GSEB std 10 science solution for Gujarati check Subject Chapters Wise::
જવાબ : It is the energy required to remove an electron from an isolated gaseous atom in its ground state. M (g) + I.E. àM+ (g) + e–
જવાબ : Lanthanoids (the fourteen elements after Lanthanum) and actinides (the fourteen elements after actinium) are called inner transition elements.
જવાબ : General electronic configuration of f-block elements =(n – 2) f1-14 (n -1) d0-1 ns2.
જવાબ : The elements of group 1 (alkali metals), group 2,(alkaline earth metals) and group 13 to 17 constitute the representative elements. They are elements of s-block and p-block.
(i)Mg or Ca (ii) S or Cl
જવાબ : (i) Ca (ii) S.
જવાબ : Horizontal rows are called periods and vertical columns are called groups.
જવાબ : Electron gain enthalpy of Cl is more negative than that of F.
Al, Si, Ba and O
જવાબ : Al and Si.
જવાબ : Ar, K+, CT, S2-, or P3- are isoelectronic with ca2+.
જવાબ : They have fully filled s-orbitals and hence have no tendency to accept an additional electron. That’s why energy is needed if an extra electron is to be added. Therefore, electron gain enthalpies of Be and Mg are positive.
જવાબ : Nitrogen has exactly half filled p-orbitals.
જવાબ : Modem Periodic Law states that physical and chemical properties of the elements are a periodic function of their atomic numbers.
જવાબ : The ionisation enthalpy of oxygen is lower than that of Nitrogen as because when we remove one electron from oxygen then it easily donates it to attain half-filled stability but in case of nitrogen, it is difficult to remove one electron as it already has half-filled stability and it would become unstable after that.
જવાબ : Ti has an atomic number of 22 and electronic configuration [Ar]3d24s2 and can show three oxidation states of +2,+3 and +4 in various compounds like TiO29(+4), Ti2O3(+3) and TiO(+2). The non-transition elements like the p block elements show variable oxidation states like in case of phosphorous. It has -3,+3 and +5.
જવાબ : (i) 72, 160 Because neon has van der Waals radii and fluorine has covalent radii. Covalent radius is always less than van der Waal’s radius, so the radius of Fluorine is 72pm and Neon is 160pm.
જવાબ : 1. They show variable oxidation states. The reducing character increases down the group and oxidising character increases along the period. 2. They have a high ionisation enthalpy than the s-block elements. 3. They usually form covalent compounds. 4. Both metals and non-metals are found in this group, but non-metals are slightly more in number.
જવાબ : (i) Carbon has the highest ionisation enthalpy. It increases from left to right across the period and decreases as we go down the group. (ii) Aluminium has the most metallic character. As we move down, the metallic character increases and decreases across the period from left to right.
જવાબ : There are 118 elements in the 7 periods of the modern periodic table. Therefore the element with atomic number 119 will lie in the 8th period of the first group and will have the outermost electronic configuration of 8s1. It belongs to group 1 and has a valency of one. The formula of its oxide will be M2O.
જવાબ : The elements having their outermost shell filled with d electrons are known as d block elements. All d block are not transition elements because it is important to have incompletely filled d orbital of the element like calcium and zinc etc.
જવાબ : Fluorine has a smaller size as compared to chlorine as a result of which the attraction outside the shell to gain electron is less. Moreover, there are inter electronic repulsions as well in the 2p orbitals which results in the less negative electron gain enthalpy.
જવાબ : The element is chlorine (Cl) with atomic number (Z) = 17.
જવાબ : Period – 7 and Group -14 Block-p.
જવાબ : The sixth period corresponds to sixth shell. The orbitals present in this shell are 6s, 4f, 5p, and 6d. The maximum number of electrons which can be present in these sub¬shell is 2 + 14 + 6 + 10 = 32. Since the number of elements in a period corresponds to the number of electrons in the shells, therefore, sixth period should have a maximum of 32 elements.
જવાબ : The basic difference in approach between Mendeleev’s Periodic Law and Modem Periodic Law is the change in basis of classification of elements from atomic weight to atomic number.
જવાબ : Mendeleev used atomic weight as the basis of classification of elements in the periodic table. He did stick to it and classify elements into groups and periods.
જવાબ : The basic theme of organisation of elements in the periodic table is to simplify and systematize the study of the properties of all the elements and millions of their compounds. This has made the study simple because the properties of elements are now studied in form of groups rather than individually.
