Nitrogen family Questions in English

(B) In warfare,$PH_3$ (phosphine) is used to create smoke screens. When $PH_3$ is released,it reacts with air to form dense white fumes of phosphoric acid $(H_3PO_4)$ and water vapor,which act as a smoke screen.

(C) The acidic character of non-metal oxides increases with the increase in the oxidation state of the central atom and the electronegativity of the central atom.
$1$. Oxidation states: $N_2O_5 (+5)$,$SO_2 (+4)$,$CO_2 (+4)$,$CO (+2)$.
$2$. $N_2O_5$ is the strongest acid because $N$ is highly electronegative and in its highest oxidation state $(+5)$.
$3$. Comparing $SO_2$ and $CO_2$: $S$ is less electronegative than $C$,but $S$ is larger and can stabilize the negative charge better in the conjugate base,or by comparing their hydration products ($H_2SO_3$ vs $H_2CO_3$),$H_2SO_3$ is a stronger acid.
$4$. $CO$ is a neutral oxide.
$5$. Therefore,the correct order is $N_2O_5 > SO_2 > CO_2 > CO$.

(C) Phosphorus exists in several allotropic forms,including white,red,and black phosphorus.
White phosphorus is the least stable due to high angular strain in the $P_4$ molecule.
Red phosphorus is more stable than white phosphorus.
Black phosphorus is the most thermodynamically stable allotrope of phosphorus under standard conditions because of its layered polymeric structure.

(C) Both $NH_3$ and $PH_3$ are hydrides of Group $15$ elements.
Both molecules have a lone pair of electrons on the central atom ($N$ and $P$ respectively).
Due to the presence of this lone pair,both act as Lewis bases,exhibiting basic nature.

(C) White phosphorus $(P_4)$ consists of four phosphorus atoms arranged at the corners of a regular tetrahedron. Each phosphorus atom is bonded to the other three phosphorus atoms by single covalent bonds. The bond angle is $60^\circ$.

(A) $1$. $PH_3$ (phosphine) is a much weaker Lewis base than $NH_3$ (ammonia) because the lone pair on $P$ is in a larger $3sp^3$ orbital,making it less available for donation compared to the $2sp^3$ orbital of $N$ in $NH_3$.
$2$. $NH_3$ is more basic than $PH_3$.
$3$. $PH_3$ is highly toxic,whereas $NH_3$ is less toxic.
$4$. The electronegativity of $N$ $(3.04)$ is higher than $P$ $(2.19)$.
$5$. Therefore,the correct statement is that $PH_3$ is less basic than $NH_3$.

(A) The acidic character of oxides of elements in a group decreases as we move down the group because the metallic character increases.
For the group $15$ elements,the oxides are $N_2O_3, P_2O_3, As_2O_3, Sb_2O_3, Bi_2O_3$.
$N_2O_3$ and $P_2O_3$ are acidic,$As_2O_3$ and $Sb_2O_3$ are amphoteric,and $Bi_2O_3$ is basic.
Among the given options,$P_2O_3$ is the most acidic because phosphorus is the least metallic (most non-metallic) element among the choices provided.

(A) Nitrogen has a small atomic size and high electronegativity,which allows it to form stable $p\pi - p\pi$ multiple bonds,leading to the formation of diatomic $N_2$ molecules.
In contrast,phosphorus has a larger atomic size,which makes the $p\pi - p\pi$ overlap ineffective and weak.
Therefore,phosphorus prefers to form single $P-P$ bonds,resulting in a tetrahedral $P_4$ structure where each phosphorus atom is bonded to three other phosphorus atoms.

(D) Nitrogen $(N_2)$ exists as a diatomic molecule with a triple bond between the two nitrogen atoms $(N \equiv N)$.
This triple bond has a very high bond dissociation energy of $941.4 \ kJ \ mol^{-1}$.
Due to this exceptionally strong bond,nitrogen is chemically inert under standard conditions.

(D) In the nitrogen family (Group $15$),metallic character increases as we move down the group due to the increase in atomic size and decrease in ionization enthalpy.
$P$ (Phosphorus) is a non-metal.
$As$ (Arsenic) and $Sb$ (Antimony) are metalloids.
$Bi$ (Bismuth) is a metal.
Therefore,$Bi$ is the most metallic element among the given options.

