C (Z = 35) belongs to period 4, p-block (Bromine, Br).
- D (Z = 37) belongs to period 5, s-block (Rubidium, Rb).
20. **Compare the first ioniza on enthalpy of elements A, C, and Si.**
- Answer:
First ioniza on enthalpy decreases in the order: C > Si > A (Group 14 elements, as ioniza on
energy decreases down the group).
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This answers the ques ons based on the paper provided.