જવાબ : Caesium is the least electronegative alkali metal as electronegativity decreases as we go down the group. Caesium is a group 1 element and lies down the group as it has the largest size due to a decrease in the effective nuclear charge.
જવાબ : Sodium atom loses one electron to form sodium cation and after the formation of a cation, the effective nuclear charge on the ion increases on the left electrons which results in a decrease in radius.
જવાબ : Metallic character decreases when we move from left to right across the period and non-metallic character increases as there is an increase in ionisation enthalpy and electron gain enthalpy along the period.
જવાબ : Lithium and beryllium are examples. Li is the first group element. It has different properties and forms of covalent compounds and nitrides. Beryllium is the first element of the second group. It has various anomalies like it forms a covalent compound with coordination number four, unlike other elements that have a coordination number 6.
જવાબ : The outermost electronic configuration of element (114Z) is
[Rn] 5f14 6d107s27p2. It has n = 7, so period → 7 It belongs to p-block so,
group number = 10 + 4 = 14.
જવાબ : The cause of periodicity in properties is the repetition of similar outer electronic configuration after certain regular intervals.
For example, all the elements of group LA i.e., alkali metals, have similar outer electronic configuration as ns1.
Where n refer to the number of outermost principal shell.
In a similar manner all the halogens i.e., elements of group VIL
જવાબ :
- As the principal quantum number for the valence shell is 4, the element is present in the 4th period.
- Since the last electron has been filled in 4s sub-shell (or orbital), the element belongs to s-block.
- As there is only one electron in the valence s-sub-shell, the element is present in group I.
જવાબ : Ionization enthalpy, among other things, depends upon the electronic configuration (number of electrons) and nuclear charge (number of protons). Since isotopes of an element have the same electronic configuration and same nuclear charge, they have same ionization enthalpy.
જવાબ : The ionisation enthalpy is for 1 mole atoms. Therefore, ground state energy of the , atoms may be expressed as E (ground state) = ( – 2.18 x 10-18 J) x(6.022 x 1023 mol-1)= -1.312 x 106 J mol-1
Ionisation enthalpy =E∞–E ground state
= 0-(-1.312 x 106mol-1)
= 1.312 x 106 J mol-1.
N3-, O2-, F–, Na+, Mg2+, Al3+
(a) What is common in them?
(b) Arrange them in order of increasing ionic radii?
જવાબ : (a) All of them are isoelectronic in nature and have 10 electrons each.
(b) In isoelectronic species, greater the nuclear charge, lesser will be the atomic or ionic radius.
Al3+ < Mg2+ < Na+ < F– < O2- < N3-
(i) F–(ii) Ar (iii) Mg2+(iv) Rb+
જવાબ : Isoelectronic species are those species (atoms/ions) which have same number of
electrons. The isoelectronic species are:
(i)Na+ (iii) Na+
(ii)K+ (iv) Sr2+
જવાબ : Atomic radius. The distance from the centre of nucleus to the outermost shell o electrons in the atom of any element is called its atomic radius. It refers to both covalen or metallic radius depending on whether the element is a non-metal or a metal.
Ionic radius. The Ionic radii can be estimated by measuring the distances between cations and anions in ionic crystals.
જવાબ : The elements in a group have same valence shell electronic configuration and hence have similar physical and chemical properties.
(i)Lawrence Berkeley Laboratory
(ii)Seaborg’s group?
જવાબ : (i) Lawrencium (Lr) with atomic number (z) = 103
(ii) Seaborgium (Sg) with atomic number (z) = 106.
જવાબ : The second electron gain enthalpy of O is +ve. This is because energy has to be supplied to convert O–(g) to O2-(g) in order to overcome the repulsive forces.
જવાબ : A cation is smaller than its parent atom because it has fewer electrons while its nuclear charge remains the same. The size of an anion will be larger than that of the parent atom because the addition of one or more electrons would result in increased repulsion among the electrons and a decrease in effective nuclear charge.
N3-, O2-, F–, Na+, Mg2+ and Al3+
(a) What is common in them?
(b) Arrange them in the order of increasing ionic radii.
જવાબ : (a) All these are isoelectronic species as they are having same number of electrons i.c., 10.
(b) As Z/e decreases, size increases so, order should be,
N3- > O2- > F – > Na+ > Mg2+ > Al3+.
જવાબ : Atomic radius: It is the distance between the centre of the nucleus and outermost shell where electrons are present.
