Advertisements
Advertisements
Question
Match the species given in Column I with the hybridisation given in Column II.
| Column I | Column II |
| (i) Boron in [B(OH)4]– | (a) sp2 |
| (ii) Aluminium in [Al(H2O)6]3+ | (b) sp3 |
| (iii) Boron in B2H6 | (c) sp3d2 |
| (iv) Carbon in Buckminsterfullerene | |
| (v) Silicon in \[\ce{SiO^{4-}4}\] | |
| (vi) Germanium in [GeCl6]2– |
Advertisements
Solution
| Column I | Column II |
| (i) Boron in [B(OH)4]– | (b) sp3 |
| (ii) Aluminium in [Al(H2O)6]3+ | (c) sp3d2 |
| (iii) Boron in B2H6 | (b) sp3 |
| (iv) Carbon in Buckminsterfullerene | (a) sp2 |
| (v) Silicon in \[\ce{SiO^{4-}4}\] | (b) sp3 |
| (vi) Germanium in [GeCl6]2– | (c) sp3d2 |
Explanation:
(i) Boron in [B(OH)4]– is sp3
(ii) Aluminium in [Al(H20)6]3+ is sp3d2 hybridized
(iii) Boron in B2H6 is sp3
(iv) Carbon in Buckminsterfullerene sp2 is hybridised.
(v) Silicon in \[\ce{SiO^{4-}4}\] is sp3
(vi) Germanium in [GeCl6]2- is sp3d2
APPEARS IN
RELATED QUESTIONS
If B–Cl bond has a dipole moment, explain why BCl3 molecule has zero dipole moment.
Suggest reasons why the B–F bond lengths in BF3 (130 pm) and `"BF"_4^(-)` (143 pm) differ.
A compound X, of boron reacts with NH3 on heating to give another compound Y which is called inorganic benzene. The compound X can be prepared by treating BF3 with Lithium aluminium hydride. The compounds X and Y are represented by the formulas.
Which of the following statements are correct. Answer on the basis of Figure.
(i) The two birdged hydrogen atoms and the two boron atoms lie in one plane;
(ii) Out of six B – H bonds two bonds can be described in terms of 3 centre 2-electron bonds.
(iii) Out of six B – H bonds four B – H bonds can be described in terms of 3 centre 2 electron bonds;
(iv) The four-terminal B – H bonds are two centre-two electron regular bonds.
Explain why the following compounds behave as Lewis acids?
AlCl3
Explain the following:
PbX2 is more stable than PbX4.
Explain the following:
Pb4+ acts as an oxidising agent but Sn2+ acts as a reducing agent.
Explain the following:
Electron gain enthalpy of chlorine is more negative as compared to fluorine.
Explain the following:
Tl (NO3)3 acts as an oxidising agent.
Match the species given in Column I with the properties mentioned in Column II.
| Column I | Column II |
| (i) \[\ce{BF^{-}4}\] | (a) Oxidation state of central atom is +4 |
| (ii) AICI3 | (b) Strong oxidising agent |
| (iii) SnO | (c) Lewis acid |
| (iv) PbO2 | (d) Can be further oxidised |
| (e) Tetrahedral shape |
Describe the general trends in the following properties of the elements in Groups 13 and 14.
Ionisation enthalpy
Describe the general trends in the following properties of the elements in Groups 13 and 14.
Metallic character
Describe the general trends in the following properties of the elements in Groups 13 and 14.
Oxidation states
Account for the following observations:
The +1 oxidation state of thallium is more stable than its +3 state.
Three pairs of compounds are given below. Identify that compound in each of the pairs which has group 13 element in more stable oxidation state. Give reason for your choice. State the nature of bonding also.
InCl3, InCl
BCl3 exists as monomer whereas AlCl3 is dimerised through halogen bridging. Give reason. Explain the structure of the dimer of AlCl3 also.
A nonmetallic element of group 13, used in making bullet proof vests is extremely hard solid of black colour. It can exist in many allotropic forms and has unusually high melting point. Its trifluoride acts as Lewis acid towards ammonia. The element exihibits maximum covalency of four. Identify the element and write the reaction of its trifluoride with ammonia. Explain why does the trifluoride act as a Lewis acid.
Boron compounds behave as Lewis acids because of their ______.
A group 13 element ‘X’ reacts with chlorine gas to produce a compound XCl3. XCl3 is electron deficient and easily reacts with NH3 to form \[\ce{Cl3X –> NH3}\] adduct; however, XCl3 does not dimerize X is ______.
