YASH CLASSES
Question:1. What is the basic theme of organization in the
periodic table?
Answer: The basic theme of organization in the periodic table is
the periodic law, which states that the properties of elements are periodic
functions of their atomic numbers. This means that elements are arranged in the
table in order of increasing atomic number, which corresponds to the number of
protons in the nucleus of an atom.
Or
The elements in the
periodic table are organized in such a manner that elements with similar
properties are placed in the same group. Since the properties of the elements
depend upon the valence shell configuration, therefore, elements of a group
generally have similar valence shell con- figuration.
Question:2. Which important property did Mendeleev use to classify the elements in his periodic table and did he stick to that?
Answer: Mendeleev used the atomic mass as the primary criterion
for the classification of elements in his periodic table. However, he also
considered the properties of elements, and when necessary, he adjusted the
order of elements to maintain consistency in the properties. This led to the
prediction of the existence and properties of undiscovered elements.
Question: 3.What is the difference between atomic mass and atomic number?
Answer:
Atomic mass is the weighted average mass of an atom
of an element, measured in atomic mass units (amu), considering the natural
abundance of its isotopes. Atomic number, on the other hand, is the total
number of protons in the nucleus of an atom and is unique to each element. The
atomic number determines the element's identity and its position in the
periodic table.
Question:4. How does electronic configuration of atoms change in the periodic table across the period and down the group?
Answer:
- Across a Period:
As you move across a period from left to right, the
atomic number increases, leading to an increase in the number of electrons.
Electrons are added to the same principal energy level, which results in the
filling of s and p orbitals in that order.
Down a Group:
As you move down a group, the atomic number
increases, and electrons are added to new principal energy levels, increasing
the size of the electron cloud. This results in the addition of new shells,
leading to a repetition of similar outer shell electron configurations within a
group.
Question:5. Describe the general trends in the following properties of elements in the periodic table. (a) Ionization enthalpy (b) Electron gain enthalpy (c) Atomic radii (d) Electronegativity
Answer: Ionization Enthalpy
Across a Period:
Increases from left to right due to increased nuclear charge, making it harder
to remove an electron.
Down a Group:
Decreases as the atomic size increases, making it easier to remove an electron
due to the increased distance from the nucleus.
- Electron Gain Enthalpy:
- Across a Period:
Becomes more negative (more exothermic) from left to right as the atomic number
increases, indicating a stronger attraction for an additional electron.
- Down a Group:
Becomes less negative as the size of the atom increases, reducing the
attraction for an additional electron.
- Atomic Radii:
- Across a
Period: Decreases from left to right due to increased effective nuclear
charge pulling electrons closer.
- Down a Group:
Increases due to the addition of new energy levels, increasing the size of the
electron cloud.
- Electronegativity:
- Across a
Period: Increases from left to right as the effective nuclear charge
increases, enhancing the ability to attract shared electrons.
- Down a Group:
Decreases as the atomic size increases, reducing the nucleus's pull-on shared
electrons.
Intext Questions and Answers
Question3.1: What is the basic theme of organization in the periodic table ?Answer: The basic theme of organization in the periodic table is the periodic law, which states that the physical and chemical properties of elements are periodic functions of their atomic numbers. This implies that elements are arranged in order of increasing atomic numbers, leading to a recurring pattern of properties.
or
The elements in the periodic table are organized in such a manner that elements with similar properties are placed in the same group. Since the properties of the elements depend upon the valence shell configuration, therefore, elements of a group generally have similar valence shell con- figuration.
Answer: Mendeleev used the atomic mass as the primary criterion for the classification of elements in his periodic table. However, he also considered the chemical properties of the elements. When there was a conflict between atomic mass order and properties, he prioritized the chemical properties to group similar elements together and left gaps for undiscovered elements.
Answer: The basic difference is that Mendeleev’s Periodic Law is based on the atomic mass of elements, whereas the Modern Periodic Law is based on the atomic number of elements. The Modern Periodic Law, proposed by Henry Moseley, resolved anomalies in Mendeleev’s table by correctly predicting the properties of elements based on their atomic numbers.
Question 3.4:On the basis of quantum numbers, justify that the sixth period of the periodic table should have 32 elements.
Answer: The sixth period of the periodic table involves the filling of the 6s, 4f, 5d, and 6p orbitals. The possible number of electrons that can fill these orbitals is calculated as follows:
- 6s: 2 electrons , - 4f: 14 electrons , - 5d: 10 electrons , - 6p: 6 electrons
Adding these gives \(2 + 14 + 10 + 6 = 32\) electrons, meaning
the sixth period can accommodate 32 elements.
