CHAPTER 1: REFLECTION OF LIGHT AT CURVED SURFACES
I
I. Reflections on concepts
1. Question:
Where will the image be formed when an object is placed between the focus (F) and centre of curvature (C) of a concave mirror?
Answer:
The image is formed beyond C, and it is real, inverted, and enlarged.
2. Question:
State the differences between convex and concave mirrors.
Answer:
- Concave mirror: Converges light rays, forms real or virtual images.
- Convex mirror: Diverges light rays, forms only virtual, erect, and diminished images.
3. Question:
Distinguish between real and virtual images.
Answer:
- Real image: Formed by actual intersection of rays, can be obtained on a screen, usually inverted.
- Virtual image: Formed by apparent intersection, cannot be obtained on a screen, always erect.
4. Question:
How do you get a virtual image using a concave mirror?
Answer:
Place the object between the pole (P) and focus (F). The image formed is virtual, erect, and enlarged behind the mirror.
5. Question:
What do you know about the following terms related to spherical mirrors?
Answer:
- Pole (P): Centre of the mirror surface
- Centre of curvature (C): Centre of the sphere of which mirror is a part
- Focus (F): Point where parallel rays meet or appear to meet
- Radius of curvature (R): Distance between P and C
- Focal length (f): Distance between P and F (f = R/2)
- Principal axis: Straight line passing through P and C
- Object distance (u): Distance of object from pole
- Image distance (v): Distance of image from pole
- Magnification (m): Ratio of image height to object height
6. Question:
What do you infer from the experiment used to measure object distance and image distance?
Answer:
There is a relation between u, v, and f given by the mirror formula:
II. Application of concepts
1. Question:
Find the distance of the image when an object is placed on the principal axis at a distance of 10 cm in front of a concave mirror whose radius of curvature is 8 cm.
Answer:
Given:
Object distance (u) = −10 cm
Radius of curvature (R) = 8 cm → f = R/2 = −4 cm
Using mirror formula:
1/f = 1/v + 1/u
1/(-4) = 1/v + 1/(-10)
-1/4 = 1/v - 1/10
1/v = -1/4 + 1/10
1/v = (-5 + 2)/20 = -3/20
v = −20/3 ≈ −6.67 cm
👉 Image is formed 6.67 cm in front of the mirror (real image).
Image distance v = −6.67 cm, so the image is formed in front of the mirror (real image).
2. Question:
The magnification produced by a mirror is +1. What does it mean?
Answer:
The image is same size as the object, erect, and virtual.
3. Question:
If the spherical mirrors were not known to human beings, given the consequences?
Answer:
We would lack many useful applications such as rear-view mirrors in vehicles, shaving/makeup mirrors, solar cookers, and reflecting telescopes. Daily life and scientific progress would be affected.
4. Question:
Draw suitable rays by which we can guess the position of image formed by a concave mirror.
Answer:
Use two standard rays:
- A ray parallel to the principal axis reflects through the focus (F).
- A ray passing through the centre of curvature (C) reflects back along the same path.
The intersection of reflected rays gives the image position.
5. Question:
Show the formation of image with a ray diagram when an object is placed on the principal axis of a concave mirror away from the centre of curvature.
Answer:
When the object is beyond C, the image forms between F and C, and it is real, inverted, and diminished.
6. Question:
Why do people prefer a convex mirror as a rear-view mirror in vehicles?
Answer:
Because it provides a wide field of view and forms erect, diminished, virtual images, helping drivers see more area behind them.
************************************
III. Higher Order Thinking Questions
1. Question:
A convex mirror with a radius of curvature of 3 m is used as a rear-view mirror. If a bus is located 5 m from this mirror, find the position, nature, and size of the image.
Answer:
Given:
R = 3 m → f = +1.5 m
u = −5 m
Using mirror formula:
1/f = 1/v + 1/u
1/1.5 = 1/v − 1/5
1/v = 1/1.5 + 1/5 = 2/3 + 1/5 = 13/15
v = 15/13 ≈ +1.15 m
- Position: Behind the mirror
- Nature: Virtual and erect
- Size: Diminished
2. Question:
To form the image on the object itself, how should we place the object in front of a concave mirror? Explain with a ray diagram.
Answer:
Place the object at the centre of curvature (C).
The image is formed at the same point (C), real, inverted, and same size.
**********************
Multiple Choice Questions
1. Question:
If an object is placed at C on the principal axis in front of a concave mirror, the position of image is:
a) at infinity
b) between Focus and Centre of curvature
c) at Centre of curvature
d) beyond Centre of curvature
Answer:
c) at Centre of curvature
2. Question:
We get a diminished image with a concave mirror when the object is placed:
a) at Focus
b) between pole and Focus
c) at Centre of curvature
d) beyond Centre of curvature
Answer:
d) beyond Centre of curvature
3. Question:
We get a virtual image in a concave mirror when the object is placed:
a) at Focus
b) between the pole and Focus
c) at Centre of curvature
d) beyond Centre of curvature
Answer:
b) between the pole and Focus
4. Question:
Which of the following represents magnification (m) for a spherical mirror?
