EXPLANATIONS
Objective Questions
11. (a) Lithium is the metal among the first ten elements as
boron is a metalloid, carbon is a non-metal and potassium is
1. (c) Dobereiner could identify only three triads from the
not in first ten elements because it’s atomic number is 19.
elements known at that time. These are
Li, Na, K; Ca, Sr, Ba; Cl, Br, I
12. (d) Oxides of metals are of basic in nature while those of
non-metals are acidic.
2. (b) Newlands’ law of octaves was applicable only to lighter
Group 1 and group 2 consists of metals. Therefore, all the
elements having atomic masses upto 40 u, i.e. upto calcium.
elements of these groups form basic oxides. The element
After calcium, every eighth element did not possess
present in group 13 and period 3 is aluminium, whose oxides
properties similar to that of the first element.
is amphoteric in nature.
3. (c) Gallium has a valency of 3. Hence, it forms an oxide
Group 16 consists of non-metals. Therefore, all the elements
having molecular formula E2
O3. In other options, valency of
of this group forms acidic oxide. Thus, an element X which
E is not 3.
form an acidic oxides belongs to group 16 and period 3.
4. (d) Na (group 1) has one, Al (group 13) has three (13-10), Si
13. (a)Q The oxide turns the red-litmus to blue but shows no
(group 14) has four (14-10) and P (group 15) has five (15-10)
effect on blue litmus. Also, the oxide is a metal oxide, thus
valence electrons. Therefore, P has maximum number of
one of its component must be a metal. Hence, it is MgO.
valence electrons, i.e. 5 (maximum among the given).
14. (c) The element A will form a basic oxide while B, C and D
will form an acidic oxide. This is because on moving along a
5. (b) The number of electrons present in the outermost orbit
period, the acidic character of the oxide increases.
of the element are known as valence electrons, while
15. (b) The most electropositive element is ‘d’ as electropositive
number of electrons used by any element in any chemical
character increases down the group and decreases along a
reaction is called its valency.
period.
Both are co-related as follows
16. (a) Both A and R are true and R is the correct explanation
(i) For first four elements in any period,
of A. According to the Dobereiner’s triads, the three
Valence electrons = Valency
elements in a traid were arranged in the order of increasing
(ii) For last four elements in any period,
atomic masses, the atomic mass of middle element was
Valency = 8 − valence electrons
roughly the average of the atomic masses of the other two
elements. So, taking Be, Mg and Ca as a triad.
6. (d) Q belongs to group II, so its valency is +2. R belongs to
15 or VA group, so its valency is −3 (as it requires 3 electrons
Elements Be Mg Ca
to complete its octet).
Atomic mass
9
24
40
Q
R
Average atomic mass of first and third element
9+
40
= 24.5
+2
-3 Charge
R
2
17. (c) A is true but R is false. All the elements of group I contains
7. (a) In long form of periodic table, atomic size decreases
one valence electron and valency as the electronic
along the period due to increase in effective nuclear charge.
configuration of Li = 2, 1, Na = 2, 8, 1 and K= 2, 8, 8, 1. All the
Thus, atomic size of B is less than of Be.
elements contain only one valence electron.
8. (d) Because along the period, atomic radii decreases as the
18. (a) Both A and R are true and R is the correct explanation of A.
atomic number increases. Thus, the correct increasing order
Noble gases are also called inert gases because they don’t need
is F < O < N.
to react with other elements to fill their outer shell octet, as
9. (c) On moving from left to right, the atomic number
they already posses full valence shell.
increases and hence, the nuclear charge increases. With the
19. (a)BothA and Rare true and Risthe correct explanationofA.
increase of nuclear charge, the force binding the electron
20. (c) A is true but R is false. Electronegativity of fluorine is
increases, so the atom lose the electrons with more difficulty
and not easily.
greater than that of oxygen, since the non-metallic character
increases along a period from left to right in the modern
10. (a) The correct match for the given item is
periodic table.
A-(iii), B-(i), C-(ii), D-(iv).
21.
(i)
(a) In 1913, Henry Moseley discovered the modern
● Be belongs to group 2 because it contains 2 valence
periodic table by modifying Mendeleev’s periodic law.
electrons.
(ii)
(b) Along the period 3, (Na) sodium can loose electron
● F belongs to group 17 because it contains 7 valence
easily because moving from left to right along a period,
electrons.
atomic radius decreases due to increase in effective
● P contains 5 electrons in it’s outermost shell, so it belongs
nuclear charge which tends to pull the valence
to group 15.
electrons closer to the nucleus and hence, tendency to
● Ar belongs to group 18 because it has full-filled electronic
lose the electrons decreases.
configuration.