Answer: The element present in the third period and the seventeenth group is Chlorine (Cl), which has the atomic number 17.
Answer: (i) The element named by Lawrence Berkeley Laboratory is Berkelium (Bk), with the atomic number 97.
(ii) The element named by Seaborg’s group is Seaborgium (Sg),
with the atomic number 106.
Answer: Elements in the same group have similar physical and chemical properties because they have the same number of valence electrons. This similar electron configuration leads to similar ways of interacting with other elements and forming chemical bonds.
Answer: The first ionization enthalpy of sodium is lower than that of magnesium because sodium (Na) has one valence electron in the 3s orbital, making it easier to remove compared to magnesium (Mg), which has two valence electrons in the 3s orbital. However, the second ionization enthalpy of sodium is higher than that of magnesium because, after the removal of the first electron, sodium attains a stable noble gas configuration (Ne), making it significantly harder to remove a second electron. In contrast, magnesium, after losing one electron, still has one electron left in the 3s orbital, which is easier to remove than breaking into a noble gas configuration.
Answer: (i) Be has a higher ionization enthalpy than B because Be has a completely filled 2s subshell (1s22s2), making it more stable and requiring more energy to remove an electron compared to B, which has the configuration (1s 2 2s 2 2p1).
(ii) O has a lower ionization enthalpy than N and F because O has a half-filled 2p subshell (1s22s22p4), making it less stable due to electron-electron repulsion compared to N, which has a stable half-filled configuration (1s22s22p3), and F, which has a high nuclear charge attracting electrons more strongly.
Answer: The term ‘isolated gaseous atom’ refers to an atom in its gaseous phase, free from any interactions with other atoms or molecules, ensuring that the measurement of ionization or electron gain enthalpy is not influenced by external forces. The term ‘ground state’ refers to the lowest energy state of an atom, meaning it is not excited, ensuring that the ionization or electron gain process occurs from the most stable and standard energy level.
Answer: To calculate the ionization enthalpy of atomic hydrogen, we need to convert the energy of a single electron into energy per mole of electrons: Energy of one electron = -2.18x 10-18J
= -1.31x 106J mol-1.
Convert J to kJ: = Thus, the ionization enthalpy of atomic hydrogen is 1312 kJ mol⁻¹.
(a) ([Ne]3s2 3p1)
(b) ([Ne]3s23p3)
(c) ([Ne]3s23p2)
(d) ([Ar]3d{10}4s24p3)
The element with the highest ionization enthalpy is option (b) [Ne]3s23p3), which has a half-filled p subshell, making it more stable than the others Half-filled and fully filled orbitals confer extra stability, resulting in a higher ionization enthalpy.
Answer: The element with the highest electronegativity in the periodic table is Fluorine (F), with an electronegativity value of 3.98 on the Pauling scale.
Answer: No, the second ionization enthalpy of element X would not be smaller than the first ionization enthalpy. Once the first electron is removed, the remaining electrons experience a higher effective nuclear charge due to the reduced electron-electron repulsion, making it harder to remove the next electron. These results.
Q.9. What does atomic radius and ionic radius really mean to you?
Ans. Atomic radius gives us idea about size of
the atom. Atomic radius may be taken as the distance between the centre of the
nucleus and the outermost shell of the atom. It can be measured either by X-ray
or by spectroscopic methods. The ionic radius tells us about size of the ion.
It is defined as the distance between centre of the nucleus and the point up to
which the ion has influence in the ionic bond.
Ans.
The atomic radii of the elements decrease in a period. The reason is that the
outer electron enters the same shell but the nuclear charge goes on increasing.
Q.
11. What do you understand by isoelectronic species? Name a species that will
be with each or ions:
(i) F- (ii) Ar (iii) Mg2+ (iv) Rb +
Ans.
Those species which have same number of electrons and belong to different
elements and have different magnitude of nuclear charge are called
isoelectronic species.
The
species that are isoelectronic:
(i)
With F- are Na+ , Mg2+ and Al3+
which contain a10 electrons each.
(ii) With Ar are K+ ,Ca2+ and Cl- which contain 18 electrons each.
(iii) With Mg2+ ,F-,Na+ ,and Al3+are which contain 10 electrons each.
(iv) With Rb+ are Kr, Sr2+ which contain 36 electrons each.
Q. 12. Consider the following species : N3-, Al3+,F-, Na+ , O 2-, Mg2+
and Al3+. (a) What is common in them?
(b) Arrange them in the order of increasing ionic radii.
Ans. All of them are isoelectronic in nature and have 10 electrons each, In isoelectronic species, greater the nuclear charge, lesser will be the atomic or ionic radius. Based upon that, the increasing order of atomic radii are:
Al3+ < Mg2+
< Na+ < F- < O 2- < N3-
Q.13. Explain are smaller and anions larger in radii
than their parent atoms ?
