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5. Periodic Classification of Elements
5.1 MAKING ORDER OUT OF CHAOS — EARLY ATTEMPTS AT
THE CLASSIFICATION OF ELEMENTS
Three Marks Question (50 words)
1. Lithium, sodium and potassium are placed in one group on the basis of their similar
properties. List three similar properties of these elements. (2016-RKPKF3H; 2014-Z1F6W74)
Ans. ● Lithium, sodium and potassium have similar valence shell electron.
● They vigorously react with oxygen.
● They form basic oxide and are alkali
in nature.
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5.2 MAKING ORDER OUT OF CHAOS
— MENDELEEV’S PERIODIC TABLE
Two Marks Question (30 words)
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1. Three elements ‘X’, ‘Y’ and ‘Z’ having atomic numbers 11, 7 and 6 respectively
react with oxygen to form their oxides.
(a) Arrange these oxides in increasing order of their basic nature.
(b) Give reason for your answer.
Ans.
11X = 2, 8, 1
7Y = 2, 5
6 Z = 2, 4
(a) Y < Z < X
(b) X is metallic in nature hence, its oxide is basic in nature. While Y, and Z are nonmetals and their oxides are acidic in nature.W
O NL
O character decreases.)
(In a period acidic character increases
E D and basic
Three Marks Questions (50 words)
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2. Describe the basic character of the
oxide of third period elements across the period.
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(2016-K4N9OOO; 2015-3K1YM0O; 2014-BG7DPSI)
Ans. (3rd period)
Element :
11Na
12Mg
13Al
Oxide
Na2O
MgO
Al2O3 SiO2
:
14Si
15P
16S
P4O10 SO2
17Cl
18Ar
Cl2O
—
Basic character decreases across a period when we move from left to right because :
(i) Metallic character decreases across a period.
(ii) Electropositive character decreases across a period.
3. Explain the term ‘periodicity’. Illustrate that the properties of the elements placed
in a group are same.
(2016-24TYHSZ; 2014-QPVS66P)
Ans. When elements are arranged in a definite pattern of their atomic number or
atomic mass, they show similar properties after regular time interval.
Due to similar outermost shell, electronic configuration or number of valence shell
electron present in the group elements show similar properties.
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e.g.,
3Li
=
Na
=
11
19K =
2, 1
2, 8, 1
2, 8, 1
Valence shell electron 1
4. (a) Amongst the following elements identify the ones that would form anions :
K, O, Na, F, Ca, Cl, Mg.
(b) Write the electronic configuration of anions identified above.
(2016-P5ZJBGS; 2015-673LH9K; 2014-GM5RD71)
Ans. (a) O, F, Cl.
(b)
K
O
=
2
8
9F = 2
17Cl = 2
L
6
7
8
M
—
—
7
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PRACTICE QUESTION
Three Marks Question
1. (a) Identify the group of the metal (M), the formula of whose chloride is :
(i) MCl
(ii) MCl2
(b) Identify the group of the non-metal (X), the formula of whose sodium salt is :
(i) Na2X (ii) NaX
(2016-2C2M4YT; 2014-DUKGXD1)
(c) Write the electronic configuration of argon.
5.3 MAKING ORDER OUT OF CHAOS — THE MODERN PERIODIC TABLE
One Mark Questions (One word or one sentence)
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1. Assess whether an element having electronic configuration 2, 8, 1 is a metal or a
non-metal ?
(2016-HDXKTMV)
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Ans. It is a metal.
2. The atomic number of an element X is 15. Identify its position in terms of group
and period in the Modern Periodic Table.
(2016-K4N9OOO, LFQMQWV)
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Ans. Element ‘X’ with atomic number
15
belongs
to
group
15 and period 3 in the
ENTERPRISES
Modern Periodic Table.
3. Predict the maximum number of valence electrons possible for atoms in first period
of periodic table.
(2014-3KFTJ28, 5ACL7YS, Z1F6W74)
Ans. Ist period
Element
H
He
Atomic No.
1
2
Electronic configuration
1
2
Valence shell electrons
1
2
∴ Maximum number of valence shell electrons = 2.
