Answer

General trends in group 15 elements

(i) Electronic configuration: All the elements in group 15 have 5 valence electrons. Their general electronic configuration is $n s^{2} n p^{3}$.

(ii) Oxidation states: All these elements have 5 valence electrons and require three more electrons to complete their octets. However, gaining electrons is very difficult as the nucleus will have to attract three more electrons. This can take place only with nitrogen as it is the smallest in size and the distance between the nucleus and the valence shell is relatively small. The remaining elements of this group show a formal oxidation state of -3 in their covalent compounds. In addition to the -3 state, $\mathrm{N}$ and $\mathrm{P}$ also show -1 and -2 oxidation states.

All the elements present in this group show +3 and +5 oxidation states. However, the stability of +5 oxidation state decreases down a group, whereas the stability of +3 oxidation state increases. This happens because of the inert pair effect.

(iii) lonization energy and electronegativity

First ionization decreases on moving down a group. This is because of increasing atomic sizes. As we move down a group, electronegativity decreases, owing to an increase in size.

(iv) Atomic size: On moving down a group, the atomic size increases. This increase in the atomic size is attributed to an increase in the number of shells.

Answer

Nitrogen is chemically less reactive. This is because of the high stability of its molecule, $\mathrm{N_2}$. In $\mathrm{N_2}$, the two nitrogen atoms form a triple bond. This triple bond has very high bond strength, which is very difficult to break. It is because of nitrogen’s small size that it is able to form $p \pi -p \pi$bonds with itself. This property is not exhibited by atoms such as phosphorus. Thus, phosphorus is more reactive than nitrogen.

Answer

General trends in chemical properties of group - 15

(i) Reactivity towards hydrogen:

The elements of group 15 react with hydrogen to form hydrides of type $\mathrm{EH_3}$, where $\mathrm{E}=\mathrm{N}, \mathrm{P}, \mathrm{As}, \mathrm{Sb}$, or Bi. The stability of hydrides decreases on moving down from $\mathrm{NH_3}$ to $\mathrm{BiH_3}$.

(ii) Reactivity towards oxygen:

The elements of group 15 form two types of oxides: $\mathrm{E_2} \mathrm{O_3}$ and $\mathrm{E_2} \mathrm{O_5}$, where $\mathrm{E}=\mathrm{N}, \mathrm{P}, \mathrm{As}, \mathrm{Sb}$, or $\mathrm{Bi}$. The oxide with the element in the higher oxidation state is more acidic than the other. However, the acidic character decreases on moving down a group.

(iii) Reactivity towards halogens:

The group 15 elements react with halogens to form two series of salts: $\mathrm{EX_3} a$ nd $\mathrm{EX_5}$. However, nitrogen does not form $\mathrm{NX_5}$ as it lacks the $d$-orbital. All trihalides (except $\mathrm{NX_3}$ ) are stable.

(iv) Reactivity towards metals:

The group 15 elements react with metals to form binary compounds in which metals exhibit -3 oxidation states.

Answer

Nitrogen is highly electronegative as compared to phosphorus. This causes a greater attraction of electrons towards nitrogen in $\mathrm{NH_3}$ than towards phosphorus in $\mathrm{PH_3}$. Hence, the extent of hydrogen bonding in $\mathrm{PH_3}$ is very less as compared to $\mathrm{NH_3}$.

Answer

An aqueous solution of ammonium chloride is treated with sodium nitrite.

$ \mathrm{NH_4} \mathrm{Cl_{(a q)}}+\mathrm{NaNO_{2(a q)}} \longrightarrow \mathrm{N_{2(g)}}+2 \mathrm{H_2} \mathrm{O_{(l)}}+\mathrm{NaCl_{(a q)}} $

$\mathrm{NO}$ and $\mathrm{HNO_3}$ are produced in small amounts. These are impurities that can be removed on passing nitrogen gas through aqueous sulphuric acid, containing potassium dichromate.

Answer

Ammonia is prepared on a large-scale by the Haber’s process.

$\mathrm{N} 2(\mathrm{~g})+3 \mathrm{H_2}(\mathrm{~g}) \longrightarrow 2 \mathrm{NH} 3(\mathrm{~g})$

Answer

Concentrated nitric acid is a strong oxidizing agent. It is used for oxidizing most metals. The products of oxidation depend on the concentration of the acid, temperature, and also on the material undergoing oxidation.

