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Periodic Properties

PERIODIC PROPERTIES 

Hints for Subjective Problems

 

LEVEL – I

2. Larger particles size and nearly perfect lattice.

  1. pπ-pπ back bonding between F and Al.
  2. (BeCl2)n formation.
  3. Inert pair effect.
  4. Successive ionization potentials have to be considered.
  5. Half filled and fulfilled shells have high stability.
  6. Metal in higher oxidation state in oxide compounds show acidic nature.
LEVEL – II

1. K is an alkali metal which can lose its electron quite easily and has got a low I.E. so that K+ formation occurs readily.

  1. I.E. of Mg is greater than Na and Al because in either case removal of an electron from valence orbital will render the ion formed, to be full-filled which has got extra stability.
  2. Atomic size, type of hybridisation, oxidation state, partial charge of an ion.
  3. s-character of sp, sp2 and sp3hybridised orbitals are to be considered.
  4. % ionic character has to be considered.
LEVEL – III
  1. This is because of increasing nuclear charge which outweighs the shielding effect of diffuse d-shell.
  2. This arises mainly from the coupling of spins of the added electron and an electron in an orbital originally only singly occupied.
  3. s > p > d > f
  4. For half filled stability of N
  5. Be+< C+3 < O+5 < F+
  6. The increased effective nuclear charge due to poor screening by inner 4f and 5 d orbitals enhances the attracting ability and hence it has got high electron affinity.
  7. In solution Cu+ disproportionate to Cu2+ and Cu.

2Cu+ → Cu + Cu2+ ΔG = –0.35 F for the reaction.

So it occurs readily.

 

Solution to Subjective Problems

 

LEVEL – I

 

1.
  1. This is probably because of larger particle size and more nearly perfect lattice.
  2. This is because of pπ-pπ back bonding between F & Al, thereby Al becomes electronically saturated hence no further attack by H2O to vacant 3pz orbital of Al
  3. This is because of (BeCl2)n formation in the solid phase. It is a chain polymer.
  4. This is due to the decrease in the size of the ions on the atomic no. increases among the isoelectronic ions.

 

  1. Due to inert pair effect, it is difficult to unpair s-electrons the hence the higher oxidation state is less prominent.

 

  1. By giving the I.P., Na is converted to Na+. Now Na+ has an additional amount of energy equal to its I.P and in order to go back to its original state – it will liberate energy and this we call as electron affinity.

 

  1. The alkali metals have one electron in excess of their octet while alkaline earth metals have 2 e’s excess of their octet. So after 1st I.P. the alkali metal attain the inert gas configuration while alkaline earth metals attain it after 2nd I.P. Now removal from a noble gas core needs high I.P., so there is a jump in the 2nd I.P. for alkali and 3rd I.P. for alkaline earth metals.

 

  1. In Be the extra electron is to be added in 2p orbital because 2s orbital is completely filled and in N, it is to be added to a half filled 2p orbital. Since half-filled and full-filled orbitals are more stable, reluctancy in accepting electron is found.

 

  1. In Mn2O7, Mn is in +7 state  i.e., highly electron deficient. So its tendency to accept electron increases and therefore  it is an acidic oxide.

 

LEVEL – II

 

  1. K is an alkali metal which can lose its electron quite easily and has got a low I.E. so that K+ formation occurs readily. But in case of Ag, d-orbitals can’t screen the nuclear charge effectively and thus valence electrons is not readily lost so that it has got high I.E. That’s why Ag is relatively inert.

 

  1. While F is formed from F then electron is added to an electronegative, neutral atom and energy (E.A.) is released by this process but even though O formation from O is energy releasing process, formation of O2- from Ois highly energy consuming. So combined effect is O2- formation from O is endothermic.

 

  1. Hint: Decreases down the group. This is because SO42- is very large in size, so it has got low hydration energy and down the group the hydration energy of cations also decreases.

 

  1. I.E. of Mg is greater than Na and Al because in either case removal of an electron from valence orbital will render the ion formed, to be full-filled which has got extra stability. (Na : 2s22p63s1 ; Al : 2s22p63s23p1 ).That’s why their I.E. is smaller than Mg whose electron has to be removed from a 3s 2 (filled) orbital. Also s-orbital lies closer to the nucleus than the p- orbital.

