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11 months agoContributor-Level 10
3.19. On moving down the group, the ionization enthalpy decreases. This is true for B and Al due to the bigger size of Al.
The ionization enthalpy of Ga is unexpectedly higher than Al because Ga contains 10d electrons in inner shell whose shielding is less effective than that of s and p electrons.
The outer electron is held fairly strongly by the nucleus. The ionization enthalpy increases slightly. A similar increase is observed from In to Tl due to presence of 14f electrons in the inner shell of Tl which have poor screening effect.
New answer posted
11 months agoContributor-Level 10
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.
New answer posted
11 months agoContributor-Level 10
3.17. 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 from Na+ requires more energy in comparison to Mg.
T
New answer posted
11 months agoContributor-Level 10
3.16. (i) In the electronic configuration 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.
Therefore, Be has higher? I H than that of B.
(ii) The electronic configuration of oxygen is 1s2 2s2 2p4. The 2p orbital contains 4 electrons out of which 2 are present in the same 2p-orbital. Due to this, the electron repulsion increases. N has stable half-filled configuration.
New answer posted
11 months agoContributor-Level 10
3.15. The ionisation enthalpy is for 1 mole atoms.
Therefore, ground state energy of theatoms may be expressed as
Eground state = (– 2.18 x 10-18 J) x (6.022 x 1023 mol-1)
= –1.312 x 106 J mol-1
Ionisation enthalpy =E? –Eground state
= 0– (–1.312 x 106mol-1)
= 1.312 x 106 J mol-1.
New answer posted
11 months agoContributor-Level 10
3.14. Significance of the 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 the term 'ground state'. Ground state of the atom represents the normal – energy state of an atom. It means electrons in a pa
New answer posted
11 months agoContributor-Level 10
3.13. A cation is obtained by removing an electron from the outermost shell. The removal of electron (s) results in decrease of the size of the resulting ion than the parent atom because it has fewer electrons while its nuclearcharge remains the same.
Anions are obtained by addition of electron (s) in the outermost shell. This results in increased repulsion among the electrons and a decrease in effective nuclear charge.
New answer posted
11 months agoContributor-Level 10
3.12. (a) All of them have 10 electrons each and are isoelectronic in nature.
(b) In isoelectronic species, higher the nuclear charge, smaller will be the atomic or ionic radius.
Al3+< Mg2+< Na+< F–< O2-< N3-
New answer posted
11 months agoContributor-Level 10
3.11. Species (atoms/ions) which have same number of electrons are called isoelectronic species. The isoelectronic species out of the given atoms/ions are:
(i) Na+ is isoelectronic to F-
(ii) K+ is isoelectronic to Ar
(iii) Na+ is isoelectronic to Mg2+
(iv) Sr2+ is isoelectronic to Rb+
New answer posted
11 months agoContributor-Level 10
3.10. Across a period, the atomic radii decrease from left to right due to increase in effective nuclear charge from left to right across a period
Within a group, atomic radius increases down the group due to continuous increases in the number of electronic shells or orbit numbers in the structure of atoms of the elements down a group.
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