$3 \mathrm{Cu}+8 \mathrm{HNO_{3 \text { (dilute) }}} \longrightarrow 3 \mathrm{Cu}\left(\mathrm{NO_3}\right)_{2}+2 \mathrm{NO}+4 \mathrm{H_2} \mathrm{O}$

$\mathrm{Cu}+4 \mathrm{HNO_{3(\text { conc. })}} \longrightarrow \mathrm{Cu}\left(\mathrm{NO_3}\right)_{2}+2 \mathrm{NO_2}+2 \mathrm{H_2} \mathrm{O}$

Answer

(1)

(2)

Answer

Hydride $\mathrm{NH_3} \mathrm{PH_3} \mathrm{AsH_3} \mathrm{SbH_3}$

$\mathrm{H}-\mathrm{M}-\mathrm{H}$ angle $107^{\circ} 92^{\circ} 91^{\circ} 90^{\circ}$

The above trend in the $\mathrm{H}-\mathrm{M}-\mathrm{H}$ bond angle can be explained on the basis of the electronegativity of the central atom. Since nitrogen is highly electronegative, there is high electron density around nitrogen. This causes greater repulsion between the electron pairs around nitrogen, resulting in maximum bond angle. We know that electronegativity decreases on moving down a group. Consequently, the repulsive interactions between the electron pairs decrease, thereby decreasing the $\mathrm{H}-\mathrm{M}-\mathrm{H}$ bond angle.

Answer

$N\left(\right.$ unlike $P$ ) lacks the $d$-orbital. This restricts nitrogen to expand its coordination number beyond four. Hence, $\mathrm{R_3} \mathrm{~N}=\mathrm{O}$ does not exist.

Answer

$\mathrm{NH_3}$ is distinctly basic while $\mathrm{BiH_3}$ is feebly basic.

Nitrogen has a small size due to which the lone pair of electrons is concentrated in a small region. This means that the charge density per unit volume is high. On moving down a group, the size of the central atom increases and the charge gets distributed over a large area decreasing the electron density. Hence, the electron donating capacity of group 15 element hydrides decreases on moving down the group.

Answer

Nitrogen exists as diatomic molecule $\left(\mathrm{N}_2\right)$ with triple bond between two $\mathrm{N}$ atoms and phosphorus as $\mathrm{P}_4$. Nitrogen can form $\mathrm{p} \pi-\mathrm{p \pi}$ multiple bonds. Phosphorous cannot form such bonds due to repulsion between non bonded electrons of inner core. In case of nitrogen, there is no such repulsion as it has only $1 \mathrm{~s}$ electrons in inner shell so that the overlap of $p$ orbitals to form pi bonds is easy.

Answer

White phosphorus
It is a soft and waxy solid. It possesses a garlic smell.	It is a hard and crystalline solid, without any smell.
It is poisonous.	It is non-poisonous.
It is insoluble in water but soluble in carbon disulphide.	It is insoluble in both water and carbon disulphide.
It undergoes spontaneous combustion in air.	It is relatively less reactive.
In both solid and vapour states, it exists as a $\mathrm{P_4}$ molecule.	It exists as a chain of tetrahedral $\mathrm{P_4}$ units.

Answer

Catenation is much more common in phosphorous compounds than in nitrogen compounds. This is because of the relative weakness of the $\mathrm{N}-\mathrm{N}$ single bond as compared to the P-P single bond. Since nitrogen atom is smaller, there is greater repulsion of electron density of two nitrogen atoms, thereby weakening the $\mathrm{N}-\mathrm{N}$ single bond.

Answer

On heating, orthophosphorus acid $\left(\mathrm{H_3} \mathrm{PO_3}\right)$ disproportionates to give orthophosphoric acid $\left(\mathrm{H_3} \mathrm{PO_4}\right)$ and phosphine $\left(\mathrm{PH_3}\right)$. The oxidation states of $\mathrm{P}$ in various species involved in the reaction are mentioned below.

$$ 4 \mathrm{H_3} \stackrel{+3}{\mathrm{P}} \mathrm{O_3} \longrightarrow 3 \mathrm{H_3} \stackrel{+5}{\mathrm{P}} \mathrm{O_4}+\stackrel{-3}{\mathrm{P}} \mathrm{H_3} $$

Answer

$\mathrm{PCl_5}$ can only act as an oxidizing agent. The highest oxidation state that $\mathrm{P}$ can show is +5 . $\mathrm{In} \mathrm{PCl_5}$, phosphorus is in its highest oxidation state (+5). However, it can decrease its oxidation state and act as an oxidizing agent.

Answer

The elements of group 16 are collectively called chalcogens.

(i) Elements of group 16 have six valence electrons each. The general electronic configuration of these elements is $n s^{2} n p^{4}$, where $n$ varies from 2 to 6 .

