Week 4

Question 1

Valency =

Answer 1

‘number of electrons in valence shell of free atom’ – ‘number of nonbonding electrons on atom in molecule’

Question 2

Define: Coordination number (CN)

Answer 2

Coordination number (CN) = number of attached atoms

Question 3

Oxidation number=

Answer 3

‘charge on compound’ – ‘charge on ligands’

Question 4

Describe the difference in structures for BX3 and AlX3 - InX3

Answer 4

BX3
Monomeric, P-X π bonding stabilises B pz orbital, BH3 exists as a dimer B2H6

AlX3 - InX3
InX3 oligomers formed (larger elements and weak π bonding for 3rd period and below)

Question 5

What is the most stable O.S for In and Tl, why?

Answer 5

Moving down group 13 Zeff is big and poor shielding by d and f orbitals… +1 O.S. becomes more stable for In and Tl

Question 6

Order in terms of Lewis acidity
BCl3
BF3
BBr3
B(OR)3

Answer 6

BBr3>BCl3>BF3>B(OR)3
Reason:
BBr3 : Poor overlap between B and Br (size mismatch) =>weak B-Br p-bond so good acceptor
B(OR)3 : Good overlap (good size mismatch), so
strong /short B-OR p- bond so poor acceptor

Question 7

Define isoelectronic

Answer 7

Isoelectronic = compounds with same number of valence electrons.

Question 8

Compare Borazine (B3N3H6) and Benzene

Answer 8

• Both isoelectronic and isostructural. Both aromatic – though borazine to a lesser extent
• p electron delocalisation uneven in borazine due to different p orbital energies of B/N
• Aromaticity prevents oligomerisation of borazine to (BHNH)∞
• Unoccupied p orbitals have more B character and occupied p orbitals have more N character
• In borazine B-N p system is polarised. So more reactive than C6H6.