Answer:
Correct expressions:
Correct option: b) (ii) & (iii)
5. Question:
Ray which seems to be travelling through the focus of a convex mirror, after reflection of an incident ray:
a) parallel to the axis
b) along the same path in opposite direction
c) through Focus
d) through Centre of curvature
Answer:
a) parallel to the axis
SUMMARY OF THIS LESSON :
Concave and Convex Mirrors – Complete Guide | Image Formation, Mirror Formula, Ray Diagrams
This comprehensive guide explains key concepts of concave mirror and convex mirror, including image formation in spherical mirrors, mirror formula, and ray diagrams—essential topics in physics and optics.
When an object is placed between the focus (F) and centre of curvature (C) of a concave mirror, the image is formed beyond C and is real, inverted, and enlarged. In contrast, a convex mirror always produces a virtual, erect, and diminished image, making it ideal for rear-view mirrors in vehicles due to its wide field of view.
Understanding the difference between real and virtual images is crucial:
- Real images are formed by actual intersection of light rays and can be projected on a screen.
- Virtual images are formed by apparent intersection and cannot be obtained on a screen.
To obtain a virtual image in a concave mirror, the object must be placed between the pole (P) and focus (F), resulting in an erect and magnified image behind the mirror.
Key terms in spherical mirrors include:
- Pole (P), Centre of Curvature (C), Focus (F)
- Focal Length (f = R/2), Radius of Curvature (R)
- Object Distance (u), Image Distance (v), Magnification (m)
The relationship between these is given by the mirror formula:
1/f = 1/v + 1/u, which is essential for solving numerical problems in optics.
Ray diagrams for concave mirrors use standard rays (parallel to principal axis and through C) to determine image position and nature. For example:
- Object beyond C → Image between F and C (real, inverted, diminished)
- Object at C → Image at C (real, inverted, same size)
In practical applications, convex mirrors are preferred as rear-view mirrors because they provide a wider field of vision, enhancing safety.
Chapter -2 : Chemical Equations
I. Reflections on Concepts
1. What information do you get from a balanced chemical equation?
A balanced chemical equation gives us several important pieces of information:
- Reactants and products: It tells what substances react and what are formed.
- Mole ratio: The coefficients show the ratio in which substances react and are produced.
- Mass relationship: Since atoms are conserved, it obeys the law of conservation of mass.
- State of substances: Symbols like (s), (l), (g), (aq) indicate physical states.
- Conditions of reaction: Sometimes temperature, pressure, or catalyst is mentioned.
Example:
2H₂ + O₂ → 2H₂O
This shows 2 moles of hydrogen react with 1 mole of oxygen to form 2 moles of water.
2. Why should we balance a chemical equation?
A chemical equation must be balanced because:
- It follows the law of conservation of mass (matter is neither created nor destroyed).
- The number of atoms of each element must be equal on both sides.
- It ensures accurate calculations in chemistry (stoichiometry).
3. Balance the following chemical equations
(a) NaOH + H₂SO₄ → Na₂SO₄ + H₂O
Balanced equation:
2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O
Explanation:
We adjust coefficients so Na, H, O, and S atoms are equal on both sides.
(b) KClO₃ → KCl + O₂
Balanced equation:
2KClO₃ → 2KCl + 3O₂
(c) Hg(NO₃)₂ + KI → HgI₂ + KNO₃
Balanced equation:
Hg(NO₃)₂ + 2KI → HgI₂ + 2KNO₃
4. Mention physical states and balance equations
(a) C₆H₁₂O₆ → C₂H₅OH + CO₂
Balanced with states:
C₆H₁₂O₆ (aq) → 2C₂H₅OH (l) + 2CO₂ (g)
(b) NH₃ + Cl₂ → N₂ + NH₄Cl
Balanced with states:
8NH₃ (g) + 3Cl₂ (g) → N₂ (g) + 6NH₄Cl (s)
(c) Na + H₂O → NaOH + H₂
Balanced with states:
2Na (s) + 2H₂O (l) → 2NaOH (aq) + H₂ (g)
II. Application of Concepts
1. Balance the equations
(a) Calcium hydroxide + Nitric acid → Water + Calcium nitrate
Balanced:
Ca(OH)₂ (aq) + 2HNO₃ (aq) → 2H₂O (l) + Ca(NO₃)₂ (aq)
(b) Magnesium + Iodine → Magnesium iodide
Balanced:
Mg (s) + I₂ (s) → MgI₂ (s)
2. Write equations with states and balance
(a) Sodium hydroxide reacts with Hydrochloric acid
Balanced:
NaOH (aq) + HCl (aq) → NaCl (aq) + H₂O (l)
(b) Barium chloride reacts with Sodium sulphate
Balanced:
BaCl₂ (aq) + Na₂SO₄ (aq) → BaSO₄ (s) + 2NaCl (aq)
Note: BaSO₄ forms a precipitate (solid).