Ionic radius : It is the distance between the nucleus and outermost shell of an ion.
જવાબ : In a group, the chemical properties of the elements remain nearly the same due to same valence shell configuration.
(i) Lawrence Berkeley Laboratory
(ii) Seaborg’s group?
જવાબ : (i) Lawrencium → 103Lr
(ii) Seaborgium → 106Sg
જવાબ : 17Cl → It belongs to the third period. So, outermost shell is n = 3. Its configuration is [Ne] 3s23p5.
Therefore, its atomic number = 17.
B Al Ga In Tl
801 577 579 558 589
How would you explain this deviation from the general trend?
જવાબ : (i) Al has lower ionization enthalpy than B due to larger size.
(ii) Ga has slightly higher ionization enthalpy than Al due to ineffective shielding by 3d electrons.
(iii) In has lower ionization enthalpy than Ga due to larger size.
(iv) Tl has higher ionization enthalpy than In due to ineffective shielding by 4f electrons.
(i) O or F
(ii) F or Cl
જવાબ : (i) O or F : F has more negative electron gain enthalpy than O due to smaller size, higher nuclear charge and greater possibility of attaining the nearest stable noble gas configuration by gaining one electron.
(ii) F or Cl : Cl has more negative electron gain enthalpy because in F the incoming electron is added to the smaller n = 2 quantum level and suffers significant repulsion from the other electrons present in this level. In Cl, the added electron goes to n = 3 quantum level and occupies a larger region of space and electron-electron repulsion experienced is far less.
જવાબ : We have to consider two factors :
(i) the attraction of electrons towards the nucleus, and (ii) the repulsion of electrons from each other. The effective nuclear charge experienced by a valence electron in an atom will be less than the actual charge on the nucleus because of shielding or screening of the valence electron from the nucleus by the intervening core electrons. As we descend the group, the outermost electron being increasingly farther from the nucleus, there is an increased shielding of the nuclear charge by the electrons in the inner levels. The increase in shielding outweighs the increasing nuclear charge and the removal of the outermost electron requires less energy down a group.
(ii) O has lower ∆iH than N and F?
જવાબ : (i) An s-electron is attracted to the nucleus more than a p-electron. In beryllium, the electron removed during the ionization is an s-electron whereas the electron removed during ionization of boron is a p-electron. The penetration of a 2s-electron to the nucleus is more than that of a 2p-electron; hence the 2 p electron of boron is more shielded from the nucleus by the inner core of electrons than the 2s electrons of beryllium. Therefore, it is easier to remove the 2p-electron from boron as compared to the removal of a 2s-electron from beryllium. Thus, boron has a smaller first ionization enthalpy than beryllium.
(ii) O has lower ionisation energy than N because N (1s2 2s2 
) has extra stable electronic configuration whereas O (1s2 2s2 ) does not. O has lower ionisation energy than F because O has larger size than F.
જવાબ : Energy of the electron in the ground state of H-atom, E1 = -2.18 x 10-18 J
Ionisation energy = E∞ – En
Ionisation enthalpy per mole of atomic hydrogen = (E∞ – E1)NA
= [0 – (- 2.18 x 10-18)] x 6.023 x 1023
= 2.18 x 6.023 x 105 J/mol = 13.13 x 105 j/mol
= 1.313 x 106 J/mol
જવાબ : In the definition of ionization enthalpy and electron gain enthalpy, isolated gaseous atom is required for comparison purposes. Ionization energy is the minimum amount of energy required to remove most loosely bound electron from an isolated atom in the gaseous state of an element so as to convert it into gaseous monovalent positive ion. Electron gain enthalpy is the energy change accompanying the process of adding an electron to a gaseous isolated atom to convert it into a negative ion, i.e., a monovalent anion.
Both the above mentioned processes are carried out on an isolated gaseous atom, which in turn is obtained from either the excitation of a ground state atom (in case the element is monoatomic) or atomisation of polyatomic elements. The force with which an electron is attracted by the nucleus is appreciably affected by presence of other atoms in the neighbourhood. Since in the gaseous state the atoms are widely separated, therefore these interatomic forces are minimum.જવાબ : Variation of atomic radii in a period :
As we move from left to right across a period, there is regular decrease in atomic radii of representative elements. This can be explained on the basis of effective nuclear charge which increases gradually in a period, i.c, electron cloud is attracted more strongly towards nucleus as the effective nuclear charge becomes more and more along the period. The increased force of attraction brings contraction in size.
