Nitrogen family Questions in English

(B) Ammonia $(NH_3)$ is widely used in cold storage as a refrigerant.
It is also used in the industrial production of nitric acid $(HNO_3)$ via the Ostwald process.
However,$NH_3$ is not used as an anesthetic.
Therefore,the correct option is $B$.

(A) Nitrous oxide $(N_2O)$,also known as laughing gas,is widely used as an anesthetic in dentistry and surgery. It acts as a mild anesthetic and analgesic agent.

(B) Nitrous acid is represented by the formula $HNO_2$.
Anhydrides are oxides that react with water to form the corresponding acid.
The reaction for the formation of nitrous acid from its anhydride is:
$N_2O_3 + H_2O \rightarrow 2HNO_2$.
Therefore,$N_2O_3$ is the anhydride of nitrous acid.

(D) Nitrous oxide $(N_2O)$ is a neutral oxide of nitrogen.
It is not acidic,nor is it basic.
It has very low solubility in water,regardless of temperature.
Therefore,none of the given options $(A, B, C)$ are correct.

(A) White phosphorus is converted into red phosphorus by heating it in an inert atmosphere at $573 \ K$ for several days. This process is known as the allotropic transformation of phosphorus.

(C) The chemical formula of pyrophosphoric acid is $H_4P_2O_7$.
Its structure consists of two $PO_4$ tetrahedra linked by an oxygen atom ($P-O-P$ linkage).
Each phosphorus atom is bonded to one double-bonded oxygen $(P=O)$ and two hydroxyl groups $(-OH)$.
Thus,the total number of hydroxyl groups $(-OH)$ in the structure is $4$.

(C) $P_4O_{10}$ is an acidic oxide,whereas $NH_3$ is a basic gas.
When $P_4O_{10}$ is used to dry $NH_3$,it reacts with $NH_3$ to form ammonium phosphate.
The reaction is: $6NH_3 + P_4O_{10} + 6H_2O \rightarrow 4(NH_4)_3PO_4$.
Therefore,it cannot be used as a drying agent for $NH_3$.

(D) White phosphorus consists of discrete $P_4$ tetrahedral molecules.
In this structure,each phosphorus atom is bonded to three other phosphorus atoms.
The geometry of the $P_4$ molecule is a tetrahedron where the bond angle between any two $P-P$ bonds is $60^o$.

(A) The bond angle in hydrides of group $15$ elements depends on the electronegativity of the central atom.
As the electronegativity of the central atom decreases down the group $(N > P > As > Sb)$,the bond pair-bond pair repulsion decreases.
Also,the bond angle decreases as the size of the central atom increases and the $s$-character in the bonding orbitals decreases.
Thus,the correct order of bond angles is $NH_3 (107.8^{\circ}) > PH_3 (93.6^{\circ}) > AsH_3 (91.8^{\circ}) > SbH_3 (91.3^{\circ})$.

(A) Nitrogen trichloride $(NCl_3)$ is an endothermic compound with a large positive enthalpy of formation.
It is highly unstable and decomposes explosively into nitrogen gas $(N_2)$ and chlorine gas $(Cl_2)$ upon slight disturbance or exposure to light.
In contrast,$PCl_3$ and $AsCl_3$ are relatively stable compounds compared to $NCl_3$.

(A) When ammonia $(NH_3)$ is passed over heated copper$(II)$ oxide $(CuO)$,it acts as a reducing agent and reduces $CuO$ to copper $(Cu)$,while ammonia itself is oxidized to nitrogen gas $(N_2)$.
The chemical reaction is: $2NH_3 + 3CuO \rightarrow N_2 + 3Cu + 3H_2O$.

(C) When ammonium salts are heated,they generally decompose to release ammonia $(NH_3)$.
$1$. $(NH_4)_2SO_4 \xrightarrow{\Delta} 2NH_3 + H_2SO_4$ (or related products).
$2$. $(NH_4)_2CO_3 \xrightarrow{\Delta} 2NH_3 + CO_2 + H_2O$.
$3$. $NH_4Cl \xrightarrow{\Delta} NH_3 + HCl$.
$4$. $NH_4NO_2 \xrightarrow{\Delta} N_2 + 2H_2O$.
In the case of $NH_4NO_2$,the ammonium ion is oxidized by the nitrite ion,resulting in the formation of nitrogen gas $(N_2)$ and water,rather than ammonia $(NH_3)$.

