Day-6

It is the part of Inorganic for Doctors

DAY-6

p-B lock E lements NCERT Based Some Key Point

Lecture-4

Vth Group (Nitrogen Family) Some important key points. 1.

We are having less reactivity of N2 as compare to P4 at room temp and this is due to triple bond in case of N2 and single bonds in case of P4 . Triple bond requires more energy as compare to single bonds for breaking.

2.

NCl 5 is not possible but PCl 5 is possible. The reason is absence of d-orbital in ‘N’ and due to this it is not having valency more than 3.

3.

PCl 3 is stable while NCl 3 is explosive and this is due to less electro-negativity difference, between N and Cl.

4.

From top to bottom in p-block elements, lower oxidation state become more stable due to inert pair effect.

49

eg. Bi +3 ® more stable than Bi +5 , Pb +2 is more stable then Pb +4 .

Inert pair effect : It is the tendency, of an S 2 electron pair to take part in bond formation in p-block elements. This is called inert pair effect means higher oxidation is not stable down the group. 5.

BiCl 3 is possible while BiCl 5 is not due to inert pair effect.

6.

In case of halides, PCl 5 is having equilibrium mixture of PCl 3 and Cl 2 on heating. This also indicates that in PCl 5 all the Cl bonds are not same. Because on heating Cl 2 is removed and this is formed from the combination of upper and lower Cl. PCl 3 + Cl 2 PCl5 Cl

Cl

b

p Cl

P a

Cl

Cl

Cl

Cl

Cl Length of bond b is more then a due to repulsive force developed by cl at the triangular side on upper and lower Cl. 7.

In case of solid PCl5 [ PCl4]+ Tetrahedral

x-ray analysis gives [Pcl6]Octahedral

structures.

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8.

Ammonia is prepared by Haber's process N2 + 3H 2 A

2NH 3 + Q (heat )

For better production of amonia. Pressure should be more and temperature should be less with Pt (platinum) catalyst and Mo as promoter. NO 2 is paramagnetic and N2 O 4 is diamagnetic. 2NO 2

Paramagnetic

A

N2 O 4

Diamagnetic

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10. NH 3 is basic in nature and basic characters decreases down the group and it is having more B.P. as compare to PH 3 , and this is due to hydrogen bonding in NH 3 , in general, B.P. increases with molecular weight. wt. BiH 3 > SbH 3 > NH 3 > ASH 3 > PH 3 (B.P. order) 11. From top to bottom reducing power increases. If a compound gives H – easily then it is good reducer. +

NH3

NH2

H-

+

BiH3

BiH2

H-

In case NH 3 after the removal of H – +Ve charge is on N i.e, +

N. Which is more electronegative as compare to Bi, Bi + is stable because +Ve charge is on metal which is stable, So remove of H – is easy While non metal N is not stable as N+ . 12. Bond angle -d

NH 3 > PH 3 > As H 3 > SbH 3 > BiH 3 N -d -dI -dI P q1

q2

Induced d = Charge

H H H H +d +d +d I +d I Since d > d’ So, more repulsion between two hydrogen in NH 3 as compare to PH 3 due to this angle increases means q1 > q2 . Q. Which is more basic in the given option. (a) CH3 - N - CH3 CH3 52

(b) Si H3- N = Si H3 Si H3

13.

‘a’ is having pyramidal shape while in case of 'b' lone pair of nitrogen goes into 'd' orbital of Si and form pp - dp bond and become planer and less basic form in nature. OH H PO 3

4

HO

P

OH

it is triprotic acid and gives 3H+

H

it is diprotic acid and gives 2H+

H

it is mono protic and gives 1H+

O OH

H3PO3 HO

P O H

H3PO2 HO

P O

H 3 PO 2 > H 3 PO 3 > H 3 PO 4 (Acidic order) Because K a of H 3 PO 2 is more than H 3 PO 3 and of H 3 PO 3 is more than H 3 PO 4 .

