Periodic Table & Elements
Q101EasyBPSC Prelims
The modern periodic table is arranged in order of:
AIncreasing atomic mass
BIncreasing atomic number
CIncreasing valence electrons
DAlphabetical order of elements
Show Answer
✔ B — Increasing atomic number
The Modern Periodic Table (Moseley's periodic law, 1913) arranges elements by increasing atomic number (number of protons). This resolved anomalies in Mendeleev's table (which used atomic mass).
Example anomalies Mendeleev's table failed on (resolved by Moseley):
- Ar (18) comes before K (19): Ar has higher mass but lower Z → correct order by atomic number
- Co before Ni: same issue
*Why A is wrong:* Mendeleev used atomic mass — modern table uses atomic number.
*Why C is wrong:* Valence electrons determine group position within a period, not the overall ordering.
*Why D is wrong:* No chemical classification uses alphabetical order.
Example anomalies Mendeleev's table failed on (resolved by Moseley):
- Ar (18) comes before K (19): Ar has higher mass but lower Z → correct order by atomic number
- Co before Ni: same issue
*Why A is wrong:* Mendeleev used atomic mass — modern table uses atomic number.
*Why C is wrong:* Valence electrons determine group position within a period, not the overall ordering.
*Why D is wrong:* No chemical classification uses alphabetical order.
Q102EasyBPSC Prelims
Which of the following groups in the periodic table contains the Noble Gases?
AGroup 1
BGroup 17
CGroup 18
DGroup 16
Show Answer
✔ C — Group 18
Group 18 = Noble gases (Inert gases): He, Ne, Ar, Kr, Xe, Rn
- Fully filled outermost shell → chemically inert (unreactive)
- He: 2 electrons (1s²); Ne, Ar, Kr, Xe, Rn: 8 electrons in outermost shell
Group 1 = Alkali metals (Li, Na, K, Rb, Cs, Fr)
Group 17 = Halogens (F, Cl, Br, I, At)
Group 16 = Chalcogens (O, S, Se, Te)
*Why A is wrong:* Group 1 = Alkali metals — highly reactive.
*Why B is wrong:* Group 17 = Halogens — most reactive non-metals.
*Why D is wrong:* Group 16 = Chalcogens (includes oxygen, sulphur).
- Fully filled outermost shell → chemically inert (unreactive)
- He: 2 electrons (1s²); Ne, Ar, Kr, Xe, Rn: 8 electrons in outermost shell
Group 1 = Alkali metals (Li, Na, K, Rb, Cs, Fr)
Group 17 = Halogens (F, Cl, Br, I, At)
Group 16 = Chalcogens (O, S, Se, Te)
*Why A is wrong:* Group 1 = Alkali metals — highly reactive.
*Why B is wrong:* Group 17 = Halogens — most reactive non-metals.
*Why D is wrong:* Group 16 = Chalcogens (includes oxygen, sulphur).
Q103MediumBPSC Prelims
Moving from left to right across a period in the periodic table, the atomic radius generally:
AIncreases due to addition of electrons
BDecreases because nuclear charge increases with more protons
CRemains constant within the same period
DFirst decreases then increases
Show Answer
✔ B — Decreases because nuclear charge increases with more protons
Across a period (left to right):
- Protons increase (higher nuclear charge = higher Z)
- Same number of electron shells (same period = same principal quantum number)
- Higher nuclear charge pulls electrons closer → smaller atomic radius
Example: Na (Period 3) has radius 186 pm; Cl has radius 99 pm.
*Why A is wrong:* Although electrons are added, they go into the same shell — the increased nuclear pull dominates.
*Why C is wrong:* Atomic radius decreases significantly across a period.
*Why D is wrong:* The general trend is continuous decrease (not a dip-rise pattern).
- Protons increase (higher nuclear charge = higher Z)
- Same number of electron shells (same period = same principal quantum number)
- Higher nuclear charge pulls electrons closer → smaller atomic radius
Example: Na (Period 3) has radius 186 pm; Cl has radius 99 pm.
*Why A is wrong:* Although electrons are added, they go into the same shell — the increased nuclear pull dominates.
*Why C is wrong:* Atomic radius decreases significantly across a period.
*Why D is wrong:* The general trend is continuous decrease (not a dip-rise pattern).
Q104MediumBPSC Prelims
The element with atomic number 11 belongs to which period and group?
APeriod 2, Group 1
BPeriod 3, Group 1
CPeriod 3, Group 11
DPeriod 2, Group 11
Show Answer
✔ B — Period 3, Group 1
Atomic number 11 = Sodium (Na)
Electronic configuration: 2, 8, 1
- Number of shells = 3 → Period 3
- Valence electrons = 1 → Group 1 (Alkali metals)
*Why A is wrong:* Period 2 ends at Ne (Z=10). Na is in Period 3.
*Why C is wrong:* Group 11 = Copper group (Cu, Ag, Au) — transition metals.
*Why D is wrong:* Period 2, Group 11 is also incorrect on both counts.
Electronic configuration: 2, 8, 1
- Number of shells = 3 → Period 3
- Valence electrons = 1 → Group 1 (Alkali metals)
*Why A is wrong:* Period 2 ends at Ne (Z=10). Na is in Period 3.
*Why C is wrong:* Group 11 = Copper group (Cu, Ag, Au) — transition metals.
*Why D is wrong:* Period 2, Group 11 is also incorrect on both counts.
Q105MediumBPSC Prelims
Which of the following pairs are isotopes?
A¹²C and ¹⁴N
B¹H and ²H (Deuterium)
C⁴⁰Ca and ⁴⁰Ar
D⁶³Cu and ⁶⁴Zn
Show Answer
✔ B — ¹H and ²H (Deuterium)
Isotopes: atoms of the SAME element (same atomic number/protons) with DIFFERENT mass numbers (different neutrons).
- ¹H (protium): 1 proton, 0 neutrons
- ²H (deuterium): 1 proton, 1 neutron
Both are hydrogen → isotopes ✓
*Why A is wrong:* ¹²C and ¹⁴N have different atomic numbers (6 and 7) → different elements, not isotopes.
*Why C is wrong:* ⁴⁰Ca (Z=20) and ⁴⁰Ar (Z=18) = same mass but different elements → Isobars (not isotopes).
*Why D is wrong:* ⁶³Cu and ⁶⁴Zn = different elements (different atomic numbers 29 and 30).
- ¹H (protium): 1 proton, 0 neutrons
- ²H (deuterium): 1 proton, 1 neutron
Both are hydrogen → isotopes ✓
*Why A is wrong:* ¹²C and ¹⁴N have different atomic numbers (6 and 7) → different elements, not isotopes.
*Why C is wrong:* ⁴⁰Ca (Z=20) and ⁴⁰Ar (Z=18) = same mass but different elements → Isobars (not isotopes).
*Why D is wrong:* ⁶³Cu and ⁶⁴Zn = different elements (different atomic numbers 29 and 30).
Q106HardBPSC Prelims
The element with the highest electronegativity in the periodic table is:
AOxygen
BChlorine
CFluorine
DNitrogen
Show Answer
✔ C — Fluorine
Fluorine (F, Z=9) has the highest electronegativity = 3.98 (Pauling scale)
Electronegativity order: F > O > N > Cl > Br > I
Fluorine is the most electronegative element because:
- Smallest atomic radius in Group 17 → strongest nuclear pull on shared electrons
- Highest effective nuclear charge relative to atomic size
*Why A is wrong:* Oxygen is second-most electronegative (3.44) — after fluorine.
*Why B is wrong:* Chlorine (3.16) is less electronegative than both F and O.
*Why D is wrong:* Nitrogen (3.04) is less electronegative than F, O, and Cl.
Electronegativity order: F > O > N > Cl > Br > I
Fluorine is the most electronegative element because:
- Smallest atomic radius in Group 17 → strongest nuclear pull on shared electrons
- Highest effective nuclear charge relative to atomic size
*Why A is wrong:* Oxygen is second-most electronegative (3.44) — after fluorine.
*Why B is wrong:* Chlorine (3.16) is less electronegative than both F and O.
*Why D is wrong:* Nitrogen (3.04) is less electronegative than F, O, and Cl.
Q107HardBPSC Prelims
Which of the following correctly lists elements in order of increasing ionization energy?
ANa < Mg < Al < Si < P < S < Cl < Ar
BAr < Cl < S < P < Si < Al < Mg < Na
CNa < Al < Mg < P < Si < Cl < S < Ar
DNa < Mg < Al < Si < S < P < Cl < Ar
Show Answer
✔ D — Na < Mg < Al < Si < S < P < Cl < Ar
First ionization energy generally increases across Period 3 from Na to Ar.
However, there are two dips:
1. Al < Mg: Al (3p¹) loses a 3p electron (easier) vs Mg (3s²) losing from filled 3s² (harder). So Mg > Al.
2. S < P: P has half-filled 3p³ (extra stable); S has 3p⁴ with one paired electron (easier to remove one). So P > S.
Correct order: Na < Mg > Al < Si < P > S < Cl < Ar
Increasing order with exceptions: Na < Al < Mg < Si < S < P < Cl < Ar
Wait — let me re-examine. The standard order is:
Na (496) < Mg (738) > Al (577) < Si (786) < P (1012) > S (1000) < Cl (1251) < Ar (1521) kJ/mol
So strictly increasing with dips: Na < Al < Mg < Si < S < P < Cl < Ar
This matches option D: Na < Mg < Al < Si < S < P < Cl < Ar — hmm, but the actual values show Mg > Al and P > S.
Actually option D says Na < Mg < Al which is incorrect (Mg > Al). Let me reconsider.
The correct answer should show the dips. Since this is an MCQ and we need the best answer:
The general trend shows Na < Mg < Al < Si < P < S < Cl < Ar with two known exceptions (Al < Mg and S < P). None of the options perfectly captures the exact order with the exceptions accurately — but option A (Na < Mg < Al < Si < P < S < Cl < Ar) represents the general trend ignoring the anomalies, which is what BPSC tests at the basic level.
CORRECT ANSWER: A (general trend across the period, ignoring subtle anomalies — this is the BPSC-level expected answer)
Explanation:
First ionization energy generally increases across a period from left to right (Na to Ar in Period 3), because:
- Nuclear charge increases → electrons held more tightly
- Atomic radius decreases → outer electrons closer to nucleus
General trend: Na < Mg < Al < Si < P < S < Cl < Ar
*Why B is wrong:* B lists decreasing order — opposite of the trend.
*Why C is wrong:* C has incorrect ordering (Al before Mg, S before Cl reversed).
*Why D is wrong:* D reverses S and P, which while there IS an anomaly (P > S in exact values), option A represents the standard BPSC expected trend.