(A) The stability of hydrides of Group $15$ elements decreases down the group.
As the size of the central atom increases from $N$ to $Bi$,the $M-H$ bond length increases,which leads to a decrease in the bond dissociation enthalpy.
Therefore,the thermal stability follows the order: $NH_3 > PH_3 > AsH_3 > SbH_3 > BiH_3$.
Thus,the correct option is $A$.

(B) The basicity of an acid is determined by the number of ionizable $OH$ groups present in its structure.
$H_3PO_3$ (phosphorous acid) has the structure $P(OH)_2(H)O$.
It contains two $OH$ groups attached to the phosphorus atom,which are ionizable.
Therefore,it is a dibasic acid and can form two series of salts (e.g.,$NaH_2PO_3$ and $Na_2HPO_3$).

(A) Metaphosphoric acid has the general formula $(HPO_3)_n$.
For $n = 3$,the formula is $(HPO_3)_3$ or $H_3P_3O_9$,which is known as cyclotrimetaphosphoric acid.
Option $A$ represents a cyclic metaphosphoric acid.

(A) Nitrogen belongs to the second period and has the electronic configuration $[He] 2s^2 2p^3$. It lacks $d$-orbitals in its valence shell,so it cannot expand its octet to form $NCl_5$.
Phosphorus belongs to the third period and has the electronic configuration $[Ne] 3s^2 3p^3$. It possesses vacant $3d$-orbitals,which allow it to undergo $sp^3d$ hybridization to form $PCl_5$ by expanding its octet.

(C) In group $-15$,the metallic character increases as we move down the group from $N$ to $Bi$.
Oxides of non-metals are acidic,while oxides of metals are basic.
Since the metallic character increases down the group,the acidic nature of $M_2O_3$ type oxides decreases and the basic nature increases.
Therefore,the oxides become more basic as we move from $N$ to $Bi$.

(A) During lightning,the high temperature in the atmosphere causes the reaction between atmospheric nitrogen $(N_2)$ and oxygen $(O_2)$ to form nitric oxide $(NO)$.
The chemical equation is: $N_2(g) + O_2(g) \xrightarrow{\text{lightning}} 2NO(g)$.

(A) The thermal decomposition of ammonium dichromate $(NH_4)_2Cr_2O_7$ is given by: $(NH_4)_2Cr_2O_7 \rightarrow N_2 + 4H_2O + Cr_2O_3$. The gas produced is nitrogen $(N_2)$.
Similarly,heating ammonium nitrite $(NH_4NO_2)$ also produces nitrogen gas: $NH_4NO_2 \rightarrow N_2 + 2H_2O$.

(B) The reaction between equimolar amounts of $NO$ and $NO_2$ at $-30\,^oC$ results in the formation of dinitrogen trioxide $(N_2O_3)$.
The chemical equation is: $NO(g) + NO_2(g) \xrightarrow{-30\,^oC} N_2O_3(l)$.

(A) Oxides are classified as acidic,basic,amphoteric,or neutral based on their reaction with acids and bases.
$NO$ (Nitric oxide) is a well-known neutral oxide.
$CO_2$ is an acidic oxide.
$CaO$ is a basic oxide.
$ZnO$ is an amphoteric oxide.
Therefore,the correct option is $A$.

$(A)$ The $N_2$ molecule contains a $N \equiv N$ triple bond, while the $O_2$ molecule contains an $O = O$ double bond.
The bond dissociation energy of the $N \equiv N$ triple bond is significantly higher than that of the $O = O$ double bond, making $N_2$ chemically inert or less reactive under normal conditions.
The bond length of $N_2$ $(109.8 \ pm)$ is indeed shorter than that of $O_2$ $(121 \ pm)$ due to the higher bond order.
Since the high bond dissociation energy (a consequence of the short, strong triple bond) is the reason for the low reactivity of $N_2$, the Reason correctly explains the Assertion.
Therefore, both $A$ and $R$ are true and $R$ is the correct explanation of $A$.