Ans. Cations (positive ions) are formed by the loss of electrons from the parent atoms. Therefore, they have smaller radii than the parent elements. On the other hand, anions are formed when the parent atoms take up one or more electrons. They have therefore, bigger radii than the parent atoms.
Q.14.
What is the significance of the terms 'isolated gaseous atom' and 'ground
state' while defining the lionization enthalpy and electron gain enthalpy?
[Hint: Requirements for comparison purposes.]
Ans- Isolated gaseous atom means an atom
in its free state when its properties are not influenced by surroundings.
Ground state means when the energy contents of the atom are minimum. In such a
state the ionization energy is the maximum which is constant. Also in this
state its electron gain enthalpy is maximum. In fact these conditions are
necessary requirements for comparison of their properties of different
elements.
Question 15 :Energy of an electron in the ground state of the
hydrogen atom is –2.18 × 10⁻¹⁸ J. Calculate the ionization enthalpy of atomic
hydrogen in terms of kJ mol⁻¹.
Answer:
To calculate the ionization enthalpy of atomic hydrogen, we need
to convert the energy of a single electron into energy per mole of electrons:
Energy of one electron =
-2.18x 10-18J
1 mole of electrons contains Avogadro’s number (\(6.022 \times
10^{23}\)) of electrons.
Ionization
enthalpy per mole: = -2.18x 10-18J x 6.022 x 1023mol-1
= -1.31x
106J mol-1.
Convert
J to kJ: = Thus, the ionization enthalpy of atomic
hydrogen is
1312 kJ mol⁻¹.
Q.
16. Among the second period elements the actual ionization enthalpies are in
the order Li < B < Br < C < O < N < F < Ne Explain why:
(1)
Be has higher ∆i H than B?
(ii)
O has lower ∆i H1
than N and F?
Ans.
(1) For answer, consult text part.
(1) ∆i H1 values
of their elements present in
(2) second period are given:
N(1402kJ mol- 1), O (1314kJ mol-
1) and F(1681kJ mol- 1)
(a)
∆i H1 of O is
expected to more than that of N but is actually lesser because the electronic
configuration of N is more symmetrical as well as stable in comparison to O.
For electronic configuration, consult text part.
(b)
∆i H1 of O is less
than that of F because the ionization enthalpy in general increases along a
period. For details, consult text part.
Q.
17. How would you explain the fact that the first lonization enthalpy of sodium
is lower than that of magnesium but its second lonization enthalpy is higher
than that of magnesium?
Ans. The electronic configuration of the two
elements and their first and second ionization enthalpies are given:
Na(Z = 11) 10[Ne] [↑] 3s1
, Mg(Z = 11) 10[Ne][↑↓] 3s2 ,
∆i H1 : 496kJ
mol- 1 ; ∆i
H1 : 737kJmol- 1
∆i H2 : 562kJmol- 1, ∆i H2 : 1450 kJ mol-
1
∆i
H1 value of Mg more than that of Na due to greater symmetry and
smaller size. But ∆i H2
value of Na is higher because Na+ has the configuration of noble gas
element neon while Mg+ does not have a symmetrical configuration.
Q.
18. What are the various factors due to which the ionization enthalpy of the
main group elements tends to decrease down a group?
Ans.
The ionization enthalpies of elements decrease down the group. It is due to the
fact that:
(i)
A new shell is added and the valence electron moves farther away.
(ii) Nuclear hold decreases.
(iii)
Screening affect increases.
Q.19. The first ionization enthalpy values (in
kJ mol- 1)of group
13 elements are: B Al
Ga In TI
801 577
579 558 589.
How would you explain this deviation from the
general trend?
Ans.
∆i H of Ga is abnormally
large. This is due to abnormally small screening effect of 3d electrons which
are filled just prior to Ga.
Q.
20. Which of the following pairs of elements would have a more negative
electron gain enthalpy : (i) O of F, (ii) For Cl.
Ans.
(i) F has a higher negative electron gain enthalpy than O. This is because F
has higher Zeff than O. (ii) F has abnormally low electron enthalpy
than Cl because of its exceptionally small size and large number of valence
electrons which cause large inter electronic repulsions.
Q.
21. Would you expect the second electron gain enthalpy of O as positive, more
negative or less negative than the first? Justify your answer.
Ans.
