Some p-Block Elements-Test Papers

 CBSE Test Paper-01

Class - 12 Chemistry (The p - Block Elements)

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1. The structure of ClF3 is

   1. Octahedral
   2. T-shaped
   3. Pyramidal
   4. Tetrahedral
2. Xenon difluoride has _____ shape.

   1. Linear
   2. Trigonal
   3. Angular
   4. Pyramidal
3. Which of the following is Paramagnetic?

   1. N2O
   2. N2O4
   3. NO
   4. N2O5
4. Which of the following is an amphoteric oxide?

   1. Cr2O3
   2. V2O5
   3. Cl2O7
   4. SnO2
5. Which gas is evolved when urea is treated with NaOH?

   1. Nitrogen
   2. Ammonia
   3. Nitrous oxide
   4. Laughing gas

6. What is the maximum covalence shown by N?

7. Explain why inspite of nearly the same electronegativity, oxygen forms hydrogen bonding while chlorine does not.

8. What happens when H3PO3 is heated?

9. Complete and Balance-

   1. P4+ 8SOCl2→$P_4+8SOC{l}_2\to$
   2. 3CH3COOH + PCl3→$3C{H}_{3}COOH+PC{l}_3\to$
   3. P4+ 10SO2Cl2→$P_4+10S{O}_{2}C{l}_2\to$
   4. POCl3+ 3H2O→$POC{l}_3+3H_{2}O\to$
   5. Sn + PCl5→$Sn+PC{l}_5\to$
   6. 4AgNO3+ 2H2O + H3PO2→$4AgN{O}_3+2H_{2}O+H_{3}P{O}_2\to$

10. Explain the bleaching action of chlorine.

11. Draw the structure of XeOF4.

12. Describe the manufacture of H2SO4 by contact process?

13. Explain why NH3 is basic while BiH3 is only feebly basic.

14. Describe the following about halogen family (Group 17 elements:

    1. Relative oxidising power.
    2. Relative strength of their hydrides.
    3. Oxyacids and their related oxidising ability.
15. How would you account for the following:

    1. NH3 is a stronger base than PH3
    2. Sulphur has a greater tendency for catenation than oxygen.
    3. F2 is a stronger oxidizing agent than Cl2

CBSE Test Paper-01
Class - 12 Chemistry (The p - Block Elements)
Solutions

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1. 2. T-shaped
      Explanation: CN=0.5(V+M-C+A) For. ClF3 CN= 5 so hybridisation is sp3d. The structure is trigonal bipyramidal.
      ClF3 has 10 electrons around the central atom. this means there are 5 electron pairs arranged in a trigonal bipyramidal shape with a 900 F-Cl-F bond angle. There are 2 equatorial lone pairs making the final structure T- shaped.
      
2. 1. Linear
      Explanation: CN=0.5(V+M-C+A) For XeF2 CN = 5 .So shape will be linear and structure will be trigonal bipyramidal. Xenon and the two fluorine atoms lie in a straight line while the three equatorial positions are occupied by three lone pairs of electrons. Hence it has a linear shape.
      
