Learning Outcomes

The learners would be enabled to

• infer the trends in the chemistry of elements of groups 15-18

• discuss the preparation, properties and uses of nitrogen and phosphorus and some other useful compounds

• discuss the preparation, properties and uses of dioxygen, ozone, chlorine and hydrochloric acid

• infer the allotropic forms of sulphur, chemistry of its important compounds and structures of its oxoacids

• infer the chemistry of interhalogens and their oxoacids

• the uses of noble gases

General principles of inorganic chemistry

Trends in the Periodic Table

Shells upto ( $\mathrm{n}-1$ ) are completely filled and differentiating electron (last filling electron) enters into np sub-shell — elements constitute what we call $p$-block elements.

General electronic configuration is $n s^{2}{ }^{11-6}$ ( $n$ varies from 2 to 7 )

where

$\mathrm{x}$ : covalent/van der Waals’ radius

$y$ : metallic character

z: heat of sublimation

p: electronegativity

q: ionisation energy $(\mathrm{N}>0, \mathrm{P}>\mathrm{S})$

r: oxidising power

s: stability of higher oxidation state within a group,

t: stability of lower oxidation state within a group

Trends in Properties of p-Block Elements in the direction of arrow

The highest oxidation state $=$ (group number- 10). Stability of this oxidation state (0.S.) decreases and that of $[(0 . S)-2]$ state increases as we move down the group-a case of inert pair effect.

Oxidation states

On descending the group, a lower oxidation state which is two units less than the highest oxidation state becomes more stable in groups 13 to 16 . This trend is due to inert pair effect. For example, the highest oxidation state for the elements of group 13 is +3 . However, in addition to +3 oxidation state, these elements also show +1 oxidation state which becomes more stable on moving down the group. In fact, for the last element, thallium, +1 oxidation state is more stable than +3 . Similarly, for group 14, the group oxidation state is +4 , but +2 oxidation state becomes more and more stable on going down the group. For example, the last element, lead +2 oxidation state is more stable than +4 oxidation state.

This trend of occurrence of oxidation state two units less than the group oxidation state is called inert pair effect and becomes more and more prominent as we move down the group.

The common oxidation states displayed-by the p-block elements are given in Table I.

Metallic and non-metallic character

The p-block contains metallic and non-metallic elements. It is very interesting to note that the nonmetals and metalloids exist only in the p-block of the periodic table. The non-metallic character increases along a period but decreases down a group. In fact the heaviest element in each p-block group is the most metallic in nature. Therefore, the elements with most metallic character are located mostly in the lower left portion while those with most non-metallic character are present at the top right portion of the periodic table. In between these, there are some elements which show characteristics of both metals and non-metals and are called metalloids. The common metalloids in p-block elements are $\mathrm{B}, \mathrm{Al}, \mathrm{Si}, \mathrm{Ge}, \mathrm{As}, \mathrm{Sb}, \mathrm{Te}, \mathrm{Po}, \mathrm{At}$. This change from non-metallic to metallic brings significant diversity in the chemistry of these elements.

In general, non-metals have higher ionization enthalpies and higher electronegativities than metals. Therefore, in contrast to metals which readily form cations, non-metals readily form anions. The compounds formed by combination of highly reactive non-metals, with highly reactive metals are generally ionic in nature because of large differences in their electronegativities. On the other hand, compounds formed between non-metals themselves are largely covalent in character because of small differences in their electronegativities. It can be understood in terms of their oxides. The oxides of non-metals are acidic or neutral whereas oxides of metals are basic in nature. The oxides of metalloids are amphoteric. Further more, the more electropositive the metal, the more basic is its oxide and the more electronegative the non-metal, the more acidic is its oxide. Therefore, in p-block elements, acidic character of the oxides increases or basic character decreases along a period. Similarly, the basic character of the oxides increases or acidic character decreases down the group.

Table 1. Common oxidation state of p-block elements

Group	$\mathbf{1 3}$	$\mathbf{1 4}$	$\mathbf{1 5}$	$\mathbf{1 6}$	$\mathbf{1 7}$	$\mathbf{1 8}$
General electronic Configuration	$\mathrm{ns}^2 \mathrm{np} \mathrm{n}^1$	$\mathrm{~ns}^2 \mathrm{np}^2$	$\mathrm{~ns}^2 \mathrm{np}^3$	$\mathrm{~ns}^2 \mathrm{nn}^4$	$\mathrm{~ns}^2 \mathrm{n} \mathrm{n}^5$	$\mathrm{~ns}^2 \mathrm{np}^6$
Group oxidation state	+3	+4	+5	+6	+7	+8
Various oxidation States	B +3 $\mathrm{Al}$ +3 $\mathrm{Ga}, \mathrm{In}, \mathrm{TI}$ $+3,+1$	C $+4,-4$ $\mathrm{Si}$ +4 $\mathrm{Ge}, \mathrm{Sn}, \mathrm{Pb}$ $+4,+2$	N $+5 \mathrm{to}-3$ $\mathrm{P}, \mathrm{As}$ $+3,+5,-3$ $\mathrm{Sb}, \mathrm{Bi}$ $+3,+5$	O $-1,-2$ $\mathrm{S}, \mathrm{Se}, \mathrm{Te}$ $-2,+2$, $+4,+6$	F -1 $\mathrm{Cl}, \mathrm{Br}, \mathrm{I}$ $-1,+1,+3$ $+5,+7$ $+2,+4$, $+6,+8$	$\mathrm{Kr}$ $+2,+4$ $\mathrm{Xe}$

Differences in behaviour of first element of each group

The first member of each group of p-block differs in many respects from its succeeding members (called congeners) of their respective groups. For example, boron shows anomalous behaviour as compared to rest of the members of the 13 group elements. The main reasons for the different behaviour of the first member as compared to other members is because of:

i) small size of the atom and its ion

ii) high electronegativity and

iii) absence of $d$-orbitals in their valence shell

These factors have significant effect on the chemistry of first element as compared to other elements (specially second). For example

a) Covalence upto four: First member of each group belongs to second period elements and have only four valence orbitals i.e., one $2 s$ and three $2 p$ orbitals available for taking part in chemical combinations. They do not have vacant $d$-orbitals in their valence shell. Therefore, they may have maximum covalence of four (using one $2 \mathrm{~s}$ and three $2 \mathrm{p}$ orbitals). In contrast, the next members belonging to third or higher periods have vacant $d$-orbitals. For example, the elements of third period of $p$-block with the electronic configuration $3 s^{2} 3 p^{1-6}$ has vacant $3 d$-orbitals lying between $3 p$ and $4 \mathrm{~s}$ energy sub-shells. Therefore, they can easily expand their octets and can show covalence above four. For example,

i) Boron forms only $\mathrm{BF} _{4}^{-}$(coordination number four) whereas aluminium forms $\mathrm{AlF} _{6}{ }^{3-}$ (coordination number six).

ii) Carbon can form only tetrahalides ( $\mathrm{CX} _{4}, \mathrm{X}=\mathrm{F}, \mathrm{Cl}, \mathrm{Br}$, I) whereas other members can form hexahalides,

$$ \mathrm{SF} _{6}, \mathrm{SiCl} _{6}^{2-} \text { etc. } $$

iii) Nitrogen forms only $\mathrm{NF} _{3}$ (upto octet) while phosphorus forms pentahalides, $\mathrm{PF} _{5}, \mathrm{PCl} _{5}$, etc.

iv) Fluorine does not form $\mathrm{FCl} _{3}$ (more than octet) while chlorine forms $\mathrm{CIF} _{3}$ (extends octet).

b) Reactivity: Due to availability of $d$-orbitlals of elements of third period, they are more reactive than elements of second period which do not have d-orbitals. For example, tetrahalides of carbon are not hydrolysed by water whereas tetrahalides of other elements of group 14 are readily hydrolysed. The hydrolysis involves the nucleophilic attack of water molecule.

c) Tendency to form multiple bonds: Because of the combined effect of smaller size and nonavailability of $d$-orbitals, the first member of each group shows tendency to form $p \pi-p \pi$ multiple bonds either with itself (such as $C=C, C \equiv C, N \equiv N, 0=0$ ) or with other members of the second period of elements (such $C=0, C \equiv N, N=0$, etc). The other members of the group do not have strong tendency to form $\pi$ - bonding. The heavier elements do form $\pi$-bonding but they involve $d$ - orbitals and form $d \pi-p \pi$ or $d \pi-d \pi$ bonding. For example, the bonds between sulphur and oxygen in oxides of sulphur $\left(\mathrm{SO} _{2}\right.$ and $\mathrm{SO} _{3}$ ) are much shorter than might be expected for a single bond. In these molecules, in addition to normal $\pi$ bond, a $\pi$ bond is also formed by the sidewise overlap of a filled $2 p$-orbital of oxygen with a vacant $3 d$-orbital on the sulphur). This is called $p \pi-d \pi$ bond and results in bringing the two atoms closer and thus accounts for shorter bond length of $\mathrm{S}-\mathrm{O}$ bond.

Because the d-orbitals are of higher energy than p-orbitals, they contribute less to the overall stability of molecules than does the $p \pi-p \pi$ bonding of second row elements. However, the coordination number in species of heavier elements may be higher in those of first element in the same oxidation state. For example, both nitrogen and phosphorus form ions in +5 oxidation state as $\mathrm{NO} _{3}{ }^{-}$(three coordination with bonding using one p-orbital of $\mathrm{N}$ ) ; $\mathrm{PO} _{4}{ } _{4}{ }^{3-}$ (having four coordination using $s, p$ and $d$ orbitals contributing to the $\pi$-bonding).

The first member of 13 group (boron) shows diagonal relationship with silicon (of group 14).

P-Block

Group 15-

The elements of nitrogen family i.e. group 15 of the periodic table are: nitrogen( $\mathrm{N})$, phosphorus( $\mathrm{P})$, arsenic (As), antimony (Sb) and bismuth (Bi)

General configuration- $n \mathrm{~s}^{2} \mathrm{np}^{5}$

The elements of group15 show a tendency to form bonds itself (catenation).

$: \overline{\mathrm{N}}=\stackrel{+}{\mathrm{N}}=\overline{\mathrm{N}}:\left(\right.$ Azide ion $\mathrm{N}^{3}$ )

Group 15 elements exhibit oxidation states -3 to +5 . On moving down the group, the stability of +5 oxidation state decreases while that of +3 oxidation state increases due to inert pair effect. The tendency to exhibit -3 oxidation state decreases on moving down from $\mathrm{P}$ to $\mathrm{Bi}$.

The maximum covalency of nitrogen is restricted to four because it does not have vacant $d$-orbitals in the outermost valence shell. Nitrogen molecule has low reactivity because it has a triple bond which has a very high bond enthalpy.

Due to absence of $d$-orbitals, nitrogen cannot form $p \pi-d \pi$ bonds while heavier elements can do so. Phosphorus and arsenic can also form $\mathrm{d} \pi-\mathrm{d} \pi$ bonds with transition elements where their compounds can act as ligands.

All the elements of group 15 form hydrides. Ammonia is the most important trihydride prepared by Haber’s process. Phosphine and hydrides of other heavier elements of the family are highly poisonous. All these hydrides are covalent in nature and have pyramidal shape.

1) Trends in properties (Physical state)

a) $\mathrm{N} _{2}$ is a gas; $\mathrm{P} _{4}$ is a solid

Nitrogen being small, can form $p \pi-p \pi$ multiple bonds $N=N$. it exists in the form of $N _{2}$ molecule. P can’t make $p \pi-p \pi$ multiple bond due to its larger size. So, it exists in the $P _{4}$ molecule.

$\mathrm{P} _{4} \rightarrow$ white $\xrightarrow{\ddot{ } \rightarrow}$ waxy solid, $\mathrm{P} _{4}$-Red $\rightarrow$ Hard solid

Red $\left(\mathrm{P} _{4}\right) \rightarrow$ Polymeric chain

High melting point, Insoluble in $\mathrm{CS} _{2}$.

2) Atomic size

Increases down the group due to addition of an extra shell at each succeeding element.

3) Ionization Enthalpy $\rightarrow$ very high

i) 15th group elements have higher I.E. than 16th group of elements since their electronic configuration is $n s^{2} n p^{3}$. (half filled)

ii) decreases down the group

4) Electron Gain Enthalpy $\mathrm{P}>\mathrm{As}>\mathrm{Sb}>\mathrm{Bi}>\mathrm{N}$

$N$ has positive electron gain enthalpy due to small size, inter electronic addition of an extra $e^{-}$ is not energetically favourable.

HYDRIDES

$\mathrm{NH} _{3} \mathrm{PH} _{3} \mathrm{AsH} _{3} \mathrm{SbH} _{3} \mathrm{BiH} _{3}$

1) Bond angles.

$\mathrm{NH} _{3}>\mathrm{PH} _{3}>\mathrm{AsH} _{3}>\mathrm{SbH} _{3}>\mathrm{BiH} _{3}$

Down the group, electronegativity of central atom decreases. This causes decrease in the bond angle.

2) Boiling point

$\mathrm{PH} _{3}<\mathrm{AsH} _{3}<\mathrm{NH} _{3}<\mathrm{SbH} _{3}<\mathrm{BiH} _{3}$

intermolecular $\mathrm{H}$ - bonding

3) Basic nature

Basic nature depends on the availability of lone pair of electrons.