4. Out of Li and K, which will have stronger metallic character and why ?
(2015-VWP5W1F; 2014-7HTEC43)
Ans. Potassium (K) will have stronger metallic character than lithium (Li) because
as we move from top to bottom in a group, the size increases which increases the ease of
liberation of electrons.
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STUDENT WORKSHEET–I
Instructions : Read the following lines written about the Modern Periodic Table and
fill in the blanks.
I am Modern Periodic Table, where elements are arranged on the basis of
their _______1. I have ______2 horizontal rows called ______3 and ________4 vertical
columns called ________5. My first period is very short with ________6 elements
and 2 and 3 periods contain ________7 elements, 4 and 5 have ________8 elements.
period 7 is _______10, with space for more
Period 6 has _______9 elements butRELIABLE
is of _____12. Atomic size of my elements
elements. Group 17 is of _____11, 18
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_____13 across the period but _____14 down the group same as the metallic
character but the non-metallic character is _____15 to that. Number of valence
electrons _____16 across the period so varies the valency but the valency of the
elements is _____17 in a group. All the properties of elements are the Periodic
function of their _____18.
Ans. 1. atomic numbers, 2. 7, 3. periods, 4. 18, 5. groups, 6. 2, 7. 8, 8. 18, 9. 18,
10. incomplete, 11. halogens, 12. noble gases, 13. decreases, 14. increases, 15. opposite,
16. vary, 17. same, 18. atomic numbers.
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1
2
a
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Group →
Period ↓
1
2
3
13
14
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STUDENT WORKSHEET–II
Following table represents, a part of the Modern Periodic Table containing first three
periods in which five elements have been represented
by the letters a, b, c, d and e (these
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are not the chemical symbols of the elements).
E
15
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16
17
18
e
c
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b
d
Q. 1. Select the letter(s) which represent alkali metal.
Ans. a and d. (Alkali metals are placed in 1st group in the Modern Periodic Table).
Q. 2. Select the letter(s) which represent noble gas(es).
Ans. c (Noble gases are placed in 18th group in the Modern Periodic Table).
Q. 3. Select the letter(s) which represent halogen(s).
Ans. e (Halogens are placed in 17th group in the Modern Periodic Table).
Q. 4. Where would you place an element ‘f ’ with electronic configuration 2,
8, 4 in the given table ?
Ans. We would place the element ‘f ’ in group 14, period 3.
(Because the number of electrons in the outermost shell is 4 and number of shells
occupied is 3.)
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8. How do Organisms Reproduce ?
8.1 DO ORGANISMS CREATE EXACT COPIES OF THEMSELVES ?
Two Marks Questions (30 words)
(2014-2FOHVZ9)
1. Explain how do organisms create an exact copy of themselves.
Ans. ● To build the copies of DNA or the genetic material, the cells use bio-chemical
reactions.
● Additional cellular apparatus along with the DNA copies are separated and so a
cell divides to give rise to two almost identical cells.
2. “Variations” are seen in the organisms. State the two main causes of variation.
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(2014-CEG0CCT)
Ans. Variations are caused by :
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(1) Change in the genetic material, i.e., DNA at the time of DNA copying.
(3) Environmental factors viz., light, temperature, nutrition, wind and water supply etc.
(3) Mutations.
Three Marks Questions (50 words)
3. Explain how do proteins control the characteristics in plants ?
(2016-8T23V6D, JW7OJ21)
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Ans. ● DNA in the cell nucleus is the information source for making proteins.
● It is the protein which leads to body design of an organism.
● If the information provided by DNA is changed, different proteins will be made.
● Different proteins will lead to altered body designs.
4. Define reproduction. How does it helps in providing stability to the population of
(2016-CBSEOS1, CBSEOS3)
species ?
Ans. Reproduction : It is a biological process
by
which
an
organism
produces another
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organism (oroffspring) similar to itself.E D
E
● The basic event in reproduction is the creation of a DNA copy.
● Using chemical reactions the cells make two copies of DNA in a reproducing cell.
● The consistency of DNA copying RELIABLE
during reproduction is important for maintenance
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of body design.
● Reproduction, therefore, is important for providing stability to the population of
species.
5. (a) What is the location of the following ?
(i) DNA in a cell (ii) Gene
(2014-BZC922Y)
(b) Expand DNA.