(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 MX2(X = halogen)
(f) the metal which can form a predominantly stable covalent halide of the formula MX (X = halogen)?
જવાબ : (a) The element V has highest first ionization enthalpy (∆ iH1) and positive electron gain enthalpy (∆egH) and hence it is the least reactive element. Since inert gases have positive∆egH, therefore, the element-V must be an inert gas. The values of ∆ i H1, ∆ iH2 and ∆egHmatch that of He.
(b) The element II which has the least first ionization enthalpy (∆ i H1) and a low negative electron gain enthalpy (∆egH) is the most reactive metal. The values of ∆ i H1, ∆ iH2 and ∆egH match that of K (potassium).
(c) The element III which has high first ionization enthalpy (∆ iH1) and a very high negative electron gain enthalpy (∆egH) is the most reactive non-metal. The values of ∆ i H1, ∆ iH2 and ∆egH match that of F (fluorine).
(d) The element IV has a high negative electron gain enthalpy (∆egH ) but not so high first ionization enthalpy (∆egH). Therefore, it is the least reactive non-metal. The values of ∆ i H1, ∆ iH2 and ∆egH match that of I (Iodine).
(e) The element VI has low first ionization enthalpy (∆ iH1) but higher than that of alkali metals. Therefore, it appears that the element is an alkaline earth metal and hence will form binary halide of the formula MX2(where X = halogen). The values of ∆ i H1, ∆ iH2 and ∆egH match that of Mg (magnesium).
(f) The element I has low first ionization (∆ iH1) but a very high second ionization enthalpy (∆ iH2), therefore, it must be an alkali metal. Since the metal forms a predominarly stable covalent halide of the formula MX (X = halogen), therefore, the alkali metal must be least reactive. The values of ∆ i H1, ∆ iH2 and ∆egH match that of Li (lithium).
(i) ns2 np4 for n = 3 (ii) (n – 1) d2 ns2 for n = 4 and (iii) (n – 2) f7 (n – 1) d1 ns2 for n = 6 in the periodic table?
જવાબ : (i) n = 3
Thus element belong to 3rd period, p-block element.
Since the valence shell contains = 6 electrons, group No = 10 + 6 = 16 configuration =1s2 2s2 2p6 3s2 3p4 element name is sulphur.
(ii) n = 4
Means element belongs to 4th period belongs to group 4 as in the valence shell (2 + 2) = 4 electrons.
Electronic configuration.=1s2 2s2 2p6 3s2 3p6 3d2 4s2, and the element name is Titanium (Ti).
(iii) n = 6
” Means the element belongs to 6th period. Last electron goes to the f-orbital, element is from f-block.
group = 3
The element is gadolinium (z = 64)
Complete electronic configuration =[Xe] 4 f7 5d1 6s2.
(a) gains an electron (b) loses an electron ?
જવાબ :
- Gain of an electron leads to the formation of an anion. The size of an anion will be larger than that of the parent atom because the addition of one or more electrons would result in increased repulsion among electrons and decrease in effective nuclear charge.
This the ionic radius of fluoride ion (F–) is 136 pm whereas atomic radius of Fluorine (F) is only 64 pm. - Loss of an electron from an atom results in the formation of a cation. A cation is smaller than its parent atom because it has fomer electrons while its nuclear charge remains the same. For example, The atomic radius of sodium (Na) is 186 pm and atomic radius of sodium ion (Na+) = 95 pm.
જવાબ : The elements of Group I have only one electron in their respective valence shells and thus have a strong tendency to lose this electron. The tendency to lose electrons 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 same order Li < Na < K < Rb < Cs. In contrast, the elements of group 17 have seven electrons in their respective valence shells and thus have strong tendency to accept one more electron to make stable configuration. It is linked with electron gain enthalpy and electronegativity. Since both of them decreases down the group, the reactivity therefore decreases.
જવાબ : On Pauling scale, the electronegativity of nitrogen, (3.0) indicates that it is sufficiently electronegative. But it is not correct to say that the electronegativity of nitrogen in all the compounds is 3. It depends upon its state of hybridisation in a particular compound, greater the percentage of s-character, more will be the electronegativity of the element. Thus, the electronegativity of nitrogen increases in moving from SP3 hybridised orbitals to SP hybridised orbitals i.e., as SP3 < SP2 < SP.