(C) Smoke screens are produced using calcium phosphide $(Ca_3P_2)$.
When $Ca_3P_2$ comes in contact with water,it undergoes hydrolysis to produce phosphine gas $(PH_3)$.
The reaction is: $Ca_3P_2 + 6H_2O \rightarrow 3Ca(OH)_2 + 2PH_3$.
Phosphine gas is spontaneously inflammable in air and produces dense white fumes of phosphoric acid $(H_3PO_4)$,which act as a smoke screen.

(A) The structure of phosphorus trioxide $(P_4O_6)$ consists of four $P$ atoms at the corners of a tetrahedron,with six $O$ atoms bridging the six $P-P$ edges. Thus,it contains $6$ $P-O-P$ bridges.
In phosphorus pentoxide $(P_4O_{10})$,each $P$ atom is also bonded to an additional terminal oxygen atom via a double bond. The core structure remains the same as $P_4O_6$,meaning it also contains $6$ $P-O-P$ bridges.
Therefore,the number of $P-O-P$ bridges in both $P_4O_{10}$ and $P_4O_6$ is $6$ and $6$ respectively.

(C) The reaction between concentrated $H_2SO_4$ and $KNO_3$ is as follows:
$2KNO_3 + H_2SO_4 \rightarrow K_2SO_4 + 2HNO_3$
Upon further heating,the $HNO_3$ decomposes to produce brown fumes of nitrogen dioxide $(NO_2)$:
$4HNO_3 \rightarrow 4NO_2 + 2H_2O + O_2$
Thus,the brown fumes are due to $NO_2$.

(B) $1$. White phosphorus $(P_4)$ reacts with $NaOH$ solution in an inert atmosphere to produce phosphine $(PH_3)$: $P_4 + 3NaOH + 3H_2O \rightarrow PH_3 + 3NaH_2PO_2$.
$2$. Red phosphorus is chemically much less reactive than white phosphorus and does not react with $NaOH$ to produce phosphine under normal conditions.
$3$. Calcium phosphide $(Ca_3P_2)$ reacts with water to produce phosphine: $Ca_3P_2 + 6H_2O \rightarrow 3Ca(OH)_2 + 2PH_3$.
$4$. Phosphonium iodide $(PH_4I)$ reacts with $NaOH$ to liberate phosphine: $PH_4I + NaOH \rightarrow PH_3 + NaI + H_2O$.
Therefore,the reaction that does not produce phosphine is heating red $P$ with $NaOH$.

(D) The reaction of phosphorus pentoxide $(P_4O_{10})$ with water is a hydrolysis reaction.
$P_4O_{10} + 6H_2O \rightarrow 4H_3PO_4$.
This product,$H_3PO_4$,is known as orthophosphoric acid.

(C) When white phosphorus $(P_4)$ is boiled with a concentrated solution of caustic soda $(NaOH)$ in an inert atmosphere,it undergoes disproportionation reaction to form phosphine $(PH_3)$ and sodium hypophosphite $(NaH_2PO_2)$.
The chemical equation is: $P_4 + 3NaOH + 3H_2O \rightarrow PH_3 + 3NaH_2PO_2$.

(D) Phosphorus pentoxide $(P_4O_{10})$ has a very strong affinity for water.
It is widely used in laboratories and industries as a powerful dehydrating agent to remove water from various organic and inorganic compounds.

(D) The stability of the $+5$ oxidation state decreases down the group due to the inert pair effect. Nitrogen cannot form $NF_5$ because it lacks $d$-orbitals. However,among the given options,$BiF_5$ is the least stable pentahalide. While $BiF_5$ does exist,it is a very strong fluorinating agent and is highly unstable compared to the others. In many contexts,$NF_5$ is considered impossible,but since it is not in the options,we look for the most unstable one. Actually,all $PF_5, AsF_5, SbF_5, BiF_5$ exist. If the question implies which one is not possible due to electronic configuration,$NF_5$ is the answer. Given the options provided,there might be a conceptual error in the question as all listed pentahalides are known. However,if we must choose,$BiF_5$ is often cited for its extreme reactivity and instability due to the inert pair effect making the $+5$ state difficult to maintain.

(A) The hydrides of group $-15$ elements are $NH_3, PH_3, AsH_3, SbH_3,$ and $BiH_3$.
These hydrides act as Lewis bases due to the presence of a lone pair of electrons on the central atom.
As we move down the group,the size of the central atom increases,and the electron density on the lone pair decreases.
Consequently,the availability of the lone pair for donation decreases,making the basic strength decrease down the group.
Therefore,$NH_3$ is the most basic hydride among them.