$\mathrm{NH} _{3}>\mathrm{PH} _{3}>\mathrm{AsH} _{3}>\mathrm{SbH} _{3}>\mathrm{BiH} _{3}$

Size of central atom increases and the availability of $I p$ of $e^{-}$for protonation decreases. So, the basic nature decreases.

4) Reducing nature

$\mathrm{NH} _{3}<\mathrm{PH} _{3}<\mathrm{AsH} _{3}<\mathrm{SbH} _{3}<\mathrm{BiH} _{3}$

Size of central atom increases and the thermal stability decreases. Ease of availability of hydrogen increases.

Ammonia forms hydrogen bonding with water molecules, therefore it is soluble in water, while other hydrides are insoluble in water.

Oxides

The oxides in the higher oxidation state of elements are more acidic than that of the oxides in the lower oxidation state.

A) Oxide of Nitrogen

a) $\mathrm{N} _{2} \mathrm{O} \rightarrow$ Laughing gas (Nitrous oxide)

$\ddot{\mathrm{N}}=\mathrm{N} \rightarrow \ddot{0}:$

Prepared by the action of any lighter metal with very dil. $\mathrm{HNO} _{3}$

$$ 4 \mathrm{Sn}(\mathrm{s})+10 \mathrm{HNO} _{3} \rightarrow 4 \mathrm{Sn}\left(\mathrm{NO} _{3}\right) _{2}+5 \mathrm{H} _{2} \mathrm{O}+\mathrm{N} _{2} \mathrm{O} $$

b) NO $\rightarrow$ Nitric oxide

$$ \ddot{\mathrm{N}} \equiv \underset{0}{0} \leftrightarrow \stackrel{+}{\mathrm{N}} \equiv \underset{0}{(-)} $$

Prepared by the action of dil. $\mathrm{HNO} _{3}$ on a heavier metal on heating.

$$ 3 \mathrm{Cu}(\mathrm{s})+8 \mathrm{HNO} _{3} \xrightarrow{\Delta} 3 \mathrm{Cu}\left(\mathrm{NO} _{3}\right) _{2}+4 \mathrm{H} _{2} \mathrm{O}+2 \mathrm{NO} $$

c) $\mathrm{N} _{2} \mathrm{O} _{3} \rightarrow$ Dinitrogen tetroxide

d) $\mathrm{NO} _{2} \rightarrow$ Nitrogen dioxide

Odd $\mathrm{e}^{-} \rightarrow$ gets excited in the visible region-Reddish Brown colour - Paramagnetic Any metal+ hot \& con. $\mathrm{HNO} _{3} \rightarrow \mathrm{NO} _{2}$

$$ \underset{\text { hot \& conc. }}{\mathrm{Cu}(\mathrm{s})}+\underset{\mathrm{HNO} _{3}}{\mathrm{Cu}\left(\mathrm{NO} _{3}\right) _{2}}+2 \mathrm{H} _{2} \mathrm{O}+2 \mathrm{NO} _{2} $$

e) $\mathrm{N} _{2} \mathrm{O} _{4}$ (dimer of $\left.\mathrm{NO} _{2}\right) \rightarrow$ dinitrogen tetraoxide

anhydride of $\mathrm{HNO} _{2} \& \mathrm{HNO} _{3}$

f) $\mathrm{N} _{2} \mathrm{O} _{5} \rightarrow$ Dinitrogen pentoxide

$\underset{\text { (conc.) }}{2 \mathrm{HNO} _{3}} \xrightarrow{\mathrm{P} _{4} \mathrm{O} _{10}} \mathrm{~N} _{2} \mathrm{O} _{5}+\mathrm{H} _{2} \mathrm{O}$

Anhydride of $\mathrm{HNO} _{3}$

B) Oxides of phosphorus

Phosphorus trioxide $\mathrm{P} _{2} \mathrm{O} _{3}$ or $\mathrm{P} _{4} \mathrm{O} _{6}$

$\mathrm{P} _{4}+3 \mathrm{O} _{2} \xrightarrow{\Delta} \mathrm{P} _{4} \mathrm{O} _{6}$

white Limited

$\mathrm{P} _{4} \mathrm{O} _{6}+6 \mathrm{H} _{2} \mathrm{O} \rightarrow 4 \mathrm{H} _{3} \mathrm{PO} _{3}$

Orthophosphorus acid

Phosphorus pentoxide

$\mathrm{P} _{4(\mathrm{~s})}+\underset{\text { excess }}{5 \mathrm{O} _{2} \xrightarrow{\Delta}} \mathrm{P} _{4} \mathrm{O} _{10}$ i) $\mathrm{P} _{4} \mathrm{O} _{10}+6 \mathrm{H} _{2} \mathrm{O} \rightarrow 4 \mathrm{H} _{3} \mathrm{PO} _{3}$

Orthophosphoric acid

ii) $\mathrm{P} _{4} \mathrm{O} _{10} \rightarrow$ dehydrating agent

$$ 2 \mathrm{HNO} _{3} \xrightarrow{\mathrm{P} _{4} \mathrm{O} _{10}} \mathrm{~N} _{2} \mathrm{O} _{5}+\mathrm{H} _{2} \mathrm{O} $$

HALIDES

A) Halides of N

• $\mathrm{NF} _{3}$

• $\mathrm{NF} _{3}$ is an endothermic species $\mathrm{N}-\mathrm{F}$ bond is stable due to inter electronic repulsion.

• $\mathrm{NF} _{3}$ does not get hydrolysed; but $\mathrm{NCl} _{3}$ gets hydrolysed :

• In $\mathrm{NF} _{3}$, both ’ $\mathrm{N}$ ’ \& ’ $\mathrm{F}$ ’ do not have vacant ’d’ orbitals to receive electrons from water molecules.

• $\mathrm{NCl} _{3}+\mathrm{H} _{2} \mathrm{O} \rightarrow \mathrm{NH} _{3}+\mathrm{HOCl}$

• Nitrogen does not form a pentahalide whereas $\mathrm{PCl} _{5}$ exists.

In ’ $\mathrm{PCl} _{5}{ }^{\prime} \rightarrow \mathrm{P} \rightarrow$ undergoes $\mathrm{sp}^{3} \mathrm{~d}$ hybridisation, but in ’ $\mathrm{N}$ ’ there is no vacant ’ $\mathrm{d}$ ’ orbital.

Maximum covalency of ’ $N$ ’ is 4.

B) Halides of P

• Trihalides of phosphorous

$\mathrm{PF} _{3}, \mathrm{PCl} _{3}, \mathrm{PBr} _{3}, \mathrm{Pl} _{3}$ does not exist

• Penthalides of phosphorous

$\mathrm{PCl} _{5} \rightarrow \mathrm{sp}^{3} \mathrm{~d}$

axial bonds are longer than the equatorial bonds due to bp-bp repulsions.

• $\mathrm{PCl} _{5}$ fumes in moist air

$$ \mathrm{PCl} _{5}+\underset{\text { White fumes }}{\mathrm{HOH}} \rightarrow \mathrm{HCl}+\mathrm{POCl} _{3} $$

• $\mathrm{PCl} _{5}$ is a covalent compound but in solid state it conducts electricity $\mathrm{PCl} _{5}$ in solid state dimerises as follows

$$ \mathrm{PCl} _{5}+\mathrm{PCl} _{5} \rightarrow\left[\mathrm{PCl} _{4}\right]^{+}\left[\mathrm{PCl} _{6}\right]^{-} $$

• Reactions of $\mathrm{PCl} _{5}$

$$ \mathrm{PCl} _{5}+4 \mathrm{H} _{2} \mathrm{O} \rightarrow 5 \mathrm{HCl}+\mathrm{H} _{3} \mathrm{PO} _{4} \text { (on complete hydrolysis) } $$

• $\mathrm{PCl} _{5}$ is good chlorinating agent.

$$ \begin{aligned} & \mathrm{PCl} _{5}+\mathrm{Zn} \rightarrow \mathrm{PCl} _{3}+\mathrm{ZnCl} _{2} \\ & \mathrm{CH} _{3} \mathrm{CH} _{2} \mathrm{OH}+\mathrm{PCl} _{5} \rightarrow \mathrm{CH} _{3} \mathrm{CH} _{2} \mathrm{Cl}+\mathrm{POCl} _{3} \end{aligned} $$

Oxides $\&$ acidic strength of the oxides.

1. Acidic strength $\propto$ oxidation state of central atom

2. Acidic strength $\propto \dfrac{1}{\text { size of central atom }}$

3. Acidic strength $\propto$ the E.N of central atom

A) Oxo-Acids of nitrogen

Nitrous acid and nitric acid are two main oxoacids of Nitrogen. Nitrous acid $\left(\mathrm{HNO} _{2}\right)$ is unstable and occurs only in aqueous solution. It acts as oxidizing and reducing agent and possesses complex forming ability, Nitric acid $\left(\mathrm{HNO} _{3}\right.$ ) forms a part of aqua regia and acts as a strong oxidizing agent. It is manufactured by Ostwald’s process.

Nitrous acid $\mathrm{HNO} _{2}$

• $\mathrm{H}-\mathrm{O}-\mathrm{N}=0$
• Prepared by the action of any nitrite with dil. $\mathrm{H} _{2} \mathrm{SO} _{4}$ or dil $\mathrm{HCl}$

$$ \mathrm{NaNO} _{2}+\mathrm{H} _{2} \mathrm{SO} _{4} \rightarrow \mathrm{HNO} _{2}+\mathrm{NaHSO} _{4} $$

• $\mathrm{HNO} _{2} \rightarrow$ Not stable

On standing it disproportionates

$$ \begin{aligned} & 3 \mathrm{HNO} _{2} \rightarrow \mathrm{HNO} _{3}+2 \mathrm{NO}+\mathrm{H} _{2} \mathrm{O} \\ & +3 \end{aligned}+5+2 $$

• Test of $\mathrm{NO} _{2}{ }^{-}$(reducing agent)

$$ \begin{aligned} & \mathrm{MnO} _{4}^{-}+\mathrm{NO} _{2}^{–}+\mathrm{H}^{+} \rightarrow \underset{\text { Colourless }}{\mathrm{Mn}^{2+}}+\mathrm{NO} _{3}^{-}+\mathrm{H} _{2} \mathrm{O} \\ & \mathrm{Cr} _{2} \mathrm{O} _{7}^{2-}+\mathrm{NO} _{2}^{-}+\mathrm{H}^{+} \rightarrow \underset{\text { orange }}{\mathrm{Cr}^{3+}}+\mathrm{NO} _{3}^{–}+\mathrm{H} _{2} \mathrm{O} \end{aligned} $$

Nitric acid $\left(\mathrm{HNO} _{3}\right)$

• Prepared by the action of hot \& conc. $\mathrm{H} _{2} \mathrm{SO} _{4}$ on a nitrate salt.

$$ \mathrm{NaNO} _{3}+\mathrm{H} _{2} \mathrm{SO} _{4} \rightarrow \mathrm{NaHSO} _{4}+\mathrm{HNO} _{3} $$

$$ \text { hot \& cone. } $$

Commercial method.

Ostwald’s method

$$ \begin{aligned} & 4 \mathrm{NH} _{3}+5 \mathrm{O} _{2} \xrightarrow[525 \mathrm{~K}]{\mathrm{Pt}} 4 \mathrm{NO}+6 \mathrm{H} _{2} \mathrm{O} \\ & 2 \mathrm{NO}+\mathrm{O} _{2} \rightarrow 2 \mathrm{NO} _{2} \\ & 2 \mathrm{NO} _{2}+\mathrm{H} _{2} \mathrm{O} \rightarrow \mathrm{HNO} _{3}+\mathrm{HNO} _{2} \end{aligned} $$

• Specific reactions of $\mathrm{HNO} _{3}$

l) Aqua regia mix of [conc. $\mathrm{HCl } \&$ [Conc. $\mathrm{HNO} _{3}$ ] in the ratio $3:$ I

It dissolves Au, Pt, etc.

$$ \begin{aligned} & 3 \mathrm{HCl}+\mathrm{HNO} _{3} \rightarrow 2 \mathrm{H} _{2} \mathrm{O}+\mathrm{NOCl}+2[\mathrm{Cl}] \\ & \mathrm{Au}+3[\mathrm{Cl}] \rightarrow \mathrm{AuCl} _{3} \\ & \mathrm{AuCl} _{3}+\mathrm{HCl} \rightarrow \mathrm{H}\left[\mathrm{AuCl} _{4}\right] \\ & \text { chloroauric acio } \\ & \mathrm{Pt}+4[\mathrm{Cl}] \rightarrow \mathrm{PtCl} _{4} \\ & \mathrm{PtCl} _{4}+2 \mathrm{HCl} \rightarrow \underset{\text { chloroplatinic acid }}{\mathrm{H} _{2}\left[\mathrm{PtCl} _{6}\right]} \end{aligned} $$

II) $\mathrm{HNO} _{3}$ is a good oxidising agent

$$ \begin{aligned} & \mathrm{P} _{4}+20 \mathrm{HNO} _{3} \rightarrow 4 \mathrm{H} _{3} \mathrm{PO} _{4}+20 \mathrm{NO} _{2}+4 \mathrm{H} _{2} \mathrm{O} \\ & \mathrm{S} _{8}+48 \mathrm{HNO} _{3} \rightarrow 8 \mathrm{H} _{2} \mathrm{SO} _{4}+48 \mathrm{NO} _{2}+16 \mathrm{H} _{2} \mathrm{O} \\ & \mathrm{C}+4 \mathrm{HNO} _{3} \rightarrow \mathrm{CO} _{2}+4 \mathrm{NO} _{2}+2 \mathrm{H} _{2} \mathrm{O} \end{aligned} $$