જવાબ : Electron gain enthalpy refers to tendency of an isolated gaseous atom to accept an additional electron to form a negative ion. Whereas electronegativity refers to tendency of the atom of an element to attract shared pair of electrons towards it in a covalent bond.
જવાબ : (i) O or F. Both O and F lie in 2nd period. As we move from O to F the atomic size decreases.
Due to smaller size of F nuclear charge increases.
Further, gain of one electron by
F —> F–
F~ ion has inert gas configuration, While the gain of one electron by
0->O–
gives CT ion which does not have stable inert gas configuration, consequently, the energy released is much higher in going from
F ->F–
than going from O —>O–
In other words electron gain enthalpy of F is much more negative than that of oxygen.
(ii) The negative electron gain enthalpy of Cl (∆ eg H = – 349 kj mol-1) is more than that of F (∆ eg H = – 328 kJ mol -1).
The reason for the deviation is due to the smaller size of F. Due to its small size, the electron repulsions in the relatively compact 2p-subshell are comparatively large and hence the attraction for incoming electron is less as in the case of Cl.
B Al Ga In Tl
801 577 579 558 589
How would you explain this deviation from the general trend?
જવાબ : The decrease in ∆iH1 value from B to Al is due to the bigger size of Al.
In Ga there is 10 3d electrons which do not screen as is done by S and P electrons. Therefore, there is an unexpected increase in the magnitude of effective nuclear charge resulting in increased ∆iH1 values. The same is with into Tl. The later has fourteen ∆f electrons with very poor shielding effect. This also increases, the effective nuclear charge thus the value of ∆iH1 increases.
જવાબ : Atomic size. With the increase in atomic size, the number of electron shells increase. Therefore, the force that binds the electrons with the nucleus decreases. The ionization enthalpy thus decreases with the increase in atomic size.
Screening or shielding effect of inner shell electron. With the addition of new shells, the number of inner electron shells which shield the valence electrons increases. As a result, the force of attraction of the nucleus for the valence electrons further decreases and hence the ionization enthalpy decreases.
Explain why
(i) Be has higher ∆iH1than B ?
(ii) O has lower ∆iH1 than N and F?
જવાબ : (i) In case of Be (1s2 2s2) the outermost electron is present in 2s-orbital while in B (1s2 2s2 2p1) it is present in 2p-orbital. Since 2s – electrons are more strongly attracted by the nucleus than 2p-electrons, therefore, lesser amount of energy is required to knock out a 2p-electron than a 2s – electron. Consequently, At of Be is higher than that ∆iH1 of B.
(ii) The electronic configuration of
N7 = 1s2 2s2 2px1 2py1 2pz1
O8 =1s2 2s2 2px1 2py1 2pz1
We can see that in case of nitrogen 2p-orbitals are exactly half filled. Therefore, it is difficult to remove an electron from N than from O. As a result ∆iH1 of N is higher than that of O.
જવાબ : Electronic configuration of Na and Mg are
Na = 1s2 2s2 2p6 3s1
Mg = 1s2 2s2 2p6 3s2
First electron in both cases has to be removed from 3s-orbital but the nuclear charge of Na (+ 11) is lower than that of Mg (+ 12) therefore first ionization energy of sodium is lower than that of magnesium.
After the loss of first electron, the electronic configuration of
Na+ = 1s2 2s2 2p6
Mg+ = 1s2 2s2 2p6 3s1
Here electron is to be removed from inert (neon) gas configuration which is very stable and hence removal of second electron requires more energy in comparison to Mg.
Therefore, second ionization enthalpy of sodium is higher than that of magnesium.
જવાબ :
- Significance of term ‘isolated gaseous atom’. The atoms in the gaseous state are far separated in the sense that they do not have any mutual attractive and repulsive interactions. These are therefore regarded as isolated atoms. In this state the value of ionization enthalpy and electron gain enthalpy are not influenced by the presence of the other atoms. It is not possible to express these when the atoms are in the ; liquid or solid state due to the presence of inter atomic forces.
- Significance of ground state. Ground state of the atom represents the normal – energy state of an atom. It means electrons in a particular atom are in the lowest energy state and they neither lose nor gain electron. Both ionisation enthalpy and I electron gain enthalpy are generally expressed with respect to the ground state ofan atom only.
જવાબ : Within a group Atomic radius increases down the group.
Reason. This is due to continuous increases in the number of electronic shells or orbit numbers in the structure of atoms of the elements down a group.
Variation across period.
