Plus Two Chemistry Notes Chapter -01 Text Exercise Part-02

12. A cubic solid is made of two elements P and Q. Atoms of Q are at the corners of the cube and P at the body-centre. What is the formula of the compound? What are the coordination numbers of P and Q? 

• Contribution by the atoms Q present at the eight corners of the cube      1/ 8= x 8 = 1 

• Contribution by the atom present at the body centre = 1 

• So P and Q present at the ratio 1 : 1 

• Coordination number of atom P and Q  = 8.  

13. Niobium crystallises in body-centred cubic structure. If density is 8.55 g cm–3, calculate atomic radius of niobium using its atomic mass 93 u.


14. If the radius of the octahedral void is rand radius of the atoms in close packing is R, derive relation between r and R. 

15. Copper crystallises into a fcc lattice with edge length 3.61 × 10–8 cm. Show that the calculated density is in agreement with its measured value of 8.92 g cm–3. 

16. Analysis shows that nickel oxide has the formula Ni0.98O1.00. What fraction of nickel exist as Ni2+ and Ni3+ ions? 

17. What is a semiconductor? Describe the two main types of semiconductors and contrast their conduction mechanism. 

Semiconductor are the substances whose conductivity lies in between those conductors and insulators. 

• N type semiconductor- when silicon or germanium crystal is doped with group 15 element like P or As the dopant atom forms four covalent bond like Si or Ge atom but fifth electron not used in bonding become  delocalised and continues its shares towards electrical conduction. 

• P type semiconductor- when silicon or germanium crystal is doped with group 13 elements like B or Al the dopant is present only with three valence electrons.  Hole is created at the place of missing fourth electron hole move throughout crystal like  positive charge give rise to electrical conductivity. 

18. Non-stoichiometric cuprous oxide, Cu2O can be prepared in laboratory. 

In this oxide, copper to oxygen ratio is slightly less than 2:1. Can you account for the fact that this substance is a p-type semiconductor?          

The ratio is less than 2:1 of Cu2O shows shows cuprous ions can replaced by cupric ions. For maintaining electrical neutrality  each and every two Cu+ ions will be replaced by one CP2+ ions thereby creating a hole. Conduction is present due to the presence of positive holes hence it is called as a p type of semiconductor.

19. Ferric oxide crystallises in a hexagonal close-packed array of oxide ions with two out of every three octahedral holes occupied by ferric ions. Derive the formula of the ferric oxide. 

• The number of oxide ions in packing =  90 

• Number of octahedral voids is =  90  

• As2/3 octahedral voids occupied by the ferric ions. No of ferric ions 2 present = 2/3×90 = 60 

• Ratio of Fe 3+ : O2 = 60: 90 = 2:3 

• Formula of ferric oxide Fe2O3. 

20. Classify each of the following as being either a p-type or a n-type semiconductor: 

(i) Ge doped with In 

(ii) Si doped with B. 

1. Ge is group 14 element and In is group 13 element. Sense electron- deficit hole is created. Hence it is a p type of semiconductor. 

2. B is group 13 element and Si is group 14 element. That will be free electrons hence it is a n type of semiconductor. 

Chemistry Notes Text Exercise Part -03 Click Here