III) On metals

i) Lighter metal + v.dil. $\mathrm{HNO} _{3} \rightarrow \mathrm{N} _{2} \mathrm{O}$

$$ 4 \mathrm{Sn}+10 \mathrm{HNO} _{3} \rightarrow 4 \mathrm{Sn}\left(\mathrm{NO} _{3}\right) _{2}+5 \mathrm{H} _{2} \mathrm{O}+\mathrm{N} _{2} \mathrm{O} $$

ii) Lighter metal + conc. $\mathrm{HNO} _{3}($ hot $) \rightarrow \mathrm{NO} _{2}$

$$ \mathrm{Zn}+4 \mathrm{HNO} _{3} \rightarrow \mathrm{Zn}\left(\mathrm{NO} _{3}\right) _{2}+2 \mathrm{H} _{2} \mathrm{O}+2 \mathrm{NO} _{2} $$

iii) Heavier metal + Conc. $\mathrm{HNO} _{3}($ hot $) \rightarrow \mathrm{NO} _{2}$

$$ \mathrm{Cu}+4 \mathrm{HNO} _{3} \rightarrow \mathrm{Cu}\left(\mathrm{NO} _{3}\right) _{2}+2 \mathrm{H} _{2} \mathrm{O}+2 \mathrm{NO} _{2} $$

iv) Heavier metal + dil $\mathrm{HNO} _{3} \rightarrow \mathrm{NO}$

$$ 3 \mathrm{Cu}+8 \mathrm{HNO} _{3} \rightarrow 3 \mathrm{Cu}\left(\mathrm{NO} _{3}\right) _{2}+4 \mathrm{H} _{2} \mathrm{O}+2 \mathrm{NO} $$

Test for $\mathrm{NO} _{3}^{-}$ion

Brown ring test

Salt + freshly prepared $\mathrm{FeSO} _{4}+$ Conc. $\mathrm{H} _{2} \mathrm{SO} _{4} \rightarrow$ Reddish brown

$\mathrm{NO} _{3}^{-}+\mathrm{Fe}^{2+} \rightarrow \mathrm{Fe}^{3+}+\mathrm{NO}$

$\mathrm{Fe}^{2+}+5 \mathrm{H} _{2} \mathrm{O}+\mathrm{NO} \rightarrow \underset{\text { (brown complex) }}{\left[\mathrm{Fe}\left(\mathrm{H} _{2}\right) \mathrm{C} _{2} \mathrm{NO}^{+2}\right.}$

From ’ $\mathrm{NO}$ ’ 11- undergoes a transfer to $\mathrm{Fe}^{2+} \&$ this electronic transition causes the brown colour.

B) Oxoacids of Phosphorus

• All the oxoacids have phosphorus atom or atoms bonded tetrahedrally to four other atoms or groups.

• These atoms contain at least one $\mathrm{P}=0$ unit and one $\mathrm{P}-\mathrm{OH}$ group. The $\mathrm{P}-\mathrm{OH}$ group is ionisable giving a proton. The number of $\mathrm{P}-\mathrm{OH}$ groups determines the basicity of the oxoacid.

• Acids which contain $\mathrm{P}-\mathrm{H}$ bonds have strong reducing properties.

• Hypophosphorus or Phosphinic acid $\left(\mathrm{H} _{3} \mathrm{PO} _{2}\right)$; Monobasic and reducing agent Orthophosphorus acid $\left(\mathrm{H} _{3} \mathrm{PO} _{3}\right)$ : Dibasic and reducing agent.

Orthophosphoric acid $\left(\mathrm{H} _{3} \mathrm{PO} _{4}\right)$ : Weak tribasic acid

• $\mathrm{H} _{3} \mathrm{PO} _{2}$ a good reducing agent,

$\mathrm{H} _{3} \mathrm{PO} _{2}+4 \mathrm{AgNO} _{3}+2 \mathrm{H} _{2} \mathrm{O} \rightarrow \mathrm{H} _{3} \mathrm{PO} _{4}+4 \mathrm{HNO} _{3}+4 \mathrm{Ag} \downarrow$

• $\mathrm{H} _{3} \mathrm{PO} _{3}$ on heating $\rightarrow$ disproportionation

$4 \mathrm{H} _{3} \mathrm{PO} _{3} \rightarrow \mathrm{PH} _{3}+3 \mathrm{H} _{3} \mathrm{PO} _{4}$

$+3 \quad-3 \quad+5$

Oxoacids of phosphorus are

i) $\mathrm{H} _{3} \mathrm{PO} _{2}$ (Hypophosphorus acid) : Reducing agent and monobasic

ii) $\mathrm{H} _{3} \mathrm{PO} _{3}$ (Orthophosphorus acid): Reducing agent and dibasic

iii) $\mathrm{H} _{3} \mathrm{PO} _{4}$ (Orthophosphoric acid) : Weak tribasic acid

iv) $\mathrm{H} _{4} \mathrm{P} _{2} \mathrm{O} _{7}$ (Pyrophosphoric acid) : It is obtained by heating to $220^{\circ} \mathrm{C}$. It is tetrabasic.

v) (Metaphosphoric acid) : It is formed by the dehydration of $\mathrm{H} _{3} \mathrm{PO} _{4}$ at $316^{\circ} \mathrm{C}$. Also exists as a trimer and is monobasic. vi) $\mathrm{H} _{4} \mathrm{P} _{2} \mathrm{O} _{6}$ (Hypophosphoric acid) : Tetrabasic

vii) $\mathrm{H} _{4} \mathrm{P} _{2} \mathrm{O} _{5}$ (Pyrophosphorous acid) : Dibasic acid

INDIVIDUAL MEMBERS

Dinitrogen $\left(\mathrm{N} _{2}\right)$

1. By heating $\left(\mathrm{NH} _{4}\right) _{2} \mathrm{Cr} _{2} \mathrm{O} _{7}$

$$ \left(\mathrm{NH} _{4}\right) _{2} \mathrm{Cr} _{2} \mathrm{O} _{7} \xrightarrow{\Delta} \mathrm{Cr} _{2} \mathrm{O} _{3}+4 \mathrm{H} _{2} \mathrm{O}+\mathrm{N} _{2} $$

2. By the action of water (or) dil acid on metal azides.

$$ 2 \mathrm{NaN} _{3} \xrightarrow{\Delta} \underset{\text { Purest } \mathrm{N} _{2} \text { sample }}{6 \mathrm{Na}+\mathrm{N} _{2}} $$

3. $\mathrm{NH} _{4} \mathrm{NO} _{2} \xrightarrow{\Delta} 2 \mathrm{H} _{2} \mathrm{O}+\mathrm{N} _{2}$

Ammonia

1) Lab. preparation

Any ammonium salt $+\mathrm{NaOH} \xrightarrow{\Delta} \mathrm{NH} _{3}$ $\mathrm{NH} _{4} \mathrm{Cl}+\mathrm{NaOH} \xrightarrow{\text { heat }} \mathrm{NaCl}+\mathrm{H} _{2} \mathrm{O}+\mathrm{NH} _{3}$

2) Haber’s process

Conditions - Acc. To Le-Chatelier’s principle

i) High pressure

ii) Low temperature (an optimum temp of $723 \mathrm{~K}$ to be maintained)

iii) Iron oxide catalyst with $\mathrm{K} _{2} \mathrm{O}$ and $\mathrm{Al} _{2} \mathrm{O} _{3}$ promoter

Other methods of preparations

$$ \begin{aligned} & \mathrm{AlN}+3 \mathrm{H} _{2} \mathrm{O} \rightarrow \mathrm{Al}(\mathrm{OH}) _{3}+\mathrm{NH} _{3} \\ & \underset{\text { urea }}{\mathrm{NH} _{2} \mathrm{CONH}}+2 \mathrm{NaOH} \rightarrow \mathrm{Na} _{2} \mathrm{CO} _{3}+2 \mathrm{NH} _{3} \\ & \mathrm{CaCl} _{2}+8 \mathrm{NH} _{3} \rightarrow \underset{\text { (Addition product) }}{\mathrm{CaCl} _{2}} .8 \mathrm{NH} _{3} \end{aligned} $$

Chemical properties of $\mathrm{NH} _{3}$

$$ \begin{aligned} & \begin{array}{l} 8 \underset{\text { (excess) }}{\mathrm{NH} _{3}}+3 \mathrm{Cl} _{2} \\ \end{array} \mathrm{N} _{2}+6 \mathrm{NH} _{4} \mathrm{Cl} \\ & \mathrm{NH} _{3}+\underset{\text { (excess) }}{3 \mathrm{Cl} _{2}} \rightarrow \mathrm{NCl} _{3}+3 \mathrm{HCl} \\ & 2 \mathrm{Na}+\mathrm{NH} _{3} \xrightarrow[\text { (sodamide) }]{575 \mathrm{~K}} 2 \mathrm{NaNH} _{2}+\mathrm{H} _{2} \\ & 2 \mathrm{NH} _{3}+\mathrm{CO} _{2} \rightarrow \underset{\text { (ammonium carbonate) }}{\left.\left[\mathrm{NH} _{2} \mathrm{COONH}\right] _{4}\right]} \rightarrow \underset{\text { (urea) }}{\mathrm{NH} _{2} \mathrm{CONH} _{2}}+\mathrm{H} _{2} \mathrm{O} \end{aligned} $$

$\mathrm{PH} _{3}$

Action of hot $\&$ conc. $\mathrm{NaOH}$ on $\mathrm{P} _{4}$ (white)

$$ \underset{\text { hot \& conc. }}{\mathrm{P} _{4}}+3 \mathrm{NaOH}+3 \mathrm{H} _{2} \mathrm{O} \rightarrow \mathrm{PH} _{3}+3 \mathrm{NaH} _{2} \mathrm{PO} _{2} $$

Action of $\mathrm{H} _{2} \mathrm{O}$ on $\mathrm{Ca} _{3} \mathrm{P} _{2}$

$\mathrm{Ca} _{3} \mathrm{P} _{2}+3 \mathrm{H} _{2} \mathrm{O} \rightarrow 3 \mathrm{Ca}(\mathrm{OH}) _{2}+2 \mathrm{PH} _{3}$

Solved Problems

1) The number of hydrogen atoms attached to phosphorus atom in hypophosporus acid is

1. zero

2. one

3. two

4. three

Show Answer

Answer: two

Hint: Structure of hypophosphorus acid:

It can be seen clearly in the structure that two atoms of hydrogen are attached to Phosphorus atom.

2) The decreasing order of bond angles from $\mathrm{NH} _{3}$ to $\mathrm{SbH} _{3}$ down the group 15 of the periodic table is due to

1. increasing $p$ character in $s p^{3}$

2. decreasing $\mid p-b p$ repulsion

3. decreasing electronegativity

4. increasing bp-bp repulsion

Show Answer

Answer: 3

Hint: As we move down the group, the size of the central atom goes on increasing and its electronegativity goes on decreasing. As a result, the bond pairs of electrons tend to lie away and away from the central atom as we move from $\mathrm{NH} _{3}$ to $\mathrm{SbH} _{3}$.

3) Which of the following statements is wrong?

1. The stability of hydrides increases from $\mathrm{NH} _{3}$ to $\mathrm{BiH} _{3}$ in group 15 of the periodic table.

2. Nitrogen cannot form $\mathrm{p} \pi-\mathrm{d} \pi$ bond.

3. Single N-N bond is weaker than the single P-P bond.

4. $\mathrm{N} _{2} \mathrm{O} _{4}$ has two resonance structures.

[AlEEE, 2011]

Show Answer

Answer: (1)

Hint: Statement 1 is wrong because as we move from $\mathrm{NH} _{3}$ to $\mathrm{BiH} _{3}$ the size of the central atom increases and therefore, its tendency to form stable covalent bond with small hydrogen atom decreases. Thus M-H bond strength decreases and therefore stability decreases.