=

14. PCl 5 cannot act as reducer because it is having highest oxidation state of P and oxidation state decreases in any process and we know that decreasing oxidation number (reduction) means increasing oxidising power. O 15. P4O6 P4O10 N2O5 P P O O P O O O O O O P P =O O= P O O N N O O OO O O O P P || O

53

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Do Yourself -7.6 Q.1. When conc. H 2SO 4 was added into an unknown salt present in a test tube, a brown gas (A) wa evolved. This gas intensified when copper turnings were also added into this test tube. On cooling, the gas ‘A’ changed into a colourless gas ‘B’. (a)

Identify the gases A and B.

(b) Write the equations for the reactions involved. Ans. The given salt is a nitrate salt. On reaction with conc. H 2 SO 4 , vapours of HNO 3 are produced when then decompose to give a brown gas. A (NO 2 ) D

2NaNO 3 + H 2SO 4 ¾¾® Na 2SO 4 + 2HNO 3 54

Sad. nitrate

(Colourless)

D

4HNO 3 ¾¾® 4NO 2 + 2H 2O + O 2 (Brown gas. A)

This gas intensified when copper turnings are added into the test tube due to the reduction of HNO 3 to give more NO 2 gas D

Cu + 4HNO 3 ¾¾® Cu(NO 3 )2 + 2NO 2 ­ + 2H 2O On cooling the gas A (NO 2 ) changes into a colourless gas B (N 2O 4 ) 2NO 2 a

colling

(A )

N 2O 4 ( B)

Q.2. NO 2 readily forms a dimer. Explain. Ans. In NO 2 , the orbital of nitrogen has an unshared electron which tends to pair up forming a dimer. Q.3.

(i) Why does PCI 3 gives fume in moistrue ? (ii) Why is Pl 5 unknown ?

Ans. (i)

PCl 3 hydrolysis in the presence of moisture giving fumes of HCl. PCl 3 + 3H 2O ® H 3PO 3 + 3HCl

(ii) P–I bonds being very large are weak. This made Pl 5 unstable. Q.4.

Explain why BiCl 3 is more stable than BiCl 5 .

Ans. The electronic configuration of Bi (Z = 83) is Xe 4 f 14 5d10 6s2 6 p3 , the

s-electrons of the valence shell are reluctant to take part in bond formation (inert pair effect). only three electrons in the p-orbital take part in the formation of BiCl 3 . Thus, BiCl 3 is more stable than BiCl 5 . Q.5. (SiH 3 )3 N is a weaker base than (CH 3 )3 N. Ans. In (SiH 3 )3 N with planar structure, the lone electron pair on the nitrogen atom is used up in pp-dp bonding with silicon atom. It is therefore, not so easily available to the attacking acid. But the lone pair electron is free in case of (CH 3 )3 N which has a pyramidal structure similar to that of NH3 . Therefore, it can be easily donated. (CH 3 )3 N is thus a stronger base than (SiH 3 )3 N. Q.6.

Why does the reactivity of nitrogen differ from that of phosphorus ? 55

Ans. Molecular nitrogen exists as a diatomic molecule (N 2 ) in which the two nitrogen atoms are linked to each other by triple bond (N º N). It is a gas at

room temperature. Multiple bonding is not possible in case of phosphorus due to its large size. It exists as P4 molecule (soild) in which P atoms are linked to one another by single covalent bond. Because of greater bond dissociation enthalpy (due to triple bond), molecular nitrogen is very less reactive as compared to phosphorus. Q.7.

Discuss the trends in chemical reactivity of group 15 elements.

Ans. Trends in chemical reactivity.