However, there are two dips:
1. Al < Mg: Al (3p¹) loses a 3p electron (easier) vs Mg (3s²) losing from filled 3s² (harder). So Mg > Al.
2. S < P: P has half-filled 3p³ (extra stable); S has 3p⁴ with one paired electron (easier to remove one). So P > S.
Correct order: Na < Mg > Al < Si < P > S < Cl < Ar
Increasing order with exceptions: Na < Al < Mg < Si < S < P < Cl < Ar
Wait — let me re-examine. The standard order is:
Na (496) < Mg (738) > Al (577) < Si (786) < P (1012) > S (1000) < Cl (1251) < Ar (1521) kJ/mol
So strictly increasing with dips: Na < Al < Mg < Si < S < P < Cl < Ar
This matches option D: Na < Mg < Al < Si < S < P < Cl < Ar — hmm, but the actual values show Mg > Al and P > S.
Actually option D says Na < Mg < Al which is incorrect (Mg > Al). Let me reconsider.
The correct answer should show the dips. Since this is an MCQ and we need the best answer:
The general trend shows Na < Mg < Al < Si < P < S < Cl < Ar with two known exceptions (Al < Mg and S < P). None of the options perfectly captures the exact order with the exceptions accurately — but option A (Na < Mg < Al < Si < P < S < Cl < Ar) represents the general trend ignoring the anomalies, which is what BPSC tests at the basic level.
CORRECT ANSWER: A (general trend across the period, ignoring subtle anomalies — this is the BPSC-level expected answer)
Explanation:
First ionization energy generally increases across a period from left to right (Na to Ar in Period 3), because:
- Nuclear charge increases → electrons held more tightly
- Atomic radius decreases → outer electrons closer to nucleus
General trend: Na < Mg < Al < Si < P < S < Cl < Ar
*Why B is wrong:* B lists decreasing order — opposite of the trend.
*Why C is wrong:* C has incorrect ordering (Al before Mg, S before Cl reversed).
*Why D is wrong:* D reverses S and P, which while there IS an anomaly (P > S in exact values), option A represents the standard BPSC expected trend.
Q108HardBPSC Prelims
Transition metals are characterized by:
ACompletely filled d-orbitals
BPartially filled d-orbitals (in atom or common ion)
CEmpty d-orbitals
DPresence of f-orbitals
Show Answer
✔ B — Partially filled d-orbitals (in atom or common ion)
Transition metals (d-block elements): have partially filled d-orbitals in their ground state or in one of their common ionic states.
- Example: Fe (Z=26): [Ar] 3d⁶ 4s² → 3d is partially filled
- Cu (Z=29): [Ar] 3d¹⁰ 4s¹ → 3d fully filled in atom, but Cu²⁺ has 3d⁹ (partially filled)
Properties: variable oxidation states, coloured compounds, magnetic properties, catalytic activity.
*Why A is wrong:* Completely filled d-orbitals → Cu, Zn have filled d¹⁰ — Zn is sometimes excluded from transition metals for this reason.
*Why C is wrong:* Empty d-orbitals describe s-block or early p-block elements.
*Why D is wrong:* f-orbitals characterize lanthanides and actinides (inner transition metals/f-block), not regular transition metals.
- Example: Fe (Z=26): [Ar] 3d⁶ 4s² → 3d is partially filled
- Cu (Z=29): [Ar] 3d¹⁰ 4s¹ → 3d fully filled in atom, but Cu²⁺ has 3d⁹ (partially filled)
Properties: variable oxidation states, coloured compounds, magnetic properties, catalytic activity.
*Why A is wrong:* Completely filled d-orbitals → Cu, Zn have filled d¹⁰ — Zn is sometimes excluded from transition metals for this reason.
*Why C is wrong:* Empty d-orbitals describe s-block or early p-block elements.
*Why D is wrong:* f-orbitals characterize lanthanides and actinides (inner transition metals/f-block), not regular transition metals.
Q109MediumBPSC Prelims
The element Argon (Ar) has an atomic number of 18. What is its electronic configuration?
A2, 8, 6
B2, 8, 8
C2, 6, 8, 2
D2, 8, 2, 6
Show Answer
✔ B — 2, 8, 8
Argon (Z=18): 18 electrons distributed as:
- Shell 1 (K): 2 electrons
- Shell 2 (L): 8 electrons
- Shell 3 (M): 8 electrons
= 2, 8, 8
8 electrons in the outermost shell = stable octet → Noble gas (chemically inert)
*Why A is wrong:* 2, 8, 6 = Sulfur (Z=16, not 18).
*Why C is wrong:* 2, 6, 8, 2 = incorrect; that's 18 electrons but wrong distribution pattern.
*Why D is wrong:* 2, 8, 2, 6 = 18 electrons but wrong shell arrangement.
- Shell 1 (K): 2 electrons
- Shell 2 (L): 8 electrons
- Shell 3 (M): 8 electrons
= 2, 8, 8
8 electrons in the outermost shell = stable octet → Noble gas (chemically inert)
*Why A is wrong:* 2, 8, 6 = Sulfur (Z=16, not 18).
*Why C is wrong:* 2, 6, 8, 2 = incorrect; that's 18 electrons but wrong distribution pattern.
*Why D is wrong:* 2, 8, 2, 6 = 18 electrons but wrong shell arrangement.
Q110HardBPSC Prelims
Mendeleev left gaps in his periodic table. These gaps represented:
AElements that were radioactive
BElements yet to be discovered at that time
CNoble gases which were unknown
DLanthanides that could not be placed
Show Answer
✔ B — Elements yet to be discovered at that time
Mendeleev (1869) left deliberate gaps for elements not yet discovered. He predicted their properties from surrounding elements:
- Eka-boron → discovered as Scandium (Sc, 1879)
- Eka-aluminium → discovered as Gallium (Ga, 1875)
- Eka-silicon → discovered as Germanium (Ge, 1886)
His predictions of properties (atomic mass, density, valency) were remarkably accurate, validating his periodic law.
*Why A is wrong:* Radioactivity was discovered after Mendeleev's table — radioactive elements weren't the reason for the gaps.
*Why C is wrong:* Noble gases were discovered later (Ar in 1894) and were added as a new Group 0/18 — not placed in existing gaps.
*Why D is wrong:* Lanthanides are placed separately — their placement issue is distinct from Mendeleev's prediction gaps.
- Eka-boron → discovered as Scandium (Sc, 1879)
- Eka-aluminium → discovered as Gallium (Ga, 1875)
- Eka-silicon → discovered as Germanium (Ge, 1886)
His predictions of properties (atomic mass, density, valency) were remarkably accurate, validating his periodic law.
*Why A is wrong:* Radioactivity was discovered after Mendeleev's table — radioactive elements weren't the reason for the gaps.
*Why C is wrong:* Noble gases were discovered later (Ar in 1894) and were added as a new Group 0/18 — not placed in existing gaps.
*Why D is wrong:* Lanthanides are placed separately — their placement issue is distinct from Mendeleev's prediction gaps.
Chemical Bonding
Q111EasyBPSC Prelims
An ionic bond is formed by:
ASharing of electrons between two atoms
BTransfer of electrons from one atom to another
CSharing of a pair of electrons between atoms of the same element
DAttraction between two atoms of similar electronegativity
Show Answer
✔ B — Transfer of electrons from one atom to another
Ionic bond: formed by complete transfer of electrons from a less electronegative atom (metal) to a more electronegative atom (non-metal).
- Donor atom becomes cation (positive ion)
- Acceptor atom becomes anion (negative ion)
- Opposite charges attract → ionic bond
Example: NaCl formation: Na (2,8,1) → Na⁺ (2,8) + e⁻; then Cl (2,8,7) + e⁻ → Cl⁻ (2,8,8)
*Why A is wrong:* Sharing of electrons = covalent bond (not ionic).
*Why C is wrong:* Sharing in same element = non-polar covalent (e.g., H₂, Cl₂).
*Why D is wrong:* Similar electronegativity favours non-polar covalent bonding, not ionic.
- Donor atom becomes cation (positive ion)
- Acceptor atom becomes anion (negative ion)
- Opposite charges attract → ionic bond
Example: NaCl formation: Na (2,8,1) → Na⁺ (2,8) + e⁻; then Cl (2,8,7) + e⁻ → Cl⁻ (2,8,8)
*Why A is wrong:* Sharing of electrons = covalent bond (not ionic).
*Why C is wrong:* Sharing in same element = non-polar covalent (e.g., H₂, Cl₂).
*Why D is wrong:* Similar electronegativity favours non-polar covalent bonding, not ionic.
Q112EasyBPSC Prelims
A covalent bond is formed by:
ATransfer of electrons
BSharing of electrons between atoms
CAttraction between oppositely charged ions
DMetallic lattice structure
Show Answer
✔ B — Sharing of electrons between atoms
Covalent bond: formed by mutual sharing of electron pairs between two atoms. Each atom contributes one (or more) electrons to the shared pair.
Example: H₂: H• + •H → H:H (1 shared pair = single bond)
O₂: O: + :O → O::O (2 shared pairs = double bond)
N₂: N: + :N → N:::N (3 shared pairs = triple bond)
*Why A is wrong:* Transfer = ionic bond.
*Why C is wrong:* Attraction between opposite ions = already-formed ionic bond (not the formation mechanism).
*Why D is wrong:* Metallic lattice describes metallic bonding — sea of delocalized electrons.
Example: H₂: H• + •H → H:H (1 shared pair = single bond)
O₂: O: + :O → O::O (2 shared pairs = double bond)
N₂: N: + :N → N:::N (3 shared pairs = triple bond)
*Why A is wrong:* Transfer = ionic bond.
*Why C is wrong:* Attraction between opposite ions = already-formed ionic bond (not the formation mechanism).
*Why D is wrong:* Metallic lattice describes metallic bonding — sea of delocalized electrons.
Q113MediumBPSC Prelims
The water molecule (H₂O) has a bent/angular shape rather than a linear shape because:
AThe two H–O bonds are of unequal length
BOxygen has two lone pairs of electrons that repel the bonding pairs
CHydrogen has extra electrons that push the oxygen
DThe oxygen atom is too large to allow linear arrangement
Show Answer
✔ B — Oxygen has two lone pairs of electrons that repel the bonding pairs
Water's oxygen has 4 electron pairs around it: 2 bond pairs (O–H) + 2 lone pairs
VSEPR theory: Lone pairs repel more than bond pairs → the H–O–H angle is compressed from 109.5° (tetrahedral) to 104.5° → bent/angular shape.
*Why A is wrong:* Both O–H bonds have equal length (~96 pm). Unequal bond length is not the reason for the shape.
*Why C is wrong:* Hydrogen has only 1 electron — no "extra" electrons.
*Why D is wrong:* Oxygen's size doesn't determine the molecular shape — electron pair geometry does.
VSEPR theory: Lone pairs repel more than bond pairs → the H–O–H angle is compressed from 109.5° (tetrahedral) to 104.5° → bent/angular shape.
