Title P BLOCK (GROUP 13 & 14).pdf ✅

 Digital www.allendigital.in [ 159 ] Inert pair effect Group-13 Group-14 Group-15 (group oxidation state) 2 1 3 ns np + 2 2 4 ns np + 2 3 5 ns np +  While moving down the group the stability of lower oxidation state (2 less than group oxidation state) progressively increases and for the last element of the group the stability of lower oxidation state increases than the group oxidation state. This is called inert pair effect. Group-13 Group-14 Group-15 B +3 C +4 N +5 Al +3 Si +4 P +5 -------------------------------------------------------------------------------- Ga +3 > +1 Ge +4 > +2 As +5 > +3 In +3 > +1 Sn +4 > +2 Sb +5 > +3 Tl +3 < +1 Pb +4 < +2 Bi + 5 < +3 Reason : As we move down the group there is presence of d & f-orbitals in inner shells which have poor shielding effect hence Zeff increases. As a result the ns2 electron pair becomes more and more tightly held to the nucleus and becomes inert /passive towards bonding.  For the last element group oxidation state is highly oxidising in nature. Examples : 1. PbCl2 is more stable than PbCl4. 2. TlCl is more stable than TlCl3. 3. GaCl3 is more stable than TlCl3. 4. SnCl4 is more stable than PbCl4. 5. Thallium (III) iodide does not exist. 6. PbF4, PbCl4 exist but PbBr4 and PbI4 does not exist. 7. Among pentahalides of Bi only BiF5 exists. F–< Cl–< Br–< I – ⎯⎯⎯⎯⎯⎯⎯⎯→ reducing nature due to strong oxidising nature of Pb+4 and strong reducing nature of Br– and I– , PbBr4 and PbI4 does not exist while Cl– and F– cannot reduce Pb+4 Illustration 1: Stability of trivalent cations of group III A will be in order. (1) Ga3+ < In3+ < T 3+ (2) Ga3+ > In3+ > T 3+ (3) Ga3+ > In3+ < T 3+ (4) T 3+ > In3+ < Ga3+ Solution: Ans. (2) p-Block (Group-13 & 14) 04 Illustrations
NEET : Chemistry [ 160 ] www.allendigital.in  Digital Illustration 2: Aluminium and Ga have the same covalent radius because of – (1) Greater shielding power of s–electrons of Ga atoms. (2) Poor shielding power of s–electrons of Ga atoms. (3) Poor shielding power of d–electrons of Ga atoms. (4) Greater shielding power of d–electrons of Ga atoms. Solution: Ans. (3) 1. Which halide is unknown ? (1) Tl(III) iodide (2) In(III) bromide (3) Ga(III) fluoride (4) B(III) chloride 2. There is a tendency for the elements of groups 13 to 15 of higher periods, particularly period 6, to form compounds with ions having a positive charge of _________ less than the group oxidation number. (1) One (2) Three (3) Four (4) Two 3. Which oxidation states are more characteristic for lead and tin ? (1) For lead + 4, for tin + 2 (2) For lead + 2, for tin + 2 (3) For lead + 4, for tin + 4 (4) For lead + 2, for tin + 4 4. Which of the following pentahalide can not be formed :- (1) PCl5 (2) AsCl5 (3) SbCl5 (4) BiCl5 Back bonding Definition : It is an additional bond which is formed between 2 adjacent bonded atoms in a covalent molecule/ion by collateral overlapping. Conditions for back bonding : (i) One bonded atom must possess vacant orbital and the other bonded atom must possess lone pair. (ii) Both bonded atoms must belong to 2nd period or one bonded atom must belong to 2nd period and other belong to 3rd period. As a result of back bonding between the bonded atoms, bond length decreases and bond energy increases. (B–F) B.O = 4/3 = 1.33 Types of back bonding : Based on type of orbitals involved in overlapping 2 types of back bonding is possible (1) p-p back bonding (2) p-d back bonding BEGINNER’S BOX-1
p-Block (Group-13 & 14)  Digital www.allendigital.in [ 161 ] (1) p-p back bonding : It is formed by the sidewise overlapping of two p orbitals. Order of strength : 2p – 2p > 2p – 3p > 2p – 4p ....... __________________________________ Size of orbital  extent of B.B.  It is used to explain following observations :- (a) Abnormal bond length and bond energy of B-F bond in BF3. (b) Lewis acidic order of Boron and Beryllium halides. BF3 < BCl3 < BBr3 < BI3 BeF2 < BeCl2 < BeBr2 < BeI2  extent of back bonding   Lewis acidic strength  (c) Hybridisation : If a lone pair participates in back bonding then it is not considered in hybridisation. Ex. B3N3H6 (inorganic benzene or borazine or borazole) Hybridisation of B as well as N is sp2 Inorganic benzene is more reactive than organic benzene as in it the bonds are polar, although overall molecule is non-polar. (d) If back bonding is present in the molecule then tendency to form dimer or polymer decreases. Ex. BF3, BeF2 (2) p-d back bonding : It is formed by the side wise overlapping between 2p and 3d orbitals. It is used to explain the following observations : (a) Hybridisation Ex. Trimethyl amine Trisilyl amine (CH3)3N (SiH3)3N • Hybridisation of N is sp3 • Hybridisation of N is sp2 • Structure is trigonal pyramidal • Structure is trigonal planar • It is a stronger Lewis base • It is a weaker Lewis base (due to presence of l.p.)
NEET : Chemistry [ 162 ] www.allendigital.in  Digital (b) Acidic strength • No back bonding in conjugate base • Back bonding present in conjugate base • Less acidic • More acidic Illustration 3: Identify the compounds showing back bonding in the following : (a) BF3 (b) BF4 – (c) O(SiH3)2 (d) O(CH3)2 Solution: (a, c) Illustration 4: Compare the given properties of followings compounds (a) BF3, BF4 – (bond length) (b) BF3, BF4 – (bond angle) (c) BF3, BCl3, BBr3, BI3 (Lewis acid strength) Solution: (a) < (b) > (c) increasing order (d) > (e) > (f) Pyramidal, Trigonal planar 1. In BF3 : (1) B-F bond has some double bond character and this bond is delocalised (2) All the B–F bonds are single covalent in nature (3) Bond energy and bond-length of B–F bond indicate its single bond character (4) All of the bonds are ionic 2. Which of the following statements is incorrect in the context of the B–F bond in BF3 :- (1) All the three B–F bond lengths are equal and each of them is shorter than the sum of the covalent radii of boron and fluorine. (2) The bond energy of the B–F bond is very high, higher than for any other single bond (3) The unusual shortness and strength of the B–F bond may be explained by a p-p interaction between boron and fluorine atoms. (4) The unusual shortness and strength of the bonds may be explained by a p-d interaction between the atoms of boron and fluorine. 3. Which of the following structures correctly represents the boron trifluoride molecule : (1) (2) (3) (4) 4. Trisilylamine [ N •• (SiH3)3] has a (1) Planar geometry (2) Tetrahedral geometry (3) Pyramidal geometry (4) None of these BEGINNER’S BOX-2 Illustrations