(i) Reactivity towards hydrogen (Formation of Hydrides). All elements of this group forms hydrides of the general formula MH3 . These hydrides are : NH 3

PH 3

Ammonia

Phosphine

AsH 3 Arsine

SbH 3 Stibine

BiH 3 Bismuthine

Characteristics of hydrides, (a) Bond angles. All hydrides are pyramidal in shape. The bond angle decreases on moving down the group due to decreases in bond pair repulsion. NH 3 (107° ) > PH 3 (94° ) > AsH 3 (92° ) SbH 3 (91° ) > BiH 3 (90° ) (b) basic strength. The decreasing order of basic strength of hydrides is as follows : NH 3 > PH 3 > AsH 3 > SbH 3 > BiH 3 (c) Boiling points of hydrides. The increasing order of boiling points is as follows: PH 3 < AsH 3 < NH 3 < SbH 3 < BiH 3 (d) Reducing character of hydrides. The increasing order of reducing character is as follows : NH 3 < PH 3 < AsH 3 < SbH 3 < BiH 3 (ii) Reactivity towards halogens (Formation of Halides). The elements of group 15 combine with halogens to form trihalides as well as pentahalides. The structure of trihalides are pyramidal in nature. All possible trihalides of N, P, As, Sb and Bi are known except NBr3 and NI3 . These trihalides, act as Lewis bases (Except NF3 ), since they have a lone pair of electrons. All trihalides are covalent in nature. The halides are volatile liquids and readily hydrolysed by water (Except NF3 and PF3 ). 56

(iii) Reactivity towards metals. All the members of the family have a tendency to form binary compounds with metals in which they exhibit-3 oxidation state. For example, (i) Calcium nitride (Ca 3N 2 ) (ii) Calcium phosphide (Ca 3P2 ) (iii) Sodium arsenide (Na 3 As) (iv) Zinc antimonide (Zn 3Sb 2 ). In addition to these, bismuth also forms a similar compund called magnesium bismuthide (Mg 3Bi 2 ). (iv) Reactivity towards oxygen (Formation of oxides). The members of the nitrogen family generally form two types of oxides i.e. trioxides and pentoxides except for nitrogen which forms a few more oxides. Q.8.

Why does NH3 form hydrogen bond while PH 3 does not ?

Ans. The N–H bond in ammonia is quite polar due to electronegativity difference of

N (3 × 0) and H (2×1). Due to polarity, intermolecular hydrogen bonding is present in the molecules of amonia. H

H d–

H

d+ d–

d+ d–

—H—N—H—N—H—N— H

H

H

On the contrary, P–H bond in phosphine is almost non-polar because both P and H atoms have almost same electronegativity (2 × 1). Thus PH3 does not show hydrogen bonding. Q.9.

Give reason : CN - ion is known but CP - is not known.

Ans. Nitrogen due to small size and higher electronegativity than phosphorus, can

form multiple bonds so CN - is formed by p-bond formation between 2p orbitals of C and N. On the other hand, CP - is not known since no p-bond formation is possible between 2p orbital of C and 3p-orbital of P (due to larger size of P). Q.10. Account for the following :

(a) Chlorine water has both oxidizing and bleaching properties. (b) H 3PO 2 and H 3PO 3 act as good reducing agents while H 3PO 4 does not. Ans. (a) Chlorine water is formed when Cl 2 is dissolved in water. Its oxidising and

bleaching action is due to nascent oxygen.

57

Cl 2 + H 2O ® [HCl + HOCl] ® 2HCl + [O] (b) H 3PO 2 and H 3PO 3 act as good reducing agents due to the presence of P–H bonds. H 3PO 4 does not act as reducing agent due to the absence of P–H bond. Q.11. Illustrate how copper metal gives different products on reaction with HNO 3 . Ans. With concentrated nitric acid.

Cu + 2HNO 3 ¾¾® CuO + 2NO 2 + H 2 O Cu + 2HNO 3 ¾¾® Cu(NO 3 )2 + H 2 O Cu + 4HNO 3 (conc. ) ¾ ¾ ¾® Cu (NO 3 )2 + 2NO 2 + 2H 2 O l Copper nitrate

With dilute nitric acid 3Cu + 2HNO 3 ¾¾® 3CuO + 2NO + H 2 Ol CuO + 2HNO 3 ¾¾® Cu(NO 3 )2 + H 2 O] ´3 3Cu + 8HNO 3 (dilute) ¾ ¾ ¾® 3Cu(NO 3 )2 + 2NO + 4 H 2 O Q.12. Can PCl 5 act as oxidising as well-as reducing agent ? Justify. Ans. Phosphorus can show a maximum oxidation state of +5 in its compounds. In

