Lecture-4
V
th Group (Nitrogen Family)
Some im por tant key points.
1.
We are having less reactivity of N
2
as compare to P
4
at
room temp and this is due to triple bond in case of N
2
and
single bonds in case of P
4
. 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.
p-B
lock E
lements
NCERT Based Some Key Point
eg. Bi
+
3
® more stable than Bi
+
5
, Pb
+
2
is more stable
then Pb
+
4
.
In ert pair ef fect :
It is the ten dency, of an S
2
elec tron pair
to take part in bond for ma tion in p-block el e ments. This is
called in ert pair ef fect means higher ox i da tion is not
sta ble 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.
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.
+
-[ PCl ]
4[Pcl ]
6 Tetrahedral OctahedralPCl
5x-ray analysis gives
In case of solid
structures.
PCl
5
PCl
3
+
Cl
2
Cl
Cl
Cl
Cl
a
Cl
Cl
Cl
Cl
b
P
p
8.
Ammonia is prepared by Haber's process
N
2
+
3
H
2
A 2
NH
3
+
Q heat
(
)
For better production of amonia.
Pressure should be more and temperature should be less
with
Pt
(plat i num) cat a lyst and
Mo
as pro moter.
9.
NO
2
is paramagnetic and N O
2 4
is diamagnetic.
2
NO
2
Paramagnetic
A N O
Diamagnetic
2 4
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MySec ret s
N
q
1
H
H
-d
-d
+d
+d
P
q
2
H
H
I
+d
I
-d
I
+d
I
-d
CH
3
CH
3
CH
3
-
N
-(a)
(b) Si
H
3
-
N =
Si H
3
Si H
3
NH
3
NH
2
H
-BiH
3
BiH
2
H
-+ +
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.
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 -
NH
3
> PH
3
> As H
3
> SbH
3
> BiH
3
Induced d = Charge
Since d > d’ So, more repulsion between two hydrogen in
NH
3
as compare to PH
3
due to this angle increases means
q > q
1
2
.
H PO
3
4
H PO
3
3
H PO
3
2
OH
P OH
HO
O
OH
P H
HO
O
H
P H
HO
O
it is triprotic acid and
+
gives 3H
it is diprotic acid and
+
gives 2H
it is mono protic and
+
gives 1H
‘a’ is having pyramidal shape while in case of 'b' lone pair
of nitrogen goes into 'd' orbital of Si and form p
p
-
d
p
bond and become planer and less basic form in nature.
13.
H PO
3
2
>
H PO
3
3
>
H PO
3
4
(Acidic order)
Because K
a
of H PO
3
2
is more than H PO
3
3
and of H
3
PO
3
is more than H PO
3
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.
15.
N
O
O
O
N
O
O
N O
2 5
P O
4 6
O
P
O
O
O
P
O O
P
P O
4 10
O
O
O
P
O
P
O
P
P
O=
=O
O
O
=
||
O
P
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My Sec ret s ... ... ... ... ... ... ... ... ... ...Q.1. When conc. H SO2 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) Iden tify the gases A and B.
(b) Write the equa tions for the re ac tions in volved.
Ans. The given salt is a nitrate salt. On reaction with conc. H SO2 4, vapours of HNO3are produced when then decompose to give a brown gas. A (NO2)
2NaNO + H SO3 2 4 ¾¾D®Na SO + 2HNO2 4 3
Sad. nitrate (Colourless)
4HNO3 ¾¾D® 4NO + 2H O + O2 2 2 (Brown gas. A)
This gas intensified when copper turnings are added into the test tube due to the reduction of HNO3 to give more NO2 gas
Cu + 4HNO3 ¾¾D®Cu(NO ) + 2NO3 2 2 + 2H O2 On cooling the gas A (NO )2 changes into a colourless gas B (N O )2 4
2NO2 N O2 4
( )A ( )B
colling
a
Q.2. NO2readily forms a dimer. Explain.
Ans. In NO2, the orbital of nitrogen has an unshared electron which tends to pair up forming a dimer.
Q.3. (i) Why does PCI3 gives fume in moistrue ? (ii) Why is Pl5 unknown ?
Ans. (i) PCl3 hydrolysis in the presence of moisture giving fumes of HCl. PCl + 3H3 2O ® H PO + 3HCl3 3
(ii) P–I bonds being very large are weak. This made Pl5 unstable.
Q.4. Explain why BiCl3 is more stable than BiCl5.
Ans. The electronic configuration of Bi (Z = 83) is Xe 4 f14 5d10 6s2 6p , the3
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 BiCl3. Thus, BiCl3 is more stable than BiCl5.
Q.5. (SiH ) N3 3 is a weaker base than (CH ) N3 3 .
