11th Standard Syllabus & Materials
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Published on: 29/04/2019
Public exam five mark questions Periodic Classification Of Elements
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1.
Calculate the ionic Radii of Na+ and F- ion in NaF crystal whose interionic distance is equal to 231 pm
2.
Given the bona lengths of H2, Cl2, C - C bond in diamond, Si - C bond in carborundum and C - CI bond in CCl4 are 0.74, 1.9, 1.54, 1.93 and 1.76A respectively, find the covalent radius of hydrogen, chlorine, carbon in diamond, silicon and carbon in CCI4.
3.
What are the factors influencing ionization enthalpy.
4.
I.E increases as we move across the period but Ionisation enthalpies (I.E) of second period of elements in the order.
Li < B < Be < C < O < N < F < Ne
Explain why?
(i) Be has higher I.E and B
(ii) O has lower I.E than N & F
5.
Calculate the effective nuclear charge of the last electron in an atom whose configuration is 1s2 2s2 2p6 3s2 3p5
6.
Calculate the effective nuclear charge experienced by the 4s electron in potassium atom.
7.
Bring out the differences between electronegativity and electron affinity
8.
What are the factors which influence the electron gain enthalpy?
9.
List out and compare the chemical properties of metals and non-metals.
10.
Give any five characteristic properties of inner transition elements.
11.
Give the characteristics of p-block elements.
12.
How do you classify elements into blocks? Give their electronic configuration.
13.
Mention Anomalies of Mendeleev's periodic table.
14.
Give the structural features of modern periodic law.
1.
d = rNa +rF-
i.e rNa + rF- = 231 pm
We know that
\(\frac { { r }_{ Na }+ }{ { r }_{ F- } } =\frac { (Zeff)F- }{ (Zeff){ Na }^{ + } } \)
(Zeff)F- = Z -Z
= 9 - 4.15
=4.85
(Zeff)Na+ = 11 - 4.15
=6.85
\(\frac { r{ Na }^{ + } }{ r{ F }^{ - } } =\frac { 4.85 }{ 6.85 } \)
\(\Rightarrow\) rNa+ = 0.71 x rF-
substituting (2) in (1)
(1) \(\Rightarrow\) 0.71 rF- + rF- = 231 pm
1.71 rF- = 231 pm
\(r{ F }^{ - }=\frac { 231 }{ 1.71 } =135.1\) pm
Substituting the value of rF- in equation (1)
rNa+ 135.1 =231
rNa+ = 95.9 pm
2.
(i) Bond length (dH-H) in hydrogen molecule = 0.74A
\(\therefore\) Covalent radius of hydrogen = \(\frac { d_{ H-H } }{ 2 } -\frac { 0.74 }{ 2 } =.37A\)
(ii) Bond length (dCI-Cl) in chlorine molecule = 1.98A
\(\therefore\) Covalent radius of chlorine = \(\frac { d_{ cl-CL } }{ 2 } =\frac { 1.98 }{ 2 } =0.99A\)
(iii) Bond length (C - C) in diamond = 1.54A
\(\therefore\) Covalent radius of carbon = \(\frac { 1.54 }{ 2 } =0.77A\)
(iv) Bond length of (dsi-c) in carborundum = 1.93A
\({ d }_{ si-c }={ r }_{ si }+{ r }_{ c }\)
\(1.93\quad =\quad { r }_{ si }+0.77\)Refer (iii)
\({ r }_{ si }=1.93-0.77=1.16A\)
(v) Bond length (dC-CI) in CCl4 = 1.76A
\({ d }_{ c-cl }={ d }_{ c }+{ d }_{ cl }\)
1.76 \(=r_{ c }+0.99\) Refer (ii)
\({ r }_{ c }=1.76-0.99=0.77A\)
3.
Factors influencing ionization enthalpy:
(i) Size of the atom:
(ii) Magnitude of nuclear charge:
(iii) Screening or shielding effect of the inner electrons:
Ionization enthalpy decreases when the shielding effect of inner electrons increases. This is because when the inner electron shells increases, the attraction between the nucleus and the outermost electron decreases.
(iv) Penetrating power of subshells s, p, d & f:
The penetration power of the electrons in various orbitals decreases in a given shell in the order: s > p > d > f.
(v) Electronic configuration:
If an atom has half-filled or completely filled sub-levels, its ionization enthalpy is higher. This is because such atoms have extra stability and hence it is difficult to remove electrons from these stable configuration.
4.
(i) 4Be - 1s2 2s2 ; 5B - 1s2 2s2 2p1
(ii) 7N - 1s2 ,2s2 ,2px1, 2py1, 2pz1
8O -1s2 ,2s2 ,2px2, 2py1, 2pz1
∴ O has lower I.E. than N.
5.
Z = 17
Z* = Z -S
= 17- [(0.35 \(\times\) No. of other electrons in nth shell) + (0.85\(\times\) No. of electrons in (n -1)th shell) + (1.00 \(\times\) total number of electrons in the inner shells)]
= 17 - [(0.35 \(\times\) 6) + (0.85 \(\times\) 8) + (1 \(\times\) 2)]
= 17 - 10.9 = 6.1
Z* = 6.1
6.