Ans. (a) (i) In the nucleus of cell, (ii) Located on the chromosomes.
(b) Deoxyribo Nucleic Acid.
PRACTICE QUESTION
Five Marks Question
1. (a) The two cells formed after DNA copying are similar but not identical. What
(2016-E0KUM11)
reason can be associated for this ?
(b) Variation though leads to certain changes in an individual but is useful for the
survival of species over time. Justify this statement.
(2016-LRI5VFM)
(c) Is the consistency of DNA copying important during reproduction ? (2015-7GEQTPL)
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9. Heredity and Evolution
9.1 ACCUMULATION OF VARIATION DURING REPRODUCTION
Three Marks Question (50 Words)
1. Variation is useful for the survival of species over long time. But the variants have
unequal chances of survival. Explain this statement.
(2016-O1AQXI6)
Or
Do all variations in a species have equal chance of survival ? Explain.
Ans. Though variations are useful for survival of species over long time but variants
have equal chances of survival due to the alteration in ecological conditions.
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● Different individuals have different
kind of advantages.
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● For example, a bacteria which can
withstand heat, will have better chances of
survival in heat than other organisms.
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PRACTICE QUESTION
Three Marks Question
1. Variations are seen when there is inheritance of characters. Explain the statement.
(2015-673LH9K)
9.2 HEREDITY
One Mark Questions (One word or one sentence)
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1. How can the chromosomes be identified ?
(2014-3KFTJ28, A14VW1K)
Ans. In human beings, the individual chromosomes are identified by their lengths,
position of centromere and banding pattern on staining.
2. What conclusion was drawn by Mendel after obtaining F2 progeny in a dihybrid
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cross ?
(2014-Q5K0ZGL, BG7DPSI)
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Ans. After obtaining progeny in F2 generation in a dihybrid cross, Mendel concluded
that when two pairs of traits are combined in a hybrid, one pair of character segregates
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independently of the other pair of character.
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3. Mendel observed a contrasting trait
in relation to position of flowers. Mention that
trait.
(2015-ARZ0O7U; 2014-B)
Ans. Terminal and Axial
↓
↓
Recessive
Dominant
Two Marks Question (30 words)
4. In a monohybrid cross a round seeded plant was crossed with a wrinkle seeded
plant. In F2 generation genotype ratio obtained was 1 : 2 : 1.
(a) How many plants were homozygous round ?
(b) How many plants were heterozygous round ?
(2015-12NQ4D2)
Ans. (a) A quarter of plants were homozygous round.
(b) Half of the plants were heterozygous round.
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10. Light — Reflection and Refraction
10.2 SPHERICAL MIRRORS
Two Marks Questions (30 words)
1. Find the radius of curvature of a spherical mirror whose focal length is of 11m.
(2015-WIPC5W2)
Mention the nature of this mirror. Give reason for your answer.
Ans. R = 2f
R = 2 × 11 m = 22 m
As the focal length is positive, it is a convex mirror.
5. List two possible ways in which aRELIABLE
concave mirror can produce a magnified image of
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an object placed in front of it. State the difference, if any, between these two images.
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Ans. ● Concave mirror produces an enlarged image when :
(a) Object is placed between pole (P) and focus (F) of mirror.
(b) Object is placed between centre of curvature (C) and focus (F) of mirror.
(c) Object is placed at focus (F).
● Yes, there is a difference between these two images. When object is placed between
pole and focus, the image obtained is virtual and erect. Whereas when object is kept
between focus and centre of curvature then image obtained is real and inverted.
6. An object is placed between focus and the pole of a concave mirror. Draw a ray
diagram and also state the position, relative size and nature of the image formed.
(2014-ZZ-104-9200)
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Ans. For diagram see Q. No. 2, Unit 10.2, Page 95
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● Position of the image — behind the
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● Relative size of the image — larger than the object
● Nature of the image — virtual and erect.
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7. How many images are seen when
two mirrors are placed in parallel ?
Ans. Infinitely many images of anENTERPRISES
object are seen when two mirrors are placed in
parallel. It happens so because each image of the object in one mirror works as an object for
the other. This phenomenon is called the multiple reflection.
