DAV CENTENARY PUBLIC SCHOOL
PASCHIM ENCLAVE
GROUP – 17 ELEMENTS
Q1.
Ans.
With what neutral molecule is ClO- isoelectronic?
ClF
Q2.
Give the formula of the noble gas species that is isostructrural with
(a) ICl4(b) IBr2(c) BrO3Try yourself. (ACCORDING TO VSEPR THEORY)
Ans.
Q3.
Ans.
Which among the following is the strongest oxidizing agent?
ClO4-, BrO4-, IO4IO4-
Q4.
Ans.
What do you mean by interhalogen compounds?
The binary compounds formed by the two different halogens are known as interhalogen compounds.
They may be represented by a general formula AXn (where n = 1, 3, 5, 7).
Q5.
Ans.
Write the one chemical method for the preparation of fluorine.
K2MnF6 + 2SbF5 → 2KSbF6 + MnF3 + ½ F2
Q6.
Ans.
Why SiF62- is known SiCl62- is not?
Size of F is smaller than Cl therefore, in SiF62- has lesser steric repulsions. Moreover, the interaction of
lone pair electrons of F with Si is stronger than that of the lone pairs of chlorine.
Q7.
Electrolysis of KBr (aq) gives Br2 at anode but that of KF (aq) does not give F2. Give reason for disparity
in behaviour.
The stantdard electrode potential of fluorine is highest, hence F - ions (from KF) cannot be oxidized to
fluorine by any oxiising agent. (Which always has lower E 0 value).
Ans.
Q8.
Ans.
Identify X in the following chemical reaction: Cl2 + 2X- → 2Cl- + X2.
X may be Br- or I-
Q9.
Ans.
Give an example of the compound in which the oxidation state of chlorine atom is +7.
Sodium perchlorate (NaClO4).
Q10.
Ans.
Arrange the hydrides of halogens in the order of increasing acid strength.
HF < HCl < HBr < HI
Q11.
Ans.
What kind of bond is expected between oxygen and fluorine in oxygen fluoride?
Covalent type of bond, because there is very small difference in the electronegativity values of F and O.
Q12.
Ans.
Arrange HClO4, HIO4 and HBrO4 in order of increasing oxidizing ability.
HClO4 < HIO4 < HBrO4 (because the standard electrode potential, XO 4- | XO3- is highest in case of
HBrO4)
Q13.
Ans.
Arrange the halogens in order of bond dissociation energies.
I2 < Br2 < F2 < Cl2
Q14.
Ans.
Name the anhydride of hypochlorous acid?
Cl2O.
Q15.
Ans.
Name the anhydride of perchloro acid.
Cl2O7.
Q16.
Which oxides of chlorine are paramagnetic?
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Ans.
ClO2 and ClO3.
Q17.
Arrange the following oxyacids in increasing order of acid strength and decreasing order of oxidizing
power. HClO4, HClO3, HClO2, HClO
HClO < HClO2 < HClO3 < HClO4 (increasing order of strength)
HClO > HClO2 > HClO3 > HClO4 (decreasing order of oxidizing power)
Ans.
Q18.
Ans.
Draw the structures of chloric (I) acid, chloric (III) acid, chloric (V) acid, chloric (VII) acid.
O
O
O
Cl
H
chloric (I) acid
Cl
H
chloric (III) acid
O
O
O
H
Cl
O
chloric (V) acid
O
O
Cl
H
O
chloric (VII) acid
Q19. Why HF is a weaker acid than HI.
Ans.
HF (aq) has extensive hydrogen bonding hence dissociation is weak. Whereas HI is weakly hydrogen
bonded and is almost fully dissociated.
Q20.
Ans.
Electron affinity of fluorine is less than that of chlorine. Why?
Electron affinity of fluorine is less than that of chlorine. The low value of electron affinity of fluorine is
probably due to the electron – electron repulsion in relatively compact 2p – orbitals of fluorine atom due
to its small size.
Q21.
What are inter halogen compounds? Why most of the inter halogens are most reactive than molecular
halogens?