The electronic configuration of K atom is
K19 = (1s2)(2s2p6)(3s23p6)4s1
Effective nuclear charge (Z*) = Z - S
Z* = 19 - [(0.85 \(\times\) No. of electrons in (n -1)th shell) + (1.00 total number of electrons in the inner shells)]
= 19-[0.85 \(\times\) (8) + (1.00 \(\times\) 10)]
Z* = 2.20
7.
| Electron gain Enthalpy | Electro negativity |
| It is the tendency of an isolated gaseous atom to attract an electron. |
It is the tendency of an atom in a molecule to attract the shared pair of electrons |
| It is measured in electron volts/atom or kcal/mole or kj/mole. |
It is a number and has no units |
| It is the property of an isolated atom. | It is property of a bonded atom. |
| An atom has an absolute value of electron gain enthalpy |
An atom has a relative value of electronegativity depending upon its bonding state. For example, sphybridized carbon is more electronegative than sp2-hybridized carbon which, in turn, is more electronegative than sp,- hybridized carbon. |
| It does not change regularly in a period or group. | It changes regularly in a period or a group |
8.
Factors influencing electron gain enthalpy:
(i) Size of the atom : Electron affinity is inversely proportional to the size of the atom. As the size of atom increases, the effective nuclear charge decreases or the nuclear attraction for adding electron decreases.
(ii) Nuclear charge: Electron affinity is directly proportional to effective nuclear charge. Greater the nuclear charge more will be the tendency to accept electrons so E.A is more.
(iii) Electronic configuration: An atom with stable electronic configuration has no tendency to gain an electron. Such atoms have zero or almost zero electron gain enthalpy.
(iv) Shielding effect: Electron affinity is inversely proportional to shielding effect. Electronic energy state, lying between nucleus and outermost state hinder the nuclear attraction for incoming electron. Therefore, greater the number of inner lying states, less will be the electron affinity.
9.
| Chemical Properties | Metals | Non-Metals |
|---|---|---|
| Oxidising and Reducing Action | Generally good reducing agents | Generally good oxidising agents |
| Nature of oxides | Metallic oxides are basic in nature | Non-metallic oxides are acidic in nature |
| Nature of Hydrides | Form unstable hydride | Form stable hydrides |
| Action with acid | Dissolve in mineral acid to form salt. | Do not react with mineral acids |
| Arrangement of valence electrons | They have one, two or three valence electrons. | They have 4,5,6, 7 valence electrons |
10.
Characteristics of f-block elements:
(i) The elements in which the last electron enters in f-orbital are called f-block elements.
(ii) This block consists of 4f block elements and 5f block elements.
Lanthanoid series (4f) 58Ce- 71Lu
Actinoid series (5f) 90Th- 103Lr
(iii) Their general electronic configuration is (n - 2) f1-14 (n - 1)d0-1 ns2.
Lanthanoid - 4f1-14 5d0-1 6s2
Actinoid - 5f1-14 6d0-1 7s2
11.
Characteristics of p-block elements:
(i) General electronic configuration is ns2np1-6 (Group 13 to 18).
(ii) They are generally non-metals but some metals and metalloids are present.
(iii) They have high electron affinity and ionization enthalpy.
(iv) Oxidising character.
(v) s-block and p-block elements are collectively called representative elements.
12.
| Elements | Belonging to Groups | Electronic Configuration | Reason for their Name |
|---|---|---|---|
| s-block | 1 and 2 | ns1 and ns2 | Valence electron enters the s-orbital |
| p-block | 13 to 18 | ns2np1 to ns2np6 | Valence electron enters the p-orbital |
| d-block | 3 to 12 | (n-1)d1-10ns(0-2) | Valence electron enters the d-orbital |
| f-block | lanthanides and Actinoids | (n-2)f1-14(n-1)d0-1ns2 | Valence electron enters the f-orbital |
13.
Anomalies of Mendeleev's periodic table:
(i) Some elements with similar properties were placed in different groups whereas some elements having dissimilar properties were placed in same group. Eg: Tellurium (127.6) was placed in VI group but iodine (127.0) was placed in VII group.
(ii) Some elements with higher atomic weights were placed before lower atomic masses in order to maintain the similar chemical nature of elements. This concept was called inverted pair of elements concept. Eg: \(_{ 27 }^{ 59 }{ Co }._{ 28 }^{ 58.7 }{ Ni }\) .
(iii) Isotopes did not find any place in Medeleev's periodic table.
(iv) Position of hydrogen could not be made clear.
(v) He did not leave any space for lanthanides and actinides which were discovered later on.
(vi) It could not explain the variable valencies of elements.
(vii) Elements with different nature were placed in one group as 'A' and 'B'; Eg: Alkali metals & coinage metals were placed together.
(viii) Diagonal and horizontal relationships were not explained.
14.
(i) According to the recommendation of IUPAC, the groups are numbered from 1 to 181A.
(ii) There are 18 vertical columns which constitute 18 groups or families.
(iii) There are 7 horizontal rows of the periodic table known as periods.
(iv) The first period contains two elements. One present in first group and the other in 18th group.
(v) Second and third periods contain 8 elements in each.
(vi) Fourth and fifth periods are completely filled as they contain 18 elements in each.
(vii) The sixth period contains 32 elements. The seventh is incomplete. Fourteen elements of both sixth and seventh periods are placed in separate panels at the bottom of the table.
(viii) This periodic table is important and useful because we can predict the properties of any element using periodic trend.
11th Standard Syllabus & Materials
11th Standard
TN 11th Tamil பீடு பெற நில் - செய்யுள் - காவடிச்சிந்து Important Questions And Answers Study Material - QB365 Set A
NEW11th Standard
TN 11th Tamil பீடு பெற நில் - உரைநடை - மலை இடப்பெயர்கள் : ஓர் ஆய்வு Important Questions And Answers Study Material - QB365 Set A
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TN 11th Tamil மாமழை போற்றுதும் - துணைப்பாடம் - யானை டாக்டர் Important Questions And Answers Study Material - QB365 Set A
NEW11th Standard
TN 11th Tamil மாமழை போற்றுதும் - செய்யுள் - ஐங்குறுநூறு Important Questions And Answers Study Material - QB365 Set A
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