PCl 5 , the oxidation state of phosphorus is +5. Since it cannot increase the same beyond +5, it cannot act as reducing agent. However, it can act as oxidising agent by undergoing a decrease in oxidation state. For example, it oxidises Ag to AgCl. 0

+5

+1

+3

2 A g + P Cl 5 ® 2 Ag Cl + P Cl 3 Q.13. BH –4 and NH +4 are isolobal. Why ?

(C.B.S.E. All India 1998)

Ans. This is because both have sp3 -hybridisation (or tetrahedral shape). Q.14. Mention an important property of hydrazine.

(C.B.S.E. All India 2000)

Ans. Hydrazine is a strong reducing agent. Q.15. Why is bond angle in PH 3 molecule lesser than that in NH 3 molecule ? (C.B.S.E. Outside Delhi 2008)

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Ans. Due to more electronegativity of N than P the bond pair of N–H bond is more closer to N as compared to that of P–H bond. So bond pair-bond pair repulsions in PH 3 are lesser than in NH 3 which leads to lesser bond angle in PH 3 than in NH 3 . Q.16. In which one of the two structures, NO+ 2 and NO 2 , the bond angle has a higher value ?

(C.B.S.E. Delhi 2008)

Ans. NO+ 2 (Because less electron pair repulsion) Q.17. Why is red phosphorus less reactive than white phosphorus ? (C.B.S.E. All India 2009)

Ans. It is because red phosphorus has a polymeric structure whereas yellow phosphorus has discrete P4 molecules and is understrain. Q.18. Nitrogen is less reactive than phosphorus. Why ? (C.B.S.E All India 2010; C.B.S.E. Delhi 2012)

Or Phosphorus is much more reactive than nitrogen. Why ? (C.B.S.E. Delhi 2009)

Ans. It is due to the presence of N º N triple bond in N 2 . As such bond enthalpy of N 2 is very high. On the other hand, phosphorus exists as P4 molecule with P–P simple bonds having lesser bond enthalpy, so it is more reactive. Q.19. The bond angles (O–N–O) are not of the same value in NO -2 and NO+ 2. Explain. (C.B.S.E. Delhi 2012)

Ans. N in NO+ 2 is sp-hybridized, therefore, its bond angle is 180°. In NO 2 , N has one

pair of electrons, so lp-bp repulsions are stronger and the bond angle decreases from 120° to 115°. Q.20. Give appropriate reason for each of the following observations :

(i) Only higher members of group 18 of the periodic table are expected to form compounds. (ii) NO 2 readily forms a dimer, whereas ClO 2 does not. (C.B.S.E. All India 2003)

Ans. (i) Higher members of group 18 i.e., Krypton and Xenon form compounds on account of their low ionisation energies. Xenon form compounds with fluorine

59

and oxygen because these elements are highly electronegative and can create unpaired electrons in xenon by excitation of electrons from fully filled 5p orbitals to empty 5d orbitals. (ii) In NO 2 the orbital of nitrogen has an unshared electron which tends to pair up, forming a dimer. Such a situation does not exist in ClO 2 . Q.21. Why does H 3PO 2 act as monobasic acid ?

(C.B.S.E. All India 2006)

Ans. H 3PO 2 (hypophosphorus acid) acts as monobasic acid because it has only one replaceable hydrogen atom present as OH group which it can easily release.

O P H

OH

H

Q.22. Bismuth is a strong oxidising agent in the pentavalent state. (C.B.S.E. Delhi 2005 S)

or Pentavalent bismuth is a strong oxidising agent.

(C.B.S.E. Delhi 2006 S)

Ans. Since that inert pair effect is very prominent in Bi, therefore, its +5 oxidation state is much less stable than its +3 oxidation state. In other words, bismuth in the pentavalent stable can easily accept two electrons and thus gets reduced to trivalent bismuth.