 

  1. This is due to a pair of electrons remaining paired in –ous form and becoming unpaired in –ic form (e.g. Sn, P, Te etc.]. But in transition metals different no. of d-electrons may take part in bonding [e.g. Fe2+, Fe3+ & Cu+, Cu2+ etc.]

 

  1. Atomic size, hybridisation mode, oxidation state, partial charge of an ion.
  2. Hybrid orbitals sp3 sp2 sp

s-character 25% 33% 50%

Electronegativity 3.68 3.92 4.67

  1. I > II > III
  2. NF2 (3.78) > Cl3 (2.96) > NH2 (2.78) > CH3(2.3)
  3. % ionic character =

As electronegativity difference is higher in HF than HI hence the order.

 

LEVEL – III

  1. For 100% ionic bond the dipole moment = 4.8 × 92 D

∴ % ionic character = = 44%

  1. This is because of increasing nuclear charge which outweighs the shielding effect of diffuse d-shell.

3 i) s, p electrons are taken to equally affected by the nucleus

  1. ii) s, p, d f electrons equally shield the electron in n, (n-1)th shell.
  2. This arises mainly from the coupling of spins of the added electron and an electron in an orbital originally only singly occupied.
  3. s > p > d > f
  4. For half filled stability of N
  5. Be+< C+3 < O+5 < F+
  6. i) Size of the nuclear charge
  7. ii) The atomic radius

iii) The screening effect of inner electron shells

  1. iv) Ellipticity of electron’s orbit
  2. II < I (1.14Å) = III (1.95Å)

 

  1. X ⎯→ X+ + e; ΔH = IE1 = a eV

X + e ⎯→ X; ΔH = –EA1 = b eV

If N/2 atoms of X lose electrons which are taken up by remaining N/2 of X to give X, then

or a – b =

∴ a – b = 8.477

Now N/2 of X lose two electrons to give X+

X ⎯→ X + e; ΔH = EA1 = + b

X ⎯→ X+ + e; ΔH = + IE1 = +a

or a + b =

a + b = 15.194

∴ a = 11.835 eV and b = 3.358 eV

  1. AlCl3 + aq. ⎯→ AlCl3(aq); ΔH = ?

ΔH = Energy released during hydration – Energy used during ionisation

= –4665 – 3 × 381 + 5137 = –671 kJ/mole

Thus formation of ions will take place.

 

  1. The increased effective nuclear charge due to poor screening by inner 4f and 5 d orbitals enhances the attracting ability and hence it has got high electron affinity.

 

  1. As we go down the period the electronegativity decreases. But after Al, due to the presence of d electrons which have minimum shielding effect, the nuclear charge increases and hence electronegativity also increases.

 

  1. Packing in crystal latice is not compact due to small size of Li+ and bigger size of . But O2– being smaller in size can stabilise Li2O.

 

  1. In solution Cu+ disproportionate to Cu2+ and Cu.

2Cu+ → Cu + Cu2+ ΔG = –0.35 F for the reaction.

So it occurs readily.

 

Solution to Objective Problems

 

LEVEL – I

1. Alred – Rochow Scale.

  1. Hybridised orbital accommodates σ bond pairs.
  2. Electronegativity of silicon <carbon.
  3. Mulliken’s scale.
  4. Periodic variation.
  5. Solvation energy > lattice energy.
  6. Octet or fulfilled shells.
  7. Overlapping.
  8. Low lattice energy.
  9. Maximum covalency less melting point.
  10. Cations have less diameter compared to atoms.
  11. More negative charge larger species.
  12. High salvation energy of Ba+2 and SO4–2.
  13. Less than 1.7.
  14. More the bond order more will be the bond energy.

 

LEVEL – II

1. High effective nuclear charge makes high ionization potential.

  1. Slater’s rule.
  2. Electrostatic repulsion is greater than the coupling energy.
  3. High φ value.
  4. Ionic potential = .
  5. I is highly polarisable.
  6. Radius values.
  7. No. of valency electrons will reduced.
  8. I.P. = IP1 + IP2
  9. Characteristic of silicon.
  10. F2 is highly oxidizing.
  11. Half filled shells.
  12. Half filled or fulfilled shells have high stability
  13. I.P. nearly equal to EA.
  14. Noble gases.

 

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