The halogens combine themselves to form a number of covalent compounds known as interhalogen
compounds. This happens due to the difference in their electronegativities. Most of the inter halogen
compounds are strong oxidizing agents. They are more reactive than halogens, because the polarity is
developed due to the large difference in their electronegativities.
Ans.
Q22.
Ans.
Q23.
Ans.
Q24.
Ans.
Why do you add KF and exclude moisture in the electrolysis of HF for the manufacture of fluorine? Give
reasons.
Addition of KF makes HF more electroyte i.e., it makes HF more conducting in molten state. Fluorine
has high affinity for hydrogen. Because of this fluorine reacts with water to liberate oxygen.
F2 + H2O → 2F- + 2H+ + ½ O2
Arrange the following according to the property mentioned against each:
(i)
HF, HCl, HBr, HI in the order of increasing acid strength.
(ii)
H2S, H2O, H2Te, H2Se in the order of decreasing thermal stability.
(iii)
HClO4, HBrO4, HIO4 in the order of increasing oxidizing ability
(i)
HF < HCl < HBr < HI
(ii)
H2S > H2O > H2Se > H2Te
(iii)
HClO4 < HIO4 < HBrO4
Arrange the following in the decreasing order of property indicated against each:
(i)
F2, Cl2, Br2, I2 (decreasing bond strength )
(ii)
As2O3, ClO2, GeO2, Ga2O3 (increasing acidity)
(iii)
NH3, PH3, AsH3, SbH3 (decreasing base strength)
(iv)
H2O, H2S, H2Se, H2Te (increasing acidic strength)
(v)
H2O, H2S, H2Se, H2Te (decreasing boiling point)
(i)
Cl2 > F2 > Br2 > I2
(ii)
Ga2O3 < As2O3 < GeO2 < ClO2
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(iii)
NH3 > PH3 > AsH3 > SbH3
(iv) H2O < H2S < H2Se < H2Te
(v) H2O > H2Te > H2Se > H2S
Q25.
Ans.
Hydrogen fluoride has higher boliling point than hydrogen chloride. Why?
Hydrogen fluoride has higher boiling point than hydrogen chloride. This is due to the presence of strong
intermolecular hydrogen bonding between H – F molecules, but there is no such strong hydrogen bonding
in H – Cl.
Q26.
Ans.
SF6 is known but SCl6 is not known. Give reason.
Due to small size of S, six large Cl atoms cannot be accommodated around S atom. But small six F atoms
can be easily accommodated around S atom to form SF6. Moreover, because of low electronegativity of
Cl, it cannot easily cause promotion of electrons in S to form S (VI).
Q27.
Ans.
State the trends observed in case of the oxidizing property of the members of the halogen family.
Since all the halogens have a strong tendency to accept electrons, they act as strong oxidizing agents.
There oxidizing power decreases from fluorine to iodine. Since fluorine is the strongest oxidizing agent
in the series, it will oxidize other halide ions to halogens in solutions or dry state.
Q28.
Why is it that anhydrous aluminium fluoride has a higher melting point than anhydrous alumimium
chloride?
Anhydrous aluminium fluoride is an ionic compound while anhydrous aluminium chloride is a covalent
compound, therefore, the melting point of anhy. AlF3 is greater than anhy. AlCl3.
Ans.
Q29.
Ans.
Draw the structures of ClO2 and Cl2O7.
O
O
O
O
O
Cl
Cl
Cl
O
O
O
O
Q30.
Ans.
ClF3 exists but FCl3 does not. Explain.
Chlorine has vacant d – orbitals, thus one p – electron on promotion to d – orbital gives 3s2 3p4 3d1
(three unpaired electrons) which forms ClF 3. Fluorine cannot form FCl3 due to the absence of d –
orbitals.
Q31.
Ans.
Despite its lower electron affinity fluorine is a stronger oxidizing agent than chlorine. Explain.
The reduction potential of fluorine is higher than the other halogens, therefore despite having lower
value of electron affinity, fluorine is a stronger oxidizing agent.
1
F2  e  aq  F  (aq)
2
Q32.
Ans.
Q33.
Ans.
Q34.
Ans.
E 0  2.87V
PCl5 is known but PI5 is not known. Why?