(ii) Oxidation state:

Asthese elements have six valence electrons $\left(n s^{2} n p^{4}\right)$, they should display an oxidation state of -2 . However, only oxygen predominantly shows the oxidation state of -2 owing to its high electronegativity. It also exhibits the oxidation state of $-1\left(\mathrm{H_2} \mathrm{O_2}\right)$, zero $\left(\mathrm{O_2}\right)$, and $+2\left(\mathrm{OF_2}\right)$. However, the stability of the -2 oxidation state decreases on moving down a group due to a decrease in the electronegativity of the elements. The heavier elements of the group show an oxidation state of $+2,+4$, and +6 due to the availability of $d$-orbitals.

(iii) Formation of hydrides:

These elements form hydrides of formula $\mathrm{H_2} \mathrm{E}$, where $\mathrm{E}=\mathrm{O}, \mathrm{S}, \mathrm{Se}, \mathrm{Te}, \mathrm{PO}$. Oxygen and sulphur also form hydrides of type $\mathrm{H_2} \mathrm{E_2}$. These hydrides are quite volatile in nature.

Answer

Oxygen is smaller in size as compared to sulphur. Due to its smaller size, it can effectively form pÃâ, $\neg$-pÃâ, $\neg$ bonds and form $\mathrm{O_2}(\mathrm{O}==\mathrm{O})$ molecule. Also, the intermolecular forces in oxygen are weak van der Wall’s, which cause it to exist as gas. On the other hand, sulphur does not form $\mathrm{M_2}$ molecule but exists as a puckered structure held together by strong covalent bonds. Hence, it is a solid.

Answer

Stability of an ionic compound depends on its lattice energy. More the lattice energy of a compound, more stable it will be.

Lattice energy is directly proportional to the charge carried by an ion. When a metal combines with oxygen, the lattice energy of the oxide involving $\mathrm{O}^{2}$ ion is much more than the oxide involving O-ion. Hence, the oxide having $\mathrm{O}^{2}$ ions are more stable than oxides having $\mathrm{O}$. Hence, we can say that formation of $\mathrm{O}^{2}$ is energetically more favourable than formation of $\mathrm{O}$ :

Answer

Freons or chlorofluorocarbons (CFCs) are aerosols that accelerate the depletion of ozone. In the presence of ultraviolet radiations, molecules of CFCs break down to form chlorine-free radicals that combine with ozone to form oxygen.

Answer

Sulphuric acid is manufactured by the contact process. It involves the following steps:

Step (i):

Sulphur or sulphide ores are burnt in air to form $\mathrm{SO_2}$.

Step (ii):

By a reaction with oxygen, $\mathrm{SO_2}$ is converted into $\mathrm{SO_3}$ in the presence of $\mathrm{V_2} \mathrm{O_5}$ as a catalyst.

$2 \mathrm{SO_{2(g)}}+\mathrm{O_2(g)} \xrightarrow{\mathrm{V_2} \mathrm{O_s}} 2 \mathrm{SO_{3(g)}}$

Step (iii):

$\mathrm{SO_3}$ produced is absorbed on $\mathrm{H_2} \mathrm{SO_4}$ to give $\mathrm{H_2} \mathrm{~S_2} \mathrm{O_7}$ (oleum).

$\mathrm{SO_3}+\mathrm{H_2} \mathrm{SO_4} \longrightarrow \mathrm{H_2} \mathrm{~S_2} \mathrm{O_7}$

This oleum is then diluted to obtain $\mathrm{H_2} \mathrm{SO_4}$ of the desired concentration.

In practice, the plant is operated at 2 bar (pressure) and $720 \mathrm{~K}$ (temperature). The sulphuric acid thus obtained is 96 $98 \%$ pure.

Answer

Sulphur dioxide causes harm to the environment in many ways:

1. It combines with water vapour present in the atmosphere to form sulphuric acid. This causes acid rain. Acid rain damages soil, plants, and buildings, especially those made of marble.

2. Even in very low concentrations, $\mathrm{SO_2}$ causes irritation in the respiratory tract. It causes throat and eye irritation and can also affect the larynx to cause breathlessness.

3. It is extremely harmful to plants. Plants exposed to sulphur dioxide for a long time lose colour from their leaves. This condition is known as chlorosis. This happens because the formation of chlorophyll is affected by the presence of sulphur dioxide.