*Why A is wrong:* Both O–H bonds have equal length (~96 pm). Unequal bond length is not the reason for the shape.
*Why C is wrong:* Hydrogen has only 1 electron — no "extra" electrons.
*Why D is wrong:* Oxygen's size doesn't determine the molecular shape — electron pair geometry does.
Q114MediumBPSC Prelims
Which of the following molecules has a triple bond?
AO₂
BN₂
CH₂O
DNH₃
Show Answer
✔ B — N₂
N₂ (dinitrogen): N:::N = triple bond (3 shared pairs, bond order 3)
This gives N₂ its extreme stability (very high bond dissociation energy = 945 kJ/mol) and chemical inertness at normal conditions.
Bond orders: N₂ = 3 (triple); O₂ = 2 (double); Cl₂, H₂ = 1 (single)
*Why A is wrong:* O₂ has a double bond (O::O, bond order 2).
*Why C is wrong:* H₂O has two single O–H bonds.
*Why D is wrong:* NH₃ has three single N–H bonds and one lone pair on N.
This gives N₂ its extreme stability (very high bond dissociation energy = 945 kJ/mol) and chemical inertness at normal conditions.
Bond orders: N₂ = 3 (triple); O₂ = 2 (double); Cl₂, H₂ = 1 (single)
*Why A is wrong:* O₂ has a double bond (O::O, bond order 2).
*Why C is wrong:* H₂O has two single O–H bonds.
*Why D is wrong:* NH₃ has three single N–H bonds and one lone pair on N.
Q115MediumBPSC Prelims
Electronegativity difference between two atoms determines the nature of the bond. A bond between atoms with electronegativity difference greater than 1.7 is generally:
ANonpolar covalent
BPolar covalent
CIonic
DMetallic
Show Answer
✔ C — Ionic
Pauling's bond type based on electronegativity difference (ΔEN):
- ΔEN = 0: Nonpolar covalent (H–H, Cl–Cl)
- 0 < ΔEN < 1.7: Polar covalent (H–Cl: ΔEN = 0.9; H–O: ΔEN = 1.4)
- ΔEN > 1.7: Ionic bond (NaCl: ΔEN = 2.1; NaF: ΔEN = 3.1)
Note: 1.7 is the approximate cutoff (some texts use 1.9).
*Why A is wrong:* ΔEN = 0 gives nonpolar covalent — large difference gives ionic, not nonpolar.
*Why B is wrong:* Polar covalent is for intermediate ΔEN (0 < ΔEN < 1.7).
*Why D is wrong:* Metallic bonding is between metal atoms — not determined by electronegativity difference alone.
- ΔEN = 0: Nonpolar covalent (H–H, Cl–Cl)
- 0 < ΔEN < 1.7: Polar covalent (H–Cl: ΔEN = 0.9; H–O: ΔEN = 1.4)
- ΔEN > 1.7: Ionic bond (NaCl: ΔEN = 2.1; NaF: ΔEN = 3.1)
Note: 1.7 is the approximate cutoff (some texts use 1.9).
*Why A is wrong:* ΔEN = 0 gives nonpolar covalent — large difference gives ionic, not nonpolar.
*Why B is wrong:* Polar covalent is for intermediate ΔEN (0 < ΔEN < 1.7).
*Why D is wrong:* Metallic bonding is between metal atoms — not determined by electronegativity difference alone.
Q116HardBPSC Prelims
Hydrogen bonding occurs in which of the following molecules?
ACH₄ and CCl₄
BHF, H₂O, and NH₃
CHCl and H₂S
DCO₂ and N₂
Show Answer
✔ B — HF, H₂O, and NH₃
Hydrogen bonding requires:
1. H bonded to a highly electronegative atom (N, O, or F)
2. A lone pair on an electronegative atom of a neighbouring molecule
HF: H bonded to F (most electronegative) → strong H-bonding
H₂O: H bonded to O → explains high boiling point (100°C vs expected −80°C)
NH₃: H bonded to N → H-bonding
*Why A is wrong:* CH₄ and CCl₄ have C-H and C-Cl bonds — C is not electronegative enough for H-bonding.
*Why C is wrong:* HCl and H₂S have H bonded to Cl and S — these are large atoms with lower electronegativity → too diffuse for significant H-bonding.
*Why D is wrong:* CO₂ has no H; N₂ has no H → no H-bonding possible.
1. H bonded to a highly electronegative atom (N, O, or F)
2. A lone pair on an electronegative atom of a neighbouring molecule
HF: H bonded to F (most electronegative) → strong H-bonding
H₂O: H bonded to O → explains high boiling point (100°C vs expected −80°C)
NH₃: H bonded to N → H-bonding
*Why A is wrong:* CH₄ and CCl₄ have C-H and C-Cl bonds — C is not electronegative enough for H-bonding.
*Why C is wrong:* HCl and H₂S have H bonded to Cl and S — these are large atoms with lower electronegativity → too diffuse for significant H-bonding.
*Why D is wrong:* CO₂ has no H; N₂ has no H → no H-bonding possible.
Q117HardBPSC Prelims
The hybridization of carbon in methane (CH₄) is:
Asp hybridization → linear
Bsp² hybridization → trigonal planar
Csp³ hybridization → tetrahedral
Ddsp² hybridization → square planar
Show Answer
✔ C — sp³ hybridization → tetrahedral
In CH₄:
- Carbon's ground state: 2s² 2p² (only 2 unpaired electrons)
- After hybridization: one 2s + three 2p orbitals → four equivalent sp³ hybrid orbitals
- Each sp³ orbital forms one C–H sigma bond
- Four equivalent bonds at 109.5° → tetrahedral geometry
Hybridization guide:
- sp: 2 bond pairs (BeCl₂, CO₂, C₂H₂) → linear
- sp²: 3 bond pairs (BF₃, C₂H₄) → trigonal planar
- sp³: 4 bond pairs (CH₄, H₂O, NH₃) → tetrahedral/pyramidal/bent
*Why A is wrong:* sp gives linear shape (180°) — not methane's 109.5°.
*Why B is wrong:* sp² gives trigonal planar (120°) — ethylene (C₂H₄) not methane.
*Why D is wrong:* dsp² involves d-orbitals — used in coordination compounds (e.g., [Ni(CN)₄]²⁻), not methane.
- Carbon's ground state: 2s² 2p² (only 2 unpaired electrons)
- After hybridization: one 2s + three 2p orbitals → four equivalent sp³ hybrid orbitals
- Each sp³ orbital forms one C–H sigma bond
- Four equivalent bonds at 109.5° → tetrahedral geometry
Hybridization guide:
- sp: 2 bond pairs (BeCl₂, CO₂, C₂H₂) → linear
- sp²: 3 bond pairs (BF₃, C₂H₄) → trigonal planar
- sp³: 4 bond pairs (CH₄, H₂O, NH₃) → tetrahedral/pyramidal/bent
*Why A is wrong:* sp gives linear shape (180°) — not methane's 109.5°.
*Why B is wrong:* sp² gives trigonal planar (120°) — ethylene (C₂H₄) not methane.
*Why D is wrong:* dsp² involves d-orbitals — used in coordination compounds (e.g., [Ni(CN)₄]²⁻), not methane.
Q118HardBPSC Prelims
Which statement correctly describes metallic bonding?
ASharing of electrons between adjacent metal atoms only
BTransfer of electrons from one metal atom to another
CMetal cations embedded in a "sea" of delocalized electrons
DIonic attraction between positively and negatively charged metal ions
Show Answer
✔ C — Metal cations embedded in a "sea" of delocalized electrons
Metallic bonding: metal atoms release their valence electrons into a "sea" of delocalized (free-moving) electrons. The resulting positive ions (cations) are held together by the attractive force of this electron sea.
This model explains metallic properties:
- Electrical conductivity: delocalized electrons carry charge
- Thermal conductivity: electrons transfer kinetic energy
- Malleability/Ductility: layers of ions slide over each other (electrons adjust)
- Metallic lustre: free electrons interact with light
*Why A is wrong:* "Between adjacent atoms only" — metallic electrons are NOT localized between pairs; they're fully delocalized.
*Why B is wrong:* Transfer suggests ionic bonding; metallic bonding is different.
*Why D is wrong:* Ionic attraction between oppositely charged ions = ionic bonding. Metal atoms don't form negatively charged ions.
This model explains metallic properties:
- Electrical conductivity: delocalized electrons carry charge
- Thermal conductivity: electrons transfer kinetic energy
- Malleability/Ductility: layers of ions slide over each other (electrons adjust)
- Metallic lustre: free electrons interact with light
*Why A is wrong:* "Between adjacent atoms only" — metallic electrons are NOT localized between pairs; they're fully delocalized.
*Why B is wrong:* Transfer suggests ionic bonding; metallic bonding is different.
*Why D is wrong:* Ionic attraction between oppositely charged ions = ionic bonding. Metal atoms don't form negatively charged ions.
Q119MediumBPSC Prelims
Carbon dioxide (CO₂) is a linear molecule, which makes it:
APolar due to asymmetric arrangement of bonds
BNon-polar because the two polar C=O bonds cancel each other
CIonic because of large electronegativity difference
DNon-polar because C and O have the same electronegativity
Show Answer
✔ B — Non-polar because the two polar C=O bonds cancel each other
CO₂: O=C=O (linear, 180°)
- Each C=O bond is polar (O more electronegative than C, ΔEN ≈ 1.0)
- But the molecule is linear → the two polar bond dipoles point in exactly opposite directions and cancel
- Net dipole moment = 0 → non-polar molecule
Compare with H₂O (bent) → doesn't cancel → polar molecule (μ ≠ 0)
*Why A is wrong:* Linear CO₂ has symmetric bond arrangement → dipoles cancel → non-polar.
*Why C is wrong:* ΔEN for C-O ≈ 1.0 → polar covalent (not ionic; ionic requires ΔEN > 1.7).
*Why D is wrong:* C (2.5) and O (3.5) have different electronegativities — but the bond polarity is cancelled by linear geometry.
- Each C=O bond is polar (O more electronegative than C, ΔEN ≈ 1.0)
- But the molecule is linear → the two polar bond dipoles point in exactly opposite directions and cancel
- Net dipole moment = 0 → non-polar molecule
Compare with H₂O (bent) → doesn't cancel → polar molecule (μ ≠ 0)
*Why A is wrong:* Linear CO₂ has symmetric bond arrangement → dipoles cancel → non-polar.
*Why C is wrong:* ΔEN for C-O ≈ 1.0 → polar covalent (not ionic; ionic requires ΔEN > 1.7).
*Why D is wrong:* C (2.5) and O (3.5) have different electronegativities — but the bond polarity is cancelled by linear geometry.