Due to small size of Cl atom, five Cl atoms can be accommodated around P atom. But I is of large size
and therefore, five I atoms cannot be accommodated around P atom. As a result, P – I bonds are weak
and prefer to form PI3 rather than PI5.
NCl3 gets readily hydrolysed while NF3 does not. Why?
In NCl3, Cl has vacant d – orbitals to accept lone pair of electrons present on oxygen atom of water
molecule.
SF6 is known but SCl6 is not known. Explain.
Due to small size of S, six large Cl atoms cannot be accommodated around S atom. But small six F atoms
can be easily accommodated around S atom to form SF6. Moreover, because of low electronegativity of
Cl, it cannot easily cause promotion of electrons in S to form S (IV).
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Q35.
Ans.
Q36.
Ans.
Halogen have maximum negative electron gain enthalpy in the respective periods of the periodic table.
Why?
The halogens have the smallest size in their respective periods and therefore, high effective nuclear
charge. Moreover, they have only one electron less than the stable noble gas configuration.
Therefore, they have strong tendency to accept one electron to acquire noble gas electronic configurations
and hence have maximum negative electron gain enthalpy in their respective periods.
Fluorine exhibits only -1 oxidation state where as other halogens exhibit positive oxidation states also
such as +1, +3, +5, +7. Why?
Fluorine is most electronegative element and cannot exhibit any positive oxidation states. On the other
hand, the other halogens are less electronegative and therefore, can exihibit positive oxidation states.
They also have vacant d – orbitals and hence can expand their octets and show +1, +3, +5, +7 oxidation
states also.
Q37.
Ans.
HCl when reacts with finely divided iron forms ferrous chloride and not ferric chloride. Why?
HCl reacts with finely divided iron and produces H2 gas.
Fe + 2HCl → FeCl2 + H2
Liberation of hydrogen prevents the formation of ferric chloride.
Q38.
Ans.
Explain why inspite of nearly same electronegativity, oxygen forms hydrogen bonding while chlorine
does not.
Oxygen has smaller size than chlorine. The smaller size of oxygen favours hydrogen bonding. In other
words, though electronegativity of Cl is same as that of O, it does not form hydrogen bonding because of
its larger size.
Q39.
Ans.
Explain why fluorine forms only one oxoacid HOF?
Due to small size and high electronegativity, fluorine cannot act as central atom in higher oxoacids.
Q40.
Ans.
Why are pentahalides more covalent than trihalides?
In pentahalides, the oxidation state is more (+5) than in trihalides (+3). As a result of higher positive
oxidation state of central atom, they have larger polarizing power and can polarize the halide ion to a
greater extent than in the corresponding trihalide. Since larger the polarization, larger is the covalent
character, therefore, pentahalides are more convalent than trihalides.
Q41.
With the increase in oxidation no. of a particular halogen atom, the acidic character of corresponding
oxoacid Increases. Explain. (HClO < HClO2 < HClO3 < HClO4)
Here oxygen is more electronegative than chlorine. The more the oxygen atom bonded to the chlorine
atom more the electrons will be pulled away from the OH bond, and the more the bond will be weakened.
Thus HClO4 requires the least energy to break the O — H bond and form H+. Hence HClO4 is the
strongest acid. The strength of acid is also explained on the basis of stability of ion formed. Thus the
stability will be
ClO4- > ClO3- > ClO2- > ClO–
Hence the acidic strength will also decreases from HClO4 to HClO.
Ans.
Q42.
Ans.
Considering the parameters such as bond dissociation enthalpy, electron gain enthalpy and hydration
enthalpy, compare the oxidizing power of F2 and Cl2.
F2, is stronger oxidizing agent than Cl2. This can be explained on the basis of bond dissociation enthalpy,
electron gain enthalpy and hydration enthalpy. The process of oxidizing behaviour may be expressed as:
½ X2 (g) → X (g) ΔHdiss.
X (g) → X- (g)
ΔHeg
X- (g) → X- (aq)
ΔHhyd
The overall tendency for the change (i.e., oxidizing behaviour) depends upon the net effect of three steps.