Bi 5+ + 2e - ® Bi 3+ Therefore, it acts as a strong oxidising agent. Q.23. Account for the following :

NH 3 is a stronger base than PH 3 . (C.B.S.E. Delhi 2008, 2009)

Ans. Due to smaller size of N than P, electron density is more on N-atom in NH 3 than in PH 3 . Thus, ammonia is a stronger base than phosphine. Q.24. Write chemical equations for the following processes :

Orthophoshorus acid is heated. Ans. 4H 3PO 3 ® 3H 3PO 4 + PH 3

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25.

Arrange the following in order of the property mentioned : PH3 , NH3 , SbH3 , AsH3 (increasing basic strength)

Ans. Increasing basic strength : SbH 3 < AsH 3 < PH 3 < NH 3 Q.26. How would you account for the following ?

NCl 3 is an endothermic compound while NF3 is an exothermic one. Ans. NCl 3 is an endothermic compound (i. e., energy is absorbed when NCl 3 is prepared this is because of following two reasons. Q.27. How would you account for the following :

The oxidising power of oxoacids of chlorine follows the order : HClO 4 < HClO 3 < HClO 2 < HClO (C.B.S.E. Foreign 2011, C.B.S.E. Delhi 2012)

Ans. Oxidising power of oxoacids of chlorine is due to the presence of oxonion in them. As the stability of oxoanion increases, As the stability of oxoanion increases, the oxidising power of anion decreases. The stability order of these oxoanions is as follows.

ClO -4 > ClO -3 > ClO -2 > ClO As such, oxidising power of these anions is in the order ClO -4 < ClO -3 < ClO -2 < ClO Q.28. How would you account for the following :

H 2S is more acidic than H 2O. (C.B.S.E. Delhi, All India 2011)

Ans. Higher acidic character of H 2S than H 2O can be explained on the basis of the

fact that bond enthalpy of H–S bond is less than the bond enthalpy of H–O bond due to greater size of sulphur. Q.29. Complete the following chemical equations :

(i) HgCl 2 + PH 3 ®

(ii) SO 3 + H 2SO 4 ®

(iii) XeF4 + H 2O ® (C.B.S.E. All India 2011)

61

Ans. (i) 3HgCl 2 + 2PH 3 ¾® Hg 3P2 + 6HCl

(ii) SO 3 + H 2SO 4 ¾® H 2S 2O7 (iii) 6XeF4 + 12H 2O ¾® 4Xe + 2XeO 3 + 24HF + 3O 2 Q.30. What is the shape of P4 ? Ans. Tetrahedral. Q.31. Which allotropic form of phosphorus is poisonous and is used to kill rats Ans. White phosphorus. Q.32. Which phosphorus causes the disease ‘‘Phossy Jaw’’ ? Ans. White phosphorus.

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Lecture-5

VIth Group (Oxygen Family) 1.

Oxygen and ozone are allotropes.

2.

Oxygen is most electronegative while sulphur is most electron affinitive, less electron affinity of ‘O’ is due to high electron density which repules coming electron

3.

Oxygen doesn't form oxyacid with fluorine, because in oxyacid, halogen is having +ve oxidation state, but fluorine is not having +ve oxidation state. In any compound, because it is most electronegative in the periodic table and having oxidation number – 1.

Ex- HClO 4 , O.N. of Cl=+7, HFO 4 O.N. of F=+7 which is not possible. 4.

Ozone form a blue color liquid on condensation.

5.

OF6 is not possible while SF6 is possible and this is due to absence of d-orbital in case of oxygen. While in case of S ‘d’ orbitals is there. So, SF 6 is possible. F

F

F

S F

F F Note: In SF6 all the S - F bonds are having equal bonds length while in case of SF4 all the S - F bonds are not equal. S a b

F F F F bonds length a > b this is due to lone pair repulsion. 63

6.