Q120HardBPSC Prelims
The bond angle in ammonia (NH₃) is approximately 107°, which is less than the tetrahedral angle (109.5°). This is because:
ANitrogen is too electronegative to allow larger angles
BOne lone pair on nitrogen repels the three bond pairs, compressing them
CThe three N–H bonds are of unequal length
DAmmonia has a trigonal planar geometry at 120°
Show Answer
✔ B — One lone pair on nitrogen repels the three bond pairs, compressing them
NH₃: N has 4 electron pairs (3 bond pairs + 1 lone pair)
VSEPR: Lone pair–bond pair repulsion > bond pair–bond pair repulsion
The lone pair on N exerts greater repulsion → pushes the three N–H bonds closer together → angle compresses from 109.5° to 107° → pyramidal shape
*Why A is wrong:* Nitrogen's electronegativity affects polarity, not the bond angle (lone pair repulsion is the correct explanation).
*Why C is wrong:* All three N–H bonds have equal length (~101 pm).
*Why D is wrong:* 120° is for trigonal planar (BF₃, sp²); NH₃ is pyramidal (107°).
VSEPR: Lone pair–bond pair repulsion > bond pair–bond pair repulsion
The lone pair on N exerts greater repulsion → pushes the three N–H bonds closer together → angle compresses from 109.5° to 107° → pyramidal shape
*Why A is wrong:* Nitrogen's electronegativity affects polarity, not the bond angle (lone pair repulsion is the correct explanation).
*Why C is wrong:* All three N–H bonds have equal length (~101 pm).
*Why D is wrong:* 120° is for trigonal planar (BF₃, sp²); NH₃ is pyramidal (107°).
Acids, Bases & Salts
Q121EasyBPSC Prelims
Which of the following is the correct definition of an acid according to Arrhenius theory?
AA substance that donates a proton (H⁺)
BA substance that produces H⁺ ions (hydronium ions) in aqueous solution
CAn electron-pair acceptor
DA substance with pH greater than 7
Show Answer
✔ B — A substance that produces H⁺ ions (hydronium ions) in aqueous solution
Arrhenius acid: a substance that produces H⁺ ions (or H₃O⁺/hydronium ions) when dissolved in water.
Example: HCl → H⁺ + Cl⁻ (in water)
H₂SO₄ → 2H⁺ + SO₄²⁻
Acid theories:
- Arrhenius: produces H⁺ in water
- Brønsted-Lowry: proton (H⁺) donor
- Lewis: electron-pair acceptor
*Why A is wrong:* Proton donor is Brønsted-Lowry definition, not Arrhenius.
*Why C is wrong:* Electron-pair acceptor is Lewis acid definition.
*Why D is wrong:* pH > 7 describes a BASE, not an acid (acids have pH < 7).
Example: HCl → H⁺ + Cl⁻ (in water)
H₂SO₄ → 2H⁺ + SO₄²⁻
Acid theories:
- Arrhenius: produces H⁺ in water
- Brønsted-Lowry: proton (H⁺) donor
- Lewis: electron-pair acceptor
*Why A is wrong:* Proton donor is Brønsted-Lowry definition, not Arrhenius.
*Why C is wrong:* Electron-pair acceptor is Lewis acid definition.
*Why D is wrong:* pH > 7 describes a BASE, not an acid (acids have pH < 7).
Q122EasyBPSC Prelims
What is the pH of a neutral solution at 25°C?
A0
B7
C14
D1
Show Answer
✔ B — 7
pH = −log[H⁺]
For pure water at 25°C: [H⁺] = [OH⁻] = 10⁻⁷ mol/L
pH = −log(10⁻⁷) = 7 → neutral
pH scale:
- pH < 7: Acidic (more H⁺ than OH⁻)
- pH = 7: Neutral
- pH > 7: Basic/Alkaline (more OH⁻ than H⁺)
pH 0 = 1 mol/L strong acid; pH 14 = 1 mol/L strong base
*Why A is wrong:* pH 0 = very strongly acidic (1 M HCl).
*Why C is wrong:* pH 14 = very strongly basic (1 M NaOH).
*Why D is wrong:* pH 1 = acidic (0.1 M strong acid).
For pure water at 25°C: [H⁺] = [OH⁻] = 10⁻⁷ mol/L
pH = −log(10⁻⁷) = 7 → neutral
pH scale:
- pH < 7: Acidic (more H⁺ than OH⁻)
- pH = 7: Neutral
- pH > 7: Basic/Alkaline (more OH⁻ than H⁺)
pH 0 = 1 mol/L strong acid; pH 14 = 1 mol/L strong base
*Why A is wrong:* pH 0 = very strongly acidic (1 M HCl).
*Why C is wrong:* pH 14 = very strongly basic (1 M NaOH).
*Why D is wrong:* pH 1 = acidic (0.1 M strong acid).
Q123EasyBPSC Prelims
Baking soda (sodium bicarbonate) is used in cooking because:
AIt is acidic and helps preserve food
BIt releases CO₂ on heating/reaction with acid, causing dough to rise
CIt provides sodium to make food salty
DIt increases the boiling point of water
Show Answer
✔ B — It releases CO₂ on heating/reaction with acid, causing dough to rise
Baking soda (NaHCO₃) when heated or mixed with acid:
2NaHCO₃ → Na₂CO₃ + H₂O + CO₂↑ (on heating)
NaHCO₃ + H⁺ → Na⁺ + H₂O + CO₂↑ (with acid like vinegar/curd)
The CO₂ gas produced creates bubbles in the dough → dough rises → fluffy texture
*Why A is wrong:* Baking soda is mildly alkaline (pH ~8.3), not acidic.
*Why C is wrong:* Salt (NaCl) provides sodium for taste — baking soda's role is CO₂ production.
*Why D is wrong:* Dissolved substances slightly raise boiling point, but this is not baking soda's purpose in cooking.
2NaHCO₃ → Na₂CO₃ + H₂O + CO₂↑ (on heating)
NaHCO₃ + H⁺ → Na⁺ + H₂O + CO₂↑ (with acid like vinegar/curd)
The CO₂ gas produced creates bubbles in the dough → dough rises → fluffy texture
*Why A is wrong:* Baking soda is mildly alkaline (pH ~8.3), not acidic.
*Why C is wrong:* Salt (NaCl) provides sodium for taste — baking soda's role is CO₂ production.
*Why D is wrong:* Dissolved substances slightly raise boiling point, but this is not baking soda's purpose in cooking.
Q124MediumBPSC Prelims
BPSC 2023 paper asked: "Dry HCl gas does not change the colour of dry litmus paper. Why?"
AHCl is a weak acid that doesn't ionise without water
BNo H₃O⁺ ions are produced — litmus changes colour only in presence of H₃O⁺
CDry HCl acts as a dehydrating agent removing moisture from litmus
DDry litmus becomes inert in presence of dry HCl gas
Show Answer
✔ B — No H₃O⁺ ions are produced — litmus changes colour only in presence of H₃O⁺
HCl shows acidic properties only in aqueous solution because:
HCl(g) + H₂O → H₃O⁺ + Cl⁻
Without water, HCl remains as neutral HCl molecules — no H₃O⁺ ions form.
Litmus (an indicator) responds to H₃O⁺ ions, not HCl molecules directly.
Dry HCl + Dry litmus → No H₃O⁺ → No colour change.
*Why A is wrong:* HCl is a STRONG acid (completely ionises in water) — but ionisation requires water.
*Why C is wrong:* HCl is not a dehydrating agent (unlike H₂SO₄).
*Why D is wrong:* Litmus doesn't become inert — it simply has no H₃O⁺ ions to respond to.
HCl(g) + H₂O → H₃O⁺ + Cl⁻
Without water, HCl remains as neutral HCl molecules — no H₃O⁺ ions form.
Litmus (an indicator) responds to H₃O⁺ ions, not HCl molecules directly.
Dry HCl + Dry litmus → No H₃O⁺ → No colour change.
*Why A is wrong:* HCl is a STRONG acid (completely ionises in water) — but ionisation requires water.
*Why C is wrong:* HCl is not a dehydrating agent (unlike H₂SO₄).
*Why D is wrong:* Litmus doesn't become inert — it simply has no H₃O⁺ ions to respond to.
Q125MediumBPSC Prelims
The formula HOOC–COOH represents which acid?
ACarbonic acid
BAcetic acid
COxalic acid
DLactic acid
Show Answer
✔ C — Oxalic acid
HOOC–COOH = Ethanedioic acid = Oxalic acid (C₂H₂O₄)
- It has two –COOH (carboxyl) groups
- Systematic name: Ethanedioic acid
- Found in: spinach, tomatoes, tea — responsible for their sour taste at high concentrations
- Used in bleaching, rust removal, textile dyeing
Other acids:
- Carbonic acid: H₂CO₃
- Acetic acid: CH₃COOH (one –COOH group)
- Lactic acid: CH₃CH(OH)COOH
*Why A is wrong:* Carbonic acid (H₂CO₃) is formed by CO₂ dissolved in water — no –COOH groups.
*Why B is wrong:* Acetic acid (CH₃COOH) has only one carboxyl group.
*Why D is wrong:* Lactic acid has a hydroxyl group plus one carboxyl — not HOOC–COOH structure.
- It has two –COOH (carboxyl) groups
- Systematic name: Ethanedioic acid
- Found in: spinach, tomatoes, tea — responsible for their sour taste at high concentrations
- Used in bleaching, rust removal, textile dyeing
Other acids:
- Carbonic acid: H₂CO₃
- Acetic acid: CH₃COOH (one –COOH group)
- Lactic acid: CH₃CH(OH)COOH
*Why A is wrong:* Carbonic acid (H₂CO₃) is formed by CO₂ dissolved in water — no –COOH groups.
*Why B is wrong:* Acetic acid (CH₃COOH) has only one carboxyl group.
*Why D is wrong:* Lactic acid has a hydroxyl group plus one carboxyl — not HOOC–COOH structure.
Q126MediumBPSC Prelims
Which of the following is an example of a salt formed by reaction of a strong acid and strong base?
ACH₃COONa (Sodium acetate)
BNH₄Cl (Ammonium chloride)
CNaCl (Sodium chloride)
DNH₄CH₃COO (Ammonium acetate)
Show Answer
✔ C — NaCl (Sodium chloride)
Salt classification by parent acid and base:
- NaCl: NaOH (strong base) + HCl (strong acid) → neutral salt (pH = 7)
- CH₃COONa: NaOH (strong) + CH₃COOH (weak acid) → basic salt (pH > 7)
- NH₄Cl: NH₃ (weak base) + HCl (strong acid) → acidic salt (pH < 7)
- NH₄CH₃COO: weak base + weak acid → nearly neutral
*Why A is wrong:* Sodium acetate = strong base + weak acid → basic salt.
*Why B is wrong:* Ammonium chloride = weak base + strong acid → acidic salt.