PRACTICE QUESTIONS
Two Marks Questions
1. Define the terms-absolute and relative refractive indices of a medium. Express each
(2016-G3CYRYZ)
in terms of velocity of light.
2. Draw ray diagram for the formation of image formed by a concave lens when an
object is placed anywhere between infinity and optical centre of the lens. State the nature
and size of image as compared to that of the object.
(2015-BKYDQ0S )
3. (i) In refraction of light through a rectangular glass slab, the emergent ray is
parallel to the direction of the incident ray. Why ?
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11. The Human Eye and the Colourful World
11.1 THE HUMAN EYE
Two Marks Questions (30 words)
1. What is meant by least distance of distinct vision ? Give the value of near point
(2016-6TKCC9E)
and far point for a normal human eye.
Ans. The minimum distance, at which objects can be seen most distinctly without
strain is called the least distance of distinct vision.
Near point, Far point : Please see Q. 4 of Unit 11.1, Page 133
2. Explain how does image of an object is communicated to the brain.
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(2015-73Z67S7; 2014-7HTEC43, S8UO9TQ)
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Ans. (1) The eye lens forms an inverted
and real image of the object on the retina.
(2) The retina is a delicate membrane having enormous number of light sensitive
cells.
(3) The light sensitive cells get activated up on illumination and generate electrical
signals.
(4) These signals are sent to the brain via the optic nerves.
3. In which of the following two cases the focal length of the eye lens will be more :
(a) When ciliary muscles of a normal eye is most relaxed.
(b) When ciliary muscles of a normal eye is in most contracted state. Explain with
reason.
(2015-24O7TPT; 2012-67018)
Or
If the ciliary muscles of a normal eye are in their
(a) most relaxed state
(b) most contracted state.
Mention in brief how focal length and power of eye lens will change in both the cases.
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(2016-739K52A)
Give reason for your answer.
E
E will be more when ciliary muscles is most relaxed.
Ans. (a) ● Focal length of the eye lens
● It is because in this state, eye lens becomes thinner, hence its focal length increases.
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(b) Focal length of the eye lens will
be least when ciliary muscles are in their most
contracted state. It is because in thisENTERPRISES
state, eye lens become thicker, hence its focal
length decreases.
Three Marks Questions (50 words)
4. Explain power of accommodation. Explain in brief the near and the far point of an
(2016-E0KUM11, GPB8I84; 2015-3IPTVL5; 2014-ENKE30R)
eye and give their values.
Ans. Power of Accommodation : It is the ability of the eye lens to adjust its focal
length.
Near Point : The nearest point upto which an eye can see the object clearly is called
the nearest point. The near point of normal human eye is at a distance of 25 cm.
Far Point : The farthest point from the eye which can be seen clearly is known as
the far point of the eye. Far point of a normal human eye is infinity.
5. (a) Define power of accommodation of the eye.
(b) Why does the power of accommodation of an eye decreases with age ? Explain.
(2016-VM3A33O)
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STUDENT WORKSHEET–I
Instructions : Match the two columns correctly :
Column I
1. The effective diameter of the circular outline of
a spherical lens
2. A transparent piece of material having two surfaces
at least one of which is curved
3. The ratio of the height of the image and the height
of the object
4. An electromagnetic wave which RELIABLE
gives sensation of vision
5. The phenomenon due to which stars
appear to twinkle
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6. It is formed when reflected or refracted rays intersect
each other virtually or actually.
Ans. 1. (c), 2. ( f ), 3. (a), 4. (e), 5. (b), 6. (d).
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Column II
(a) Magnification
(b) Atmospheric
refraction
(c) Aperture
(d) Image
(e) Light
( f ) Lens
SECTION – B
Questions Based on Practicals
EXPERIMENT 1
To study the following properties of acetic acid (ethanoic acid) :
(i) Odour
(ii) Solubility in water
(iii) Effect on litmus
(iv) Reaction with sodium bicarbonate
One Mark Questions
1. The gas released on addition of sodium carbonate to acetic acid is :
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(a) H2
(b) CO
(c) CO2
(d) O2
ENTERPRISES
Ans. (c) Carbon dioxide gas is released.