PRACTICE QUESTIONS

Question 1- The number of $\mathrm{P}-0-\mathrm{P}$ bonds in cyclic metaphosphoric acid is

a) zero

b) two

c) three

d) four

Show Answer

Answer:- c

Question 2- Ammonia can be dried by

a) conc. $\mathrm{H} _{2} \mathrm{SO} _{4}$

b) $\mathrm{P} _{4} \mathrm{O} _{10}$

c) $\mathrm{CaO}$

d) anhydrous $\mathrm{CaCl} _{2}$

Show Answer

Answer:- d

Question 3- Polyphosphates are used as water softening agents because they

a) form soluble complexes with anionic species

b) precipitate anionic species

c) form soluble complexes with cationic species

d) precipitate cationic species

Show Answer

Answer:- c

Question 4- The bonds present in $\mathrm{N} _{2} \mathrm{O} _{5}$ are

a) only ionic

b) covalent and coordinate

c) only covalent

d) covalent and ionic

Show Answer

Answer:- b

Question 5- Which one of the following is the strongest base?

a) $\mathrm{AsH} _{3}$

b) $\mathrm{NH} _{3}$

c) $\mathrm{PH} _{3}$

d) $\mathrm{SbH} _{3}$

Show Answer

Answer:- b

Question 6- Which of the following oxides of nitrogen is a coloured gas?

a) $\mathrm{N} _{2} \mathrm{O}$

b) $\mathrm{NO}$

c) $\mathrm{N} _{2} \mathrm{O} _{4}$

d) $\mathrm{NO} _{2}$

Show Answer

Answer:- d

Question 7- Amongst the trihalides of nitrogen, which one is least basic?

a) $\mathrm{NF} _{3}$

b) $\mathrm{NCl} _{3}$

c) $\mathrm{NBr} _{3}$

d) $\mathrm{NI} _{3}$

Show Answer

Answer:- a

Question 8- The reaction of $P _{4}$ with $X$ leads selectively to $P _{4} O _{6}$. The $X$, is

a) dry $\mathrm{O} _{2}$

b) a mixture of $\mathrm{O} _{2}$ and $\mathrm{N} _{2}$

c) moist $\mathrm{O} _{2}$

d) $\mathrm{O} _{2}$ in the presence of aqueous $\mathrm{NaOH}$

Show Answer

Answer:- b

Question 9- Extra pure $\mathrm{N} _{2}$ can be obtained by heating

a) $\mathrm{NH} _{3}$ with $\mathrm{CuO}$

b) $\mathrm{NH} _{4} \mathrm{NO} _{3}$

c) $\left(\mathrm{NH} _{4}\right) _{2} \mathrm{Cr} _{2} \mathrm{O} _{7}$

d) $\mathrm{Ba}\left(\mathrm{N} _{3}\right) _{2}$

Show Answer

Answer:- d

Question 10- Which ordering of compounds is according to the decreasing order of the oxidation state of nitrogen? .

a) $\mathrm{HNO} _{3}, \mathrm{NO}, \mathrm{NH} _{4} \mathrm{Cl}, \mathrm{N} _{2}$

b) $\mathrm{HNO} _{3}, \mathrm{NO}, \mathrm{N} _{2}, \mathrm{NH} _{4} \mathrm{Cl}$

c) $\mathrm{HNO} _{3}, \mathrm{NH} _{4} \mathrm{Cl}, \mathrm{NO}, \mathrm{N} _{2}$

d) $\mathrm{NO}, \mathrm{HNO} _{3}, \mathrm{NH} _{4} \mathrm{Cl}, \mathrm{N} _{2}$

Show Answer

Answer:- b

Group 16 elements

The elements oxygen (0), Sulphur (S), Sellenium (Se), Tellurium (Te) and Po (Polonium) are members of group 16 (Chalcogens). The general valence shell electronic configuration of these elements is $n s^{2} n p^{4}$.

All the elements of this group show allotropy. Oxygen exists in two forms, $\mathrm{O} _{2}$ and $\mathrm{O} _{3}$. Sulphur exist in several non metallic forms i.e. Rhombic, Monoclinic and Plastic sulphur.

The oxidation state of oxygen is -2 in most compounds (except peroxides) and the tendency of these elements to show -2 oxidation state decreases down the group.

Oxygen does not show positive oxidation state except in $\mathrm{OF} _{2}$. The other elements show oxidation states $+2,+4,+6$. Oxidation state decreases due to inert pair effect.

GENERAL TRENDS IN PHYSICAL PROPERTIES

1. Size increase on moving down the group and ionization enthalpy decreases.
2. Electron gain enthalpy

$\mathrm{S}>\mathrm{Se}>\mathrm{Te}>\mathrm{P} 0>0$

‘S’ has largest - E.G.E. value because oxygen has lower E.G.E due its small size and inter electronic repulsion.

3. Electronegativity

$0>\mathrm{S}>\mathrm{Se}>\mathrm{Te}>\mathrm{Po}$ (regular trend)

4. Catenation property

This property is maximum with ’ $\mathrm{S}$ ‘.

5. The stability of +6 state decreases and +4 increases due to the inert pair effect.

Chemical Properties

Hydrides:

All the elements of group 16 from hydrides of the general formula $\mathrm{H} _{2} \mathrm{E}$ where $\mathrm{E}$ is the element belonging to group 16. Following are some of the characteristics of these hydrides:

$$ \mathrm{H} _{2} \mathrm{O}, \mathrm{H} _{2} \mathrm{~S}, \mathrm{H} _{2} \mathrm{Se}, \mathrm{H} _{2} \mathrm{Te} $$

a) Boiling point

$$ \mathrm{H} _{2} \mathrm{~S}<\mathrm{H} _{2} \mathrm{Se}<\mathrm{H} _{2} \mathrm{Te}<\mathrm{H} _{2} \mathrm{O} $$

b) Bond angles

$$ \mathrm{H} _{2} \mathrm{O}>\mathrm{H} _{2} \mathrm{~S}>\mathrm{H} _{2} \mathrm{Se}>\mathrm{H} _{2} \mathrm{Te} $$

c) Acidic strength or reducing nature

$$ \mathrm{H} _{2} \mathrm{Te}>\mathrm{H} _{2} \mathrm{Se}>\mathrm{H} _{2} \mathrm{~S}>\mathrm{H} _{2} \mathrm{O} $$

Oxides : These elements form monoxides ( $\mathrm{MO})$, dioxides $\left(\mathrm{MO} _{2}\right)$. Ozone $\left(\mathrm{O} _{3}\right.$ ) and sulphur dioxide $\left(\mathrm{SO} _{2}\right)$ are gases while selenium dioxide $\left(\mathrm{SeO} _{2}\right)$ is solid. Reducing property of dioxide decreases from $\mathrm{SO} _{2}$ to $\mathrm{TeO} _{2} ; \mathrm{SO} _{2}$ is reducing while $\mathrm{TeO} _{2}$ is an oxidizing agent. Besides $\mathrm{MO} _{2}$ type oxides sulphur, selenium and tellurium also form $\mathrm{MO} _{3}$ type oxides $\left(\mathrm{SO} _{3}, \mathrm{SeO} _{3}, \mathrm{TeO} _{3}\right)$. Both types of oxides are acidic in nature. Increasing order of acidic nature of oxides is $\mathrm{TeO} _{3}<\mathrm{SeO} _{3}<\mathrm{SO} _{3}$

Oxo-acids of sulphur

1) Sulphurous acid series

a) $\mathrm{H} _{2} \mathrm{SO} _{3} \quad \mathrm{~S}(\mathrm{IV})$ sulphurous acid

b) $\mathrm{H} _{2} \mathrm{~S} _{2} \mathrm{O} _{5} \quad \mathrm{~S}(\mathrm{~V})$ disulphurous acid or

$\hspace{2cm}\mathrm{S}$ (III) pyrosulpurous acid

c) $\mathrm{H} _{2} \mathrm{~S} _{2} \mathrm{O} _{4} \quad \mathrm{~S}(\mathrm{III})$ dithionous acid

2) Sulphuric acid series

a) $\mathrm{H}_2 \mathrm{SO}_4 \quad \mathrm{~S}(\mathrm{VI})$ sulphuric acid

b) $\mathrm{H}_2 \mathrm{~S}_2 \mathrm{O}_3 \begin{array}{ll} & \mathrm{~S}(\mathrm{IV}) \\ & \mathrm{S}(-\mathrm{II})\end{array}$ thiosulphuric acid

c) $\mathrm{H}_2 \mathrm{~S}_2 \mathrm{O}_7 \quad \mathrm{~S}(\mathrm{VI})$ pyrosulphuric acid

3) Peroxo acid series

a) $\mathrm{H} _{2} \mathrm{SO} _{5} \quad \mathrm{~S}(\mathrm{VI})$ $\underset{\text{(Caro’s acid)}}{\text{peroxomonosulphuric acid}}$

b) $\mathrm{H} _{2} \mathrm{~S} _{2} \mathrm{O} _{8} \quad \mathrm{~S}(\mathrm{VI}) \underset{\text{(Marshall’s acid)}}{\text{peroxodisulphuric acid}}$

OXIDES OF SULPHUR

$\mathrm{SO} _{2}$ (anhydride of $\mathrm{H} _{2} \mathrm{SO} _{3}$ )

$\mathrm{SO} _{3}$ (anhydride of $\mathrm{H} _{2} \mathrm{SO} _{4}$ )

$\mathrm{SO} _{2}$ is produced in large scale

$$ \begin{aligned} & \mathrm{S}+\mathrm{O} _{2} \xrightarrow{\Delta} \mathrm{SO} _{2} \\ & 2 \mathrm{H} _{2} \mathrm{~S}+3 \mathrm{O} _{3} \rightarrow 3 \mathrm{H} _{2} \mathrm{O}+2 \mathrm{SO} _{2} \\ & 2 \mathrm{ZnS}+3 \mathrm{O} _{2} \rightarrow 2 \mathrm{ZnO}+2 \mathrm{SO} _{2} \end{aligned} $$

Properties of $\mathrm{SO} _{2}$

a) As a reducing agent (with $\mathrm{Cr} _{2} \mathrm{O} _{7}{ } _{7}^{2-}$ )

$$ \begin{aligned} & \mathrm{Cr} _{2} \mathrm{O} _{7}^{2-}+14 \mathrm{H}^{+}+6 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H} _{2} \mathrm{O} \\ & \mathrm{SO} _{2}+2 \mathrm{H} _{2} \mathrm{O} \rightarrow \mathrm{SO} _{4}^{2-}+4 \mathrm{H}^{+}+2 \mathrm{e}^{-} \end{aligned} $$

b) It reduces $\mathrm{MnO} _{4}^{-}$in acid medium.

$$ \begin{aligned} & \mathrm{MnO} _{4}^{-}+16 \mathrm{H}^{+}+5 \mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2+}+8 \mathrm{H} _{2} \mathrm{O} \\ & \mathrm{SO} _{2}+2 \mathrm{H} _{2} \mathrm{O} \rightarrow \mathrm{SO} _{4}^{2-}+4 \mathrm{H}+2 \mathrm{e}^{-} \end{aligned} $$

Properties of $\mathrm{SO} _{3}$

Saturated with $\mathrm{H} _{2} \mathrm{SO} _{4}$ is called oleum.

$$ \mathrm{H} _{2} \mathrm{SO} _{4}+\mathrm{SO} _{3} \rightarrow \mathrm{H} _{2} \mathrm{~S} _{2} \mathrm{O} _{7} $$

At room temperature $\mathrm{SO} _{3}$ is a solid.

$\mathrm{H} _{2} \mathrm{SO} _{4}$

$\mathrm{H} _{2} \mathrm{SO} _{4}$ is manufactured by the Contact process.

$2 \mathrm{SO} _{2}+\mathrm{O} _{2} \rightleftharpoons \stackrel{\mathrm{V} _{2} \mathrm{O} _{5}}{\rightleftharpoons} 2 \mathrm{SO} _{3}+\Delta \mathrm{H}$

$\mathrm{SO} _{3}+\mathrm{H} _{2} \mathrm{SO} _{4} \rightarrow \mathrm{H} _{2} \mathrm{~S} _{2} \mathrm{O} _{7}$

$\mathrm{H} _{2} \mathrm{~S} _{2} \mathrm{O} _{7}+\mathrm{H} _{2} \mathrm{O} \rightarrow 2 \mathrm{H} _{2} \mathrm{SO} _{4}$

$\mathrm{H} _{2} \mathrm{SO} _{4}$ is an oxidizing agent

$$ \begin{aligned} & \mathrm{SO} _{4}^{2-}+4 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{SO} _{2}+2 \mathrm{H} _{2} \mathrm{O} \\ & \mathrm{Cu}+2 \mathrm{H} _{2} \mathrm{SO} _{4} \xrightarrow{\Delta} \mathrm{CuSO} _{4}+\mathrm{SO} _{2}+2 \mathrm{H} _{2} \mathrm{O} \\ & \mathrm{S}+2 \mathrm{H} _{2} \mathrm{SO} _{4} \rightarrow 2 \mathrm{H} _{2} \mathrm{O}+3 \mathrm{SO} _{2} \\ & 2 \mathrm{HBr}+\mathrm{H} _{2} \mathrm{SO} _{4} \rightarrow \mathrm{Br} _{2}+\mathrm{SO} _{2}+2 \mathrm{H} _{2} \mathrm{O} \\ & \mathrm{C}+2 \mathrm{H} _{2} \mathrm{SO} _{4} \rightarrow \mathrm{CO} _{2}+2 \mathrm{H} _{2} \mathrm{O} \end{aligned} $$

$\mathrm{H} _{2} \mathrm{SO} _{4}$ is a dehydrating agent.

$$ \begin{aligned} & \mathrm{C} _{6} \mathrm{H} _{12} \mathrm{O} _{6} \xrightarrow{\text { Conc. } \mathrm{H} _{2} \mathrm{SO} _{4}} 6 \mathrm{H} _{2} \mathrm{O}+6 \mathrm{C} \\ & \mathrm{C} _{12} \mathrm{H} _{22} \mathrm{O} _{11} \xrightarrow{\text { Conc. } \mathrm{H} _{2} \mathrm{SO} _{4}} 11 \mathrm{H} _{2} \mathrm{O}+12 \mathrm{C} \end{aligned} $$

HALIDES

The elements of group 16 form a number of monohalides, dihalides, tetrahalides and hexahalides. Whereas $\mathrm{SF} _{6}$ is trigonal bipyramidal. Further, $\mathrm{SF} _{4}$ acts as a Lewis base and easily undergoes hydrolysis but $\mathrm{SF} _{6}$ does not undergo hydrolysis. Since it is sterically protected by 6 fluorine atoms.