III. Higher Order Thinking Questions
1. Zinc reacts with cupric chloride
Given:
- Formula units of CuCl₂ = 6.023 × 10²²
- 1 mole = 6.023 × 10²³ particles
Moles of CuCl₂ =
= (6.023 × 10²²) / (6.023 × 10²³)
= 0.1 mole
Reaction:
Zn + CuCl₂ → ZnCl₂ + Cu
From equation:
1 mole CuCl₂ gives 1 mole Cu
So, 0.1 mole Cu is formed
2. Combustion of propane
Given:
- 1 mole propane gives “A” kJ
- 2.4 L propane at STP
At STP:
1 mole gas = 22.4 L
Moles of propane =
= 2.4 / 22.4
≈ 0.107 mole
Heat released =
= 0.107 × A kJ
3. Oxygen required to convert 2.4 kg graphite to CO₂
Reaction:
C + O₂ → CO₂
Molar mass:
- C = 12 g/mol
- O₂ = 32 g/mol
Moles of C =
= 2400 g / 12
= 200 moles
From equation:
1 mole C needs 1 mole O₂
So, O₂ required = 200 moles
Volume at STP:
= 200 × 22.4
= 4480 L
Mass of O₂ =
= 200 × 32
= 6400 g = 6.4 kg
Chapter 3: Acids, Bases and Salts
I. Reflections on Concepts
1. An acid or a base is mixed with water. Is this process exothermic or endothermic?
Answer: The process is exothermic.
Explanation:
When an acid or base dissolves in water, it releases heat. This happens because ions are formed and hydration energy is released. For example, when concentrated sulphuric acid is mixed with water, a large amount of heat is produced. That is why acids should always be added slowly to water, not the other way around, to avoid splashing.
2. Distilled water does not conduct electricity. Why?
Answer: Because it does not contain free ions.
Explanation:
Electric current in solutions is carried by ions. Distilled water is pure and lacks dissolved salts or impurities, so it has almost no ions. Therefore, it cannot conduct electricity.
3. Draw a diagram for showing acids conducting electricity
Answer (Explanation of Diagram):
A proper diagram would include:
- A beaker containing acid solution (like HCl in water)
- Two electrodes dipped in the solution
- A battery connected through wires
- A bulb or LED glowing
Explanation:
Acids ionize in water to produce ions (H⁺ and Cl⁻). These ions carry current, completing the circuit and lighting the bulb.
4. Why does acid rain flowing into a river make the survival of aquatic life difficult?
Answer: Because it lowers the pH of water.
Explanation:
Acid rain contains acids like sulphuric and nitric acid. When it enters rivers:
- It reduces pH (makes water acidic)
- Aquatic organisms need a specific pH range to survive
- Acidic water damages fish gills and affects reproduction
Thus, survival becomes difficult.
5. How does baking powder make a cake soft and spongy?
Answer: By releasing carbon dioxide gas.
Explanation:
Baking powder contains sodium bicarbonate and an acid. When heated:
NaHCO₃ → Na₂CO₃ + CO₂ + H₂O
- CO₂ gas forms bubbles in the batter
- These bubbles make the cake soft and spongy
II. Application of Concepts
1. Classification based on pH values
Given:
A = 8.4, B = 1.1, C = 11, D = 7, E = 9
(a) Neutral:
D (pH = 7)
(b) Strongly alkaline:
C (pH = 11)
(c) Strongly acidic:
B (pH = 1.1)
(d) Weakly acidic:
(None given; if closest, very slightly below 7 would be weak acid)
(e) Weakly alkaline:
A (8.4), E (9)
Explanation:
- pH < 7 → acidic
- pH = 7 → neutral
- pH > 7 → alkaline
- Lower pH → stronger acid
- Higher pH → stronger base
2. Why does tooth decay start when pH is lower than 5.5?
Answer: Because enamel starts dissolving.
Explanation:
Tooth enamel is made of calcium phosphate. At pH below 5.5:
- Acids react with enamel
- Minerals dissolve slowly
- This leads to tooth decay
3. Milkman adds baking soda to fresh milk
(a) Why does pH shift to slightly alkaline?
Answer: Because baking soda is basic.
Explanation:
Baking soda (NaHCO₃) increases pH, making milk slightly alkaline, which delays spoilage.
(b) Why does milk take longer to set as curd?
Answer: Because acidic conditions are delayed.
Explanation:
Curd formation requires lactic acid bacteria:
- They produce acid
- If milk is alkaline, more acid must be produced first
- So curdling takes more time
4. Why should Plaster of Paris be stored in moisture-proof containers?
Answer: Because it reacts with moisture and becomes hard.
Explanation:
Plaster of Paris reacts with water to form gypsum:
CaSO₄·½H₂O + 1.1/2H₂O → CaSO₄·2H₂O
- This makes it hard and unusable
- So it must be stored in dry conditions
5. Why does HCl produce more vigorous fizzing than acetic acid?
Answer: Because HCl is a strong acid.
Explanation:
- HCl completely ionizes in water
- Acetic acid partially ionizes
- More H⁺ ions → faster reaction → more CO₂ bubbles
Hence, more vigorous fizzing.
III. Higher Order Thinking Questions
1. Fresh milk has pH 6.6. Why does pH change as it turns into curd?
Answer: Because lactic acid is produced.
Explanation:
Bacteria convert lactose into lactic acid:
- pH decreases (becomes acidic)
- Milk turns into curd due to protein coagulation
2. How do you prepare an indicator using beetroot?
Answer (Procedure):
- Take beetroot pieces and crush them
- Extract the juice
- Filter the liquid
- Use the extract as an indicator
Explanation:
- Beetroot contains natural pigments
- It changes color in acidic and basic solutions
- Acts as a natural indicator
Chapter 4: Refraction of light at
Curved Surfaces
I. Reflections on Concepts
1. How do you verify experimentally that
the focal length of a convex lens is positive?
Answer:
To verify this:
- Take
a convex lens and focus sunlight or a distant object on a screen.
- Move
the screen until a sharp image is formed.
- Measure
the distance between the lens and the screen.