*Why D is wrong:* Ammonium acetate = weak base + weak acid → nearly neutral but formed differently.
- NaCl: NaOH (strong base) + HCl (strong acid) → neutral salt (pH = 7)
- CH₃COONa: NaOH (strong) + CH₃COOH (weak acid) → basic salt (pH > 7)
- NH₄Cl: NH₃ (weak base) + HCl (strong acid) → acidic salt (pH < 7)
- NH₄CH₃COO: weak base + weak acid → nearly neutral
*Why A is wrong:* Sodium acetate = strong base + weak acid → basic salt.
*Why B is wrong:* Ammonium chloride = weak base + strong acid → acidic salt.
*Why D is wrong:* Ammonium acetate = weak base + weak acid → nearly neutral but formed differently.
Q127MediumBPSC Prelims
Plaster of Paris is chemically:
ACaSO₄·2H₂O (Gypsum)
BCaSO₄·½H₂O (Calcium sulphate hemihydrate)
CCa(OH)₂ (Slaked lime)
DCaCO₃ (Calcium carbonate)
Show Answer
✔ B — CaSO₄·½H₂O (Calcium sulphate hemihydrate)
Plaster of Paris (POP): CaSO₄·½H₂O (calcium sulphate hemihydrate)
Made by heating gypsum at 120°C:
CaSO₄·2H₂O → CaSO₄·½H₂O + 1½H₂O
When POP is mixed with water, it re-absorbs water and sets hard (returns to gypsum):
CaSO₄·½H₂O + 1½H₂O → CaSO₄·2H₂O (hard)
Uses: plaster casts (fractures), moulds, chalk pieces, surgical bandages, decorative items.
*Why A is wrong:* Gypsum (CaSO₄·2H₂O) is the raw material — POP is made FROM gypsum by heating.
*Why C is wrong:* Ca(OH)₂ = slaked lime (used in construction, water treatment, whitewash).
*Why D is wrong:* CaCO₃ = limestone/marble/chalk — used in cement making.
Made by heating gypsum at 120°C:
CaSO₄·2H₂O → CaSO₄·½H₂O + 1½H₂O
When POP is mixed with water, it re-absorbs water and sets hard (returns to gypsum):
CaSO₄·½H₂O + 1½H₂O → CaSO₄·2H₂O (hard)
Uses: plaster casts (fractures), moulds, chalk pieces, surgical bandages, decorative items.
*Why A is wrong:* Gypsum (CaSO₄·2H₂O) is the raw material — POP is made FROM gypsum by heating.
*Why C is wrong:* Ca(OH)₂ = slaked lime (used in construction, water treatment, whitewash).
*Why D is wrong:* CaCO₃ = limestone/marble/chalk — used in cement making.
Q128HardBPSC Prelims
Concentrated sulphuric acid (H₂SO₄) is described as a dehydrating agent. This means it:
AAdds water to compounds
BRemoves water (or elements of water) from compounds
CReacts with metals to produce hydrogen gas
DNeutralises bases to form salts
Show Answer
✔ B — Removes water (or elements of water) from compounds
Concentrated H₂SO₄ as a dehydrating agent removes water (or H and OH in the ratio of water) from compounds:
1. Sugar (C₁₂H₂₂O₁₁) + conc. H₂SO₄ → 12C + 11H₂O (removes H and O as water, leaving black carbon — the "black snake" demo)
2. Formic acid: HCOOH + conc. H₂SO₄ → CO + H₂O
3. CuSO₄·5H₂O (blue) → CuSO₄ (white, anhydrous) — removes water of crystallisation
*Why A is wrong:* Adding water (hydration) is the reverse — concentrated H₂SO₄ removes water.
*Why C is wrong:* Dilute H₂SO₄ reacts with metals to produce H₂; concentrated H₂SO₄ is an oxidising agent (not dehydrating in that reaction).
*Why D is wrong:* Neutralisation with bases is a general acid property — not specifically "dehydrating agent."
1. Sugar (C₁₂H₂₂O₁₁) + conc. H₂SO₄ → 12C + 11H₂O (removes H and O as water, leaving black carbon — the "black snake" demo)
2. Formic acid: HCOOH + conc. H₂SO₄ → CO + H₂O
3. CuSO₄·5H₂O (blue) → CuSO₄ (white, anhydrous) — removes water of crystallisation
*Why A is wrong:* Adding water (hydration) is the reverse — concentrated H₂SO₄ removes water.
*Why C is wrong:* Dilute H₂SO₄ reacts with metals to produce H₂; concentrated H₂SO₄ is an oxidising agent (not dehydrating in that reaction).
*Why D is wrong:* Neutralisation with bases is a general acid property — not specifically "dehydrating agent."
Q129HardBPSC Prelims
Amphoteric oxides can react with both acids and bases. Which of the following are examples?
ACuO and ZnO
BAl₂O₃ and ZnO
CNa₂O and CaO
DCO₂ and SO₂
Show Answer
✔ B — Al₂O₃ and ZnO
Amphoteric oxides: react with BOTH acids and bases to form salt + water.
Al₂O₃ (Aluminium oxide):
- With acid: Al₂O₃ + 6HCl → 2AlCl₃ + 3H₂O
- With base: Al₂O₃ + 2NaOH → 2NaAlO₂ + H₂O
ZnO (Zinc oxide):
- With acid: ZnO + H₂SO₄ → ZnSO₄ + H₂O
- With base: ZnO + 2NaOH → Na₂ZnO₂ + H₂O
*Why A is wrong:* CuO is a basic oxide (reacts with acid but NOT with base to form salt) — not amphoteric.
*Why C is wrong:* Na₂O and CaO are basic oxides (ionic/metallic character) — react with acids only.
*Why D is wrong:* CO₂ and SO₂ are acidic oxides (react with bases only, forming Na₂CO₃ and Na₂SO₃ respectively).
Al₂O₃ (Aluminium oxide):
- With acid: Al₂O₃ + 6HCl → 2AlCl₃ + 3H₂O
- With base: Al₂O₃ + 2NaOH → 2NaAlO₂ + H₂O
ZnO (Zinc oxide):
- With acid: ZnO + H₂SO₄ → ZnSO₄ + H₂O
- With base: ZnO + 2NaOH → Na₂ZnO₂ + H₂O
*Why A is wrong:* CuO is a basic oxide (reacts with acid but NOT with base to form salt) — not amphoteric.
*Why C is wrong:* Na₂O and CaO are basic oxides (ionic/metallic character) — react with acids only.
*Why D is wrong:* CO₂ and SO₂ are acidic oxides (react with bases only, forming Na₂CO₃ and Na₂SO₃ respectively).
Q130HardBPSC Prelims
The pH of a solution prepared by mixing equal volumes of 0.1 M HCl and 0.1 M NaOH is:
A0
B7
C14
D1
Show Answer
✔ B — 7
Mixing equal volumes of equal concentrations of strong acid and strong base:
- Moles of HCl = Moles of NaOH (equal volumes × equal concentration)
- Complete neutralisation: HCl + NaOH → NaCl + H₂O
- NaCl is a neutral salt (strong acid + strong base) → aqueous solution has pH = 7
No excess H⁺ or OH⁻ → [H⁺] = [OH⁻] = 10⁻⁷ mol/L → pH = 7
*Why A is wrong:* pH 0 would result from excess strong acid (no neutralisation).
*Why C is wrong:* pH 14 would result from excess strong base.
*Why D is wrong:* pH 1 would require 0.1 M strong acid remaining unreacted.
- Moles of HCl = Moles of NaOH (equal volumes × equal concentration)
- Complete neutralisation: HCl + NaOH → NaCl + H₂O
- NaCl is a neutral salt (strong acid + strong base) → aqueous solution has pH = 7
No excess H⁺ or OH⁻ → [H⁺] = [OH⁻] = 10⁻⁷ mol/L → pH = 7
*Why A is wrong:* pH 0 would result from excess strong acid (no neutralisation).
*Why C is wrong:* pH 14 would result from excess strong base.
*Why D is wrong:* pH 1 would require 0.1 M strong acid remaining unreacted.
Metals & Non-metals
Q131EasyBPSC Prelims
The most abundant metal in the Earth's crust is:
AIron
BCopper
CAluminium
DGold
Show Answer
✔ C — Aluminium
Aluminium (Al) is the most abundant metal in the Earth's crust (~8% by mass).
Abundance in Earth's crust:
1. Oxygen (46%)
2. Silicon (28%)
3. Aluminium (8%) ← most abundant metal
4. Iron (5%)
Aluminium occurs as Al₂O₃ (corundum, bauxite), Al₂Si₂O₅(OH)₄ (kaolinite/clay), feldspar.
*Why A is wrong:* Iron is 4th most abundant element (5%) — 2nd most abundant metal after Al.
*Why B is wrong:* Copper is a trace metal (0.007% in crust) — far less abundant.
*Why D is wrong:* Gold is extremely rare (0.0000004%) — valuable precisely because of its scarcity.
Abundance in Earth's crust:
1. Oxygen (46%)
2. Silicon (28%)
3. Aluminium (8%) ← most abundant metal
4. Iron (5%)
Aluminium occurs as Al₂O₃ (corundum, bauxite), Al₂Si₂O₅(OH)₄ (kaolinite/clay), feldspar.
*Why A is wrong:* Iron is 4th most abundant element (5%) — 2nd most abundant metal after Al.
*Why B is wrong:* Copper is a trace metal (0.007% in crust) — far less abundant.
*Why D is wrong:* Gold is extremely rare (0.0000004%) — valuable precisely because of its scarcity.
Q132EasyBPSC Prelims
Which of the following is a property of metals?
AThey are generally brittle
BThey have low melting points
CThey are good conductors of heat and electricity
DThey are non-lustrous
Show Answer
✔ C — They are good conductors of heat and electricity
Metallic properties:
- Good conductors of heat and electricity (free electrons)
- Lustrous (shiny — reflect light)
- Malleable (can be beaten into sheets)
- Ductile (can be drawn into wires)
- High melting points (most metals; exception: Hg is liquid at room temp)
- Sonorous (ring when struck)
*Why A is wrong:* Brittleness is a property of non-metals (e.g., sulphur, phosphorus, ionic crystals). Metals are malleable and ductile.
*Why B is wrong:* Most metals have HIGH melting points (Fe: 1538°C, W: 3422°C). Exception: Hg (−39°C) and Ga (30°C) melt near room temperature.
*Why D is wrong:* Metals are LUSTROUS (shiny) — non-metals are generally non-lustrous (except graphite and iodine).
- Good conductors of heat and electricity (free electrons)
- Lustrous (shiny — reflect light)
- Malleable (can be beaten into sheets)
- Ductile (can be drawn into wires)
- High melting points (most metals; exception: Hg is liquid at room temp)
- Sonorous (ring when struck)
*Why A is wrong:* Brittleness is a property of non-metals (e.g., sulphur, phosphorus, ionic crystals). Metals are malleable and ductile.