Individual members

• Dioxygen $\left(\mathrm{O} _{2}\right)$

Preparation of dioxygen

a) By the decomposition of oxygen rich compounds

$$ \begin{aligned} & 2 \mathrm{KMnO} _{4} \rightarrow \mathrm{K} _{2} \mathrm{MnO} _{4}+\mathrm{MnO} _{2}+\mathrm{O} _{2} \\ & 2 \mathrm{KClO} _{3} \xrightarrow{\mathrm{MnO} _{2}} 2 \mathrm{KCl}+3 \mathrm{O} _{2} \\ & 2 \mathrm{KNO} _{3} \rightarrow 2 \mathrm{KNO} _{2}+\mathrm{O} _{2} \end{aligned} $$

b) By heating peroxides $\&$ higher oxides

$$ \begin{aligned} & 2 \mathrm{HgO} \xrightarrow{\Delta} 2 \mathrm{Hg}+\mathrm{O} _{2} \\ & 2 \mathrm{Ag} _{2} \mathrm{O} \xrightarrow{\Delta} 4 \mathrm{Ag}+\mathrm{O} _{2} \\ & 3 \mathrm{MnO} _{2} \xrightarrow{\Delta} \mathrm{Mn} _{3} \mathrm{O} _{4}+\mathrm{O} _{2} \\ & 2 \mathrm{~Pb} _{3} \mathrm{O} _{4} \xrightarrow{\Delta} 6 \mathrm{PbO}+\mathrm{O} _{2} \\ & 2 \mathrm{~Pb} _{3} \mathrm{O} _{4} \xrightarrow{\Delta} 6 \mathrm{PbO}+\mathrm{O} _{2} \\ & 2 \mathrm{BaO} _{2} \xrightarrow{\Delta} 2 \mathrm{BaO}+\mathrm{O} _{2} \end{aligned} $$

Laboratory preparation of dioxygen

i) $2 \mathrm{Na} _{2} \mathrm{O} _{2}+2 \mathrm{H} _{2} \mathrm{O} \rightarrow 4 \mathrm{NaOH}+\mathrm{O} _{2}$

ii) $2 \mathrm{KMnO} _{4}+3 \mathrm{H} _{2} \mathrm{SO} _{4} \rightarrow \mathrm{K} _{2} \mathrm{SO} _{4}+2 \mathrm{MnSO} _{4}+3 \mathrm{H} _{2} \mathrm{O}+5[0]$

iii) $2 \mathrm{KMnO} _{4}+5 \mathrm{Na} _{2} \mathrm{O} _{2}+8 \mathrm{H} _{2} \mathrm{SO} _{4} \rightarrow \mathrm{K} _{2} \mathrm{SO} _{4}+5 \mathrm{Na} _{2} \mathrm{SO} _{4}+2 \mathrm{MnSO} _{4}+8 \mathrm{H} _{2} \mathrm{O}+5 \mathrm{O} _{2}$

- Ozone

1) Natural formation

uv - rays

$\mathrm{3O} _{2} \rightleftharpoons 2 \mathrm{O} _{3}$

2) Preparation of ozone

$3 \mathrm{O} _{2} \xrightarrow[\text { discharge }]{\text { electric }} 2 \mathrm{O} _{3} \Delta \mathrm{H}=+284.5 \mathrm{~kJ} / \mathrm{mol}$

Chemical properties of $\mathrm{O} _{3}$

1. Decomposition $2 \mathrm{O} _{3} \xrightarrow{473 \mathrm{~K}} 3 \mathrm{O} _{2}$
2. Ozone is an oxidising agent

$$ \mathrm{O} _{3} \rightarrow \mathrm{O} _{2}+0 $$

Some examples

$$ \begin{aligned} & \mathrm{PbS}+4 \mathrm{O} _{3} \rightarrow \mathrm{PbSO} _{4}+4 \mathrm{O} _{2} \\ & 2 \mathrm{HCl}+\mathrm{O} _{3} \rightarrow \mathrm{H} _{2} \mathrm{O}+\mathrm{Cl} _{2}+\mathrm{O} _{2} \\ & \mathrm{KNO} _{2}+\mathrm{O} _{3} \rightarrow \mathrm{KNO} _{3}+\mathrm{O} _{2} \\ & 2 \mathrm{FeSO} _{4}+\mathrm{H} _{2} \mathrm{SO} _{4}+\mathrm{O} _{3} \rightarrow \mathrm{Fe} _{2}\left(\mathrm{SO} _{4}\right) _{3}+\mathrm{H} _{2} \mathrm{O}+\mathrm{O} _{2} \end{aligned} $$

Detection of Ozone

The sample of air containing ozone is treated with excess of $\mathrm{KI} _{(a q)}$. The formed $\mathrm{I} _{2}$ is volumetrically titrated with $\mathrm{Na} _{2} \mathrm{~S} _{2} \mathrm{O} _{3}$. From the volume of $\mathrm{Na} _{2} \mathrm{~S} _{2} \mathrm{O} _{3}, \mathrm{O} _{3}$ present in the sample of air can be detected.

Solved questions

1) Which form of sulphur show paramagnetism?

a) $\mathrm{S} _{8}$

b) $\mathrm{S} _{6}$

c) $\mathrm{S} _{2}$

d) none of these

Show Answer

Answer: c

Hint: In vapour state sulphur partly exists as $S _{2}$ molecules which has 2 unpaired electrons in the antibonding $\pi^{\star}$ orbitals like $\mathrm{O} _{2}$ and hence exhibits paramagnetism.

2) Which of the following is liquid?

a) $\mathrm{H} _{2} \mathrm{~S}$

b) $\mathrm{H} _{2} \mathrm{Se}$

c) $\mathrm{H} _{2} \mathrm{Te}$

d) $\mathrm{H} _{2} \mathrm{O}$

Show Answer

Answer: d

The electronegativity difference is maximum in case of $\mathrm{H} \& \mathrm{O}$ thus $\mathrm{H}$-bonding takes place between the molecules hence making water a liquid.

PRACTICE QUESTIONS

Question 1- There is no $S$ - $S$ bond is

a) $\mathrm{S} _{2} \mathrm{O} _{4}{ }^{2-}$

b) $\mathrm{S} _{2} \mathrm{O} _{5}{ }^{2}$

c) $\mathrm{S} _{2} \mathrm{O} _{3}{ }^{2-}$

d) $\mathrm{S} _{2} \mathrm{O} _{7}{ }^{2-}$

Show Answer

Answer:- d

Question 2- The correct order of acidic strength is

a) $\mathrm{Cl} _{2} \mathrm{O} _{7}>\mathrm{SO} _{2}>\mathrm{P} _{4} \mathrm{O} _{10}$

b) $\mathrm{CO} _{2}>\mathrm{N} _{2} \mathrm{O} _{5}>\mathrm{SO}_3$

c) $\mathrm{Na} _{2} \mathrm{O}>\mathrm{MgO}>\mathrm{Al} _{2} \mathrm {O} _{3}$

d) $\mathrm{K} _{2} \mathrm{O}>\mathrm{CaO}>\mathrm{MgO}$

Show Answer

Answer:- a

Question 3- Amongst $\mathrm{H} _{2} \mathrm{O}, \mathrm{H} _{2} \mathrm {~S}, \mathrm{H} _{2} \mathrm{Se}$ and $\mathrm{H} _{2} \mathrm{Te}$, the one with the highest boiling point is

a) $\mathrm{H} _{2} \mathrm{O}$ because of hydrogen bonding

b) $\mathrm{H} _{2}$ Te because of higher molecular weight

c) $\mathrm{H} _{2} \mathrm{~S}$ because of hydrogen bonding

d) $\mathrm{H} _{2} \mathrm{Se}$ because of lower lower molecular weight

Show Answer

Answer:- a

Question 4- Aqueous solution of $\mathrm{Na} _{2} \mathrm{~S} _{2} \mathrm{O} _{3}$ on reaction with $\mathrm{Cl} _{2}$ gives

a) $\mathrm{Na} _{2} \mathrm{~S} _{4} \mathrm{O} _{6}$

b) $\mathrm{NaHSO} _{4}$

c) $\mathrm{NaCl}$

d) $\mathrm{NaOH}$

Show Answer

Answer:- b

Question 5- The number of $\mathrm{S}-\mathrm{S}$ bonds in sulphur trioxide
trimer, $\left(\mathrm{S} _{3} \mathrm{O} _{9}\right)$ is

a) three

b) two

c) one

d) zero

Show Answer

Answer:- d

Question 6- Which of the following is not oxidised by $\mathrm{O} _{3}$ ?

a) $\mathrm{KI}$

b) $\mathrm{FeSO} _{4}$

c) $\mathrm{KMnO} _{4}$

d) $\mathrm{K} _{2} \mathrm{MnO} _{4}$

Show Answer

Answer:- c

Question 7- Identify, the correct order of acidic strength of $\mathrm{CO} _ {2}, \mathrm{CuO}, \mathrm{CaO}, \mathrm{H} _{2} \mathrm{O}$

a) $\mathrm{CaO}<\mathrm{CuO}<\mathrm{H} _{2} \mathrm{O}<\mathrm{CO} _{2}$

b) $\mathrm{H} _{2} \mathrm{O}<\mathrm{CuO}<\mathrm{CaO}<\mathrm{CO} _{2}$

b) $\mathrm{CaO}<\mathrm{H} _{2} \mathrm{O}<\mathrm{CuO}<\mathrm{CO} _{2}$

d) $\mathrm{H} _{2} \mathrm{O}<\mathrm{CO} _{2}<\mathrm{CaO}<\mathrm{CuO}$

Show Answer

Answer:- a

Question 8- Identify, the correct order of solubility of $\mathrm{Na} _{2} \mathrm{~S}, \mathrm{CuS}$ and $\mathrm{ZnS}$ in aqueous medium

a) $\mathrm{CuS}>\mathrm{ZnS}>\mathrm{Na} _{2} \mathrm{~S}$

b) $\mathrm{ZnS}>\mathrm{Na} _{2} \mathrm{~S}>\mathrm{CuS}$

c) $\mathrm{Na} _{2} \mathrm{~S}>\mathrm{CuS}>\mathrm{ZnS}$

d) $\mathrm{Na} _{2} \mathrm{~S}>\mathrm{ZnS}>\mathrm{CuS}$

Show Answer

Answer:- d

Question 9- Which of the following has $-0-0$ - linkage?

a) $\mathrm{H} _{2} \mathrm{~S} _{2} \mathrm{O} _{6}$

b) $\mathrm{H} _{2} \mathrm{~S} _{2} \mathrm{O} _{8}$

c) $\mathrm{H} _{2} \mathrm{~S} _{2} \mathrm{O} _{3}$

d) $\mathrm{H} _{2} \mathrm{~S} _{4} \mathrm{O} _{6}$

Show Answer

Answer:- b

Question 10- A gas that cannot be collected over water is

a) $\mathrm{N} _{2}$

b) $\mathrm{O} _{2}$

c) $\mathrm{SO} _{2}$

d) $\mathrm{PH} _{3}$

Show Answer

Answer:- c

Group 17 elements

Fluorine (F), Chlorine (Cl), Bromine (Br), lodine (I) and Astatine (At) are members of halogen family (Group 17). The general valence shell electronic configuration of these elements is $n s^{2} n p^{5}$. They all are nonmetallic and are collectively known as halogens.

Trends in Properties

1) Physical state

2) Electronic configutation

$n s^{2}, n p^{5}\left(n s^{2} ; n p _{x}^{2}, n p _{y}^{2}, n p _{z}{ }^{1}\right)$

Largest negative E.GE.

3) Ionization enthalpy (Endothermic) $\Delta \mathrm{H}=+\mathrm{ve}$

Halogens have very high I. E values. They need just one electron for stable configuration. Removal of an electron is energetically favourable.

4) Electron gain enthalpy [Exothermic] $\Delta \mathrm{H}=-$ ve

Halogens have very large negative E.G E value; as they need just one e- for stable configuration.

$\mathrm{Cl}$	$ > \hspace{1cm} \mathrm{F} \hspace{1cm} > \hspace{1cm} \mathrm{Br} \hspace{1cm} > \hspace{1cm} \mathrm{I}$
Largest	small in size
negative	Inter electronic repulsion
E. G. E	Addition of an extra electron is not energetically favourable

5) Halogens are good oxidising agents

They have $n s^{2}, n^{5}$ configuration. They need just one $e^{-}$for stable configuration They tend to give an $e^{-}$and reduce.

“F” has less negative $\mathrm{E}$. G. $\mathrm{E}$ than “Cl” despite $\mathrm{F} _{2}$ is the strongest oxidising agent. Why?

$\begin{array}{llll} \mathrm{X}_2 & \rightarrow & \mathrm{X}+\mathrm{X} & \text { Bond dissociation energy } \\ \mathrm{X}^{-}+\mathrm{e}^{-} & \rightarrow & \mathrm{X}^{-} & \text {E. G E } \\ \mathrm{X}^{-}+\mathrm{aq} & \rightarrow & \mathrm{X}^{-}(\mathrm{aq}) & \text { Hydration energy } \end{array}$

• $\mathrm{F}$ is smallest, it has maximum hydration energy.