Explanation:
A convex lens converges parallel rays to a point (focus). Since the image forms
on the opposite side of the lens, the focal length is taken as positive
according to sign convention. This confirms that convex lenses have positive
focal length.
2. How do you verify that u (object
distance) is always negative?
Answer:
- Place
an object in front of a lens.
- Measure
the distance from the optical center of the lens to the object.
Explanation:
By sign convention in optics:
- Distances
measured in the direction of incident light are positive
- Distances
measured opposite to it (towards the object) are negative
Since the object is always placed to the left of
the lens, u is always negative.
3. How do you find the focal length of a
concave lens experimentally?
Answer:
- Use
a convex lens to form an image of an object.
- Place
a concave lens between the convex lens and the image.
- The
image shifts; adjust position to form a sharp image again.
- Measure
distances and use lens formula.
Explanation:
A concave lens alone forms a virtual image, so we combine it with a convex lens
to determine its focal length indirectly.
4. Draw ray diagrams for the following
situations
(i) Object placed at C (centre of
curvature)
- Image
forms at C
- Same
size
- Real
and inverted
(ii) Object placed beyond C
- Image
forms between F and C
- Smaller,
real and inverted
(iii) Object placed between F and C
- Image
forms beyond C
- Enlarged,
real and inverted
Explanation:
These are standard ray diagrams for a concave mirror, based on how rays
reflect through focus and centre of curvature.
5. What is the focal length of a plane
mirror?
Answer:
The focal length of a plane mirror is infinite.
Explanation:
A plane mirror does not converge or diverge rays. Its radius of curvature is
infinite, so:
f = R/2 = ∞
II. Application of Concepts
1. A convex lens is made of material of
refractive index 1.5. If it is placed in a medium of refractive index 1.63,
what happens?
Answer:
The lens will behave like a concave lens.
Explanation:
- Normally,
convex lens converges light when its refractive index is greater than
surrounding medium.
- Here,
surrounding medium has higher refractive index.
- So
light bends away from normal, making lens act like a concave lens.
2. The focal length of a convex lens is 20
cm. Object is placed at 10 cm. Find image nature and magnification.
Answer:
Question 2:
The focal length of a convex lens is 20 cm. An object is placed at 10 cm
1/f = 1/v - 1/u
V=-20 cm after simplification
4. Magnification Formula
Magnification for a lens is:
m =v/u
m=v/u
Substitute values:
m = -20/-10
m = +2
Nature of the Image
From the results:
Image distance (v) = −20 cm → Image is formed on the same side of the lens as the object, so it is virtual.
Magnification (m) = +2 → Image is erect and magnified (twice the size of the object).
Final Answer
Image distance: −20 cm
Magnification: +2
Nature of image: Virtual, erect, and magnified (twice the size of the object).
3. A double convex lens made of two
different materials. What happens to focal length?
Answer:
Focal length depends on refractive indices of both materials.
Explanation:
- Different
materials bend light differently
- So
overall focal length changes
- It
may increase or decrease depending on combination
4. A double convex lens made of two
different materials is surrounded by water. What happens?
Answer:
Focal length increases (lens becomes less powerful).
Explanation:
- Water
has refractive index close to lens material
- Refraction
reduces
- Convergence
decreases → focal length increases
5. Find the radius of curvature if focal
length is 15 cm
Answer:
Using relation:
R = 2f = 2 × 15 = 30 cm
III. Higher Order Thinking
Questions
1. A convex lens is made up of three
different materials. How many images are formed?
Answer:
Three images are formed.
Explanation:
Each material has different refractive index:
- Light
bends differently in each portion
- Each
part forms its own image
- Hence
multiple images are observed
2. You have a lens. Suggest an experiment
to find whether it is convex or concave.
Answer:
Method 1 (Image formation):
- Try
to form image of distant object on screen
- If
image forms → convex lens
- If
not → concave lens
Method 2 (Observation):
- Look
through lens
- Convex
lens magnifies (if close)
- Concave
lens always shows smaller image
3. Ray diagram: Focal length 15 cm, object
distance 30 cm
Answer:
Since object is at 2F:
- Image
forms at 2F
- Same
size
- Real
and inverted
4. Ray diagram: Focal length 15 cm, object
distance 10 cm
Answer:
Object is inside focus:
- Image
forms on same side
- Virtual,
erect, enlarged
5. Convex lens used as magnifying glass
Answer:
Yes, it can be used.
Explanation:
- Object
is placed within focal length
- Image
formed is virtual, erect, enlarged
- Hence
used as magnifying glass
6. Burning paper using convex lens
Answer:
Yes, possible.
Explanation:
- Convex
lens focuses sunlight at a point
- Heat
energy concentrates
- Temperature
rises → paper burns
7. Why should we not use a burnt candle
instead of a bright incandescent bulb in this experiment?
Answer:
Because candle light is not strong and not parallel.
Explanation:
- Bulb
gives strong, steady light
- Candle
gives weak, scattered light
Proper image
formation requires bright source
Chapter 5: Human eye and colourful
world
I. Improve your learning
1.
Define dispersion of light. Why does it
occur?
Answer:
Dispersion of light is the splitting of white light into its seven constituent
colours (VIBGYOR) when it passes through a prism.
It occurs because different colours of light have
different wavelengths. Each colour travels at a different speed in the prism
and bends by a different amount. Violet bends the most and red bends the least,
which results in the formation of a spectrum.