Answer

The general electronic configuration of halogens is $n p^{5}$, where $n=2-6$. Thus, halogens need only one more electron to complete their octet and to attain the stable noble gas configuration. Also, halogens are highly electronegative with low dissociation energies and high negative electron gain enthalpies. Therefore, they have a high tendency to gain an electron. Hence, they act as strong oxidizing agents.

Answer

Fluorine forms only one oxoacid i.e., HOF because of its high electronegativity and small size.

Answer

Both chlorine and oxygen have almost the same electronegativity values, but chlorine rarely forms hydrogen bonding. This is because in comparison to chlorine, oxygen has a smaller size and as a result, a higher electron density per unit volume.

Answer

Uses of $\mathrm{ClO_2}$ :

(i) It is used for purifying water.

(ii) It is used as a bleaching agent.

Answer

Almost all halogens are coloured. This is because halogens absorb radiations in the visible region. This results in the excitation of valence electrons to a higher energy region. Since the amount of energy required for excitation differs for each halogen, each halogen displays a different colour.

Answer

(i) $\mathrm{Cl_2}+\mathrm{H_2} \mathrm{O} \longrightarrow \underset{\text{Hydrochloric acid}}{\mathrm{HCl}}+ \underset{\text{Hypochlorous acid}}{\mathrm{HOCl}}$

(ii) $2 \mathrm{~F_{2(g)}}+2 \mathrm{H_2} \mathrm{O_{(l)}} \longrightarrow 4 \mathrm{H_{(a q)}^{+}}+4 \mathrm{~F_{(a q)}^{-}}+\mathrm{O_{2(g)}}+4 \mathrm{HF_{(a q)}}$

Answer

(i) $\mathrm{Cl_2}$ can be prepared from $\mathrm{HCl}$ by Deacon’s process.

$4 \mathrm{HCl}+\mathrm{O_2} \xrightarrow{\mathrm{CuCl_2}} 2 \mathrm{Cl_2}+2 \mathrm{H_2} \mathrm{O}$

(ii) $\mathrm{HCl}$ can be prepared from $\mathrm{Cl_2}$ on treating it with water.

$ \mathrm{Cl_2}+\mathrm{H_2} \mathrm{O} \longrightarrow \underset{\text{Hydrochloric acid}}{\mathrm{HCl}}+ \underset{\text{Hypochlorous acid}}{\mathrm{HOCl}} $

Answer

Neil Bartlett initially carried out a reaction between oxygen and $\mathrm{PtF_6}$. This resulted in the formation of a red compound, $\mathrm{O_2}^{+}\left[\mathrm{PtF_6}\right]^{-}$.

Later, he realized that the first ionization energy of oxygen $(1175 \mathrm{~kJ} / \mathrm{mol})$ and $\mathrm{Xe}(1170 \mathrm{~kJ} / \mathrm{mol})$ is almost the same. Thus, he tried to prepare a compound with $\mathrm{Xe}$ and $\mathrm{PtF_6}$. He was successful and a red-coloured compound, $\mathrm{Xe}^{+}\left[\mathrm{PtF_6}\right]^{-}$was formed.

Answer

Let the oxidation state of $p$ be $x$

(i) $\mathrm{H_3} \mathrm{PO_3}$

$3+x+3(-2)=0$

$3+x-6=0$

$x-3=0$

$x=+3$

(ii) $\mathrm{PCl_3}$

$x+3(-1)=0$

$x-3=0$

$x=+3$

(iii) $\mathrm{Ca_3} \mathrm{P_2}$

$3(+2)+2(x)=0$

$6+2 x=0$

$2 x=-6$

$x=-3$

(iv) $\mathrm{Na_3} \mathrm{PO_4}$

$3(+1)+x+4(-2)=0$

$3+x-8=0$

$x-5=0$

$x=+5$

(v) $\mathrm{POF_3}$

$x+(-2)+3(-1)=0$

$x-5=0$

$x=+5$

Answer

(i) $4 \mathrm{NaCl}+\mathrm{MnO_2}+4 \mathrm{H_2} \mathrm{SO_4} \longrightarrow \mathrm{MnCl_2}+4 \mathrm{NaHSO_4}+2 \mathrm{H_2} \mathrm{O}+\mathrm{Cl_2}$

(ii) $ \mathrm{Cl_2}+\mathrm{NaI} \longrightarrow 2 \mathrm{NaCl}+\mathrm{I_2} $

Answer

$\mathrm{XeF_2}, \mathrm{XeF_4}$ and $\mathrm{XeF_6}$ are obtained by a direct reaction between $\mathrm{Xe}$ and $\mathrm{F} 2$. The condition under which the reaction is carried out determines the product.