• Order of Bond Dissociation energy

$ \mathrm{Cl} _{2}>\mathrm{Br} _{2}>\mathrm{F} _{2}>\mathrm{I} _{2} $

$\mathrm{F} \rightarrow$ small size;

Inter electronic repulsion, So, “F-F” bond is relatively weaker.

These two factors overcome E. G. E

6) Halogens are coloured

$\mathrm{F} _{2} \rightarrow$ greenish yellow

$\mathrm{C1} _{2} \rightarrow$ pale green

$\mathrm{Br} _{2} \rightarrow$ reddish brown

$\mathrm{I} _{2} \rightarrow$ violet

Halogens $\rightarrow \mathrm{ns}^{2}, \mathrm{np} _{\mathrm{x}}^{2}, \mathrm{np} _{\mathrm{y}}{ }^{2}, \mathrm{np} _{\mathrm{z}}^{1}$

[ $n p _{2}^{\prime} e^{-}$gets excited in visible reason]

In halogen molecules, the e- gets excited from the lower molecular orbital within the visible region. In case of I, the size of atom is large. So, electrons absorb energy from lower energy region. So, the complementary colour is emitted.

7) Electronegativity

i) Electronegativity $\rightarrow \mathrm{F}>\mathrm{Cl}>\mathrm{Br}>\mathrm{I}$

ii) Electron gain enthalpy $\rightarrow \mathrm{Cl}>\mathrm{F}>\mathrm{Br}>\mathrm{I}$

$\quad$ Large negative electron gain enthalpy

iii) Bond dissociation enthalpy $\rightarrow \mathrm{Cl} _{2}>\mathrm{Br} _{2}>\mathrm{F} _{2}>\mathrm{I} _{2}$

By considering all the 3 factors, the reactivity order is concluded.

8) Hydrogen halides:

• Hydrogen fluoride is a low boiling liquid (due to hydrogen bonding) while $\mathrm{HCl}, \mathrm{HBr}$ and $\mathrm{HI}$ are gases.

• $\mathrm{HF}>\mathrm{HCl}>\mathrm{HBr}>\mathrm{HI}$ (decreasing order of ionic character)

• $\mathrm{HF}<\mathrm{HCl}<\mathrm{HBr}<\mathrm{HI}$ (increasing order of bond length, reducing character)

• $\mathrm{HF}>\mathrm{HCl}>\mathrm{HBr}>\mathrm{HI}$ (decreasing order of bond dissociation enthalpy, thermal stability)

• $\mathrm{HI}>\mathrm{HBr}>\mathrm{HCl}>\mathrm{HF}$ (decreasing order of acidic strength)

• $\mathrm{HF}$ is not stored in glass vessels as it reacts with $\mathrm{SiO} _{2}$ of glass.

• $\mathrm{HF}>\mathrm{HI}>\mathrm{HBr}>\mathrm{HCl}$ (Trend of B.pt.)

9) Reducing Nature

$\mathrm{HI}>\mathrm{HBr}>\mathrm{HCl}>\mathrm{HF}$

Acidic nature $\propto$ Electronegativity of central atom

10) Oxoacids of halogens

• Fluorine forms one oxoacid HOF (hypofluorous acid)- at ice temperature. The rest of the halogens form four series of oxoacids, $\mathrm{HXO}, \mathrm{HXO} _{2}, \mathrm{HXO} _{3}$ and $\mathrm{HXO} _{4}$.

$\mathrm{HOCl}(+1)$ hypo chlorus acid

$\mathrm{HClO} _{2}(+3)$ chlorus acid

$\mathrm{HClO} _{3}(+5)$ chloric acid

$\mathrm{HClO} _{4}(+7)$ per chloric acid

Acidic strength

i) Per Halic acid with the same halogen atom ; but different oxidation state

$$ \mathrm{HClO} _{4}>\mathrm{HClO} _{3}>\mathrm{HClO} _{2}>\mathrm{HOCl} $$

oxidation state

$$ \begin{aligned} & +7 \quad+5 \quad+3 \quad+1 \\ & \mathrm{HClO} _{4}>\mathrm{HClO} _{3}>\mathrm{HClO} _{2}>\mathrm{HOCl} \\ & \\ & \text { Conjuate base}-\mathrm{ClO} _{4}^{-}>\underset{\text { Perchlorate }}{\mathrm{ClO} _{3}^{-}}>\underset{\text { chlorate }}{\mathrm{ClO} _{2}^{-}}>\underset{\text { chlorite }}{\mathrm{ClO} _{2}^{-}}>\underset{\text { hyperchlorite }}{\mathrm{ClO}} \end{aligned} $$

a) Acidic strength $\propto$ stability of conjugate base

b) Acidic nature $\propto$ Electronegativity of the central atom

c) Acidic strength $\propto$ oxidation state of the central atom

11) Inter halogen compounds

The compounds of one halogen with the other are called interhalogens or interhalogen compounds. The main reason for their formation is the large electronegativity and the size differences between the different halogens. Taking $A$ as the less electronegative and $B$ as the more electronegative halogen, they are divided into the following four types. The less electronegative halogen $(A)$ is always written first.

$A B$	$\mathrm{AB}_3$	$\mathrm{AB}_5$	$\mathrm{AB}_7$
$\mathrm{ClF}$	$\mathrm{ClF}_3$	$\mathrm{BrF}_5, \mathrm{IF}_5$	$\mathrm{IF}_7$
$\mathrm{BrF}, \mathrm{BrCl}, \mathrm{ICl}$ $\mathrm{IBr} , \mathrm{IF}$	$\mathrm{IF}_3, \mathrm{ICI}_3$

These interhalogen compounds are unstable and more reactive

a) General properties:

i) Largest halogen always serves the central atom

ii) The highest interhalogen compound i.e. $\mathrm{IF} _{7}$ is obtained with iodine, the largest halogen attached to the smallest one

iii) The bonds in interhalogen compounds are essentially covalent due to little electronegativity difference between different halogens.

iv) Thermal stability decreases as the size difference decreases and increases as the polarity of the bond increases. Thus CIF is thermally more stable as compared to IBr.

v) They ionize in solution or in the liquid state.

$$ 2 \mathrm{ICl} \rightleftharpoons \mathrm{I}^{+}+\mathrm{ICl} _{2}^{-} \quad ; \quad 2 \mathrm{ICl} _{3} \rightleftharpoons \mathrm{ICl} _{2}^{+}+\mathrm{ICl} _{4}^{-} $$

vi) Hydrolysis of interhalogen compounds always produces a halide ion derived from smaller halogen and oxyhalide derived from larger halogen

$$ \mathrm{ICI}+\mathrm{H} _{2} \mathrm{O} \longrightarrow \mathrm{Cl}^{-}+\mathrm{OI}^{-}+2 \mathrm{H}^{+} \quad ; \quad \mathrm{BrF} _{5}+3 \mathrm{H} _{2} \mathrm{O} \longrightarrow 5 \mathrm{~F}^{-}+\mathrm{BrO} _{3}^{-}+6 \mathrm{H}^{+} $$

vii) They are strong oxidising agents and are diamagnetic in nature.

viii) Largest number of interhalogens are formed by fluorine due to its smaller size and higher electronegativity or oxidising power.

ix) They are more reactive than the component halogens (except fluorine) due to weakness of the covalent bond between two dissimilar electronegative elements.

b) Structure: Interhalogen compounds

i) of the type $A B$, i.e. $I C I, I B r$, IF etc. are linear

ii) of the type $\mathrm{AB} _{3}$ i.e. $\mathrm{IF} _{3}, \mathrm{CIF} _{3}, \mathrm{BrF} _{3}$ have distorted trigonal bipyramidal (dsp3-hybridlization) structures or $\mathrm{T}$ shape due to two lone pairs in equatorial positions. $\mathrm{ICl} _{3}$ dimeric, $\mathrm{I} _{2} \mathrm{Cl} _{6}$ and has a planar structure.

iii) of the types $\mathrm{AB} _{5}$ ie. $\mathrm{BrF} _{5}, \mathrm{IF} _{5}$ have distorted octahedral ( $\mathrm{d}^{2} \mathrm{sp}^{3}$-hybridizatipn) shapes or square pyramidal due to a lone pair in one of the axial positions.

iv) of the type $\mathrm{AB} _{7}$ i.e. $\mathrm{IF} _{7}$, have pentagonal bipyramidal ( $\mathrm{d}^{3} \mathrm{sp}^{3}$-hybridization) structures.

Reactions with alkalies:

With cold and dilute $\mathrm{NaOH}, \mathrm{F} _{2}$ gives $\mathrm{OF} _{2}$ : while with hot and conc. $\mathrm{NaOH}$, it gives $\mathrm{O} _{2}$.

$$ 2 \mathrm{~F} _{2}+2 \mathrm{NaOH} \xrightarrow{\text { cold }} 2 \mathrm{NaF}+\mathrm{OF} _{2}+\mathrm{H} _{2} \mathrm{O} ; 2 \mathrm{~F} _{2}+4 \mathrm{NaOH} \xrightarrow{\text { hot }} 4 \mathrm{NaF}+2 \mathrm{H} _{2} \mathrm{O}+\mathrm{O} _{2} $$

Other halogens form hypohalites $\left(\mathrm{XO}^{-}\right)$with cold dilute $\mathrm{NaOH}$ solution and halates $\left(\mathrm{XO} _{3}^{-}\right)$with hot and conc. $\mathrm{NaOH}$ solution.

$2 \mathrm{NaOH}$ (dil.) $+\mathrm{X} _{2} \xrightarrow{\text { cold }} \mathrm{NaXO}+\mathrm{NaX}+\mathrm{H} _{2} \mathrm{O}$ where $\mathrm{X}=\mathrm{Cl}$, Br or I

$6 \mathrm{NaOH}$ (conc.) $+3 \mathrm{X} _{2} \xrightarrow{\text { hot }} \mathrm{NaXO} _{3}+5 \mathrm{NaX}+3 \mathrm{H} _{2} \mathrm{O}$ where $\mathrm{X}=\mathrm{Cl}$, Br or I.

Polyhalide Ions

Halogens or interhalogens combine with halide ions to form polyhalide ions. The most common example of polyhalide ion formation is furnished by the increase in solubility of iodine in water in the presence of $\mathrm{KI}$ which is due to the formation of tri-iodide ion,

$$ \mathrm{I}^{-}+\mathrm{I} _{2} \rightarrow \mathrm{I} _{3} $$

Many other examples of polyhalides ions are i) $\mathrm{Cl} _{3}^{-}, \mathrm{Br} _{3}^{-}, \mathrm{ICl} _{2}^{-}, \mathrm{IBr} _{2}^{-}$including $\mathrm{I} _{3}^{-}$. In these ions, one of the halogen atoms (In case of similar atoms) or halogen atom larger in size undergoes $\mathrm{sp}^{3} \mathrm{~d}$ - hybridization giving a linear shape with three lone pairs at equatorial positions.

ii) $\mathrm{Cl} _{3}{ }^{+}, \mathrm{Br} _{3}{ }^{+}, \mathrm{I} _{3}{ }^{+}, \mathrm{ICl} _{2}{ }^{+}, \mathrm{IBr} _{2}{ }^{+}$. Here we find central atom $\mathrm{sp}^{3}$ hybridized giving a bent shape with two lone pairs of electrons on the central atom

iii) $\mathrm{ICl} _{4}^{-}, \mathrm{BrF} _{4}^{-}, \mathrm{I} _{5}^{-}$. Here central atom involves $\mathrm{sp}^{3} \mathrm{~d}^{2}$ hybridization giving square planar shape with two lone pairs of electrons on axial positions.

iv) $\mathrm{ICl} _{4}^{+} \mathrm{BrF} _{4}^{+}, \mathrm{I} _{5}^{+}$. In these ions central atom involves $\mathrm{sp}^{3} \mathrm{~d}^{2}$ hybridization giving, a distorted tetrahedrai structure with one lone pair of electrons on equatorial position. v) $\mathrm{I} _{7}, \mathrm{IF} _{6}$ : The central atom $\mathrm{I}$ undergoes $\mathrm{sp}^{3} \mathrm{~d}^{3}$ hybridization giving a distorted octahedral structure with one lone pair of electrons.

vi) $\mathrm{I} _{7}^{+}$, Here central $\mathrm{I}$ atom involves $\mathrm{sp}^{3} \mathrm{~d}^{2}$ hybridization giving an octahedral structure.

Fluorine due to its highest electronegativity (and only - I oxidation state) does not form polyhalide ions where it acts as a central atom.

Anomalous behaviour of fluorine.

Fluorine differs from rest of the elements of its family due to (a) its small size (b) highest electronegativity, (c) low bond dissociation energy and (d) absence of d-orbitals in the valence shell. The main points of difference are:

i) Fluorine is most reactive of all the halogens due to lower value of $\mathrm{F}-\mathrm{F}$ bond dissociation energy $\left(\mathrm{F} _{2}=158, \mathrm{Cl} _{2}=243, \mathrm{Br} _{2}=192\right.$ and $\left.\mathrm{I} _{2}=151 \mathrm{~kJ} \mathrm{~mol}^{-1}\right)$. ii) Being the most electronegative element, it shows only an oxidation state of -1 and does not show any positive oxidation states due to absence of $d$-orbitals in its valence shell. Other halogens show positive oxidation states of $+1,+3,+5$ and +7 .

iii) Due to small atomic size and high electronegativity of $\mathrm{F}, \mathrm{HF}$ undergoes strong $\mathrm{H}$-bonding while other halogen acids do not.