2CH3COOH + Na2CO3 → 2CH3COONa + CO2 + H2O
2. Pure acetic acid is known as glacial acetic acid because
(a) it is transparent like glass.
(b) it freezes in winter.
(c) it is colourless.
(d) it is found in glaciers.
Ans. (b) The melting point of pure ethanoic acid is 290 K and hence it often freezes
during winter in cold climate. This is why it is also called as glacial acetic acid.
3. A bottle containing a solution was left open by mistake in the laboratory. As soon
as, Rahul entered the laboratory, he got the smell of vinegar. Rahul concluded that the
bottle surely contained :
(a) Sodium hydroxide
(b) Acetic acid
(c) Hydrochloric acid
(d) Sodium bicarbonate
Ans. (b) Acetic acid smells like vinegar.
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4. CH3COOH is a
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E
(a) weak acid.
(b) strong acid.
(c) weak base.
(d) strong base.
Ans. (a) Acetic acid does not get ionized completely in aqueous solution. So, it is a
weak acid.
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CH3COOH CH3COO– + H+
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5. When a pinch of sodium bicarbonate is added to acetic acid, a gas is evolved which
(I) has dense white fumes.
(II) extinguishes a burning splinter.
(III) turns lime water milky.
(2015-12NQ4D2; 2014-3KFTJ28)
The correct observation is :
(a) (I) and (II)
(b) (II) and (III)
(c) (I) and (III)
(d) (I), (II) and (III)
Ans. (b) When sodium bicarbonate reacts with acetic acid, carbon dioxide gas is
evolved which turns lime water milky and extinguishes a burning splinter.
6. On adding acetic acid to a solid X kept in a test tube, a student observed that a
colourless and odourless gas Y evolves which turns lime water milky. On the basis of this
(2014-7HTEC43, U5K0ZGL)
information, the conclusion which can be drawn is :
(a) Solid X is sodium hydroxide and the gas Y is carbon dioxide.
(b) Solid X is sodium hydrogen carbonate and the gas Y is carbon monoxide.
(c) Solid X is sodium hydroxide and the gas Y is carbon monoxide.
(d) Solid X is sodium hydrogen carbonate and the gas Y is carbon dioxide.
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PRACTICE PAPERS
Practice Paper – 1 (Solved)
SCIENCE
CLASS – X
Time : 3 Hours
Maximum Marks : 90
General Instructions :
(i) The question paper comprises of two Sections A and B. You are to attempt both the sections.
(ii) All questions are compulsory.
(iii) There is no choice in any of the questions.
(iv) All questions of Section–A and all questions of Section–B are to be attempted separately.
(v) Question numbers 1 to 3 in Section–A are one mark questions. These are to be answered
ENTERPRISES
in one word or in one sentence.
(vi) Question numbers 4 to 6 in Section–A are two marks questions. These are to be answered
in about 30 words each.
(vii) Question numbers 7 to 18 in Section–A are three marks questions. These are to be answered
in about 50 words each.
(viii) Question numbers 19 to 24 in Section–A are five marks questions. These are to be answered
in about 70 words each.
(ix) Question numbers 25 to 33 in Section–B are multiple choice questions based on practical
skills. Each question is a one mark question. You are to select one most appropriate
response out of the four provided to you.
(x) Question numbers 34 to 36 in Section–B are based on practical skills. Each question is
two marks question. These are to be answered in about 30 words each.
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SECTION–A
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DO NLO
1. State maximum covalence of carbon.
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Ans. 4
2. Which element has only two shells
which are completely filled ?
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Ans. Neon
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3. Name the part of the eye which controls the amount of light entering the eye.
Ans. Pupil
4. Give one example of each :
(i) Metal having valency 2.
(ii) Non-metal having valency 2.
(iii) Element with completely filled outermost shell.
(iv) Element with three shells, having 4 electrons in the outermost shell.
Ans. (i) Magnesium
(ii) Oxygen
(iii) Nickel
(iv) Silicon
5. “The chromosome number of the sexually producing parents and their offspring is
the same.” Justify this statement.
Ans. In sexual reproduction, both the gametes (male and female) contain half the
number of chromosomes (haploid or n) and by the fusion of these gametes, the zygote have
full set (diploid 2n) chromosomes.
Contd...
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