English
Karnataka Board PUCPUC Science Class 11

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]

Advertisements
Advertisements

Question

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.

Options

  • sp3, tetrahedral

  • sp3, square planar

  • sp3d2, octahedral

  • dsp2, square planar

MCQ
Advertisements

Solution

sp3, tetrahedral

Explanation:

Boron has die electronic configuration: 1s2 2s2 2px1 2p0 y 2p0z

In the excited state, 2s-orbital electrons are impaired and one electron is shifted to a p-orbital. Now, hybridisation occurs between one s-and three p-orbitals to give sp3 hybridisation and tetrahedral geometry.

shaalaa.com
  Is there an error in this question or solution?
Chapter 11: The p-block Elements - Multiple Choice Questions (Type - I) [Page 134]

APPEARS IN

NCERT Exemplar Chemistry Exemplar [English] Class 11
Chapter 11 The p-block Elements
Multiple Choice Questions (Type - I) | Q 3 | Page 134

RELATED QUESTIONS

How can you explain higher stability of BClas compared to TlCl3?


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.


Dry ice is ______.


Cement, the important building material is a mixture of oxides of several elements. Besides calcium, iron and sulphur, oxides of elements of which of the group (s) are present in the mixture?


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


When BCl3 is treated with water, it hydrolyses and forms [B[OH]4] only whereas AlCl3 in acidified aqueous solution forms [Al(H2O)6]3+ ion. Explain what is the hybridisation of boron and aluminium in these species?


Explain the following:

Boron does not exist as B3+ ion.


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.


Identify the compounds A, X and Z in the following reactions:

\[\ce{A + 2HCl + 5H2O -> 2NaCl + X}\]


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

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–  

Account for the following observations:

Though fluorine is more electronegative than chlorine yet BF3 is a weaker Lewis acid than BCl3 


Boron fluoride exists as BF3 but boron hydride doesn’t exist as BH3. Give reason. In which form does it exist? Explain its structure.


Boron compounds behave as Lewis acids because of their ______.


Taking stability as the factor, which one of the following represents the correct relationship?


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×