Ans. In (SiH ) N3 3 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 ) N3 3 which has a pyramidal structure similar to that of NH3. Therefore, it
can be easily donated. (CH ) N3 3 is thus a stronger base than (SiH ) N3 3 .
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 ºN). 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) Re ac tiv ity to wards hy dro gen (For ma tion of Hy drides). All el e ments of this group forms hy drides of the gen eral for mula MH3. These hy drides are : NH3 PH3 AsH3 SbH3 BiH3 Ammonia Phosphine Arsine Stibine Bismuthine
Char ac ter is tics of hy drides,
(a) Bond an gles. All hy drides are py ram i dal in shape. The bond an gle de creases on mov ing down the group due to de creases in bond pair re pul sion.
NH3(107° >) PH3(94° >) AsH3(92°)SbH3(91° >) BiH3(90°)
(b) ba sic strength. The de creas ing or der of ba sic strength of hy drides is as fol lows : NH > PH > AsH > SbH > BiH3 3 3 3 3
(c) Boil ing points of hy drides. The in creas ing or der of boil ing points is as fol lows:
PH < AsH < NH < SbH < BiH3 3 3 3 3
(d) Re duc ing char ac ter of hy drides. The in creas ing or der of re duc ing char ac ter is as follows :
NH < PH < AsH3 3 3< SbH3 < BiH3
(ii) Re ac tiv ity to wards halo gens (For ma tion of Halides). The el e ments of group 15 com bine with halo gens to form trihalides as well as pentahalides. The struc ture of trihalides are py ram i dal in na ture. All pos si ble trihalides of N, P, As, Sb and Bi are known ex cept NBr3 and NI3. These trihalides, act as Lewis bases (Ex cept NF3), since they have a lone pair of elec trons. All trihalides are co va lent in na ture. The halides are vol a tile liq uids and readily hy dro lysed by wa ter (Ex cept NF3 and PF3).
(iii) Reactivity to wards met als. All the mem bers of the fam ily have a ten dency to form bi nary com pounds with met als in which they ex hibit-3 ox i da tion state. For ex am ple,
(i) Calcium nitride (Ca N )3 2 (ii) Calcium phosphide (Ca P )3 2 (iii) Sodium arsenide (Na As)3 (iv) Zinc antimonide (Zn Sb )3 2 .
In addition to these, bismuth also forms a similar compund called magnesium bismuthide (Mg Bi ).3 2
(iv) Re ac tiv ity to wards ox y gen (For ma tion of ox ides). The mem bers of the ni tro gen fam ily gen er ally form two types of ox ides i.e. tri ox ides and pentoxides ex cept for ni tro gen which forms a few more ox ides.
Q.8. Why does NH3 form hydrogen bond while PH3 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.
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 PO3 2 and H PO3 3 act as good reducing agents while H PO3 4 does not. Ans. (a) Chlorine water is formed when Cl2 is dissolved in water. Its oxidising and
bleaching action is due to nascent oxygen.
—H—N—H—N—H—N—d– d+ d– d+ d–
H H H
Cl + H O2 2 ®[HCl + HOCl]®2HCl + [O]
(b) H PO3 2 and H PO3 3 act as good reducing agents due to the presence of P–H bonds. H PO3 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
HNO3.
Ans. With con cen trated ni tric acid.
Cu + 2HNO3 ¾ ®¾ CuO + 2NO + H O2 2 Cu + 2HNO3 ¾ ®¾ Cu(NO3)2 + H O2
Cu + 4HNO3 conc. Cu NO )3 2 NO + 2
Copper nitrate
2
( ) ¾¾¾® ( +2 H O2 l
With dilute nitric acid
3Cu + 2HNO3 ¾ ®¾ 3CuO + 2NO + H O2 l CuO + 2HNO3 ¾ ®¾ Cu(NO3)2 + H O] 32 ´
3Cu + 8HNO3(dilute) ¾¾¾®3Cu NO( 3)2 +2NO + H O4 2
Q.12. Can PCl5 act as oxidising as well-as reducing agent ? Justify.
Ans. Phosphorus can show a maximum oxidation state of +5 in its compounds. In PCl5, 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.
2 A g + P Cl0 +5 5 ® 2 Ag Cl + P Cl+1 +3 3
Q.13. BH4– and NH4+ 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 PH3 molecule lesser than that in NH3 molecule ?
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 PH3 are lesser than in NH3 which leads to lesser bond angle in PH3 than in NH3.
Q.16. In which one of the two structures, NO+2 and NO2-, 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
Phos pho rus is much more re ac tive than ni tro gen. Why ?
(C.B.S.E. Delhi 2009) Ans. It is due to the presence of N ºN triple bond in N2. As such bond enthalpy of
N2 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 NO2- 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 NO2-, 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 mem bers of group 18 of the pe ri odic ta ble are ex pected to
form com pounds.
