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The p-Block Elements

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Chemistry Part II

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Class 10 Class 12

Describe what happens (give chemical equations only) when:
(i) carbon monoxide is treated with chlorine. 
(ii) carbon monoxide is passed through heated NaOH under pressure. 
(iii) Vapours of carbon monoxide are passed over nickel and
(iv) carbon monoxide is passed through heated ferric oxide. 


space left parenthesis straight i right parenthesis space Phosgene space gas space is space produced
space space space space space space space space CO space space plus Cl subscript 2 space space rightwards arrow space space space stack COCl subscript 2 with Phosgene below

space left parenthesis ii right parenthesis space Sodium space formate space is space formed.
space space space space space space space space space space CO space plus space NaOH space rightwards arrow from 6 space atm to 575 space straight K of space HCOONa with Sodium space formate below

space left parenthesis iii right parenthesis space Nickel space carbonyl space is space formed.
space space space space space space space space Ni space plus space 4 CO space rightwards arrow space space stack Ni left parenthesis CO right parenthesis subscript 4 with Nickel space carbonyl below

space left parenthesis iv right parenthesis space Carbon space dioxide space is space produced.
space space space space space space space space space space space 3 CO space plus space Fe subscript 2 straight O subscript 3 space space rightwards arrow space space space 2 Fe space plus space 3 CO subscript 2
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Aluminium trifluoride is insoluble in anhydrous HF but dissolves on the addition of NaF. Aluminium trifluoride precipitates out of the resulting solution when gaseous BFis bubbled through. Give reasons. 


(i) Anhydrous HF, being a covalent compound and is strongly H-bonded, therefore it does not give ions. Hence AlF3 does not dissolve in HF. On the other hand, NaF is an ionic compound and gives F– ions and hence AlF3 dissolves in NaF forming soluble complex
           bold AlF subscript bold 3 bold plus bold 3 bold NaF bold rightwards arrow stack bold Na subscript bold 3 open square brackets bold AlF subscript bold 6 close square brackets with bold Sodium bold space bold hexafluoroaluminate bold left parenthesis bold III bold right parenthesis bold space bold left parenthesis bold soluble bold space bold complex bold right parenthesis below

(ii) Borax has much higher tendency to form complexes than aluminium because of its smaller size and higher electronegativity. Hence when gaseous BF3 is bubbled through the resulting solution. AlF3 gets precipitated. bold 3 bold BF subscript bold 3 bold space bold plus bold space bold Na subscript bold 3 bold left square bracket bold AlF subscript bold 6 bold right square bracket bold space bold rightwards arrow bold space bold space stack bold 3 bold Na open square brackets bold BF subscript bold 4 close square brackets bold space bold plus bold space bold AlF subscript bold 3 bold left parenthesis bold s bold right parenthesis with stack bold Sod bold. bold space bold tetrafluoroboarte bold space bold left parenthesis bold III bold right parenthesis with bold left parenthesis bold soluble bold space bold complex bold right parenthesis below below

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Suggest a reason why the B - F bond lengths in BF3 (130 pm) and BF subscript 4 superscript minus (143 pm) differ?
Or
Why B - F bond length in BF3 is smaller than the expected value?


In BF3, boron is sp2 hybridised and therefore BF3 is a planar molecule. It has a vacant 2p-orbital. F-atom has three lone pairs of electrons. In BF3 molecule, one 2p-orbital of fluorine atom overlaps sidewise with empty 2p-orbtial of boron to form pπ space minus space pπ back bonding (back donation) in which the lone pair is transferred from F to B as shown.



As a result of this back bonding (or black donation), the B-F bond acquires some double bond character. 
On the other hand in open square brackets BF subscript 4 close square brackets to the power of minus ion,  boron is sp3 hybridised and therefore open square brackets BF subscript 4 close square brackets to the power of minus is a tetrahedral molecule. B in open square brackets BF subscript 4 close square brackets to the power of minus ion does not have vacant p-orbital available to accept the electrons donated by the F atom. Hence open square brackets BF subscript 4 close square brackets to the power of minus ion, B -F is a purely single bond. Since double bonds are shorter than single bonds, therefore B-F bond length in BF3 is shorter (130 pm) than B-F bond length (143 pm) in  [BF4].

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When metal X is treated with sodium hydroxide, a white precipitate (A) is obtained, which is soluble in excess of NaOH to give soluble complex (B). Compound (A) is soluble in dilute HCl to form compound (C).
The compound (A) when heated strongly gives (D) which is used to extract metal. Identify (X), (A), (B), (C) and (D). Write suitable equations to support their identities. 