3. 3. NO
      Explanation: NO is paramagnetic. Its due unpaired electron present in antibonding molecular orbital.
4. 1. Cr2O3
      Explanation: Higher oxidation state oxides are acidic while lower oxidation state oxides are basic. Intermediate oxidation state oxides are amphoteric. In other words, it behaves acidic with bases and as basic with acids.
      The oxidation state of Cr in Cr2O3 is +3, Vanadium in V2O5 is +5, Cl in Cl2O7 is +7 and Sn in SnO2 is +4.
5. 2. Ammonia
      Explanation: Urea on reaction with NaOH liberates ammonia.
      NH2CONH2 + 2NaOH →$\to$ Na2CO3 + 2NH3↑$\uparrow$
6. Nitrogen shows a maximum covalence of +4 because only four orbitals in its valence, i.e. one s and three p- orbitals are available for bonding in Nitrogen.
7. Both chlorine and oxygen have almost the same electronegativity values, but chlorine rarely forms hydrogen bonding. This is because in comparison to chlorine, oxygen has a smaller size and as a result, a higher electron density per unit volume.
8. Orthophosphoric acid, when heated gives phosphine and orthophosphoric acid as follows:
   4H3PO3Orthophorphloricacid−→−heatPH3phosphine+H3PO4Orthophorphloricacid$\underset{\text{acid}}{\underset{\text{Orthophorphloric}}{4H_{3}P{O}_3}}\stackrel{heat}{\to }\underset{\text{phosphine}}{P{H}_3}+\underset{\text{acid}}{\underset{\text{Orthophorphloric}}{H_{3}P{O}_4}}$
9. 1. P4+8SOCl2→4PCl3+4SO2+2S2Cl2$P_4+8SOC{l}_2\to 4PC{l}_3+4S{O}_2+2S_{2}C{l}_2$
   2. 3CH3COOH+PCl3→3CH3COCl+H3PO3$3C{H}_{3}COOH+PC{l}_3\to 3C{H}_{3}COCl+H_{3}P{O}_3$
   3. P4+10SO2Cl2→4PCl5+10SO2$P_4+10S{O}_{2}C{l}_2\to 4PC{l}_5+10S{O}_2$
   4. POCl3+ 3H2O→H3PO4+ 3HCl$POC{l}_3+3H_{2}O\to H_{3}P{O}_4+3HCl$
   5. Sn+PCl5→SnCl4+2 PCl3$Sn+PC{l}_5\to SnC{l}_4+2PC{l}_3$
   6. 4AgNO3+2H2O+H3PO2→4Ag+4HNO3+H3PO4$4AgN{O}_3+2H_{2}O+H_{3}P{O}_2\to 4Ag+4HN{O}_3+H_{3}P{O}_4$
10. The bleaching action of chlorine water is due to its tendency to give nascent oxygen so that the substance gets oxidized.
    Cl2+H2O→[HCl+HClO(yellow)]→2HCl(colourless)+2[O]Nascent${\mathrm{C}\mathrm{l}}_2+{\mathrm{H}}_2\mathrm{O}\to \left[\mathrm{H}\mathrm{C}\mathrm{l}+\underset{(\mathrm{y}\mathrm{e}\mathrm{l}\mathrm{l}\mathrm{o}\mathrm{w})}{\mathrm{H}\mathrm{C}\mathrm{l}\mathrm{O}}\right]\to \underset{\left(\mathrm{c}\mathrm{o}\mathrm{l}\mathrm{o}\mathrm{u}\mathrm{r}\mathrm{l}\mathrm{e}\mathrm{s}\mathrm{s}\right)}{2\mathrm{H}\mathrm{C}\mathrm{l}}+\underset{\mathrm{N}\mathrm{a}\mathrm{s}\mathrm{c}\mathrm{e}\mathrm{n}\mathrm{t}}{2\left[\mathrm{O}\right]}$
    So, it helps to bleach the given substance. i.e.
    Coloured substances + [O] →$\to$ Colourless substance.
11. Structure of XeOF4 square pyramidal as shown below:
    
12. Contact process: It involves the following steps:
    1. Burning of sulphur or sulphide ores in presence of oxygen to produce SO2.
       S+ O2→$\to$SO2
       4FeS2 + 11O2 →$\to$ 2Fe2O3 + 8SO2
    2. SO2 is reacted with O2 in presence of V2O5 as a catalyst to form SO3. 2SO2 + O2 ↔V2O5$\stackrel{V_2{O}_5}{\leftrightarrow }$ 2SO3,
       Δ$\mathrm{\Delta }$rH = -196.6 kJ mol-1 .The plant is operated at a pressure of 2 bar and a temperature of 720 K.
    3. SO3 is absorbed in H2SO4 to give oleum which is H2S2O7. SO3 + H2SO4 →$\to$ H2S2O7.
    4. Dilution of oleum with water gives H2SO4 of required concentration. H2S2O7 + H2O →$\to$ 2H2SO4
       The sulphuric acid obtained by contact process is 96 - 98% pure.
       
13. NH3 is distinctly basic while BiH3 is feebly basic. Nitrogen has a small size due to which the lone pair of electrons is concentrated in a small region. This means that the charge density per unit volume is high. On moving down a group, the size of the central atom increases and the charge gets distributed over a large area decreasing the electron density. Hence, the electron donating capacity of group 15 element hydrides decreases on moving down the group.
14. The decreasing order of the property for the elements of Group 17 is given below:
    1. F2 > Cl2 > Br2 > I2 is decreasing order to oxidising power.
    2. HI > HBr > HCl > HF is decreasing order of strength of an acid.
    3. The order of oxidising power of different oxyacids is given below:
       HClO >HClO2 >HClO3 >HClO4
       also; HOI > HOBr > HOCl
       HOF does not exist at room temperature.
15. 1. Due to the presence of lone pair of electrons on the centre atom both NH3 and PH3 are Lewis Bases. When NH3 or PH3 accepts a proton, an additional N - H or P - H bond is formed.
       H3N+ H+→$\to$NH4+,H3P + H+→$\to$PH4+
       Due to smaller size of N than P, N - H bond thus formed is much stronger than P - H bond. As a result NH3 has more tendency than PH3 to accept a proton. Therefore, NH3 is a stronger base than PH3.
    2. The property of catenation depends upon the strength of the element - element bond. Since sulphur S - S bond strength is much more than O - O bond strength. So sulphur has greater tendency for catenation than oxygen.
    3. Since F2 has smaller size than Cl2 and there is absence of d-orbital in fluorine, that's why F2 is stronger oxidizing agent than Cl2.