For oxygen atom:
O(g)+e0(g);
(AH) =-141kJ mol¹ 0(g)e02(g); (AH) = +780kJ mol- 1. The first
electron gain enthalpy of oxygen is negative because energy is released when a
gaseous atom accepts an electron to form monovalent anion. The second electron
gain enthalpy is positive because energy is needed to over-come the force of
repulsion between monovalent an- ion and the second incoming electron.
Q.
22. What is the basic difference between the term’s electron gain enthalpy and
electronegativity?
Ans.
Electron gain enthalpy is the amount of energy released when an electron is added
to the neutral gaseous atom. But electronegativity is the power of an element
to attract covalent electron towards it.
Q.
23. How would you react to the statement that the electronegativity of N on
Pauling scale is 3-0 in all the nitrogen compounds?
Ans.
The electronegativity of an element on Pauling scale is a relative value w. r.
t. a standard reference and thus it remains same in all its com- pounds.
Q.
24. Describe the theory associated with the radius of an atom as it :
(a) Gains an electron, (b) Loses an electron.
Ans. This is explained in term of change in
effective nuclear charge (Zeff) of an element.
(a)
Zeff decreases on gaining an electron and thus electron could expand
and its radius increases.
(b)
Zeff increases on losing an
electron and thus electron cloud contracts and its radius decreases.
Q.
25. Would you expect the first ionization enthalpies for two isotopes of the
same element to be the same or different? Justify your answer.
Ans.
Two isotopes of the same element have same number of electrons,
same nuclear charge, same atomic radii.
Therefore, they are expected to
same
ionization enthalpies.
same
ionization enthalpies.
Q.
26. What are the major differences between metals and non-metals?
Ans. The major differences between metals and non-metals are:
Metals -
1.
Malleability: Can be hammered into
thin sheets.
2. Ductility:
Can be stretched into thin wires.
3.
Conductivity:
Good conductors of electricity and heat.
4.
Luster: Have a shiny
appearance.
5.
Density:
Generally have high densities.
6. Reactivity: Tend to lose electrons to
form positive ions (cations).
7.
Electron configuration:
Typically have 1-3 electrons in their outermost energy level.
Nonmetals-
1.
Brittleness: Break when
hammered or stretched.
2.
Insulators: Poor conductors of
electricity and heat.
3.
Dullness: Have a dull
appearance.
4.
Low density: Generally have
low densities.
5. Reactivity: Tend to gain electrons to
form negative ions (anions).
6.
Electron configuration:
Typically have 4-8 electrons in their outermost energy level.
Additionally, metals tend to:
-
Be solids at room temperature (except mercury)
-
Have high melting and boiling points
-
Be good reducing agents
Nonmetals tend to:
-
Be gases or solids at room temperature
-
Have low melting and boiling points
-
Be good oxidizing agents
the
periodic table tower the Ing questions:
Q.27.Use the periodic table to answer the following
questions :
(a) Identify an element with free electro the outer sub shell (f).
(b) Identify an element that would tend to lose two electrons.
(c) Identify an element that would tend to gain two electrons.
(d) Identify the group having metal, non-me liquid as well as gas
at the room temperature.
Ans. (a) All the elements of 17th group (Halogen fam have five electrons in outermost of subshell family) ns2 np5 electronic configuration. So they have five electrons in outermost shell.
(b) Lower members of 2nd group e.g.
Mg , Ca, Sr, Ba and Ra.
(c) Upper members of 16th group e.g. , O and S.
(d) First group contains H which is gas and nоn-metal Li, Na, K, Rb solid metals and Co which is liquid metal (MP 28°C).
Q. 28. The increasing order group 1 elements is Li < Na < K < Rb < c Whereas that among group 17 elements is F Cl > Br > I Explain.
Ans. The elements of group-1 have a strong tendency to lose electron. The tendency to lose electron in turn, depends upon the ionization enthalpy Since the ionization enthalpy decreases down the group, therefore the reactivity of group-1 elements increases in the order: Li < Na < K < Rb <Cs.
On the other hand, the elements of group- 17, have a strong tendency to accept electron. The tendency to accept electrons, in turn, depends upon their electrode potential. Since the electrode potentials of group-17 elements decrease on descending the group, therefore, their reactivities decrease in the
order: F > CI > Br > I
Q.
29. Write the general outer electronic configuration of s, p, d-and f-block
elements.
Ans.
s-Block elements: ns1-2 ,
p-Block elements: ns2p1-6
,
d-Block
elements: (n-1)d1-10ns0-2 ,
f-Block elements: (n-2)f 1-2(n-1)d0-1ns2
,
Q.
30. Assign the position of the element having outer electronic configuration:
(i) n s2np4 for n
= 3 (ii) (n - 1)d2 ns2 for n = 4 and
(iii)
(n - 2) f 7 s1-2 ,(n-1) d1ns2 for n
= 6 in the periodic table.