The given data suggests that the metal X must be aluminium because it reacts with NaOH to first give a white ppt. (A) which dissolves is excess of NaOH to receive a soluble complex (B).

   stack bold 2 bold Al with bold left parenthesis bold X bold right parenthesis below bold space bold plus bold space bold 3 bold NaOH bold space bold rightwards arrow bold space bold space bold space stack bold Al bold left parenthesis bold OH bold right parenthesis subscript bold 3 with bold Aluminium bold space bold hydroxide below bold space bold plus bold space bold 3 bold Na to the power of bold plus
stack bold Al bold left parenthesis bold OH bold right parenthesis subscript bold 3 with bold left parenthesis bold A bold right parenthesis below bold space bold plus bold space bold NaOH with bold Excess below bold space bold space bold rightwards arrow bold space bold space bold space stack bold Na to the power of bold plus open square brackets bold Al bold left parenthesis bold OH bold right parenthesis subscript bold 4 close square brackets to the power of bold minus with bold Sodium bold space bold tetrahydroxoaluminate bold left parenthesis bold III bold right parenthesis below
Since compound Al(OH)3 [A] is amphoteric in nature, it reacts with dil. HCl to form (C) which must be AlCl3

space stack bold Al bold left parenthesis bold OH bold right parenthesis subscript bold 3 with bold left parenthesis bold A bold right parenthesis below bold space bold plus bold space bold 3 bold HCl bold space bold rightwards arrow bold space bold space stack bold AlCl subscript bold 3 with bold left parenthesis bold C bold right parenthesis below bold space bold plus bold space bold 3 bold H subscript bold 2 bold O

Further (A) on heating gives (D) which is usded to extract metal, it shows that (D) must be alumina left parenthesis Al subscript 2 straight O subscript 3 right parenthesis.

space stack bold 2 bold Al bold left parenthesis bold OH bold right parenthesis subscript bold 3 with bold left parenthesis bold A bold right parenthesis below bold space bold space bold rightwards arrow with bold Heat on top bold space stack bold Al subscript bold 2 bold O subscript bold 3 with bold left parenthesis bold D bold right parenthesis below bold space bold plus bold space bold 3 bold H subscript bold 2 bold O
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Discuss the pattern of variation in the oxidation states of B(Boron) to Tl(Thallium).
Or
What is inert pair effect? Illustrate it with reference to Boron family. 


Inert pair effect:The inert pair effect represents the reluctance of the valence electrons to take part in the chemical combination due to their penetration in the nucleus of heavy elements.
B and Al do not exhibit inert pair effect due to the absence of d – or f-electrons. As a result, they show an oxidation state of +3 only due to the presence of two electrons in the s– and one electron in the p-orbital of the valence shell.

straight B presubscript 5 colon space space space 1 straight s squared space 2 straight s squared space 2 straight p to the power of 1
Al presubscript 13 colon space 1 straight s squared 2 straight s squared 2 straight p to the power of 6 3 straight s squared 3 straight p to the power of 1

On the other hand, the elements from Ga to Tl contain only d and f-electrons and hence show oxidation states of +1 and  +3 due to inert pair effect. 
As we move down the group, the stability of +3 oxidation state decreases and that of +1 oxidation state increases. This means that as we move down the group, the tendency of the electrons of the valence shell to participate in bond formation decreases. In other words, ns2 electron pair in Ga, In and Tl tends to remain paired. This is called inert pair effect. Because of inert pair effect, only the electron of thallium takes parts in bonding with the atoms of the other elements. Thus, monovalent compounds of thallium are stable.

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Explain structures of diborane. Write the structures of diborane and explain the nature of bonding in it ?


Electronic configuration of boron (Z=5) in the excited state is 1 straight s squared space 2 straight s to the power of 1 space 2 straight p subscript straight x superscript 1 space space 2 straight p subscript straight y superscript 1. There are three half filled orbitals in its valence shell. If each boron atom in B2H6 forms three covalent bonds, 14 electrons are required (six B-H bonds and one B– B bond). But there are only 12 electrons (six from two boron atoms and six from hydrogen atoms). So B2H6molecule is short of 2 electrons. Therefore, it cannot have a structure similar to that of C2H6 (ethane).
The electron diffraction studies have shown bridged structure for diborane.



In this structure, the four terminal hydrogen atoms (shown by thick lines) and two boron atoms lie in one plane while other two hydrogen atoms, one lying above and the other lying below this plane and hence are called bridge hydrogens. The above structure of diborane depicts that there are two types of hydrogen atoms. Four of them are of one type which is used in making four normal covalent bonds (two centre electron pair bonds) with boron. The remaining two form bridges between two boron atoms through three centre electron pair bonds. Three centre electron pair bond is a bond involving three atoms and only two electrons.
 
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