First Ionisation energy of 6th group element is less than 5th group and this is due to half filled subshell of 5th group. N O 1s2 2s2 2p3 1s2 2s2 2p4 last subshell 2P3

2P4

N is having more IE then O due to more stable half filled subshell of 'N'. N+

O+

In case of N+ and O + more stable is O + it is due to half filled subshell of O + If N and O + then. In both the cases number of electrons are 7 but in case of O + is number of protons are ’8’. So more attraction on electrons In O + there due to which remove of electons will not be easy and more ionisation energy. 7.

H 2 O is having Hydrogen bonding while H 2 S is not having hydrogen bonding and due to hydrogen bonding H 2 O is liquid while H 2 S is a gas and boiling point of H 2 O is more than H 2 S.

8.

Isonisation energy (I.E.) of 7th group is more as compare to 6th group and this is due to more number of protons in

64

7th group as compare to 6th and having more attraction on electrons. 9.

Due to hydrogen bonding, thermal stability is as H 2 O > H 2 S > H 2 Se

10. In case of melting and boiling point we are having order as It depends upon molecular weight mainly but in case of H 2 O B.P. is more than H 2 S due to Hydrogen bonding H 2 O > H 2 Te > H 2 Se > H 2 S 11. Bond angle and dipole moment is as H 2 O > H 2 S > H 2 Se > H 2 Te –d' –d' -d -d S O q2 H H +d +d +d' +d' In case of O—H bond more polorisation is there and due to this charge develop on O—H in H 2 O is d while in case of S—H it is d' and d is more than d' due to which more repulsion between two ‘H’ atoms in H 2 O as compare to in H 2S H

q1

H

12. Sulphur is having allotropic forms. Ex- Plastic sulphur, orthorohmbic, monoclinic etc. 13. ClO 2 is strong oxidising agent and used as bleaching agent for paper industry and as a germicide. 14. 6th group elements are not colored but 7th group elements are having color, In case of 7th it is due to absorbtion of 65

energy from visible light by the molecules for the excitation of outer electron to higher energy levels and when this electron or electrons return to lower energy level then it radiates the light which is in visible region. 15. SO 2 is air pollutant and this is the cause of acid rain because the combination of SO 2 with water in presence of oxygen gives H 2 SO 4 an acid. H 2 O + SO 2 +

1 O 2 ® H 2 SO 4 2

16. SO is good oxidiser and trimerise. as 3 So ¾ 3 ¾® S 3O 9 3 and the structure of trimer is as. O O S O O O O

S

S O

O O

17. In case of contact process for the preparation of H 2 SO 4 low temperature, high pressure and presence of V2 O 5 (as a catalyst). 18. Ozone is good oxidiser and a bleaching agent and having resonating structure with bond order 3/2. 19. Sulphur is having catenation property better than Se because in case of Se, bond is weak due to large size of Se and less weak overlapping of orbitals. 20. SF6 is not hydrolysed because S is surrounded by 6F and it is not easy to approach S by H 2 O . 66

21. SF6 is colorless, odovrless and non toxic gas at room temp and due to its complete octet, it is inert towards the reaction. 22. H 2 S is good reducer and can reduce SO 2 to ‘S’ So 2 + H 2 S ¾ ¾® H 2 O + S 23. All the acids are stored in glass bottle, but HF is the only acid which is stored in steel vessel or in wax bottle because HF reacts with silicate of glass to give a waxy compound and due to this impression on glass is there. This is known as etcing of glass. Na2 SiO 3 + HF ¾ ¾®Na2 SiF6 + H 2 O Waxy compound

24. Oxygen is paramagnetic in nature due to two unpaired electron in antibonding site and also attracted by the magnet.

Anti bonding electrons

P4

P4 Bonding electrons

25. SF6 is there but not SH 6 . It is due to low polarisation between ‘S’ and ‘H’ or high promotion of electron by fluorine.