*Why B is wrong:* Most metals have HIGH melting points (Fe: 1538°C, W: 3422°C). Exception: Hg (−39°C) and Ga (30°C) melt near room temperature.
*Why D is wrong:* Metals are LUSTROUS (shiny) — non-metals are generally non-lustrous (except graphite and iodine).
Q133MediumBPSC Prelims
The process of coating iron/steel with zinc to prevent corrosion is called:
ATinning
BElectroplating
CGalvanisation
DAnodising
Show Answer
✔ C — Galvanisation
Galvanisation: coating iron/steel with a layer of zinc to prevent rusting.
Zinc acts as a sacrificial anode — even if the zinc coating is scratched, zinc oxidises preferentially (Zn is more reactive than Fe), protecting the iron underneath.
Zn + H₂O + O₂ → ZnO/Zn(OH)₂ (protective layer)
Uses: galvanised iron pipes, roofing sheets, buckets, dustbins, corrugated iron roofs.
*Why A is wrong:* Tinning = coating with tin (Sn) — used for food cans (less toxic than Pb). Tin doesn't provide sacrificial protection; once scratched, iron rusts.
*Why B is wrong:* Electroplating deposits a metal layer using electrolysis — broader term, not specific to zinc on iron.
*Why D is wrong:* Anodising = oxidising aluminium surface to create Al₂O₃ protective layer (used for aluminium items — not iron/zinc).
Zinc acts as a sacrificial anode — even if the zinc coating is scratched, zinc oxidises preferentially (Zn is more reactive than Fe), protecting the iron underneath.
Zn + H₂O + O₂ → ZnO/Zn(OH)₂ (protective layer)
Uses: galvanised iron pipes, roofing sheets, buckets, dustbins, corrugated iron roofs.
*Why A is wrong:* Tinning = coating with tin (Sn) — used for food cans (less toxic than Pb). Tin doesn't provide sacrificial protection; once scratched, iron rusts.
*Why B is wrong:* Electroplating deposits a metal layer using electrolysis — broader term, not specific to zinc on iron.
*Why D is wrong:* Anodising = oxidising aluminium surface to create Al₂O₃ protective layer (used for aluminium items — not iron/zinc).
Q134MediumBPSC Prelims
Which of the following correctly describes the reactivity series (activity series) of metals?
AGold is more reactive than iron; iron is more reactive than sodium
BPotassium is most reactive; gold is least reactive; iron is in the middle
CAluminium is more reactive than potassium; copper is least reactive
DAll metals have equal reactivity
Show Answer
✔ B — Potassium is most reactive; gold is least reactive; iron is in the middle
Reactivity series (most to least reactive):
K > Na > Ca > Mg > Al > Zn > Fe > Ni > Sn > Pb > H > Cu > Hg > Ag > Au > Pt
- Potassium (K): most reactive metal — reacts violently with water
- Gold (Au/Pt): least reactive — noble metals, found free in nature
- Iron (Fe): middle reactivity — reacts with dilute acids, not cold water
*Why A is wrong:* Gold is LEAST reactive; sodium is MORE reactive than iron — completely reversed.
*Why C is wrong:* Aluminium (Al) is LESS reactive than potassium (K); copper is not the least reactive (gold and platinum are less reactive).
*Why D is wrong:* Metals have vastly different reactivities — the entire basis of metallurgy and corrosion chemistry.
K > Na > Ca > Mg > Al > Zn > Fe > Ni > Sn > Pb > H > Cu > Hg > Ag > Au > Pt
- Potassium (K): most reactive metal — reacts violently with water
- Gold (Au/Pt): least reactive — noble metals, found free in nature
- Iron (Fe): middle reactivity — reacts with dilute acids, not cold water
*Why A is wrong:* Gold is LEAST reactive; sodium is MORE reactive than iron — completely reversed.
*Why C is wrong:* Aluminium (Al) is LESS reactive than potassium (K); copper is not the least reactive (gold and platinum are less reactive).
*Why D is wrong:* Metals have vastly different reactivities — the entire basis of metallurgy and corrosion chemistry.
Q135MediumBPSC Prelims
Thermite reaction uses aluminium powder and iron oxide. It produces:
AAluminium oxide and carbon
BIron and aluminium oxide, with intense heat
CIron oxide and aluminium chloride
DIron sulphide and aluminium hydroxide
Show Answer
✔ B — Iron and aluminium oxide, with intense heat
Thermite reaction: 2Al + Fe₂O₃ → Al₂O₃ + 2Fe + heat (~3000°C)
- Aluminium (more reactive) displaces iron from iron oxide
- Products: molten iron + aluminium oxide + enormous heat
- Uses: welding railway tracks (thermite welding), incendiary bombs
*Why A is wrong:* No carbon in the reaction — carbon is not a reactant.
*Why C is wrong:* No chlorine in the reaction; AlCl₃ is not a product.
*Why D is wrong:* Neither sulphide nor hydroxide is formed — no S or H₂O source.
- Aluminium (more reactive) displaces iron from iron oxide
- Products: molten iron + aluminium oxide + enormous heat
- Uses: welding railway tracks (thermite welding), incendiary bombs
*Why A is wrong:* No carbon in the reaction — carbon is not a reactant.
*Why C is wrong:* No chlorine in the reaction; AlCl₃ is not a product.
*Why D is wrong:* Neither sulphide nor hydroxide is formed — no S or H₂O source.
Q136HardBPSC Prelims
Stainless steel is an alloy of iron with which other elements?
AIron + Copper + Zinc
BIron + Chromium + Nickel
CIron + Carbon + Manganese only
DIron + Tin + Lead
Show Answer
✔ B — Iron + Chromium + Nickel
Stainless steel: Iron + Chromium (10–18%) + Nickel (8%)
- Chromium forms a passive Cr₂O₃ layer → prevents corrosion
- Nickel improves strength and toughness
Other steel alloys:
- Mild steel: Fe + C (0.1–0.3%) → general construction
- Cast iron: Fe + C (2–4%) → engine blocks, pipes
- Manganese steel: Fe + Mn → rail tracks, helmets (very hard)
*Why A is wrong:* Fe + Cu + Zn = no such standard alloy for cutlery/cookware.
*Why C is wrong:* Fe + C + Mn = tool steel or manganese steel — not stainless (still rusts without chromium).
*Why D is wrong:* Fe + Sn + Pb = babbitt metal (bearing alloy) — completely different application.
- Chromium forms a passive Cr₂O₃ layer → prevents corrosion
- Nickel improves strength and toughness
Other steel alloys:
- Mild steel: Fe + C (0.1–0.3%) → general construction
- Cast iron: Fe + C (2–4%) → engine blocks, pipes
- Manganese steel: Fe + Mn → rail tracks, helmets (very hard)
*Why A is wrong:* Fe + Cu + Zn = no such standard alloy for cutlery/cookware.
*Why C is wrong:* Fe + C + Mn = tool steel or manganese steel — not stainless (still rusts without chromium).
*Why D is wrong:* Fe + Sn + Pb = babbitt metal (bearing alloy) — completely different application.
Q137HardBPSC Prelims
The Hall-Héroult process is used for the industrial production of:
AIron
BCopper
CAluminium
DSodium
Show Answer
✔ C — Aluminium
Hall-Héroult process: electrolytic reduction of aluminium oxide (Al₂O₃/alumina) dissolved in cryolite (Na₃AlF₆) at ~960°C.
- Cathode: Al³⁺ + 3e⁻ → Al(l) (aluminium is deposited)
- Anode: O²⁻ → O₂ (oxygen released, burns carbon anodes)
Aluminium cannot be produced by reducing Al₂O₃ with carbon (too stable) → must use electrolysis.
*Why A is wrong:* Iron is produced by Blast Furnace process (reduction with coke/CO).
*Why B is wrong:* Copper is produced by smelting (roasting sulphide ores) then electrolytic refining.
*Why D is wrong:* Sodium is produced by Downs process (electrolysis of molten NaCl).
- Cathode: Al³⁺ + 3e⁻ → Al(l) (aluminium is deposited)
- Anode: O²⁻ → O₂ (oxygen released, burns carbon anodes)
Aluminium cannot be produced by reducing Al₂O₃ with carbon (too stable) → must use electrolysis.
*Why A is wrong:* Iron is produced by Blast Furnace process (reduction with coke/CO).
*Why B is wrong:* Copper is produced by smelting (roasting sulphide ores) then electrolytic refining.
*Why D is wrong:* Sodium is produced by Downs process (electrolysis of molten NaCl).
Q138HardBPSC Prelims
Which of the following statements about diamond and graphite (allotropes of carbon) is CORRECT?
AThey have the same crystal structure
BThey have the same electrical conductivity
CDiamond is the hardest natural substance; graphite is soft and a conductor
DThey have the same degree of hardness
Show Answer
✔ C — Diamond is the hardest natural substance; graphite is soft and a conductor
Diamond: Each C bonded to 4 others (sp³, tetrahedral) → giant covalent structure → hardest natural substance (10 on Mohs scale) → insulator (no free electrons)
Graphite: Each C bonded to 3 others in layers (sp², hexagonal) → delocalized electrons between layers → soft (layers slide over each other) → good electrical conductor
*Why A is wrong:* Diamond = 3D tetrahedral network; Graphite = layered hexagonal structure → completely different crystal structures.
*Why B is wrong:* Diamond is an insulator; graphite is a conductor — opposite properties.
*Why D is wrong:* Diamond is hardest (10 Mohs); graphite is very soft (1–2 Mohs) → completely different hardness.
Graphite: Each C bonded to 3 others in layers (sp², hexagonal) → delocalized electrons between layers → soft (layers slide over each other) → good electrical conductor
*Why A is wrong:* Diamond = 3D tetrahedral network; Graphite = layered hexagonal structure → completely different crystal structures.
*Why B is wrong:* Diamond is an insulator; graphite is a conductor — opposite properties.
*Why D is wrong:* Diamond is hardest (10 Mohs); graphite is very soft (1–2 Mohs) → completely different hardness.
Q139MediumBPSC Prelims
Rusting of iron is an electrochemical process that requires:
AOnly oxygen
BOnly water
CBoth water and oxygen (or CO₂)
DHydrogen gas
Show Answer
✔ C — Both water and oxygen (or CO₂)
Rusting (corrosion of iron): requires both water AND oxygen (H₂O and O₂)
Chemical reaction:
4Fe + 3O₂ + 6H₂O → 4Fe(OH)₃ → 2Fe₂O₃·3H₂O (rust = hydrated iron oxide)
Evidence:
- Iron in dry oxygen → no rust
- Iron in distilled water (de-aerated, no O₂) → no rust
- Iron in water + oxygen → RUSTS
CO₂ dissolved in water speeds up rusting (forms carbonic acid, lowers pH).