Chromyl chloride

When solid metal chloride is heated with conc. $\mathrm{H} _{2} \mathrm{SO} _{4}$ in presence of solid $\mathrm{K} _{2} \mathrm{Cr} _{2} \mathrm{O} _{7}$ in a dry test tube, deep red vapours of chromyl chloride are evolved

$$ \begin{gathered} \mathrm{NaCl}+\mathrm{H} _{2} \mathrm{SO} _{4} \rightarrow \mathrm{NaHSO} _{4}+\mathrm{HCl} \\ \mathrm{K} _{2} \mathrm{Cr} _{2} \mathrm{O} _{7}+2 \mathrm{H} _{2} \mathrm{SO} _{4} \rightarrow \underset{\text { Chromyl Chloride }}{2 \mathrm{KHSO} _{4}+2 \mathrm{CrO} _{3}+\mathrm{H} _{2} \mathrm{O}} \\ \mathrm{CrO} _{3}+2 \mathrm{HCl} \rightarrow \underset{\text { CrO }}{\mathrm{CrO} _{2} \uparrow}+\mathrm{H} _{2} \mathrm{O} \end{gathered} $$

When these vapours are passed through $\mathrm{NaOH}$ solution, the solution becomes yellow due the formation of sodium chromate

$$ \mathrm{CrO} _{2} \mathrm{Cl} _{2}+4 \mathrm{NaOH} \underset{\text { Yellow colour }}{\mathrm{Na} _{2} \mathrm{CrO} _{4}}+2 \mathrm{NaCl}+2 \mathrm{H} _{2} \mathrm{O} $$

The yellow solution is neutralised with acetic acid and on addition of lead acetate gives a yellow precipitate of lead chromate

$$ \mathrm{Na} _{2} \mathrm{CrO} _{4}+\mathrm{Pb}\left(\mathrm{CH} _{3} \mathrm{COOH}\right) _{2} \rightarrow \underset{\text { Yellow ppt }}{\mathrm{PbCrO} _{4}}+2 \mathrm{CH} _{3} \mathrm{COONa} $$

18th Group of Elements

Properties	$\mathrm{He}$	$\mathrm{Ne}$	$\mathrm{Ar}$	$\mathrm{Kr}$	$X e$	$\operatorname{Re}$
Atomic number	2	10	18	36	54	86
Atomic mass $\left(\mathrm{g} \mathrm{mol}^{-1}\right)$	4.00	20.18	39.95	83.80	131.30	222
Atomic radius (pm) (van der Waals’ radius)	120	160	190	200	220	-
First lonisation enthalpy $\Delta \mathrm{H}\left(\mathrm{kJ} \mathrm{mol}^{-1}\right)$	2372	2080	1520	1351	1170	1037
Electron gain enthalpy $\left(\mathrm{kJ} \mathrm{mol}^{-1}\right)$	+48	+116	+96	+96	+77	+68
Melting point/K	-	25	84	116	161	202
Boiling point/K	4.2	27.1	87.3	120	165	211
Relative abundance (ppm)	5.2	18.2	93.4	1.1	0.09	Traces

ii) Electronic config. $n s^{2} n p^{6}$ except $\mathrm{He} \rightarrow \mathrm{IS}^{2}$

iii) Ionization enthalpy $\rightarrow 18^{\text {th }}$ group elements have very high I.E. Already they have achieved stable config. removal of an $e^{-}$is not energetically favourable.

’ $\mathrm{Xe}$ ’ is capable of forming certain compounds only with ’ $\mathrm{F} ’ \& 0 $.

Both $\mathrm{F} \& 0 \rightarrow$ Highly E. N.

Clathrates compounds are formed when $\mathrm{Ar}, \mathrm{Kr}$ and $\mathrm{Xe}$ are passed through organic liquids such as phenol, hydroquinone etc. under pressure and solution is crystallised. The atoms of inert gases are trapped into the crystal lattice of organic molecules.

3) Some compounds of Xenon:

a) $\mathrm{XeO} _{3}$ : pyramidal,

b) $\mathrm{XeO} _{4}$ : Tetrahedral

c) $\mathrm{XeOF} _{2}: \mathrm{T}$-shaped

d) $\mathrm{XeOF} _{4}:$ Square pyramidal

e) $\mathrm{XeF} _{2}$ - Linear

f) $\mathrm{XeF} _{4}$ - Sq. Planar

g) $\mathrm{XeF} _{6}$ - Distorted octanedron

Diagonal Relationship: The similarity in the properties of some of the elements of second period to diagonally opposite elements of third period is called diagonal relationship.

Inert pair effect: The reluctance of s-electrons to take part in bond formation. This is observed in the lower elements of each group (from Gr No-13 to Gr No-16).

Covalent character of an ionic bond: In ionic bonds due to small size and high charge density the cation pulls the electron density of the anion towards itself thus leading to the development of covalent character. This is called the polarization effect.

Fajan’s rule: Higher the charge density of the cation, greater is its polarising power. Larger the size of the anion greater will be its polarizahility.

Hydration energy: The amount of energy released when one mole of ions are completely hydrated

iv) E. G. E $\rightarrow$ Noble gases have +ve E. G. E

They have achieved stable configuration. So, addition of an extra e- is not favourable.

v) $E N \downarrow $ decreases as the size increases

vi) Nature of liquification

Ease of liquification increases as the size of molecules and hence the V.W.F also increases

Neils Bartlett $\rightarrow$ He prepared the first compound of $\mathrm{Xe}$.

$ \mathrm{O} _{2}+\mathrm{PtF} _{6} \xrightarrow{\Delta}\left[\mathrm{O} _{2}\right]^{+}\left[\mathrm{PtF} _{6}\right]^{-} $

Both $\mathrm{O} _{2} \& \mathrm{Xe}$ have similar E.N and similar 1st IE.

$\mathrm{Xe}+\mathrm{PtF} _{6} \xrightarrow{\Delta}[\mathrm{Xe}]^{+}\left[\mathrm{PtF} _{6}\right]^{-}$

First compound of Xe prepared in the lab.

Structure of compounds of $\mathrm{Xe}$.

1) $\mathrm{XeF} _{2}$

$ \begin{aligned} & H=\dfrac{1}{2}(V+M-C+A) \\ & =\dfrac{1}{2}(8+2-0+0) \\ & =5 \rightarrow s p^{3} d \end{aligned} $

$\mathrm{XeF} _{2} \rightarrow$ Linear

(In $\mathrm{sp}^{3} \mathrm{~d}$, the Ip of $\mathrm{e}^{-}$occupy equitional position)

2) $\mathrm{XeF} _{4}$

$\begin{aligned} & H=\dfrac{1}{2}(8+4-0+0) \\ & =6 \rightarrow \mathrm{sp}^3 \mathrm{~d}^2 \end{aligned}$

(In $\mathrm{sp}^{3} \mathrm{~d}^{2}$ hybridisation, the Ip of $\mathrm{e}^{-}$occupy only the axial position)

3) $\mathrm{XeOF} _{2}$

$ \begin{aligned} & \mathrm{H}=\dfrac{1}{2}(8+2) \\ & =5 \rightarrow \mathrm{Sp}^{3} \mathrm{~d} \end{aligned} $

4) $\mathrm{XeOF} _{4} \rightarrow \mathrm{sp}^{3} \mathrm{~d}^{2}$

Square pyramidal

5) $\mathrm{XeO} _{2} \mathrm{~F} _{2} \rightarrow \mathrm{sp}^{3} \mathrm{~d}$

Bent sea-saw model

6) $\mathrm{XeO} _{2} \mathrm{~F} _{4} \rightarrow \mathrm{sp}^{3} \mathrm{~d}^{2}$

Octahedral or square bipyramidal

7) $\mathrm{XeO} _{3}$

$ \begin{aligned} & \mathrm{H}=\dfrac{1}{2}(8+0-0+0) \\ & =4 \rightarrow \mathrm{sp}^{3} \end{aligned} $

8) $\mathrm{XeF} _{6}$

$ \begin{aligned} & H=\dfrac{1}{2}(8+6-0+0) \\ & =7 \rightarrow \mathrm{sp}^{3} \mathrm{~d}^{3} \end{aligned} $

Expected structure $\rightarrow$ pentagonal bipyramidal

Typical reactions of ’ $X$ e’ compounds

1. Preparation of $\mathrm{XeF} _{2}, \mathrm{XeF} _{4} \& \mathrm{XeF} _{6}$

$$ \begin{aligned} & \mathrm{Xe}+\mathrm{F} _{2} \xrightarrow[\text { 1 bar }]{673 \mathrm{~K}} \mathrm{XeF} _{2} \\ & \\ & \mathrm{Xe}+2 \mathrm{~F} _{2} \xrightarrow{8 \text { bar }} \text { } \mathrm{XeF} _{4} \\ & \\ & \mathrm{Xe}+3 \mathrm{~F} _{2} \xrightarrow[60-70 \mathrm{bar}]{573 \mathrm{~K}} \mathrm{XeF} _{6} \end{aligned} $$

2. Hydrolysis of xenon flouride

i) $\mathrm{XeF} _{2}+\underset{\text { complete }}{\mathrm{H} _{2} \mathrm{O}} \longrightarrow \mathrm{Xe}+\mathrm{HF}+\mathrm{O} _{2}$

ii) $\mathrm{XeF} _{4}+\underset{\text { complete }}{\mathrm{H} _{2} \mathrm{O}} \longrightarrow \mathrm{Xe}+\mathrm{XeO} _{3}+\mathrm{HF}+0$

iii) $\mathrm{XeF} _{4}+\underset{\text { partial }}{\mathrm{H} _{2} \mathrm{O}} \longrightarrow 2 \mathrm{HF}+\mathrm{XeOF} _{2}$

iv) $\mathrm{XeF} _{6}+\underset{\text { completely }}{3 \mathrm{H} _{2} \mathrm{O}} \longrightarrow \mathrm{HF}+\mathrm{XeO} _{3}$

v) $\mathrm{XeF} _{6}+\underset{\text { semipartial }}{2 \mathrm{H} _{2} \mathrm{O}} \longrightarrow 4 \mathrm{HF}+\mathrm{XeOF} _{2}$

vi) $\mathrm{XeF} _{6}+\mathrm{H} _{2} \mathrm{O} \longrightarrow 2 \mathrm{HF}+\mathrm{XeOF} _{4}$

Solved Examples

Question 1. Identify the incorrect statement:

1. Silicon reacts with $\mathrm{NaOH}(\mathrm{aq})$ in the presence of air to give $\mathrm{Na} _{2} \mathrm{SiO} _{3}$ and $\mathrm{H} _{2} \mathrm{O}$.

2. $\mathrm{Cl} _{2}$ reacts with excess of $\mathrm{NH} _{3}$ to give $\mathrm{N} _{2}$ and $\mathrm{HCl}$.

3. $\mathrm{Br} _{2}$ reacts with hot and strong $\mathrm{NaOH}$ solution to form $\mathrm{NaBr}, \mathrm{NaBrO} _{3}$ and $\mathrm{H} _{2} \mathrm{O}$.

4. Ozone reacts with $\mathrm{SO} _{2}$ to give $\mathrm{SO} _{3}$.

[AIEEE, 2007]

Show Answer

Answer: (3)

The correct equation for this reaction is:

$3 \mathrm{Br} _{2}+6 \mathrm{NaOH} \rightarrow 5 \mathrm{NaBr}+\mathrm{NaBrO} _{3}+3 \mathrm{H} _{2} \mathrm{O}$

Question 2. In which of the following arrangements, the sequence is not strictly according to the property written against it?

1. $\mathrm{CO} _{2}<\mathrm{SiO} _{2}<\mathrm{SnO} _{2}<\mathrm{PbO} _{2}$ (increasing oxidizing power)

2. $\mathrm{HF}<\mathrm{HCl}<\mathrm{HBr}<\mathrm{HI}$ (increasing acid strength)

3. $\mathrm{NH} _{3}<\mathrm{PH} _{3}<\mathrm{AsH} _{3}<\mathrm{SbH} _{3}$ (increasing basic strength)

4. $\mathrm{B}<\mathrm{C}<\mathrm{O}<\mathrm{N}$ (increasing first ionisation enthalpy)

[AlEEE, 2009]

Show Answer

Answer: (3)

Option (3) is incorrect because as we move down the group, the size of the central atom increases. Due to this, availability of lone pair of electrons decreases. Hence the basic character decreases. Thus the correct order is $\mathrm{NH} _{3}>\mathrm{PH} _{3}>\mathrm{AsH} _{3}>\mathrm{SbH} _{3}$

Question 3. Which of the following reactions of xenon compounds is not feasible?