67

26. Ozone remove fluidity of mercury and this is known as tailing of mercury. When ’ O 3 ’ reacts with Hg then solid Hg2 O is formed. Hg + O 3 ¾ ¾® Hg2 O + O 2 Solid

27. Sulphur in vapour state having paramagnetic nature. 28. SF6 is more stable than SF4 and this is due to complete orbital filling of SF6 while in case of SF4 , 2 orbitals are incomplete and due to these two incomplete orbitals SF4 is more reactive than SF6 . 29. If Excess of So 2 reacts with NaOH then product formed is NaHSO 3 . NaOH + SO 2 ¾ ¾®NaHSO 3 excess

30. SO 3 is oxidiser, H 2 S is reducer and SO 2 is having both oxidising and reducing nature. 31. H 2 SO 4 is dehydrating agent H SO

2 4 HCOOH ¾ ¾¾¾ ® H 2 O + CO

COOH H SO 2 4 ¾ ¾¾¾ ® H 2 O + CO + CO 2 | COOH 32. H 2 SO 4 burns sugar H SO

2 4 C 12 H 22 O 11 ¾ ¾¾¾ ® C + H 2O

¾® H 2 S2 O 7 is known as oleum. 33. H 2 So 4 + So 3 ¾

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Do Yourself -7.7 Q.1.

Of PH 3 and H 2S which is more acidic and why ?

Ans. In H 2S, S–H bond is weaker due to larger size of sulphur. So this bond breaks

easily to give H+. Thus H 2S is more acidic. Q.2. SF6 is well know compound but SCl 6 is not known. Explain. (C.B.S.E. Delhi 2005 C)

Ans. Six floride ions can be accomadated around sulphur because of smaller size of

florine. On the other hand, Cl – ion is large in size so six Cl – ions cannot be accomadated around sulphur as there will be interelectronic repulsions. Q.3.

Bleaching of flowers by chlorine is permanent while by sulphur dioxide is temporary. Explain. (C.B.S.E. Delhi 2004 S)

Ans. Bleaching action of chlorine is due to oxidation and hence bleaching is permanent.

Cl 2 + H 2O ® 2HCl + [O] Coloured material + [O] ® Colourless On the other hand, SO 2 bleaches coloured material by reduction and hence bleaching is temporary because when the bleached colourless material is exposed to air, it gets oxidised and the colour is restored. SO 2 + 2H 2O ® 2H 2SO 4 + 2[H] Aerial

Coloured material + [H] ® colourless material ¾ ¾ ¾ ¾® Coloured material oxidation

Q.4.

Why does sulphur in the vapour state exhibit paramagnetic behaviour (C.B.S.E. All India 2006, 2008)

Ans. In the vapour state sulphur exists as diatomic (S 2 ). Due to the presence of unpaired electrons in the antibonding p-orbitals, it exhibits paramagnetic character.

69

Q.5.

Account for the following : NH 3 is a stronger base than PH 3 .

(C.B.S.E. Delhi 2008, 2009)

Ans. Due to smaller size of N than P, electron density is more on N-atom in NH 3 than in PH 3 . Thus, ammonia is a stronger base than phosphine. Q.6.

Explain the following facts giving appropriate reason in each case : (i)

NF3 is an exothermic compound whereas NCl 3 is not.

(ii) All the bonds in SF4 are not equivalent.

(C.B.S.E. All India 2012)

Ans. (i) Since NF3 is very unstable compound. Due to very high repulsive fore between F — F atom & N—F atom, it is an exothermic compound where as this all thing happen in NCl 3 but with less repulsive fore that’s why NCl 3 is not

16 5p m

(ii) SF4 has trigonal bipyramidal structure with lone pair in the equatorial position. This lone pair in the equatorial position. This lone pair repels the axial bond pair greater than the equtorial bond pair. So the bonds in SF4 are not equivalent.

S

F F

m 155p

F F Q.7.

How would you account for the following : (i) Sulphur vapour exhibits paramagnetic behaviour. (ii) Red phosphorus is less reactive than white phosphorus. (C.B.S.E. Foreign)

Ans. (i) Sulphur in vapour state contains S2 molecules. Due to the presence of two unpaired electrons, S2 molecules are paramagnetic in nature.