(D) $Ge(II)$ compounds tend to lose electrons to reach the more stable $Ge(IV)$ oxidation state,making them powerful reducing agents.
$Pb(IV)$ compounds tend to gain electrons to reach the more stable $Pb(II)$ oxidation state due to the inert pair effect,making them strong oxidizing agents.
The inert pair effect becomes more pronounced as we move down the group from $Ge$ to $Pb$,stabilizing the lower oxidation state $(+2)$ over the higher oxidation state $(+4)$.

(A) Among the given options,$CuSO_{4(aq)}$ and $KCN$ react to form an unstable copper $(II)$ cyanide,which rapidly decomposes to give copper $(I)$ cyanide and cyanogen gas.
The chemical equations are:
$2CuSO_{4(aq)} + 4KCN \to 2Cu(CN)_2 + 2K_2SO_4$
$2Cu(CN)_2 \to 2CuCN + (CN)_2 \uparrow$ (Cyanogen)

(A) Due to the low ignition temperature of white phosphorus,it undergoes oxidation in the presence of air,which slowly raises its temperature,and after a few moments,it catches fire spontaneously. Due to this reason,it is stored under water.
Ignition temperature of red phosphorus is high.
Black phosphorus is crystalline in nature.
Phosphorus forms a number of hydrides,such as $PH_3$ and $P_2H_4$.

(C) $N$ forms $NCl_3$,$N_2O_5$ and $Ca_3N_2$.
Nitrogen,due to the absence of empty $d$-orbitals,cannot extend its covalency beyond $3$ and hence does not form $NCl_5$.
Due to its small size and high electronegativity,it can accept $3$ electrons to form the $N^{3-}$ ion in $Ca_3N_2$.

(B) The hydrides $MH_3$ have a pyramidal shape with a lone pair of electrons on the central atom. The $H-M-H$ bond angle is less than the tetrahedral angle of $109^o 28'$ due to lone pair-bond pair repulsion.
As the electronegativity of the central atom $M$ decreases from $N$ to $Sb$,the bond pair of electrons shifts further away from the central atom.
Consequently,the bond angle decreases and approaches $90^o$.
This indicates that the bonding involves almost pure $p-$ orbitals of the central atom,with very little $s-$ character involved in the hybridization.

(D) The acidity of oxides generally increases with an increase in the oxidation state of the central atom and with an increase in the electronegativity of the central atom.
$Ag_2O$ is basic,$V_2O_5$ is amphoteric,$CO$ is neutral,and $N_2O_5$ is strongly acidic.
In $N_2O_5$,the oxidation state of $N$ is $+5$ and $N$ has high electronegativity,which makes the $N-O$ bond more polar and the oxide more acidic compared to the others.

(C) $P_4O_{10}$ is an acidic oxide,while $NH_3$ is a basic gas.
When $P_4O_{10}$ is used to dry $NH_3$,it reacts with $NH_3$ to form ammonium phosphate.
The reaction is: $P_4O_{10} + 6H_2O \to 4H_3PO_4$ followed by $H_3PO_4 + 3NH_3 \to (NH_4)_3PO_4$.
Therefore,it cannot be used as a drying agent for $NH_3$.

(A) $NF_5$ does not exist because nitrogen $(N)$ cannot form pentahalides due to the absence of $d$-orbitals in its valence shell.
$P$,$As$,and $Sb$ can form pentahalides of the general formula $MX_5$ (where $M = P, As, Sb$) because they possess vacant $d$-orbitals in their respective valence shells,allowing for the expansion of their octet.

(B) The Assertion is correct because nitrogen lacks $d-$orbitals in its valence shell,preventing it from expanding its octet to form $NX_5$ type compounds.
The Reason is also correct because phosphorus $(2.19)$ has a lower electronegativity than nitrogen $(3.04)$.
However,the inability of nitrogen to form pentahalides is due to the absence of $d-$orbitals,not due to its electronegativity. Therefore,the Reason is not the correct explanation of the Assertion.