(iii)
(d) Rb has the highest metallic character because down
3. Yes, Dobereiner’s triads also exist in the columns of
the group, the effective nuclear charge decreases as the
Newlands’ octaves, e.g. lithium (Li), sodium (Na) and
atomic radius increases. So, the outermost electrons are
potassium (K) constitute a Dobereiner’s triads. Now, if we
farther away from nucleus which can be lost easily.
consider Li as the first element, then the eighth element
Hence, metallic character increases down the group.
from it is Na and if we consider Na as the first element, then
(iv)
(b) Electropositive nature increases down the group
the eight element from it is K.
due to increase in atomic size.
Similarly, Dobereiner’s triad consisting of the elements
(v)
(d) Along the period 4, Ca is the most electronegative
beryllium (Be), magnesium (Mg) and calcium (Ca) is also
atom because along a period, non-metallic character
included in the column of Newlands’ octaves.
increases and so electronegativity also increases.
Thus, Dobereiner’s triads are included in the columns of
22.
(i)
(a)H and Phave complete octet.So,theyare noble gases.
Newlands’ octaves.
(ii)
(c) G and O have 7 electrons in their outermost shell.
4.
(i) Here, the elements are arranged in the order of
So, they belongs to halogen group.
increasing atomic masses, so according to Newlands’
law of octaves there is a repetition of every eighth
(iii)
(c)D has6electrons.So,itselectronicconfigurationis2,4.
element as compared to the given element. The two
L has 14 electrons. So, its electronic configuration is
sets of elements which have similar properties are
2, 8, 4.
Set I → F, Cl
∴Both have 4 valency.
Set II → Na, K
(iv)
(d) A has one valence electron. So, its valency is 1.
F and Cl are first and eighth element in the above
It can form A+ ion.
sequence, therefore, they have similar properties.
G has seven valence electron. It needs one electron to
Although Na and K have similar properties but they are
complete its octet. So, A can make ionic bond with G
not related as first and eighth element in the above
and form stable compound, AG.
sequence.
(v)
(b) B has 4 electrons. So, its electronic configuration is
(ii) The given sequence is according to Newlands’ law of
2, 2 and hence, have 2 valence electrons.
octaves represented as
J has 12 electrons. So, it’s electronic configuration is 2,
F Na Mg Al Si P S Cl Ar K
8, 2 so, it also contains 2 valence electrons.
5.
● This law was applicable only upto calcium. After calcium,
∴ Both B and J contains 2 valence electrons.
every eighth element did not possess the same properties
similar to that of the first.
Subjective Questions
● Newland assumed that there were only 56 elements existed
1. All the elements discovered at that time could not be
in nature and no more elements would be discovered in the
classified into triads, only a limited number of elements
future. But, later on, several new elements were discovered,
could be arranged in such triads.
whose properties did not fit into the law of octaves.
e.g. The three elements nitrogen (N), phosphorus (P) and
● In order to fit elements into his table, Newlands’ adjusted two
arsenic (As) have similar properties. Therefore, they should
elements in the same slot and also put some unlike elements
be regarded to form a triad.
under the same column.
However, the actual mass of the middle element P(31.04) is
e.g. Cobalt and nickel are in the same slot and these are
much lower than the average (44.454) of the atomic masses
placed in the same column as fluorine, chlorine and bromine
of nitrogen (14.4) and arsenic (74.94). Therefore, these three
which have very different properties than these elements.
elements do not constitute a Dobereiner’s triad inspite of
Iron, which resembles cobalt and nickel in properties, has
their similar chemical properties.
been placed far away from these elements. Hence, Newlands’
2.
(i) Na, Si and Cl have different properties, therefore, they
law of octaves worked well with lighter elements only.
do not form Dobereiner’s triad even though the atomic
mass of the middle atom (Si) is approximately the
6. The criteria used by Mendeleev were:
average of the atomic masses of Na and Cl, i.e.
(i) The arrangement of elements in increasing order of
Na (23); Si (28); Cl (35)
atomic masses.
(ii) Similarity in chemical properties of the elements.
23
+
35
58
Atomic mass of Si =
=
=
29
7. In Mendeleev’s periodic table, cobalt (Co) with a higher
2
2
atomic mass of 58.93 u is placed before nickel (Ni) due to the
(ii) Be, Mg and Ca have many similar properties and also
following reasons :
the atomic mass of the middle element Mg is
(i) The properties of cobalt are similar to those of rhodium
approximately the average of the atomic masses of Be
(Rh) and iridium (Ir) (same group) and
and Ca, i.e.