(ii) NO2 readily forms a dimer, whereas ClO2 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
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 NO2 the orbital of nitrogen has an unshared electron which tends to pair up, forming a dimer. Such a situation does not exist in ClO2.
Q.21. Why does H PO3 2 act as monobasic acid ?
(C.B.S.E. All India 2006) Ans. H PO3 2 (hypophosphorus acid) acts as monobasic acid because it has only one
replaceable hydrogen atom present as OH group which it can easily release.
Q.22. Bismuth is a strong oxidising agent in the pentavalent state.
(C.B.S.E. Delhi 2005 S)
or
Pentavalent bis muth is a strong oxi dis ing 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.
Bi5++2e- ®Bi3+ Therefore, it acts as a strong oxidising agent.
Q.23. Account for the following :
NH3 is a stronger base than PH3.
(C.B.S.E. Delhi 2008, 2009) Ans. Due to smaller size of N than P, electron density is more on N-atom in NH3
than in PH3. Thus, ammonia is a stronger base than phosphine.
Q.24. Write chemical equations for the following processes : Orthophoshorus acid is heated.
Ans. 4H PO3 3 ® 3H PO + PH3 4 3 O P H H OH
25. Arrange the following in order of the property mentioned :
PH3,NH3,SbH3,AsH3 (in creas ing ba sic strength)
Ans. Increasing basic strength : SbH < AsH < PH < NH3 3 3 3
Q.26. How would you account for the following ?
NCl3 is an endothermic compound while NF3 is an exothermic one.
Ans. NCl3 is an endothermic compound (i e. ., energy is absorbed when NCl3is 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 < HClO < HClO < HClO4 3 2
(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 > ClO > ClO > ClO4- 3- 2- -As such, oxidising power of these anions is in the order
ClO < ClO4- 3- < ClO2-< ClO
-Q.28. How would you account for the following :
H S2 is more acidic than H O2 .
(C.B.S.E. Delhi, All India 2011) Ans. Higher acidic character of H S2 than H O2 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 + PH2 3 ® (ii) SO + H SO3 2 4 ®
(iii) XeF + H O4 2 ®
Ans. (i) 3HgCl + 2PH2 3 ¾®Hg P + 6HCl3 2 (ii) SO + H SO3 2 4 ¾® H S O2 2 7
(iii) 6XeF + 12H O4 2 ¾® 4Xe + 2XeO + 24HF + 3O3 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
V
Ith 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.
OF
6
is not possible while SF
6
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.
Note:
In SF
6
all the S F
-
bonds are having equal bonds length
while in case of SF
4
all the S F
-
bonds are not equal.
bonds length a > b this is due to lone pair repulsion.
F F
F
S
F
b
a
F
F
F
F
F
F
S
6.
First Ionisation energy of
6th
group el e ment is less than
5th
group and this is due to half filled subshell of 5th
group.
N is having more IE then O due to more stable half filled
subshell of 'N'.
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 O
2
is having Hydrogen bonding while H S
2
is not
having hydrogen bonding and due to hydrogen bonding
H O
2
is liquid while H S
2
is a gas and boiling point of
H O
2
is more than H S
2
.
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
N
32P
2
1s
2s
2
2p
3
1s
2
2s
O
2
2
p4
42P
last subshell
+N
O
+7th group as compare to 6th and having more attraction
on electrons.
9.
Due to hydrogen bonding, thermal stability is as
-H O -H S -H Se
2
>
2
>
2
10.
In case of melting and boiling point we are having order
as
-It depends upon molecular weight mainly but in case of
H O
2
B.P. is more than H S
2
due to Hydrogen bonding
H O H Te H Se H S
2
>
2
>
2
>
2
11.
Bond angle and dipole moment is as
-H O -H S -H Se -H Te
2
>
2
>
2
>
2
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 O
2
as compare to in
H S
2
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
O
q
1
H
H
-d
-d
+d
S
q
2
H
'
+d
+d
H
'
'
–d
–d
'
+d
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 SO
2
4
an acid.
H O SO
2
2
1
O
2
H SO
2
4
2
+
+
®
16.
SO
3
is good oxidiser and trimerise. as 3
So
3
¾ ®
¾
S O
3 9
and the structure of trimer is as.
17.
In case of contact process for the preparation of H SO
2
4
low
temperature, high pressure and presence of V O
2 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.
SF
6
is not hydrolysed because S is surrounded by 6F and
it is not easy to approach S by H O
2
.
S
O
S
O
O
S
O
O
O
O
O
O
21.
SF
6
is colorless, odovrless and non toxic gas at room temp
and due to its complete octet, it is inert towards the
reaction.
22.
H S
2
is good reducer and can reduce SO
2
to ‘S’
So
2
+
H S
2
¾ ®
¾
H O S
2
+
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.
Na SiO
HF
Na SiF
H O
Waxy compound
2
3
+
¾ ®
¾
2
6
+
2
24.