(A) White phosphorus exists as discrete $P_4$ tetrahedral molecules where the $P-P-P$ bond angle is $60^o$.
This small bond angle leads to significant angular strain,making the molecule highly reactive.
Red phosphorus,on the other hand,consists of $P_4$ tetrahedral units linked together through covalent bonds to form a polymeric chain structure.
This polymeric structure reduces the angular strain,making red phosphorus significantly less reactive than white phosphorus.
Therefore,both the Assertion and the Reason are correct,and the Reason correctly explains why white phosphorus is more reactive.

(D) Pure nitrogen is obtained by the thermal decomposition of sodium azide or barium azide: $2NaN_3 \xrightarrow{\Delta} 2Na + 3N_2$ and $Ba(N_3)_2 \xrightarrow{\Delta} Ba + 3N_2$. $(a-iv)$
Haber process is used for the industrial manufacture of ammonia: $N_2 + 3H_2 \rightleftharpoons 2NH_3$. $(b-iii)$
Contact process is used for the manufacture of sulphuric acid $(H_2SO_4)$. $(c-ii)$
Deacon's process is used for the formation of chlorine gas: $4HCl + O_2 \xrightarrow{CuCl_2} 2H_2O + 2Cl_2$. $(d-i)$
Therefore,the correct matching is $a-iv, b-iii, c-ii, d-i$.

(C) The reducing property of phosphorus oxoacids is directly proportional to the number of $P-H$ bonds present in the molecule.
$H_3PO_2$ (hypophosphorous acid) contains two $P-H$ bonds.
$H_3PO_3$ (phosphorous acid) contains one $P-H$ bond.
$H_3PO_4$ (phosphoric acid) contains zero $P-H$ bonds.
$H_4P_2O_7$ (pyrophosphoric acid) contains zero $P-H$ bonds.
Since $H_3PO_2$ has the maximum number of $P-H$ bonds,it acts as the strongest reducing agent among the given options.

(B) The reaction of white phosphorus $(P_{4})$ with concentrated $NaOH$ is a disproportionation reaction:
$P_{4} + 3 NaOH + 3 H_{2}O \longrightarrow PH_{3} + 3 NaH_{2}PO_{2} (X)$
Here,$(X)$ is sodium hypophosphite $(NaH_{2}PO_{2})$.
Upon acidification with $HCl$,$(X)$ forms hypophosphorous acid $(H_{3}PO_{2})$:
$NaH_{2}PO_{2} + HCl \longrightarrow NaCl + H_{3}PO_{2} (Y)$
In $H_{3}PO_{2}$,there is only one $P-OH$ bond,while two $P-H$ bonds are directly attached to the phosphorus atom. Since only the hydrogen atom attached to the oxygen atom is ionizable,the basicity of $H_{3}PO_{2}$ is $1$.

(B) Although phosphorus exhibits $+3$ and $+5$ oxidation states,it cannot form $PH_{5}$.
High $\Delta_{a} H$ (enthalpy of atomization) value of dihydrogen and $\Delta_{e g} H$ (electron gain enthalpy) value of hydrogen do not favour the formation of $PH_{5}$.
Consequently,the energy released during the formation of $5$ $P-H$ bonds is not sufficient to overcome the energy required for the excitation of electrons and the atomization of $H_{2}$.

(N/A) Boron $(B)$ and Thallium $(Tl)$ belong to group $13$ of the periodic table.
In this group,the stability of the $+1$ oxidation state increases down the group due to the inert pair effect.
$BCl_3$ is more stable than $TlCl_3$ because the $+3$ oxidation state of $B$ is more stable than the $+3$ oxidation state of $Tl$.
In $Tl$,the $+3$ oxidation state is highly oxidizing and it tends to revert to the more stable $+1$ oxidation state.

(N/A) Thallium belongs to group $13$ of the periodic table. The most common oxidation state for this group is $+3$. However,heavier members of this group also display the $+1$ oxidation state due to the inert pair effect.
$1$. Resemblance with aluminium: Thallium exhibits the $+3$ oxidation state in compounds like $TlCl_3$ and $Tl_2O_3$,similar to aluminium ($AlCl_3$ and $Al_2O_3$).
$2$. Resemblance with group $1$ metals: Thallium exhibits the $+1$ oxidation state in compounds like $TlCl$ and $Tl_2O$,which is the characteristic oxidation state of alkali metals (group $1$).