(ii) The less reactivity of red phosphorus in comparison to white phosphorus is due to the polymeric nature of red phosphorus. Q.8.

How would you account for the following : (i) SF6 is kinetically inert.

(C.B.S.E. Delhi, All India )

Ans. (i) SF6 is kinetically stable because :

(a) In SF6 , S is sterically protected by the six F-atoms, and 70

(b) Due to the presence of six S—F bonds in SF6 it is co-ordinatively saturated. Q.9.

Complete the following reaction equations : (C.B.S.E. Foreign )

P4 + NaOH + H 2 O ¾® Ans.

P4 + 3NaOH + 3H 2 O ¾¾® PH 3 + 3NaH 2 PO 2

Q.10. Explain H 2S is less acidic than H 2 Te. Ans. Due to decrease in bond (E-M) dissociation enthalpy down the group, acidic character increases. Q.11. What happens when sulphur dioxide gas is passed through an aqueous solutions of a Fe (III) salt ? Ans.

2Fe 3+ + SO 2 + 2H 2O ¾® 2Fe 2+ + SO 24 - + 4H+

Q.12. How would you account for the following :

(i) H 2S is more acidic than H 2O. (C.B.S.E. All India 2010)

(ii) Both O 2 and F2 stabilize high oxidation states but the ability of oxygen to stabilize the higher state exceeds that of fluorine. (C.B.S.E. All India 2011, C.B.S.E. Delhi 2012)

Ans. (i) Higher acidic character of H 2S than H 2O can be explained on the basis of the fact that bond enthalpy of H–S bond is less than the bond enthalpy of H–O bond due to greater size of sulphur.

(ii) Oxygen stabilizes higher oxidation state better than fluorine due to its ability to form multiple bond. Q.13.

Ans.

Name the element of group 16 which has maximum tendency for catenation. (P.S.B. 1999) Sulphur.

Q.14. Why the elements of oxygen family are called as chalcogens ? Ans. Chalcogens means ‘‘ore-forming element’’ and most of the ores of metals are available as their oxides or sulphides. Q.15. Covalency of oxygen never exceeds 2. Why ? Ans. This is because of the absence of d-orbitals in oxygen.

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Q.16. What is oleum ? Ans. Concentrated H 2SO 4 saturated with SO 3 is called oleum.

H 2SO 4 + SO 3 ¾® H 2S 2O7 Q.17. What is the shape of SO 3 molecule ? Ans. Triangular. Q.18. What happens when SO 3 is passed through water ? Ans. It dissolves SO 3 to give H 2SO 4 .

SO 3 + H 2O ¾® H 2SO 4 Q.19. Why conc. H 2SO 4 is not used for drying of H 2S gas ? Ans. Conc. H 2SO 4 is an oxidising agent and oxidises H 2S to sulphur.

H 2S + H 2SO 4 ¾® 2H 2O + SO 2 + S Q.20. What happens when sulphuric acid is added to sugar ? Ans. Sugar gets charred. H SO

4 C 12H 22O11 ¾ ¾2 ¾¾ ® 12C + 11H 2O

Q.21. What is Marshall’s acid ? Ans. It is H 2S 2O 8 (peroxodisulphuric acid). Q.22. H2 S acts only as reducing agent while SO2 can act both as a reducing

agent and an oxidising agent. Ans. In H2 S, the oxidation state of sulphur is -2. It can only increase its oxidation

state but cannot decrease it. Therefore, it can act as a reducing agent. In SO2 , the oxidiation state of sulphur is +4. It is in a position to undergo a decrease as well as an increase in the oxidation state. Thus, SO2

can act both as an

oxidising agent and a reducing agent. Q.23. Arrange the following is directed :

H2 O, H2 S, H2 Se, H2 Te (decreasing order of boiling point). Ans. H2 O > H2 Te > H2 Se > H2 S

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This content is the part of book ‘Inorganic for Doctors’ with video lectures and solutions by Er. Dushyant Kumar(B.Tech., IIT-Roorkee)