$ \underset{\text{(Excess)}}{\mathrm{Xe_{(g)}}}+\mathrm{F_{2(g)}} \xrightarrow{673 \mathrm{~K}, \text { 1 bar }} \mathrm{XeF_{2(\mathrm{~s})}} $

$ \underset{\text{(1:5 ratio)}}{\mathrm{Xe_{(g)}}}+2 \mathrm{~F_{2\left(g\right)}} \xrightarrow{873 \mathrm{~K}, 7 \text { 7 bar }} \mathrm{XeF_{4(s)}} $

$ \underset{\text{(1:20 ratio)}}{\mathrm{Xe_{(g)}}}+3 \mathrm{~F_{2(g)}} \xrightarrow{573 \mathrm{~K} \cdot 60-70 \mathrm{bar}} \quad \mathrm{XeF_{6(s)}} $

Answer

$\mathrm{CIO}$ is isoelectronic to CIF. Also, both species contain 26 electrons in all as shown.

Total electrons $\mathrm{CIO}=17+8+1=26$

In $\mathrm{CIF}=17+9=26$

CIF acts like a Lewis base as it accepts electrons from $\mathrm{F}$ to form $\mathrm{CIF_3}$.

Answer

(i) $\mathrm{XeO_3}$ can be prepared in two ways as shown.

$ 6 \mathrm{XeF_4}+12 \mathrm{H_2} \mathrm{O} \longrightarrow 4 \mathrm{Xe}+2 \mathrm{XeO_3}+24 \mathrm{HF}+3 \mathrm{O_2} \\ \mathrm{XeF_6}+3 \mathrm{H_2} \mathrm{O} \longrightarrow \mathrm{XeO_3}+6 \mathrm{HF} $

(ii) $\mathrm{XeOF_4}$ can be prepared using $\mathrm{XeF_6}$.

$ \mathrm{XeF_6}+\mathrm{H_2} \mathrm{O} \longrightarrow \mathrm{XeOF_4}+2 \mathrm{HF} $

Answer

(i) Bond dissociation energy usually decreases on moving down a group as the atomic size increases. However, the bond dissociation energy of $\mathrm{F} 2$ is lower than that of $\mathrm{Cl} 2$ and $\mathrm{Br} 2$. This is due to the small atomic size of fluorine. Thus, the increasing order for bond dissociation energy among halogens is as follows:

$\mathrm{I_2}<\mathrm{F_2}<\mathrm{Br_2}<\mathrm{Cl_2}$

(ii) $\mathrm{HF}<\mathrm{HCl}<\mathrm{HBr}<\mathrm{HI}$

The bond dissociation energy of $\mathrm{H}-\mathrm{X}$ molecules where $\mathrm{X}=\mathrm{F}, \mathrm{Cl}, \mathrm{Br}, \mathrm{I}$, decreases with an increase in the atomic size. Since $\mathrm{H}-\mathrm{I}$ bond is the weakest, $\mathrm{HI}$ is the strongest acid.

(iii) $\mathrm{BiH_3} < \mathrm{SbH_3}<\mathrm{AsH_3}<\mathrm{PH_3}<\mathrm{NH_3}$

On moving from nitrogen to bismuth, the size of the atom increases while the electron density on the atom decreases. Thus, the basic strength decreases.

Answer

$\mathrm{NeF_2}$ does not exist.

Answer

(i)

$\mathrm{XeF_4}$ is isoelectronic with $\mathrm{ICl_4}^{-}$and has square planar geometry.

(ii)

XeF ${ _2}$ is isoelectronic to ${ }^{\mathrm{IBr_2}^{-}}$and has a linear structure.

(iii)

$\mathrm{XeO_3}$ is isostructural to $\mathrm{BrO_3}^{-}$and has a pyramidal molecular structure.

Answer

Noble gases do not form molecules. In case of noble gases, the atomic radii corresponds to van der Waal’s radii. On the other hand, the atomic radii of other elements correspond to their covalent radii. By definition, van der Waal’s radii are larger than covalent radii. It is for this reason that noble gases are very large in size as compared to other atoms belonging to the same period.

Answer

Uses of neon gas:

(i) It is mixed with helium to protect electrical equipments from high voltage.

(ii) It is filled in discharge tubes with characteristic colours.

(iii) It is used in beacon lights.

Uses of Argon gas:

(i) Argon along with nitrogen is used in gas-filled electric lamps. This is because Ar is more inert than $\mathrm{N}$.

(ii) It is usually used to provide an inert temperature in a high metallurgical process.

(iii) It is also used in laboratories to handle air-sensitive substances.