As energy is required to dissociate or convert molecular halogen into atomic halogen, the enthalpy
change for this step is positive. On the other hand, energy is released in step (II) and (III), therefore,
enthalpy change for these steps is negative. Now although fluorine has less negative electron gain
enthalpy, yet ist is stronger oxidizing agent because of low enthalpy of dissociation and very high
enthalpy of hydration. In other words, large amount of energy required in step (I) overweigh the smaller
energy released in step (II) for fluorine. As a result, the ΔH overall is more negative for F2 than for Cl2.
Hence, F2 is stronger oxidizing agent than Cl2.
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IMPORTANT REACTIONS OF GROUP – 17
1.
The relative oxidizing power of halogens can be illustrated by their reactions with water. Fluorine
oxidizes water to oxygen whereas chlorine and bromine react with water to form corresponding
hydrohalic and hypohalous acids. The reaction of iodine with water is non – spontaneous. In fact, I- can
be oxidized by oxygen in acidic medium; just the reverse of the reaction observed with fluorine.

 4H+ + 4F- + O2
X2 + H2O 
(where X = Cl or Br)
 HX + HOX
+
4I + 4H + O2 
 2I2 + 2H2O
2F2 + 2H2O
2.
Laboratory preparation of Cl2
By heating manganese dioxide with conc. HCl
MnO2 + 4HCl 
 MnCl2 + Cl2 + 2H2O
However, a mixture of common salt and conc. H2SO4 is used in place of HCl.
4NaCl + MnO2 + 4H2SO4 
 MnCl2 + 4NaHSO4 + 2H2O + Cl2
By the action of HCl on potassium permanganate.
2KMnO4 +16HCl 
 2KCl + 2MnCl2 + 8H2O + 5Cl2
Manufacturing of Cl2
Deacon’s process :- By the oxidation of hydrogen chloride gas by atmospheric oxygen in the presence of
CuCl2 (catalyst) at 723 K.
4HCl + O2
CuCl2

 2Cl2 + 2H2O
723 K
Electrolytic process :- Chlorine is obtained by the electrolysis of brine (conc. NaCl solution). Chlorine is
liberated at anode. It is also obtained as a by – product in many chemical industries.
3.
With excess ammonia, chlorine gives nitrogen and ammonium chloride whereas with excess chlorine,
nitrogen trichloride (explosive) is formed.
8 NH 3  3Cl2  6 NH 4Cl  N 2
( excess )
Cl  NH 3  NCl3  3HCl
2
( excess )
4.
With cold and dilute alkalies chlorine produces a mixture of chloride and hypochlorite but with hot and
concentrated alkalies it gives chloride and chlorate.
2 NaOH  Cl2  NaCl  NaOCl  H 2O
( Cold & Dilute )
6 NaOH  3Cl2  5NaCl  NaClO3  3H 2O This reaction is an example of disproportionation
( Hot & Conc.)
5.
reaction.
With dry slaked lime Cl2 gives bleaching powder
2Ca(OH)2 + 2Cl2  Ca(OCl)2 + CaCl2 + 2H2O
6.
The composition of bleaching powder is Ca(OCl)2 . CaCl2 . Ca(OH)2 . 2H2O
7.
Chlorine water on standing loses its yellow colour due to the formation of HCl and HOCl. Hypochlorous
acid (HOCl) so formed, gives nascent oxygen which is responsible for oxidizing and bleaching properties
of chlorine.
8.
Oxidising action of chlorine
2FeSO4 + H2SO4 + Cl2  Fe2(SO4)3 + 2HCl
Na2SO3 + Cl2 + H2O  Na2SO4 + 2HCl
 H2SO4 + 2HCl
SO2 + 2H2O + Cl2
I2 + 6H2O + 5Cl2  2HIO3 + 10HCl
9.
Laboratory preparation of HCl
NaCl + H2SO4
420 K

 NaHSO4 + HCl
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NaHSO4 + NaCl 
 Na2SO4 + HCl
HCl gas can be dried by passing through concentrated sulphuric acid but HI can’t.
823K
10.
Interhalogen compounds are very useful fluorinating agents. ClF3 and BrF3 are used for the production
of UF6 in the enrichment of
U(s) + 3ClF3 (l)
U 92235 .
 UF6 (g) + 3ClF (g)
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