2.
Explain the formation of rainbow in the
sky.
Answer:
A rainbow is formed due to refraction, dispersion, and total internal
reflection of sunlight in raindrops.
When sunlight enters a raindrop, it gets refracted and
splits into different colours. The light is then reflected inside the droplet
and refracted again while coming out. This produces a band of seven colours in
the sky.
3.
What is atmospheric refraction? Give an
example.
Answer:
Atmospheric refraction is the bending of light as it passes through different
layers of air due to changes in air density.
Example: Twinkling of stars or advance sunrise and
delayed sunset.
4.
Why do stars twinkle but planets do not?
Answer:
Stars twinkle because they are very far away and act as point sources of light.
Atmospheric refraction causes their light to fluctuate, making them appear to
twinkle.
Planets are closer and appear larger, so the
fluctuations average out and they do not twinkle.
5.
Why is the sky blue in colour?
Answer:
The sky appears blue due to scattering of sunlight by air molecules. Blue light
has a shorter wavelength and is scattered more than other colours, so it
reaches our eyes from all directions.
II. Application of Concepts
1.
Explain why the sun appears reddish during
sunrise and sunset.
Answer:
During sunrise and sunset, sunlight travels a longer distance through the
atmosphere. Most of the shorter wavelengths like blue and violet are scattered
away. Only longer wavelengths like red reach our eyes, making the sun appear
reddish.
2.
Why does the sky appear dark to
astronauts?
Answer:
In space, there is no atmosphere to scatter sunlight. Without scattering, light
does not spread in all directions, so the sky appears dark or black.
3.
What is Tyndall effect? Give an example.
Answer:
The Tyndall effect is the scattering of light by small particles in a medium,
which makes the path of light visible.
Example: Sunlight passing through a dusty room or car
headlights visible in fog.
4.
Why do we see colours in soap bubbles or
oil films?
Answer:
This is due to interference of light. Light waves reflected from different
surfaces combine and produce different colours depending on thickness and
angle.
III. Multiple Choice Questions
1.
Which phenomenon causes splitting of white
light?
Answer: Dispersion
2.
Which colour deviates the most in a prism?
Answer: Violet
3.
Which effect explains scattering of light?
Answer: Tyndall effect
4.
Why does the sun look red at sunset?
Answer: Due to scattering
of shorter wavelengths
5.
What causes twinkling of stars?
Answer: Atmospheric
refraction
I. Improve your learning
1. Reflections of concepts
1(a). What information does the electronic
configuration of an atom provide?
Answer:
Electronic configuration shows the distribution of electrons in different
shells, subshells, and orbitals of an atom.
It provides information about:
- Number
of electrons in each shell
- Valency
of the atom
- Chemical
reactivity
- Position
of the element in the periodic table
For example, the configuration of oxygen (2,6) shows
it needs 2 electrons to complete its octet.
1(b). Rainbow is an example for continuous
spectrum. Explain.
Answer:
A rainbow forms when sunlight is dispersed into its constituent colours. It
shows a continuous range of colours (VIBGYOR) without any gaps.
Since all wavelengths are present smoothly from red to
violet, it is called a continuous spectrum.
2. How are shells different from Bohr’s
orbits?
Answer:
- Shells
represent energy levels (K, L, M, N) where electrons are present.
- Bohr’s
orbits describe fixed circular paths around
the nucleus where electrons move.
Thus, shells are energy levels, while Bohr’s orbits
are the paths corresponding to those energy levels.
3. Explain the significance of the four
quantum numbers in predicting the positions of an electron in an atom.
Answer:
The four quantum numbers describe the exact state of an electron:
1.
Principal quantum number (n):
Indicates energy level or shell
2.
Azimuthal quantum number (l):
Indicates subshell (s, p, d, f)
3.
Magnetic quantum number (m):
Indicates orientation of orbital
4.
Spin quantum number (s):
Indicates direction of spin (+½ or −½)
Together, they completely describe the position and
behavior of an electron.
4. How are weaving (n+l) method in writing
electronic configuration?
Answer:
The (n + l) rule helps determine the order of filling orbitals.
- Orbitals
with lower (n + l) value fill first
- If
(n + l) is same, orbital with lower n fills first
Example order:
1s < 2s < 2p < 3s < 3p < 4s < 3d
This rule is also known as the Aufbau principle.
5. Why electronic shell is a higher energy
level than L?
Answer:
Energy of shells increases as we move away from the nucleus.
Order: K < L < M < N
Since M shell is farther from the nucleus than L
shell, it has higher energy.
II. Applications of concepts
1(a). How many maximum number of electrons
can be accommodated in a principal energy level?
Answer:
The formula is:
Maximum electrons = 2n²
Example:
- n
= 1 → 2 electrons
- n
= 2 → 8 electrons
- n
= 3 → 18 electrons
1(b). How many maximum number of electrons
can be accommodated in a sub shell?
Answer:
Each subshell can hold:
- s
= 2 electrons
- p
= 6 electrons
- d
= 10 electrons
- f
= 14 electrons
1(c). How many maximum number of electrons
can be accommodated in an orbital?
Answer:
Each orbital can hold a maximum of 2 electrons with opposite spins.
1(d). How many sub shells are present in a
principal energy level?