*Why A is wrong:* Oxygen alone (dry) → no rusting (confirmed by experiments).
*Why B is wrong:* Water alone (deoxygenated) → no rusting.
*Why D is wrong:* Hydrogen is not involved in rusting — it's the product of some corrosion reactions, not a requirement.
Chemical reaction:
4Fe + 3O₂ + 6H₂O → 4Fe(OH)₃ → 2Fe₂O₃·3H₂O (rust = hydrated iron oxide)
Evidence:
- Iron in dry oxygen → no rust
- Iron in distilled water (de-aerated, no O₂) → no rust
- Iron in water + oxygen → RUSTS
CO₂ dissolved in water speeds up rusting (forms carbonic acid, lowers pH).
*Why A is wrong:* Oxygen alone (dry) → no rusting (confirmed by experiments).
*Why B is wrong:* Water alone (deoxygenated) → no rusting.
*Why D is wrong:* Hydrogen is not involved in rusting — it's the product of some corrosion reactions, not a requirement.
Q140HardBPSC Prelims
The process of extracting a pure metal from its ore using electricity is called:
ASmelting
BCalcination
CElectrolytic reduction (Electrolysis)
DFroth flotation
Show Answer
✔ C — Electrolytic reduction (Electrolysis)
Electrolytic reduction: using electric current to decompose molten ore or aqueous salt solution to obtain the pure metal at the cathode.
Used for: Al (from Al₂O₃), Na (from molten NaCl), Mg, Ca, Cu (electrolytic refining)
M^n+ + ne⁻ → M (at cathode)
Other metallurgy processes:
- Smelting: heating ore with reducing agent (coke) → Fe, Pb, Sn
- Calcination: heating ore without air → removes CO₂ (CaCO₃ → CaO + CO₂)
- Roasting: heating ore in excess air → removes sulphur (sulphide ores → oxides)
- Froth flotation: ore concentration step (not extraction) — sulphide minerals float on foam
*Why A is wrong:* Smelting uses carbon/coke as reducing agent, not electricity — for moderately reactive metals.
*Why B is wrong:* Calcination decomposes carbonates/hydroxides — a preliminary step, not final extraction.
*Why D is wrong:* Froth flotation concentrates ore (removes gangue) but doesn't extract pure metal.
Used for: Al (from Al₂O₃), Na (from molten NaCl), Mg, Ca, Cu (electrolytic refining)
M^n+ + ne⁻ → M (at cathode)
Other metallurgy processes:
- Smelting: heating ore with reducing agent (coke) → Fe, Pb, Sn
- Calcination: heating ore without air → removes CO₂ (CaCO₃ → CaO + CO₂)
- Roasting: heating ore in excess air → removes sulphur (sulphide ores → oxides)
- Froth flotation: ore concentration step (not extraction) — sulphide minerals float on foam
*Why A is wrong:* Smelting uses carbon/coke as reducing agent, not electricity — for moderately reactive metals.
*Why B is wrong:* Calcination decomposes carbonates/hydroxides — a preliminary step, not final extraction.
*Why D is wrong:* Froth flotation concentrates ore (removes gangue) but doesn't extract pure metal.
Carbon & Organic Chemistry
Q141EasyBPSC Prelims
The functional group –OH (hydroxyl group) is present in:
AAldehyde
BKetone
CAlcohol
DEther
Show Answer
✔ C — Alcohol
Functional groups:
- Alcohol: –OH (hydroxyl) — R–OH (e.g., CH₃OH = methanol, C₂H₅OH = ethanol)
- Aldehyde: –CHO (carbonyl at chain end) — R–CHO (e.g., HCHO = formaldehyde)
- Ketone: –C=O– (carbonyl between carbons) — R–CO–R' (e.g., CH₃COCH₃ = acetone)
- Ether: –O– (oxygen between carbons) — R–O–R' (e.g., CH₃OCH₃ = dimethyl ether)
*Why A is wrong:* Aldehyde has –CHO (carbonyl + H at end), not –OH hydroxyl.
*Why B is wrong:* Ketone has C=O between carbon atoms — no –OH.
*Why D is wrong:* Ether has C–O–C linkage, not –OH.
- Alcohol: –OH (hydroxyl) — R–OH (e.g., CH₃OH = methanol, C₂H₅OH = ethanol)
- Aldehyde: –CHO (carbonyl at chain end) — R–CHO (e.g., HCHO = formaldehyde)
- Ketone: –C=O– (carbonyl between carbons) — R–CO–R' (e.g., CH₃COCH₃ = acetone)
- Ether: –O– (oxygen between carbons) — R–O–R' (e.g., CH₃OCH₃ = dimethyl ether)
*Why A is wrong:* Aldehyde has –CHO (carbonyl + H at end), not –OH hydroxyl.
*Why B is wrong:* Ketone has C=O between carbon atoms — no –OH.
*Why D is wrong:* Ether has C–O–C linkage, not –OH.
Q142EasyBPSC Prelims
Methane (CH₄) is the simplest alkane. Which of the following correctly describes alkanes?
AContain double bonds between carbon atoms
BContain triple bonds between carbon atoms
CContain only single bonds between carbon atoms (saturated)
DContain alternating single and double bonds
Show Answer
✔ C — Contain only single bonds between carbon atoms (saturated)
Alkanes (saturated hydrocarbons): CₙH₂ₙ₊₂ — contain ONLY single C–C and C–H bonds
- CH₄ (methane), C₂H₆ (ethane), C₃H₈ (propane), C₄H₁₀ (butane)
- Saturated = maximum hydrogen content for given number of carbons
Alkenes (CₙH₂ₙ): contain one double bond (C=C)
Alkynes (CₙH₂ₙ₋₂): contain one triple bond (C≡C)
Benzene/Arenes: alternating single and double bonds (aromatic)
*Why A is wrong:* Double bonds = alkenes (unsaturated).
*Why B is wrong:* Triple bonds = alkynes.
*Why D is wrong:* Alternating single-double = benzene ring (aromatic compounds).
- CH₄ (methane), C₂H₆ (ethane), C₃H₈ (propane), C₄H₁₀ (butane)
- Saturated = maximum hydrogen content for given number of carbons
Alkenes (CₙH₂ₙ): contain one double bond (C=C)
Alkynes (CₙH₂ₙ₋₂): contain one triple bond (C≡C)
Benzene/Arenes: alternating single and double bonds (aromatic)
*Why A is wrong:* Double bonds = alkenes (unsaturated).
*Why B is wrong:* Triple bonds = alkynes.
*Why D is wrong:* Alternating single-double = benzene ring (aromatic compounds).
Q143MediumBPSC Prelims
The fermentation of glucose by yeast produces:
AMethane and water
BEthanol and carbon dioxide
CAcetic acid and oxygen
DPropanol and nitrogen
Show Answer
✔ B — Ethanol and carbon dioxide
Alcoholic fermentation:
C₆H₁₂O₆ (glucose) → 2C₂H₅OH (ethanol) + 2CO₂↑
Catalyst: yeast (contains enzyme zymase/invertase)
Conditions: anaerobic (no oxygen), ~35°C
This is the basis of:
- Beer/wine making (grain/grape fermentation)
- Ethanol production for industrial and fuel use
- Bread making (CO₂ makes dough rise)
*Why A is wrong:* Methane + water is produced by anaerobic digestion of complex organic matter (biogas production), not simple glucose fermentation.
*Why C is wrong:* Acetic acid (vinegar) is produced by further oxidation of ethanol by Acetobacter bacteria (aerobic) — not the primary fermentation product.
*Why D is wrong:* Propanol and nitrogen — no fermentation pathway produces propanol from glucose fermentation.
C₆H₁₂O₆ (glucose) → 2C₂H₅OH (ethanol) + 2CO₂↑
Catalyst: yeast (contains enzyme zymase/invertase)
Conditions: anaerobic (no oxygen), ~35°C
This is the basis of:
- Beer/wine making (grain/grape fermentation)
- Ethanol production for industrial and fuel use
- Bread making (CO₂ makes dough rise)
*Why A is wrong:* Methane + water is produced by anaerobic digestion of complex organic matter (biogas production), not simple glucose fermentation.
*Why C is wrong:* Acetic acid (vinegar) is produced by further oxidation of ethanol by Acetobacter bacteria (aerobic) — not the primary fermentation product.
*Why D is wrong:* Propanol and nitrogen — no fermentation pathway produces propanol from glucose fermentation.
Q144MediumBPSC Prelims
Saponification is the process by which soap is made. It involves:
AHeating oils/fats with concentrated H₂SO₄
BHeating oils/fats with concentrated NaOH (or KOH) solution
CFermenting vegetable oils with yeast
DReacting oils with HCl to form fatty acids
Show Answer
✔ B — Heating oils/fats with concentrated NaOH (or KOH) solution
Saponification: Triglyceride (fat/oil) + NaOH → Glycerol + Sodium fatty acid salt (soap)
Fat (ester) + 3NaOH → Glycerol + 3R-COONa (sodium soap)
- NaOH → hard soap (sodium salt of fatty acid) — bar soap
- KOH → soft soap/liquid soap (potassium salt) — shaving cream
The soap molecule: hydrophobic tail (R, hydrocarbon) + hydrophilic head (–COO⁻Na⁺) → forms micelles → removes grease
*Why A is wrong:* Heating with H₂SO₄ gives hydrolysis → glycerol + free fatty acids (not soap).
*Why C is wrong:* Fermentation of oils doesn't produce soap — wrong chemistry entirely.
*Why D is wrong:* Reacting with HCl would neutralise the alkaline conditions needed for saponification.
Fat (ester) + 3NaOH → Glycerol + 3R-COONa (sodium soap)
- NaOH → hard soap (sodium salt of fatty acid) — bar soap
- KOH → soft soap/liquid soap (potassium salt) — shaving cream
The soap molecule: hydrophobic tail (R, hydrocarbon) + hydrophilic head (–COO⁻Na⁺) → forms micelles → removes grease
*Why A is wrong:* Heating with H₂SO₄ gives hydrolysis → glycerol + free fatty acids (not soap).
*Why C is wrong:* Fermentation of oils doesn't produce soap — wrong chemistry entirely.
*Why D is wrong:* Reacting with HCl would neutralise the alkaline conditions needed for saponification.
Q145MediumBPSC Prelims
Ethylene (C₂H₄) is used to ripen fruits artificially. Ethylene belongs to which class of hydrocarbons?
AAlkanes (saturated)
BAlkynes (triple bond)
CAlkenes (double bond)
DAromatic (benzene ring)
Show Answer
✔ C — Alkenes (double bond)
Ethylene (Ethene, C₂H₄): H₂C=CH₂ — contains one C=C double bond → Alkene (unsaturated hydrocarbon)
As a plant hormone: ethylene triggers fruit ripening (climacteric fruits: banana, mango, tomato)
Industrial uses: production of polyethylene (plastic), ethanol (by hydration), vinyl chloride (PVC)
*Why A is wrong:* Alkanes (CₙH₂ₙ₊₂) have only single bonds; ethylene has formula C₂H₄ = CₙH₂ₙ → alkene.