Ans- Period Group
(i) 3 2 + 4 + 10 = 16
(ii)
4 2+ 2=
16
(iii)
6 3
Q.31.
The first (∆i H1) and the second (∆r H2)
ionization enthalpies (in kJ mol¹) and the ( ∆eg H) electron gain
enthalpy (in kJ mol-1 ) of a few elements are given below:
Which of the above elements is likely to be :
(a) The least reactive element ?
(b) The most reactive metal ?
(c) The most reactive non-metal?
(d) The least reactive non-metal?
(e) The metal which can form a stable binary halide of the formula M*X_{2} (X = halogen)?
(f)
The metal which can form a predominantly stable covalent halide of the formula
MX (X = halogen)?
Ans. (a) The element V has highest first ionization enthalpy and positive electron gain enthalpy and hence it is likely to be the least reactive elements.
(b) The element II which has the least first ionization enthalpy and a low negative electron gain enthalpy is the most reactive metal.
(c) The element III which has high first ionization enthalpy and a very high negative electron gain enthalpy is likely to be the most reactive non-metal.
(d) The elements IV have a high negative electron gain enthalpy but not so high first ionization enthalpy. Therefore, it is the least reactive non-metal.
(e) The element VI has low values for first and second ionization enthalpies.
therefore it appears that the elements are
an alkaline earth metal and will
form binary halide of the formula MX2. metal and hence will form binary halide of the formula MX2.
(f) The element A has low first ionization enthalpy a very high second ionization enthalpy, therefore it must be an alkali metal with smaller atomic number and is likely to form a predominantly stable covalent halide of the formula MX.
Q.
32. Predict the formulae of the stable binary com- pounds that would be formed
by the combination of the following pairs of elements:
(a) Lithium and oxygen
(b) Magnesium and nitrogen
(c) Aluminum and iodine
(d) Silicon and oxygen
(e) Phosphorus and fluorine
(f) Element 71 and fluorine.
Ans. (a) Li2O, (b) Mg3N2 , (c) AlI3, (d) SiO2, (e) PF3 or PF5, (f) LuF3.
Q. 33. In the modern periodic table, the period indicates the value of:
(a) Atomic number
(b) Atomic mass
(c) Principal quantum number
(d) Azimuthal quantum number.
Ans. (c).
Q. 34. Which of the following statements related to the modern periodic table is incorrect:
(a) The p-block has 6 columns because a maxi- mum of 6 electrons can occupy all the orbital in a p-shell.
(b) The d-block has 8 columns, because a maxi- mum of 8 electrons can occupy all the orbital in a d-subshell.
(c) Each block contains a number of columns equal to the number of electrons that can occupy that subshell.
(d) The block indicates value of azimuthal quantum number (1) for the last subshell that received electrons in building up the electronic configuration.
Ans. (b).
Q. 35. Anything that influences the valence electrons will affect the chemistry of the element. Which one of the following factors does not affect the valence shell:
(a) Valence principal quantum number (n)
(b) Nuclear charge (Z)
(c) Nuclear mass
(d) Number of core electrons.
Ans. (c).
Q. 36. The size of iso-electronic species F, Ne and Nat is affected by:
(a) Nuclear charge (Z)
(b) Valence principal quantum number (n)
(c) Electron-electron interaction in the outer orbitals
(d) None of the factors because their size is the same.
Ans. (a).
Q. 37. Which one of the following statements is in correct in relation to ionization enthalpy :
(a) Ionization enthalpy increases for each successive electron.
(b) The greatest increase in ionization enthalpy is experienced on removal of electron from core noble gas configuration.
(c) End of valence electrons is marked by a big jump in ionization enthalpy.
(d) Removal of electron from orbitals bearing lower n value is easier than from orbital having higher n value.
Ans. (d).
Q. 38. Considering the elements B, Al, Mg and K, the correct order of their metallic character is:
(a) B > Al > Mg > K
(b) Al > Mg > B > K
(c) Mg > Al > K > B
(d) K > Mg > Al > B
Ans. (d).
Q. 39. Considering the elements B, C, N, F and Si, the correct order of their non-metallic character is:
(a)
B > C > Si > N > F .
(b) Si > C > B > N > F
(c) F > N > C > B > Si
(d) F > N > C > Si > B
Ans. (c).
Q. 40. Considering the elements F, Cl, O and N, the correct order of their chemical reactivity in terms of oxidizing property is :
(a) F > CI > 0 > N
(b) F > 0 > Cl > N
(c) CI > F > 0 > N
(d) C > F > N > Cl
Ans. (b).