Answer:
Number of subshells = n
Example:
- n
= 1 → 1 subshell (s)
- n
= 2 → 2 subshells (s, p)
- n
= 3 → 3 subshells (s, p, d)
1(e). How many spin orientations are
possible for an electron in an orbital?
Answer:
There are two spin orientations:
+½ (clockwise) and −½ (anticlockwise)
2(a). How is the outer most shell
determined?
Answer:
The outermost shell is the shell with the highest principal quantum number
(n) in the electronic configuration.
2(b). How many electrons are there in the
outermost shell?
Answer:
Count the electrons present in the outermost shell from the configuration.
These are called valence electrons.
2(c). What is the atomic number of the
element?
Answer:
Atomic number = Total number of electrons in a neutral atom.
2(d). Write the electronic configuration
of the element.
Answer:
Electronic configuration is written using Aufbau principle.
Example: 1s² 2s² 2p⁶ 3s²
3. Which electronic configuration of
nitrogen atom does not support Hund’s rule?
Answer:
Hund’s rule states that electrons occupy orbitals singly before pairing.
Incorrect configuration:
Pairing electrons in one orbital before filling others in 2p subshell.
Correct configuration:
Each 2p orbital has one electron before pairing.
4. Write the four quantum numbers for the
valence electron of sodium (Na).
Answer:
Electronic configuration of Na = 1s² 2s² 2p⁶ 3s¹
Quantum numbers:
- n
= 3
- l
= 0 (s orbital)
- m
= 0
- s
= +½
5(a). An electron has quantum numbers.
Which orbital does it belong to?
Answer:
From given values (n = 2, l = 1):
- l
= 1 indicates p orbital
- n
= 2 indicates second shell
So, electron belongs to 2p orbital.
5(b). Write the four quantum numbers for
the electron.
Answer:
Possible values:
- n
= 2
- l
= 1
- m
= −1, 0, or +1
- s
= +½ or −½
6. The wavelength of radio wave is 1 cm.
Find its frequency.
Answer:
Formula:
v = νλ
Where:
v = speed of light = 3 × 10⁸ m/s
λ = 1 cm = 0.01 m
ν = v / λ
ν = (3 × 10⁸) / (0.01)
ν = 3 × 10¹⁰ Hz
Chapter 7:
Classification of Elements and Periodic Table
. Improve your learning
1(a). How did Mendeleev classify the
elements based on their atomic masses?
Answer:
Mendeleev arranged elements in increasing order of their atomic masses.
While arranging, he noticed that elements with similar chemical properties
appeared at regular intervals. Based on this observation, he proposed the Periodic
Law, which states that the properties of elements are periodic functions
of their atomic masses.
He grouped elements with similar properties into the
same vertical columns (groups), leaving gaps for undiscovered elements.
1(b). How did he rectify the drawbacks of
classification?
Answer:
Mendeleev corrected several drawbacks by:
- Leaving
gaps for undiscovered elements
- Predicting
their properties accurately (like Eka-aluminium → Gallium)
- Rearranging
certain elements based on chemical properties instead of strict atomic
mass order
2. What are the limitations of Mendeleev’s
periodic table?
Answer:
The main limitations are:
- Position
of hydrogen was not fixed clearly
- Isotopes
could not be placed properly (same element, different masses)
- Some
elements were not in correct order of atomic mass
- No
proper explanation for periodicity
3. Define the modern periodic law. Discuss
the merits of the long form of periodic table.
Answer:
Modern Periodic Law:
The physical and chemical properties of elements are periodic functions of
their atomic numbers.
Merits of modern periodic table:
- Elements
are arranged based on atomic number, removing earlier confusion
- Isotopes
occupy the same position
- Proper
classification into groups and periods
- Explains
periodic trends like atomic size, valency, and reactivity
4. Explain why elements are classified
into s, p, d and f blocks.
Answer:
Elements are classified based on the subshell in which the last electron
enters:
- s-block:
last electron enters s-orbital
- p-block:
last electron enters p-orbital
- d-block:
last electron enters d-orbital (transition elements)
- f-block:
last electron enters f-orbital (inner transition elements)
This classification helps in understanding their
properties easily.
5. Complete the table (Electronic
configuration and properties).
Answer (Filled Table):
- Group
number: 1 → Valency: 1
- No.
of valence electrons: 2 → Valency: 2
- Metal
or non-metal: Metals → Usually 1 to 3 valence
electrons
- Element
family: Noble gases → Valency: 0
6. Complete the table (Maximum electrons
in shells)
Answer:
|
Period number |
Shells present |
Maximum electrons (2n² rule) |
Total elements |
|
1 |
K |
2 |
2 |
|
2 |
K, L |
8 |
8 |
|
3 |
K, L, M |
18 |
8 |
|
4 |
K, L, M, N |
32 |
18 |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
7. Complete the table (Blocks in periodic
table)
Answer:
|
Period |
Total elements |
s-block |
p-block |
d-block |
f-block |
|
1 |
2 |
2 |
0 |
0 |
0 |
|
2 |
8 |
2 |
6 |
0 |
0 |
|
3 |
8 |
2 |
6 |
0 |
0 |
|
4 |
18 |
2 |
6 |
10 |
0 |
|
5 |
18 |
2 |
6 |
10 |
0 |
|
6 |
32 |
2 |
6 |
10 |
14 |
|
7 |
32 |
2 |
6 |
10 |
14 |
II. Applications of Concepts
1(a). Arrange the elements A, B, C, D, E
in increasing atomic radius.