Atomic Radii. From left to right across a period atomic radii generally decreases due
to increase in effective nuclear charge from left to right across a period.
જવાબ : Ionisation enthalpy is the energy required by an isolated and gaseous atom in its ground state to remove an electron. The effective nuclear charge due to the screening effect, the valence electrons are shielded by the inner core electrons. This effective nuclear charge is less than the actual charge present on the atom. Penetrated orbitals: It is difficult to remove an electron from the orbitals that are closer to the nucleus and are penetrated towards the nucleus. The order of penetration is given by s>p>d>fStability of orbitals: Half-filled and filled orbitals have a high ionisation enthalpy as they don’t want to lose their stability. Across a period the ionisation enthalpy increases along the period. Down the group, the ionisation enthalpy decreases.
જવાબ : The factors affecting electron gain enthalpy and the trend in its variation in the periodic table are:
1)ATOMIC SIZEAs we go down the group electron gain enthalpy decreases as the distance of the nucleus from the outermost shell increases which decreases its tendency to gain electron and electron gain enthalpy becomes less negative. 2)EFFECTIVE NUCLEAR CHARGEAs we go from left to right in a period the effective nuclear charge increases and when we move down the group it decreases which results in the attraction of electrons from the outermost shell 3) ELECTRONIC CONFIGURATIONThe tendency to gain electrons depends upon the stability of an element. Elements having completely or half-filled stable orbitals have a very low tendency to gain electron thus they have very low electron gain enthalpy.TRENDSAcross a period the electron gain enthalpy becomes more negative. Down the group, the electron gain enthalpy becomes less negative.જવાબ : Exothermic reaction- The reaction in which heat is evolved is called an exothermic reaction.
for example, Cao + CO2→ CaCO3 ΔH=-178kJmol-1 Endothermic reaction- The reaction in which heat is absorbed is called an endothermic reaction. for example, 2NH3 → 3N2+ H2 ΔH=918kJmol-1જવાબ : Sodium attains stable configuration when it loses on an electron from its outermost shell. That is why its first ionisation enthalpy is less than magnesium. But in case of second ionisation enthalpy magnesium has one electron in its outermost shell and to attain stability it loses that easily as compared to sodium which is already stable.
જવાબ : ACIDIC OXIDES
SO2, B2O3 are acidic oxides and p block elements. The reaction of SO2 with water SO2 + H2O → H2SO3 The reaction of B2O3 with water B2O3 +3 H2O → 2H3BO3 Acidic oxides are those oxides that form acids after reacting with water. BASIC OXIDES Cao, BaO, Ti2O form basic oxides The reaction of Ti2O with water. BaO + H2O → Ba(OH)2 REACTION OF CaO WITH WATER Cao + H2O → Ca(OH)2 Basic oxides are those oxides that form bases after reacting with water AMPHOTERIC OXIDES Zinc oxides and aluminium oxides are the two amphoteric oxides. Reaction of ZnO with water: ZnO + 2H2O + 2NaOH → Na3Zn[OH]4 + H2 ZnO +2HCl → ZnCl2 + H2O Reaction of Al2O3 with water: Al2O3 (s) + 6 NaOH(aq) + 3H2O(l) → Na3 [Al(OH)6] (aq) Al2O3 (s) + 6HCl(aq) + 9H2O(l) → 2[Al(H2O)]3+(aq) + 6Cl-જવાબ : (i) S< P< O< N is the increasing order of the first ionisation enthalpy.
As we go down the group the ionisation enthalpy decreases and as we go along the period then it increases but in case of oxygen and nitrogen due to half-filled stability of 2 p orbitals of nitrogen it has a higher ionisation enthalpy than oxygen. (ii) Pજવાબ : (a) As we move from left to right in a period, the size of the atoms decreases due to the increase in the effective nuclear charges on the outermost electron. As a result of which electronegativity of elements increases on moving from left to right in the periodic table.
(b) As we go down the group the atomic size increases which result in the increase in the distance of the electrons in the outermost shell as a result of which effective nuclear charge decreases. This results in a decrease of ionisation enthalpy.
જવાબ : The ionisation enthalpy of elements varies across period and group. The ionisation enthalpy decreases down a group and increases as we move from left to right in a period.
• Effective nuclear charge on the outermost electrons. • Electron-electron repulsion force. • Stability of the element due to half-filled and filled orbitals are some of the parameter affected.There are No Content Availble For this Chapter
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