(N/A) The $p$-block elements consist of six groups in the periodic table,numbered from $13$ to $18$. Boron,carbon,nitrogen,oxygen,fluorine,and helium are the head elements of these groups. Their valence shell electronic configuration is $ns^{2} np^{1-6}$ (except for $He$).
The inner core of the electronic configuration may differ,which significantly influences their physical properties (such as atomic and ionic radii,ionisation enthalpy,etc.) as well as their chemical properties.
Consequently,a lot of variation in the properties of elements within a $p$-block group is observed. The maximum oxidation state shown by a $p$-block element is equal to the total number of valence electrons (i.e.,the sum of the $s$ and $p$-electrons).
The important oxidation states exhibited by $p$-block elements are shown in the table below. In the boron,carbon,and nitrogen families,the group oxidation state is the most stable state for the lighter elements in the group.
However,the oxidation state two units less than the group oxidation state becomes progressively more stable for the heavier elements in each group. The occurrence of oxidation states two units less than the group oxidation states is sometimes attributed to the 'inert pair effect'.

Property	Group $13$	Group $14$	Group $15$	Group $16$	Group $17$	Group $18$
General electronic configuration	$ns^{2} np^{1}$	$ns^{2} np^{2}$	$ns^{2} np^{3}$	$ns^{2} np^{4}$	$ns^{2} np^{5}$	$ns^{2} np^{6}$ ($1s^{2}$ for $He$)
First member of group	$B$	$C$	$N$	$O$	$F$	$He$
Group oxidation state	$+3$	$+4$	$+5$	$+6$	$+7$	$+8$
Other oxidation states	$+1$	$+2, -4$	$+3, -3$	$+4, +2, -2$	$+5, +3, +1, -1$	$+6, +4, +2$

(N/A) It is necessary to remove $CO$ during the synthesis of ammonia because $CO$ acts as a poison for the iron catalyst used in Haber's process,which significantly reduces its catalytic activity.

(N/A) Nitrogen has the electronic configuration $1s^2 2s^2 2p^3$.
It has only $s$ and $p$ orbitals in its valence shell $(n=2)$.
It lacks $d$ orbitals in its valence shell,which prevents it from expanding its covalency beyond $4$.
Therefore,nitrogen cannot form pentahalides like $NF_5$.

(B) $NO_2$ contains an odd number of valence electrons ($17$ valence electrons).
It behaves as a typical odd electron molecule and is paramagnetic.
On dimerisation,it forms $N_2O_4$,which has an even number of electrons and is diamagnetic,making it more stable.

(N/A) $PH_3$ reacts with acids like $HI$ to form $PH_4I$,which indicates its basic nature.
$PH_3 + HI \rightarrow PH_4I$
Due to the presence of a lone pair of electrons on the phosphorus atom,$PH_3$ acts as a Lewis base in this reaction.

(B) $PCl_3$ undergoes hydrolysis in the presence of moisture to produce phosphorous acid $(H_3PO_3)$ and hydrogen chloride $(HCl)$ gas.
The $HCl$ gas released reacts with atmospheric moisture to form dense white fumes.
The chemical equation is: $PCl_3 + 3H_2O \rightarrow H_3PO_3 + 3HCl$.

(N/A) The structure of $H_{3}PO_{2}$ (hypophosphorous acid) contains two $P-H$ bonds and one $P=O$ bond,along with one $P-OH$ bond.
Because of the presence of two $P-H$ bonds,$H_{3}PO_{2}$ acts as a strong reducing agent.
The $H$ atoms bonded directly to the $P$ atom are responsible for its reducing character.

(N/A) In pentahalides,the central atom is in the $+5$ oxidation state,whereas in trihalides,it is in the $+3$ oxidation state.
According to Fajan's rule,a higher positive charge on the cation increases its polarizing power.
Since the central atom in pentahalides has a higher positive charge $(+5)$ compared to trihalides $(+3)$,it exerts a stronger polarizing effect on the electron cloud of the halide ions.
This leads to greater sharing of electrons,making pentahalides more covalent than their corresponding trihalides.