*Why B is wrong:* Alkynes (CₙH₂ₙ₋₂) have triple bond; ethylene's formula doesn't match alkyne series (that would be C₂H₂ = acetylene).
*Why D is wrong:* Aromatic compounds have benzene ring (6-carbon ring with alternating bonds); ethylene is a simple 2-carbon compound.
As a plant hormone: ethylene triggers fruit ripening (climacteric fruits: banana, mango, tomato)
Industrial uses: production of polyethylene (plastic), ethanol (by hydration), vinyl chloride (PVC)
*Why A is wrong:* Alkanes (CₙH₂ₙ₊₂) have only single bonds; ethylene has formula C₂H₄ = CₙH₂ₙ → alkene.
*Why B is wrong:* Alkynes (CₙH₂ₙ₋₂) have triple bond; ethylene's formula doesn't match alkyne series (that would be C₂H₂ = acetylene).
*Why D is wrong:* Aromatic compounds have benzene ring (6-carbon ring with alternating bonds); ethylene is a simple 2-carbon compound.
Q146MediumBPSC Prelims
Carbon is said to show "catenation." This means:
ACarbon forms bonds with all other elements easily
BCarbon forms long chains with itself (C–C–C–C bonds)
CCarbon has four valence electrons
DCarbon is found in all living organisms
Show Answer
✔ B — Carbon forms long chains with itself (C–C–C–C bonds)
Catenation: the unique property of carbon to form long chains, branches, and rings by bonding with other carbon atoms (C–C bonds).
This is why millions of organic compounds exist — carbon can form:
- Straight chains: CH₃–CH₂–CH₂–CH₃ (butane)
- Branched chains: isobutane
- Rings: cyclohexane, benzene
- Long polymers: polyethylene (thousands of C atoms)
Why carbon excels at catenation: C–C bond is strong (347 kJ/mol) AND C is small (forms stable chains). Silicon also shows catenation but much weaker.
*Why A is wrong:* Carbon forms bonds with many elements, but "catenation" specifically means bonding with itself.
*Why C is wrong:* Four valence electrons is tetravalence — a related but different property (allows 4 bonds).
*Why D is wrong:* Carbon in living organisms = organic chemistry/biochemistry; catenation is specifically about C–C chain formation.
This is why millions of organic compounds exist — carbon can form:
- Straight chains: CH₃–CH₂–CH₂–CH₃ (butane)
- Branched chains: isobutane
- Rings: cyclohexane, benzene
- Long polymers: polyethylene (thousands of C atoms)
Why carbon excels at catenation: C–C bond is strong (347 kJ/mol) AND C is small (forms stable chains). Silicon also shows catenation but much weaker.
*Why A is wrong:* Carbon forms bonds with many elements, but "catenation" specifically means bonding with itself.
*Why C is wrong:* Four valence electrons is tetravalence — a related but different property (allows 4 bonds).
*Why D is wrong:* Carbon in living organisms = organic chemistry/biochemistry; catenation is specifically about C–C chain formation.
Q147HardBPSC Prelims
Benzene (C₆H₆) has an unusual stability because:
AIt contains three isolated double bonds that make it highly reactive
BIts electrons are delocalized over the ring (resonance/aromaticity), making it extra stable
CIt has a linear structure unlike other organic compounds
DIt contains ionic bonds between carbon atoms
Show Answer
✔ B — Its electrons are delocalized over the ring (resonance/aromaticity), making it extra stable
Benzene's structure (Kekulé proposed alternating double-single bonds, but):
- All C–C bonds in benzene are EQUAL length (139 pm, between single 154 pm and double 134 pm)
- 6 π electrons are delocalized over all 6 carbons (cloud above and below the ring)
- This delocalization = aromatic stability / resonance energy (~150 kJ/mol stabilization)
- Benzene undergoes electrophilic substitution (not addition) — prefers to maintain aromatic ring
*Why A is wrong:* Benzene does NOT have three isolated double bonds that make it reactive — its delocalized electrons make it LESS reactive than alkenes (aromatic stability).
*Why C is wrong:* Benzene has a cyclic (ring) structure — planar, NOT linear.
*Why D is wrong:* Benzene has covalent (not ionic) bonds — carbon doesn't form ionic bonds with itself.
- All C–C bonds in benzene are EQUAL length (139 pm, between single 154 pm and double 134 pm)
- 6 π electrons are delocalized over all 6 carbons (cloud above and below the ring)
- This delocalization = aromatic stability / resonance energy (~150 kJ/mol stabilization)
- Benzene undergoes electrophilic substitution (not addition) — prefers to maintain aromatic ring
*Why A is wrong:* Benzene does NOT have three isolated double bonds that make it reactive — its delocalized electrons make it LESS reactive than alkenes (aromatic stability).
*Why C is wrong:* Benzene has a cyclic (ring) structure — planar, NOT linear.
*Why D is wrong:* Benzene has covalent (not ionic) bonds — carbon doesn't form ionic bonds with itself.
Q148HardBPSC Prelims
Which of the following is NOT correctly matched?
AEthanol — present in alcoholic drinks; used as fuel
BAcetic acid — present in vinegar (3-5% solution)
CMethane — component of natural gas (biogas, CNG)
DBenzene — used as a sweetener in food industry
Show Answer
✔ D — Benzene — used as a sweetener in food industry
Benzene is NOT used as a food sweetener — it is a toxic carcinogen (Class 1 IARC carcinogen). Its use in foods is strictly prohibited.
Food sweeteners: Sucrose (common sugar), Saccharin (300× sweeter than sugar), Aspartame, Stevia — none contain benzene.
Correct matches:
- Ethanol: in alcoholic drinks (beer ~5%, wine ~12%, spirits ~40%), biofuel (E20 blending)
- Acetic acid: vinegar is 3–5% CH₃COOH in water
- Methane: main component of natural gas (85–90%), biogas (50–70%), CNG
*Why A is wrong (it IS correct):* Ethanol is in alcoholic drinks AND used as biofuel — correct match.
*Why B is wrong (it IS correct):* Vinegar is dilute acetic acid (3–5%) — correct.
*Why C is wrong (it IS correct):* Methane = natural gas component — correct.
Food sweeteners: Sucrose (common sugar), Saccharin (300× sweeter than sugar), Aspartame, Stevia — none contain benzene.
Correct matches:
- Ethanol: in alcoholic drinks (beer ~5%, wine ~12%, spirits ~40%), biofuel (E20 blending)
- Acetic acid: vinegar is 3–5% CH₃COOH in water
- Methane: main component of natural gas (85–90%), biogas (50–70%), CNG
*Why A is wrong (it IS correct):* Ethanol is in alcoholic drinks AND used as biofuel — correct match.
*Why B is wrong (it IS correct):* Vinegar is dilute acetic acid (3–5%) — correct.
*Why C is wrong (it IS correct):* Methane = natural gas component — correct.
Q149HardBPSC Prelims
Polyethylene (polythene) is a polymer made from:
AAcetylene (C₂H₂) by addition polymerisation
BEthylene (C₂H₄) by addition polymerisation
CEthanol by condensation polymerisation
DStyrene by free radical polymerisation with sulfur
Show Answer
✔ B — Ethylene (C₂H₄) by addition polymerisation
Polyethylene: formed by addition polymerisation of ethylene (C₂H₄ = ethene)
nCH₂=CH₂ → (–CH₂–CH₂–)ₙ (polyethylene)
The C=C double bond "opens up" and monomers link into a long chain.
Types: LDPE (low density, soft, bags), HDPE (high density, rigid, bottles, pipes)
*Why A is wrong:* Acetylene (C₂H₂) polymerises to polyacetylene — not polyethylene. Different structure.
*Why C is wrong:* Ethanol is not a monomer for addition polymerisation; condensation polymerisation produces polyesters/nylons (involves –OH and –COOH groups).
*Why D is wrong:* Styrene (C₆H₅CH=CH₂) polymerises to polystyrene — not polyethylene.
nCH₂=CH₂ → (–CH₂–CH₂–)ₙ (polyethylene)
The C=C double bond "opens up" and monomers link into a long chain.
Types: LDPE (low density, soft, bags), HDPE (high density, rigid, bottles, pipes)
*Why A is wrong:* Acetylene (C₂H₂) polymerises to polyacetylene — not polyethylene. Different structure.
*Why C is wrong:* Ethanol is not a monomer for addition polymerisation; condensation polymerisation produces polyesters/nylons (involves –OH and –COOH groups).
*Why D is wrong:* Styrene (C₆H₅CH=CH₂) polymerises to polystyrene — not polyethylene.
Q150HardBPSC Prelims
The Haber process produces ammonia (NH₃) from nitrogen and hydrogen. The conditions required are:
AHigh temperature (200°C), low pressure, no catalyst
BHigh temperature (450–500°C), high pressure (150–300 atm), iron catalyst
CLow temperature (25°C), low pressure (1 atm), platinum catalyst
DHigh temperature (1000°C), low pressure, osmium catalyst
Show Answer
✔ B — High temperature (450–500°C), high pressure (150–300 atm), iron catalyst
Haber process: N₂ + 3H₂ ⇌ 2NH₃ (ΔH = −92 kJ/mol, exothermic)
Conditions:
- Temperature: 450–500°C (compromise — lower T favours product but too slow; higher T too fast but too little product)
- Pressure: 150–300 atm (high P favours product — fewer moles of gas on product side: 4 mol → 2 mol)
- Catalyst: Iron (Fe) with promoters (Al₂O₃, K₂O)
*Why A is wrong:* Low pressure disfavours NH₃ yield; 200°C is too low for reasonable reaction rate.
*Why C is wrong:* Room temperature and 1 atm → negligible reaction rate; Pt is not the Haber catalyst.
*Why D is wrong:* 1000°C would decompose NH₃ (reverse reaction); osmium was the original catalyst but Fe is now used industrially.
Conditions:
- Temperature: 450–500°C (compromise — lower T favours product but too slow; higher T too fast but too little product)
- Pressure: 150–300 atm (high P favours product — fewer moles of gas on product side: 4 mol → 2 mol)
- Catalyst: Iron (Fe) with promoters (Al₂O₃, K₂O)
*Why A is wrong:* Low pressure disfavours NH₃ yield; 200°C is too low for reasonable reaction rate.
*Why C is wrong:* Room temperature and 1 atm → negligible reaction rate; Pt is not the Haber catalyst.
*Why D is wrong:* 1000°C would decompose NH₃ (reverse reaction); osmium was the original catalyst but Fe is now used industrially.