Answer:
Atomic radius decreases across a period (left to right).
So, order will be:
E < D < C < B < A
1(b). Which has the lowest melting point?
Answer:
Element with weakest intermolecular forces (generally noble gases or non-metals
at right end) has the lowest melting point.
1(c). Which shows maximum electropositive
character?
Answer:
Electropositive character increases down a group and towards left.
So, the element on the left-most bottom has maximum electropositivity.
2. Identify the element based on
electronic configuration.
Answer:
Given configuration: 1s² 2s² 2p⁶ 3s² 3p¹
- Atomic
number = 13
- Element
= Aluminium
3. Calculate atomic number based on given
data.
Answer:
Using given equation and solving:
Atomic number = 17
Element is Chlorine.
Chapter 8: Chemical Bonding
I: Reflections on concepts
1. Explain the difference between the valence
electrons and the co-valency of an element. (AS₁)
Valence electrons are the electrons in the outermost
shell of an atom. These are the electrons that are involved in forming chemical
bonds. Covalency (or co-valency) is the number of covalent bonds an atom can
form. It represents the number of electrons an atom shares with other atoms to
achieve a stable electron configuration, typically an octet.
2. Which chemical compound has the following Lewis
notation: (AS₁)
- Fig-02
H shows a Lewis structure where element Y has three lone pairs and forms
one bond, and element X forms one bond and has one lone pair.
a) How many valence electrons does element Y have?
Element Y has 7 valence electrons (6 from the three lone pairs and 1 shared in
the bond).
b) What is the valency of element Y? Element Y has a
valency of 1 (it forms one bond).
c) What is the valency of element X? Element X has a
valency of 1 (it forms one bond).
d) How many covalent bonds are there in the molecule?
There is 1 covalent bond in the molecule.
e) To which groups the elements X and Y belong? (AS₁)
- Since
Y has 7 valence electrons and forms 1 bond, it belongs to Group 17
(halogens).
- Since
X forms 1 bond and from the common representation of X with a single
electron and dot, it's likely representing a hydrogen atom (H), which
would belong to Group 1.
3. How bond energies and bond lengths of molecule help
us in predicting their chemical properties? Explain with examples. (AS₁)
Bond energy is the amount of energy required to break
a specific bond in one mole of gaseous molecules. Bond length is the average
distance between the nuclei of two bonded atoms in a molecule.
These two factors are inversely related: generally,
shorter bonds have higher bond energies. Together, they provide crucial
insights into a molecule's chemical properties:
- Stability
and Reactivity: Higher bond energy indicates a stronger bond, meaning more
energy is needed to break it. Molecules with strong bonds are generally
more stable and less reactive (e.g., N₂ has a very high bond energy due to
its triple bond, making it quite unreactive). Lower bond energy implies
weaker bonds, leading to higher reactivity (e.g., C-C single bonds are
weaker than C=C double bonds, influencing organic reaction pathways).
- Molecular
Structure and Geometry: Bond lengths influence the overall shape and size
of a molecule. For instance, the shorter C=C bond in ethene compared to
the C-C bond in ethane significantly affects their molecular geometry and
chemical behavior.
- Spectroscopic
Properties: Bond lengths and energies are directly related to vibrational
frequencies observed in infrared (IR) spectroscopy, which helps in
identifying functional groups and molecular structure.
4. Draw simple diagrams to show the arrangement of
valence electrons in the following compounds: (AS₁)
a) Calcium oxide (CaO) Calcium (Ca) is in Group 2,
having 2 valence electrons. Oxygen (O) is in Group 16, having 6 valence
electrons. Calcium transfers 2 electrons to oxygen, forming Ca²⁺ and O²⁻ ions.
b) Water (H₂O) Oxygen (O) has 6 valence electrons and
Hydrogen (H) has 1 valence electron. Oxygen forms two single covalent bonds
with two hydrogen atoms and has two lone pairs.
c) Chlorine (Cl₂) Each Chlorine (Cl) atom has 7
valence electrons. They form a single covalent bond, each sharing one electron
pair and having three lone pairs.
5. Represent each of the following atoms using Lewis
notation: (AS₁)
a) Beryllium (Be): Group 2, 2 valence electrons
b) Calcium (Ca): Group 2, 2 valence electrons
c) Lithium (Li): Group 1, 1 valence electron
d) Bromine (Br): Group 17, 7 valence electrons
e) Calcium chloride (CaCl₂): Calcium (Group 2) loses 2
electrons, and two Chlorine (Group 17) atoms each gain 1 electron.
f) Carbon dioxide (CO₂): Carbon (Group 14) has 4
valence electrons and Oxygen (Group 16) has 6. Carbon forms two double bonds
with two oxygen atoms, with each oxygen having two lone pairs.
Q:
How does Lewis dot structure helps in understanding bond
formation between atoms? (AS₁)
Lewis
dot structures are diagrams that show the valence electrons of atoms within a
molecule. They depict these electrons as dots around the atomic symbol, and
shared electron pairs (covalent bonds) as lines or pairs of dots between atoms.
Lewis structures help in understanding bond formation by:
- Visualizing
Valence Electrons: They clearly show the number of valence electrons
available for bonding for each atom.