(ii) The properties of nickel are similar to those of
Be (9); Mg (24); Ca (40)
palladium (Pd) and platinum (Pt) (same group).
9
+ 40
49
Atomic mass of Mg =
=
= 24.5
8. Eka-silicon is germanium (Ge). It lies in group 4 of the
2
2
Mendeleev’s periodic table and thus, has a valency of 4.
Therefore, they form Dobereiner’s triad.
∴ The formula of its chloride is GeCl4.
Eka-aluminium is gallium (Ga). It lies in group 3 of the
average of the atomic masses of other two elements,
Mendeleev’s periodic table and thus, has a valency of 3.
e.g.
∴ The formula of its chloride is GaCl3.
Elements
Cl
Br
I
9. Two main characteristics of Mendeleev’s periodic table are :
Atomic mass
35.5
80
127
(i) It consists of 8 vertical columns, called groups and
Average atomic mass
35.5 + 127
= 81.25
6 horizontal rows, called period.
of first and that elements
2
(ii) In every period, elements are arranged in increasing
(iii) Modern periodic law The physical and chemical
order of their atomic masses.
propertiesare a periodicfunctionoftheir atomicnumber.
Name of elements of second period are lithium, beryllium,
13. Atomic number of A =19
boron, carbon, nitrogen, oxygen, fluorine.
Electronic configuration is 2, 8, 8, 1.
10. Electronic configuration of an element decides its position
Hence, element A is metal potassium (K) and
in modern periodic table.
Atomic number of B = 17.
If we take an example of sodium (Na), which has
atomic number =11, i.e. it’s electronic configuration = 2, 8,1
Electronic configuration is 2, 8, 7.
As Na contains 1 electron in its outermost shell, this means
It is a non-metal, chlorine (Cl).
that it belongs to group 1 and sodium contains 3 shells so, it
So, the electron dot structure of KCl is
belongs to period number 3.
××
××-
K
+ ×Cl×
K+
×
Cl×
∴ We can conclude that,
2, 8, 8, 1
××
××
Group number = Number of valence electrons
2, 8, 7
2, 8, 8
2, 8, 8
Potassium
Chlorine
Potassium chloride
(When valence electrons are 1 and 2)
and group number =10 + valence electrons
The bond formed between K+ and Cl− is ionic bond and
(When valence electrons are 3 and above)
formula of the product formed K+ Cl− or KCl.
Period number = Number of shells in which electrons are filled.
14. Since, species X has 12 protons and 12 electrons, it is
11. Hydrogen occupies a unique position in the modern
electrically neutral. Since, species Y has 12 protons and
periodic table due to the following reasons
10 electrons, therefore, it has two units positive charge.
(i) Both hydrogen and alkali metals have similar outer
The electronic configuration of the two species are
electronic configuration as both have one electron in
Species X
Species Y
the valence shell. Therefore, some of the properties of
K L M
K L
hydrogen are similar to those of alkali metals and hence,
2
8
2
2
8
it can be placed in group 1 alongwith alkali metals.
Since, species X has three shells while species Y has two
(ii) Both hydrogen and halogens have similar outer
shells, therefore, species Y has smaller radius than species X.
electronic configuration (both have one electron less
15.
(i) Atomic radii decreases along a period from left to right
than the nearest inert gas configuration). Therefore,
due to increase in nuclear charge. Li, Be, F and N
some of the properties of hydrogen are similar to those
belong to same period. Thus, the atomic radii of Li, Be,
of halogens and hence, it can be placed in group 17
F and N increases in the order:
alongwith halogens.
F < N < Be < Li
(iii) In some properties, it differs from both hydrogen and
halogens, e.g. the oxide of hydrogen, i.e. H Ois neutral
(ii) Atomic radii increase in a group from top to bottom due
2
to the corresponding increase in the number of filled
but the oxides of alkali metals (i.e. Na O, K O
2
2
etc.) are
electronic shells. Cl, At, Br, I belong to same group. Thus,
basic while those of halogens (i.e. Cl O , Br O , I O
2
7
2
5
2
5
atomic radii of Cl, At, Br and I increase in the order:
etc.) are acidic.
Cl < Br < I < At
12.
