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
Suggest reasons why the B–F bond lengths in BF3 (130 pm) and `"BF"_4^(-)` (143 pm) differ.
How would you explain the lower atomic radius of Ga as compared to Al?
Write a balanced equation for Al + NaOH → ?
The geometry of a complex species can be understood from the knowledge of type of hybridisation of orbitals of central atom. The hybridisation of orbitals of central atom in [Be(OH)4]– and the geometry of the complex are respectively.
Which of the following oxides is acidic in nature?
The exhibition of highest co-ordination number depends on the availability of vacant orbitals in the central atom. Which of the following elements is not likely to act as central atom in \[\ce{MF^{3-}6}\]?
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?
BCl3
Aluminium dissolves in mineral acids and aqueous alkalies and thus shows amphoteric character. A piece of aluminium foil is treated with dilute hydrochloric acid or dilute sodium hydroxide solution in a test tube and on bringing a burning matchstick near the mouth of the test tube, a pop sound indicates the evolution of hydrogen gas. The same activity when performed with concentrated nitric acid, reaction doesn’t proceed. Explain the reason.
Explain the following:
Boron does not exist as B3+ ion.
Describe the general trends in the following properties of the elements in Groups 13 and 14.
Atomic size
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.
TlCl3, TlCl
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.
AlCl3 , AlCl
BCl3 exists as monomer whereas AlCl3 is dimerised through halogen bridging. Give reason. Explain the structure of the dimer of AlCl3 also.
Boron fluoride exists as BF3 but boron hydride doesn’t exist as BH3. Give reason. In which form does it exist? Explain its structure.
Which one of the following is the correct statement?