- Illustrating Octet
Rule: They help predict how atoms will share or transfer electrons to
achieve a stable electron configuration, typically an octet (eight valence
electrons) or a duet (for hydrogen).
- Identifying Bond
Types: By showing electron transfer (for ionic) or sharing (for covalent),
they visually represent how different types of bonds are formed.
- Predicting Number
of Bonds: They allow us to predict the number of bonds an atom will form
to satisfy the octet rule.
- Identifying Lone
Pairs: They show unshared pairs of electrons (lone pairs), which are
crucial for determining molecular geometry and reactivity.
- Understanding
Molecular Geometry (implicitly): While not directly showing geometry, the
arrangement of bonds and lone pairs in a correct Lewis structure is the
basis for VSEPR theory, which predicts molecular shapes.
Chapter 9: Electric Current
I. Reflections on concepts
**1. Explain how electrons flow causes electric current with Lorentz–Drude theory of electrons.
Answer:
According to the Lorentz–Drude model, a metal contains a large number of free electrons that move randomly due to thermal energy. In the absence of an electric field, their motion is random and no net current flows. When an electric field is applied, these electrons experience a force opposite to the field direction and acquire a small average drift velocity. This systematic drift of electrons superimposed on their random motion constitutes electric current. Collisions with lattice ions limit their गति, giving rise to resistance.
**2. Write the difference between potential difference and emf.
Answer:
- Potential Difference (V): The work done per unit charge between two points in a circuit; it represents energy used by components.
- EMF (Electromotive Force): The total energy supplied per unit charge by a source (like a battery).
- Key difference: EMF is the cause (energy supplied), while potential difference is the effect (energy used).
**3. How can you verify that the resistance of a conductor is temperature dependent?
Answer:
Set up a circuit with a conductor, battery, ammeter, and voltmeter. Measure resistance at room temperature using Ohm’s law. Then heat the conductor (e.g., using a flame or hot water) and measure resistance again. The observed change in resistance (usually an increase for metals) confirms temperature dependence.
**4. Explain how electric shock takes place.
Answer:
Electric shock occurs when current passes through the human body. If a person comes in contact with a live wire, a potential difference is established across the body, allowing current to flow. This interferes with nerve signals and muscle control, and can damage tissues or organs depending on current magnitude and duration.
**5. Draw a circuit diagram in which two resistors A and B are connected in series with a battery and a voltmeter is connected to measure the potential difference across the resistor A.
Answer:
In the circuit:
- Connect resistors A and B in series with a battery.
-
Place a voltmeter in parallel across resistor A.
This ensures the voltmeter measures only the potential drop across A.
**6. In the figure Q-6 the potential at A is ______ when the potential at B is zero.
Answer:
Given current = 1 A and resistance = 5 Ω,
Using Ohm’s law: V = IR = 1 × 5 = 5 V
So, potential at A = +5 V (with respect to B = 0 V).
II. Applications of concepts
**1. Explain overloading of household circuit.
Answer:
Overloading occurs when too many appliances draw current from a single circuit, exceeding its safe limit. This causes excessive heating of wires, which may damage insulation and lead to fire hazards.
**2. Why do we use fuses in household circuits?
Answer:
A fuse is a safety device that melts when current exceeds a safe value, breaking the circuit. It protects appliances and wiring from damage due to overload or short circuit.
**3. Two bulbs have ratings 100 W, 220 V and 60 W, 220 V. Which one has the greater resistance?
Answer:
Using :
- For 100 W bulb:
- For 60 W bulb:
Since power is smaller, resistance is larger.
Thus, the 60 W bulb has greater resistance.
**4. Why do we consider tungsten as a suitable material for making the filament of a bulb?
Answer:
Tungsten is used because it has:
- Very high melting point
- High resistivity
- Ability to glow (incandescence) without melting
- Good mechanical strength at high temperatures
**5. Are the head lights of a car connected in series or parallel? Why?
Answer:
They are connected in parallel so that:
- Each gets full voltage
- If one fails, the other still works
**6. Why should we connect electric appliances in parallel in a household circuit? What happens if they are connected in series?
Answer:
Parallel connection ensures:
- Same voltage across all appliances
- Independent operation
If connected in series:
- Voltage divides among appliances
- Devices won’t work properly
- Failure of one breaks the entire circuit
**7. If the resistance of your body is 10000 Ω, what would be the current that flows in your body when you touch the terminals of a 12 V battery?
Answer:
Using Ohm’s law:
So, current = 1.2 mA
III. Higher Order Thinking Questions
**1. Imagine that you have three resistors of 3 Ω each. How many resultant resistances can be obtained by connecting these in different ways? Draw their relevant diagrams.
Answer:
Possible combinations:
- All in series: 9 Ω
- All in parallel: 1 Ω
- Two in series + one parallel: 2 Ω
- Two in parallel + one series: 4.5 Ω
So, 4 different equivalent resistances are possible.
**2. A house has 3 tube lights, two fans and a television… Find the cost of electric energy used in 30 days at ₹3.00 per kWh.
Answer:
Power consumption:
- 3 tube lights: 3 × 40 = 120 W
- 2 fans: 2 × 80 = 160 W
-
TV: 100 W
Total = 380 W = 0.38 kW
Time used daily = 5 h
Energy per day = 0.38 × 5 = 1.9 kWh
For 30 days = 1.9 × 30 = 57 kWh
Cost = 57 × 3 = ₹171
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