Oxygen is paramagnetic in nature due to two unpaired
electron in antibonding site and also attracted by the
magnet.
25.
SF
6
is there but not SH
6
. It is due to low polarisation
between ‘S’ and ‘H’ or high promotion of electron by
fluorine.
Anti bonding electrons
4
P
P
426.
Ozone remove fluidity of mercury and this is known as
tailing of mercury. When ’
O
3
’
reacts with Hg then solid
Hg
2
O is formed.
Hg O
Hg O O
Solid
+
3
¾ ®
¾
2
+
2
27.
Sulphur in vapour state having paramagnetic nature.
28.
SF
6
is more stable than SF
4
and this is due to complete
orbital filling of SF
6
while in case of SF
4
, 2 orbitals are
incomplete and due to these two incomplete orbitals SF
4
is
more reactive than SF
6
.
29.
If Excess of So
2
reacts with NaOH then product formed is
NaHSO
3
.
NaOH SO
NaHSO
excess
+
2
¾ ®
¾
3
30.
SO
3
is oxidiser, H
2
S is reducer and SO
2
is having both
oxidising and reducing nature.
31.
H SO
2
4
is dehydrating agent
HCOOH
H SO
2 4H O CO
2
¾
¾¾¾
®
+
|
COOH
COOH
H O CO CO
H SO
2 42
2
¾
¾¾¾
®
+
+
32.
H SO
2
4
burns sugar
C H O
12 22 11
¾
H SO
¾¾¾
2 4®
C H O
+
2
33.
H So
2
4
+
So
3
¾ ®
¾
H S O
2 2 7
is known as oleum.
Q.1. Of PH3 and H S2 which is more acidic and why ?
Ans. In H S2 , S–H bond is weaker due to larger size of sulphur. So this bond breaks easily to give H+. Thus H S2 is more acidic.
Q.2. SF6 is well know compound but SCl6 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 + H O2 2 ® 2HCl + [O]
Coloured material +[O]® Colourless
On the other hand, SO2 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 + 2H O2 2 ® 2H SO + 2[H]2 4
Coloured material +[H]® colourless material Aerial
oxidation
¾¾ ¾¾® Coloured material
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.
Q.5. Account for the following :
NH3 is a stronger base than PH3. (C.B.S.E. Delhi 2008, 2009) Ans. Due to smaller size of N than P, electron density is more on N-atom in NH3
than in PH3. 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 NCl3 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 NCl3 but with less repulsive fore that’s why NCl3 is not (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.
Q.7. How would you account for the following :
(i) Sul phur vapour ex hib its para mag netic be hav iour.
(ii) Red phosphorus is less re ac tive than white phos pho rus.
(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
F S F F F 155pm 165p m
(b) Due to the presence of six S—F bonds in SF6it is co-ordinatively saturated.
Q.9. Complete the following reaction equations :
P + NaOH + H O 4 2 ¾® (C.B.S.E. Foreign )
Ans. P4 +3NaOH+3H O2 ¾ ®¾ PH + 3NaH PO3 2 2
Q.10. Explain H S2 is less acidic than H Te2 .
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. 2Fe3++ SO + 2H O2 2 ¾® 2Fe2++ SO42-+ 4H+
Q.12. How would you account for the following : (i) H S2 is more acidic than H O2 .
(C.B.S.E. All India 2010)
(ii) Both O2 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 S2 than H O2 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. Name the element of group 16 which has maximum tendency for
catenation. (P.S.B. 1999)
Ans. 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.
Q.16. What is oleum ?
Ans. Concentrated H SO2 4 saturated with SO3 is called oleum. H SO + SO2 4 3 ¾®H S O2 2 7
Q.17. What is the shape of SO3 molecule ? Ans. Triangular.
Q.18. What happens when SO3 is passed through water ? Ans. It dissolves SO3 to give H SO2 4.
SO + H O3 2 ¾®H SO2 4
Q.19. Why conc. H SO2 4 is not used for drying of H S2 gas ?
Ans. Conc. H SO2 4 is an oxidising agent and oxidises H S2 to sulphur. H S + H SO2 2 4 ¾® 2H O + SO + S2 2
Q.20. What happens when sulphuric acid is added to sugar ? Ans. Sugar gets charred.
C H O12 22 11 ¾H SO¾ ¾2 ¾4 ®12C + 11H O2
Q.21. What is Marshall’s acid ?
Ans. It is H S O2 2 8 (peroxodisulphuric acid).
Q.22. H S2 acts only as reducing agent while SO2 can act both as a reducing agent and an oxidising agent.
Ans. In H S2 , 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 :
H O H S H Se H Te2 , 2 , 2 , 2 (decreasing order of boiling point). Ans. H O2 >H Te2 >H Se2 >H S2