1. $\mathrm{XeO} _{3}+6 \mathrm{HF} \rightarrow \mathrm{XeF} _{6}+3 \mathrm{H} _{2} \mathrm{O}$

2. $3 \mathrm{XeF} _{4}+6 \mathrm{H} _{2} \mathrm{O} \rightarrow 2 \mathrm{Xe}+\mathrm{XeO} _{3}+12 \mathrm{HF}+1.5 \mathrm{O} _{2}$

3. $2 \mathrm{XeF} _{2}+2 \mathrm{H} _{2} \mathrm{O} \rightarrow 2 \mathrm{Xe}+4 \mathrm{HF}+\mathrm{O} _{2}$

4. $\mathrm{XeF} _{6}+\mathrm{RbF} \rightarrow \mathrm{Rb}\left[\mathrm{XeF} _{7}\right]$

[AIEEE, 2009]

Show Answer

Answer: (1)

The reaction is not feasible because $\mathrm{XeF} _{6}$ formed is hydrolysed in presence of moisture to form $\mathrm{XeO} _{3}$

PRACTICE QUESTIONS

Question 1- Which of the following represents the correct order of increasing pKa values of the given acids?

a) $\mathrm{HClO} _{4}<\mathrm{HNO} _{3}<\mathrm{H} _{2} \mathrm{CO} _{3}<\mathrm{B}(\mathrm{OH}) _{3}$

b) $\mathrm{HNO} _{3}<\mathrm{HClO} _{4}<\mathrm{B}(\mathrm{OH}) _{3}<\mathrm{H} _{2} \mathrm{CO} _{3}$

c) $\mathrm{B}(\mathrm{OH}) _{3}<\mathrm{H} _{2} \mathrm{CO} _{3}<\mathrm{HClO} _{4}<\mathrm{HNO} _{3}$

d) $\mathrm{HClO} _{4}<\mathrm{HNO} _{3}<\mathrm{B}(\mathrm{OH}) _{3}<\mathrm{H} _{2} \mathrm{CO} _{3}$

Show Answer

Answer:- a

Question 2- In which case, the order of acidic strength is not correct?

a) $\mathrm{HI}>\mathrm{HBr}>\mathrm{HCl}$

b) $\mathrm{HIO} _{4}>\mathrm{HBrO} _{4}>\mathrm{HClO} _{4}$

c) $\mathrm{HClO} _{4}>\mathrm{HClO} _{3}>\mathrm{HClO} _{2}$

d) $\mathrm{HF}>\mathrm{H} _{2} \mathrm{O}>\mathrm{NH} _{3}$

Show Answer

Answer:- b

Question 3- The relative basic character of the following is

a) $\mathrm{ClO}^{-}<\mathrm{ClO} _{2}^{-}<\mathrm{ClO} _{3}^{-}<\mathrm{ClO} _{4}^{-}$

b) $\mathrm{ClO} _{4}^{-}<\mathrm{ClO} _{3}^{-}<\mathrm{ClO} _{2}^{-}<\mathrm{ClO}^{-}$

c) $\mathrm{ClO} _{3}^{-}<\mathrm{ClO} _{4}^{-}<\mathrm{ClO} _{2}^{-}<\mathrm{ClO}^{-}$

d) $\mathrm{ClO} _{2}^{-}<\mathrm{ClO}^{-}<\mathrm{ClO} _{3}^{-}<\mathrm{ClO} _{4}^{-}$

Show Answer

Answer:- b

Question 4- The correct order of acidic strength is

a) $\mathrm{K} _{2} \mathrm{O}>\mathrm{CaO}>\mathrm{MgO}$

b) $\mathrm{CO} _{2}>\mathrm{N} _{2} \mathrm{O} _{5}>\mathrm{SO} _{3}$

c) $\mathrm{Na} _{2} \mathrm{O}>\mathrm{MgO}>\mathrm{Al} _{2} \mathrm{O} _{3}$

d) $\mathrm{Cl} _{2} \mathrm{O} _{7}>\mathrm{SO} _{2}>\mathrm{P} _{4} \mathrm{O} _{10}$

Show Answer

Answer:- d

Question 5- The correct order of bond energy is

a) $\mathrm{Cl} _{2}>\mathrm{Br} _{2}>\mathrm{F} _{2}>\mathrm{I} _{2}$

b) $\mathrm{Cl} _{2}>\mathrm{F} _{2}>\mathrm{Br} _{2}>\mathrm{I} _{2}$

c) $\mathrm{I} _{2}>\mathrm{Br} _{2}>\mathrm{Cl} _{2}>\mathrm{F} _{2}$

d) $\mathrm{I} _{2}>\mathrm{Br} _{2}>\mathrm{F} _{2}>\mathrm{Cl} _{2}$

Show Answer

Answer:- a

Question 6- $\mathrm{HClO} _{4}, \mathrm{HNO} _{3}$ and $\mathrm{HCl}$ are all very strong acids in aqueous solution. In glacial acetic acid medium, their acid strength varies in the order as

a) $\mathrm{HClO} _{4}>\mathrm{HNO} _{3}>\mathrm{HCl}$

b) $\mathrm{HNO} _{3}>\mathrm{HClO} _{4}>\mathrm{HCl}$

c) $\mathrm{HCl}>\mathrm{HClO} _{4}>\mathrm{HNO} _{3}$

d) $\mathrm{HCl}>\mathrm{HNO} _{3}>\mathrm{HClO} _{4}$

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Answer:- a

Question 7- The electronegativity follows the order

a) $\mathrm{F}>\mathrm{O}>\mathrm{Cl}>\mathrm{Br}$

b) $\mathrm{F}>\mathrm{Cl}>\mathrm{Br}>0$

c) $\mathrm{O}>\mathrm{F}>\mathrm{Cl}>\mathrm{Br}$

d) $\mathrm{Cl}>\mathrm{F}>\mathrm{O}>\mathrm{Br}$

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Answer:- a

Question 8- In the oxoacids of chlorine $\mathrm{Cl}-0$ bond contains

a) $\mathrm{d} \pi-\mathrm{d} \pi$ Bonding

b) $\mathrm{p} \pi-\mathrm{d} \pi$ Bonding

c) $p \pi-p \pi$ Bonding

d) none of the above

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Answer:- b

Question 9- The correct order of electron gain enthalpy values $(\Delta \mathrm{H})$ of the halogen atoms is

a) $\mathrm{F}<\mathrm{Cl}<\mathrm{Br}<\mathrm{I}$

b) $\mathrm{Cl}<\mathrm{F}<\mathrm{Br}<\mathrm{I}$

c) l $<\mathrm{Br}<\mathrm{F}<\mathrm{Cl}$

d) $\mathrm{Cl}<\mathrm{Br}<\mathrm{I}<\mathrm{F}$

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Answer:- c

Question 10- Which of the following sets has strongest tendency to form anions?

a) Ga, In, TI

b) $\mathrm{Na}, \mathrm{Mg}, \mathrm{Al}$

c) N, O, F

d) $\mathrm{V}, \mathrm{Cr}, \mathrm{Mn}$

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Answer:- c

PRACTICE QUESTIONS (OBJECTIVE TYPE)

Question 1- The element which liberated $\mathrm{O} _{2}$ from water is

a) $\mathrm{P}$

b) $\mathrm{N}$

c) $\mathrm{F}$

d) 1

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Answer:- d

Question 2- When, $\mathrm{I} _{2}$ is dissolved in $\mathrm{CCl} _{4}$ the colour that results is

a) Brown

b) Violet

c) Colourless

d) Bluish green

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Answer:- b

Question 3- Ozonised Oxygen can be obtained from $\mathrm{H} _{2} \mathrm{O}$ by the action of

a) Conc. $\mathrm{H} _{2} \mathrm{SO} _{4}$

b) $\mathrm{KMnO} _{4}$

c) $\mathrm{MnO} _{4}^{-}$

d) $\mathrm{F} _{2}$

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Answer:- c

Question 4- Halogen which can be prepared from caliche is

a) $\mathrm{Cl} _{2}$

b) $\mathrm{Br} _{2}$

c) $\mathrm{I} _{2}$

d) $\mathrm{F} _{2}$

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Answer:- b

Question 5- Which one of the following elements can have both positive and negative oxidation state?

a) $\mathrm{F}$

b) 1

c) $\mathrm{Li}$

d) $\mathrm{He}$

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Answer:- b

Question 6- $\mathrm{HI}$ can be prepared by all the following methods, except

a) $\mathrm{PI} _{3}+\mathrm{H} _{2} \mathrm{O}$

b) $\mathrm{KI}+\mathrm{H} _{2} \mathrm{SO} _{4}$

c) $\mathrm{H} _{2}+\mathrm{I} _{2}$

d) $\mathrm{I} _{2}+\mathrm{H} _{2} \mathrm{~S}$

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Answer:- d

Question 7- Which among the following is paramagnetic?

a) $\mathrm{Cl} _{2} \mathrm{O}$

b) $\mathrm{ClO} _{2}$

c) $\mathrm{Cl} _{2} \mathrm{O} _{7}$

d) $\mathrm{Cl} _{2} \mathrm{O} _{6}$

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Answer:- d

Question 8- Which of the following pairs will give chlorine gas most quickly, upon reaction?

a) $\mathrm{HCl}$ and $\mathrm{KMnO} _{4}$

b) $\mathrm{NaCl}$ and $\mathrm{H} _{3} \mathrm{PO} _{4}$

c) $\mathrm{NaCl}$ and $\mathrm{MnO} _{2}$

d) $\mathrm{CaCl} _{2}$ and $\mathrm{Br} _{2}$

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Answer:- c

Question 9- Which one of the following oxyacids of chlorine is the least oxidizing in nature?

a) $\mathrm{HOCl}$

b) $\mathrm{HClO} _{2}$

c) $\mathrm{HClO} _{3}$

d) $\mathrm{HClO} _{4}$

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Answer:- d

Question 10- Which of the following bonds is the strongest?

a) $\mathrm{F}-\mathrm{F}$

b) $\mathrm{Cl}-\mathrm{Cl}$

c) $1-1$

d) $\mathrm{Br}-\mathrm{Br}$

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Answer:- a

Question 11- In the clathrates of xenon with water, the nature of bonding between xenon and water molecule is

a) Covalent

b) Hydrogen bonding

c) Co-ordinate

d) Dipole-induced dipole interaction

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Answer:- a

Question 12- The last member of the family of inert gases is

a) Argon

b) Radon

c) Xenon

d) Neon

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Answer:- b

Question 13- The coloured discharge tubes for an advertisement mainly contain

a) Xenon

b) Helium

c) Neon

d) Argon

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Answer:- c

Question 14- $\mathrm{XeF} _{4}$ on partial hyrolysis produces

a) $\mathrm{XeF} _{2}$

b) $\mathrm{XeOF} _{2}$

c) $\mathrm{XeOF} _{4}$

d) $\mathrm{XeO} _{3}$

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Answer:- b

Question 15- Which element out of $\mathrm{He}, \mathrm{Ar}, \mathrm{Kr}$ and $\mathrm{Xe}$ forms least number of compounds?

a) $\mathrm{He}$

b) $\mathrm{Ar}$

c) $\mathrm{Kr}$

d) $\mathrm{Xe}$

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Answer:- a

Question 16- Which of the following is the correct sequence of the noble gases in their group in the periodic table?

a) $\mathrm{Ar}, \mathrm{He}, \mathrm{Kr}, \mathrm{Ne}, \mathrm{Rn}, \mathrm{Xe}$

b) $\mathrm{He}, \mathrm{Ar}, \mathrm{Ne}, \mathrm{Kr}, \mathrm{Xe}, \mathrm{Rn}$

c) $\mathrm{He}, \mathrm{Ne}, \mathrm{Kr}, \mathrm{Ar}, \mathrm{Xe}, \mathrm{Rn}$

d) $\mathrm{He}, \mathrm{Ne}, \mathrm{Ar}, \mathrm{Kr}, \mathrm{Xe}, \mathrm{Rn}$

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Answer:- d

Question 17- Which of the following noble gases does not have an octet of electrons in its outermost shell?

a) Neon

b) Radon

c) Argon

d) Helium

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Answer:- d

Question 18- The value of ionization potential for inert gases is

a) Zero

b) Low

c) High

d) Negative

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Answer:- c

Question 19- The elements which occupy the peak’s ionization energy curve are

a) $\mathrm{Na}, \mathrm{K}, \mathrm{Rb}, \mathrm{Cs}$

b) $\mathrm{Na}, \mathrm{Mg}, \mathrm{Cl}, \mathrm{I}$

c) $\mathrm{Cl}, \mathrm{Br}, \mathrm{I}, \mathrm{F}$

d) $\mathrm{He}, \mathrm{Ne}, \mathrm{Ar}, \mathrm{Kr}$

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Answer:- b

Question 20- Sea divers go deep in the sea water with a mixture of which of the following gases?

a) $\mathrm{O} _{2}$ and $\mathrm{He}$

b) $\mathrm{O} _{2}$ and $\mathrm{Ar}$

c) $\mathrm{O} _{2}$ and $\mathrm{CO} _{2}$

d) $\mathrm{CO} _{2}$ and $\mathrm{Ar}$

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Answer:- a

Question 21- Gradual addition of electronic shells in the noble gases causes a decrease in their

a) Ionization energy

b) Atomic radius

c) Boiling Point

d) Density

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Answer:- a

Question 22- Molecular shapes of $\mathrm{SF} _{4}, \mathrm{CF} _{4}$ and $\mathrm{XeF} _{4}$ are

a) The same, with 2, 0 and 1 lone pairs of electrons respectively

b) The same, with 1, 1 and 1 lone pairs of electrons respectively

c) Different, with 0, 1 and 2 lone pairs of electrons respectively

d) Different with 1,0 and 2 lone pairs of electrons respectively

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Answer:- d