(i)
16. Since, the element ‘X’ of group 15 exists as a
Mendeleev’s periodic table
Modern periodic table
diatomic molecule and combines with hydrogen at 773 K
In the Mendeleev’s
In modern periodic table,
in presence of a catalyst to form ammonia which has a
periodic table, the
the elements are arranged in
characteristic smell, therefore, the element ‘X’ is
elements were arranged
the increasing order of their
nitrogen (N).
in increasing order of
atomic number.
773 K
their atomic masses.
N
2
+
3H
2
⎯
⎯⎯→
2NH
3
Nitrogen
Hydrogen
Catalyst
Ammonia
This table consists of 8
This contains 18 groups and
(diatomic molecule)
(pungent smell)
groups and 6 periods.
7 periods.
(i) The atomic number of nitrogen is 7. So, its electronic
(ii) Dobereiner arrange the elements with similar
configuration is 2, 5. Thus, it has five valence electrons.
properties into groups having three elements each and
(ii) Nitrogen has 5 valence electrons. Therefore, it needs 3
named these groups as triads.
more electrons to complete its octet.
He showed that when the three elements in a triad
To do so, it shares three of its electrons with three
were arranged in the order of increasing atomic masses,
electrons of the other nitrogen atom to form a diatomic
the atomic mass of the middle element was roughly the
molecule of N2 gas.
Thus, three covalent bonds are formed between two
(Q Valency of group 2 element (A ) is 2 and that of group
nitrogen atoms and each nitrogen atom is left with one
17 element (B) is 1).
lone pair of electrons.
∴
A B
N
+ N
N N or N N
2
1
⇒ AB2
Two nitrogen atoms
Nitrogen molecule
20.
The electronic configurations of the given elements
(iii) Electron dot structure for ammonia is as follows :
are as follows
H
H
3Li→2,1
×
19 K→2,8,8,1
N
+
3H×
N
×
H or
N
H
×
12 Mg→2,8,2
H
H
6C→2,4
Nitrogen atom
Three hydrogen
Ammonia molecule
13 Al→2,8,3
atoms
S→ 2, 8, 6
16
In NH3 molecule, there are three N—H single
(i)
Element belong to same group are Li and K as they
covalent bonds and one lone pair of electrons on the
both contain one electron in their outermost shell.
nitrogen atom.
(ii)
Element which has tendency lose two electrons is
17. Since, both the ions consists of same number of
magnesium as it cantains 2 electrons its outermost
electrons and has +1 and −1 charges, hence the ions
shell.
should belong to group 1A (cation, i.e. A+) and group VII A
(iii)
Element which prefer sharing of electron to complete
(anion, i.e. B−).
its octet is carbon due to its small size and strong C⎯ C
Same number of electrons indicates that their electronic
bond.
configuration is same as that of a noble gas whose
(iv)
Most metallic element is potassium. Elements of group
atomic number lie between that of the two elements A
1 are metallic in nature as they readily loose their one
and B.
valence electron.
Dividing the molecular weight (which is sum of atomic
(v)
Non-metals form acidic oxides. Among the given
masses of A and B), we get the rough idea about the atomic
elements, S is a non-metal. Thus, it forms most acidic
74.5
mass of the noble gas which is
= 37.25, i.e. nearest to
oxide.
2
(vi)
Aluminium belongs to group 13 as it contains
argon (Ar - 40).
3 elements in its outermost shell.
Hence, A is K (group IA, 4th period) and element B is Cl
(group VII A, 3rd period).
21.
(i)
Element ‘D’ is a metal with valency two.
Because, the group number of an element having upto
18.
(i) Element A has 3 valence electrons, therefore, its
two valence electrons is equal to the number of valence
valency is 3 and thus belongs to group 13 (3 + 10). As
electrons.
such, it could be any one of the following elements : B,
Al, Ga, In or Tl.
(ii)
‘C’ is the least reactive element. Because, it belongs to
group 18. Group 18 (Noble gases) are least reactive due
(ii) Element B has 4 valence electrons, therefore, its
zero valency of group 18 elements.
valency is 4 and it belongs to group 14 (4 + 10). The
element B could be any one of the following : C, Si, Ge,
(iii)
‘E’ has a smaller atomic radius than ‘D’ because ‘E’ is
Sn or Pb.
on the right side of the modern periodic table.
(iii) Element C has two valence electrons, therefore, its
Across the period, atomic size/radius decreases on
valency is 2 and it belongs to group 2. The element C
moving left to right. This is due to an increase in
could be any one of the following : Be, Mg, Ca, Sr, Ba
nuclear charge which tends to pull the valence
or Ra.
electrons closer to the nucleus and reduces the size of
the atoms.
19.
(i) The valency of the group 1 elements is 1 and that of
oxygen is 2.
22.
(i)
Metalloids as these elements show the properties of
both the metals and non-metals.
Na O
(ii)
The list of metalloids alongwith their atomic number is
1
2
as follows:
∴ Molecular formula of oxide = Na O
B (5), Si(14), Ge (32), As (33), Sb (51), Te (52) and Po (84).
2
(iii)
These elements are located in groups 13, 14, 15 and 16.
(ii) The valency of group 13 element is 3 and that of halide
is 1.
23.
(i)
Thallium has the most metallic character. Metallic
character increases down in a group.
Al
Cl
(ii)
Boron has the highest electronegativity because
3
1
electronegativity decreases down a group.
∴ Molecular formula of halide =AlCl3
(iii)
Less metallic in character, because on moving across a
(iii) Molecular formula of compound formed = AB2
period, metallic nature decreases.
24. Oxygen is a member of group VI A in Mendeleev’s periodic
27.
(i)
Noble gases are the elements which have completely
table. Its valency is
2. Similarly, the valencies of all the
filled shells. The noble gas with two shells (K, L) is Ne
elementsgivencanbe predicted from their respective group.
having atomic number 10 and electronic configuration
This can help in writing the formula of their oxides.
KL both of the shells are completely filled.
28
(i) Potassium (K) is a member of group IA. Its valency
(ii)
Electronic configuration 2, 8, 2 suggests that atomic
is 1. Therefore, the formula of its oxide is K O.
number is 12 (2 + 8 + 2). Magnesium (Mg) has atomic
2
(ii) Carbon (C) is a member of group IVA. Its valency is 4.
number 12.
Therefore, the formula of its oxide is C O
2
4orCO2.
(iii)
The element with three shells and four electrons in
(iii) Aluminium (Al) belongs to group IIIA and its valency is
the valence shell will have electronic configuration
3. Therefore, the formula of its oxide is Al O .
KL M. The atomic number of this element is
2
3
28
4
(iv) Silicon (Si) is present in group IVA after carbon. Its
14
(2+8+4) so it will belong to group 14. Hence, it is
valency is also 4. Therefore, the formula of its oxide is
silicon (Si).
2
Si O
4 orSiO2.
(iv)
Element with two shells and 3 electrons in the valence
(v) Barium (Ba) belongs to group IIA and its valency is 2.
shell will exist in second period and will have the
Therefore, the formula of its oxide is Ba O
2
2 orBaO.
electronic configurationKL.
25.
The atomic number of this element will be 5 (2, 3). So,
Mendeleev’s periodic table
Modern periodic table
it will be boron (B).
The properties of elements
The properties of elements are
(v)
The element has two shells. We know that first shell
are the periodic functions of
the periodic functions of their
can have only 2 electrons, so according to the question
their atomic mass.
atomic number.
there will be 4 electrons (double the number of
electrons in the first shell) in the valence shell. The
It has 8 groups.
It has 18 groups.
K L, so the atomic
electronic configuration will be
24
There is no place for
Isotopes of an element are
number is 6. Hence, the element is carbon (C).
isotopes of an element.
assigned the same place with
28.
(i)
The electronic configuration of element X with atomic
their respective elements as
they have the same atomic
number 17 is 2, 8, 7. Since, it has 7 valence electrons.
Therefore, it lies in group 17 (10
+
7
). Further, since in
number.
element X, third shell is being filled, it lies in third
No fixed position was given
Hydrogen is given a special
period. In other words, X is chlorine.
to hydrogen in this periodic
position in modern periodic
The electronic configuration of element Ywith atomic
table.
table.
number 20 is 2, 8, 8, 2. Since, it has 2 valence electrons,
Inert gases were not known Inert gases have been placed at
it lies in group 2. Further, since in element Y, fourth
at the time of Mendeleev.
the end of period in group 18.
shell is being filled, it lies in 4th period. In other words,
26.
(i) Since, the element lies in group 2, it must be an
Y is calcium.
alkaline earth metal. Since, it lies in the third period, it
(ii)
Since, element X (i.e. Cl) has seven electrons in the
must be magnesium (Mg).
valence shell and needs one more electron to complete
(ii) Atomic number of Mg is 12, therefore, its electronic
its octet. Therefore, it is a non-metal. Further, the
K L M
element Y has two electrons in the valence shell that
configuration is 2,
8,
2
can be easily lost to achieve the stable electronic
(iii) When Mg burns in the presence of air, it forms a basic
configuration of the nearest inert gas, therefore, it is a
oxide, MgO.
metal.
Heat
2Mg (s) + O (g)
2
⎯
⎯
2MgO (s)
Magnesium Oxygen
Magnesium oxide
(iii)
Since, element Y (i.e. Ca) is a metal, therefore, its oxide
(i.e. CaO) must be basic in nature. Further, metals and
(iv) When MgO is dissolved in water, it forms magnesium
non-metals form ionic compounds, therefore, the
hydroxide.
nature of bonding in calcium oxide is ionic.
2MgO (s)
+
2
2HO(l)
⎯→
2Mg(OH) (
2 aq
)
Magnesium oxideWater
Magnesium hydroxide
(iv)
Electronic configuration of20Ca = 2, 8, 8, 2 [valence
electrons = 2], electronic configuration of17Cl = 2, 8, 7
(v) Mg has 2 valence electrons [as electronic configuration
[valence electrons = 7]. The electron dot structure of
of12 Mg = [2, 8, 2]. Oxygen has 6 valence electrons [as
divalent metal halide,
electronic configuration of8O = [2, 6]. Electron dot
××
structure for the formation of magnesium oxide.
×
-
×Cl
×
××
××
×
2-
Ca
+
××
Ca2+
×Cl
××
Heat
××
i.e. CaCl2
×
×
×Cl×
××
2
Mg
+
O
×
Mg2+
O×
×
–
××
××
××
××
×
Magnesium oxide
2, 8, 8, 2
2, 8, 7
or Y2+
×X
2, 8, 2
2, 6
×××
Calcium
Chlorine
2
Magnesium
Oxygen
Y
X
Calcium chloride
29.
(i)
(a) Lithium and potassium, due to presence of same
(iv)
As the element belongs to group-15, i.e., it’s valence
number of valence electrons.
electrons are 5 and valency is 3, so it has the electronic
configuration = 2, 8, 5.
(b) N → 2, 5
P→ 2, 8, 5
(v)
At present, 118 elements are known to us. All these
N is more electronegative element as
have different properties. Out of these 118 elements,
electronegativity decrease on moving down the
only 94 are naturally occurring.
group.
31.
(i)
The atomic number of element B is 9 and number of
(c) Both fluorine and chlorine form their hydrides on
reacting with hydrogen.
valence electrons are 7, i.e. it belongs to group 17.
Valency = 8 − valence electrons
2
H +F
2
⎯→
2HF
=8−7 =1
2
H + Cl
2
⎯→
2HCl
Hence, the valency of B is 1.
K L M
(ii)
(a) The electronic configuration of E
=
(ii)
Electronic configuration of element C is 2, 8, 7, i.e. it
2,
8,
2
has 7 valence electrons and valency is −1 (as it requires
Hence, it should be placed in 3rd period and
1 electron to complete it’s octet) and electronic
group II A.
configuration of element D is 2, 8, 8, 2, i.e. it has 2
(b) The element E is magnesium (Mg).
valence electrons and valency.
(c) Mg loses 2 electrons to form noble gas (Ne) with
C
D
K L
configuration as 2,
8
–1
+2 Charge
(d) Since, Mg is electropositive, it will form basic
2
C D
oxide.
This forms a divalent compound (C D)
2
(e) The formula of chloride is ECl2 or MgCl2.
(iii)
The atomic number of element E is 36, so the
10
2
6
30.
(i)
The element whose atomic number is 17 and belongs to
electronic configuration is [Ar]3d
4s
4p
. Therefore,
third period is chlorine. If we go down the group, i.e. in
this element belongs to 4th period as last electron goes
fifth period, number of shells increases as18e− are
in 4th shell and have zero number of valence electrons.
increased in consecutive periods.
Hence, the element is krypton.
Therefore, the atomic number of the element
belonging to same group and present in fifth period is
(iv)
The atomic number of element A is 3 and electronic
2
1
configuration is1s
2s
. As last electron enters in 2nd
17 +18 +18 = 53.
shell, so, it belongs to period 2.
(ii)
Hydrogen (H) and helium (He).
(v)
The element that belongs to group 17 is ‘C’ because
(iii)
The number of valence shell electrons increases by one
unit as the atomic number increases by one unit on
this group contain those elements which have 7
electrons in their outermost shell and element C also
moving from left to right in a period. Therefore, the
atoms of different elements with same number of shells
contains 7 valence electrons. Therefore, option (b) is
correct.
are placed in the same period.