(C) When excess water is added to a solution of $BiCl_3$,it undergoes hydrolysis to form a white precipitate of bismuth oxychloride $(BiOCl)$.
The chemical reaction is as follows:
$BiCl_3 + H_2O \rightarrow BiOCl(s) + 2HCl(aq)$

(B) As we move down the group from $N$ to $Sb$,the electronegativity of the central atom $(M)$ decreases.
Due to the decrease in electronegativity,the central atom prefers to use its pure $p$-orbitals for bonding with hydrogen atoms,rather than undergoing $sp^3$ hybridization.
Since $p$-orbitals are oriented at $90^o$ to each other,the $H-M-H$ bond angle in the hydrides ($NH_3$ to $SbH_3$) decreases and approaches $90^o$.

(C) The enthalpy of vaporization depends on the strength of intermolecular forces.
In $PH_3$ and $AsH_3$,the intermolecular forces are primarily weak van der Waals forces,which increase with increasing molecular mass $(PH_3 < AsH_3)$.
However,$NH_3$ molecules exhibit strong intermolecular hydrogen bonding.
Due to this strong hydrogen bonding,$NH_3$ has a significantly higher enthalpy of vaporization compared to $PH_3$ and $AsH_3$.
Therefore,the correct increasing order is $PH_3 < AsH_3 < NH_3$.

(B) $1$. Hydrolysis of $NCl_3$ yields $NH_3$ and $HOCl$ $(NCl_3 + 3H_2O \rightarrow NH_3 + 3HOCl)$. This statement is correct.
$2$. $NH_3$ is more stable than $PH_3$ because the $N-H$ bond energy is higher than the $P-H$ bond energy due to the smaller size of the $N$ atom. Thus,the statement that $NH_3$ is less stable than $PH_3$ is incorrect.
$3$. $NH_3$ is a weaker reducing agent than $PH_3$ because the $P-H$ bond is weaker and breaks more easily to release hydrogen. This statement is correct.
$4$. $NO$ is paramagnetic in the gaseous state due to an unpaired electron,but in the solid state,it exists as a dimer $(N_2O_2)$,which is diamagnetic. This statement is correct.

(B) Phosphine $(PH_3)$ is prepared in the laboratory by heating white phosphorus with concentrated sodium hydroxide solution $(NaOH)$ in an inert atmosphere of $CO_2$.
The chemical reaction is:
$P_4 + 3NaOH + 3H_2O \rightarrow PH_3 + 3NaH_2PO_2$.

(C) The structure of $P_4O_6$ consists of a $P_4$ tetrahedron where each phosphorus atom is bonded to three oxygen atoms.
Each oxygen atom acts as a bridge between two phosphorus atoms ($P$-$O$-$P$ linkage).
Therefore,each phosphorus atom is bonded to $3$ oxygen atoms.

(B) The hydrolysis of phosphorus trichloride $(PCl_3)$ with water is given by the following chemical equation:
$PCl_3 + 3H_2O \rightarrow H_3PO_3 + 3HCl$.
In this reaction,$PCl_3$ reacts with water to form phosphorous acid $(H_3PO_3)$ and hydrogen chloride $(HCl)$.

(A) In the industrial production of phosphorus,calcined phosphate rock (calcium phosphate) is heated with coke and sand in an electric furnace at $1773 \ K$.
The chemical reaction is:
$2Ca_3(PO_4)_2 + 6SiO_2 + 10C \rightarrow 6CaSiO_3 + 10CO + P_4$.
Here,$P_4$ (white phosphorus) is the main product,while $CaSiO_3$ (calcium silicate) is formed as a slag.
Among the given options,phosphorus is the correct product.

(B) The reaction of ammonia with chlorine depends on the relative amounts of the reactants.
When ammonia is in excess,the reaction is: $8NH_3 + 3Cl_2 \rightarrow 6NH_4Cl + N_2$.
When chlorine is in excess,the reaction is: $NH_3 + 3Cl_2 \rightarrow NCl_3 + 3HCl$.
Since the question specifies excess chlorine,the products are nitrogen trichloride $(NCl_3)$ and hydrogen chloride $(HCl)$.

(D) The basicity of an oxoacid of phosphorus is determined by the number of $P-OH$ groups present in its structure.
$1$. Orthophosphoric acid $(H_3PO_4)$ has $3$ $P-OH$ groups,so it is tribasic.
$2$. Hypophosphorous acid $(H_3PO_2)$ has $1$ $P-OH$ group,so it is monobasic.
$3$. Metaphosphoric acid $(HPO_3)_n$ is generally monobasic per unit.
$4$. Pyrophosphoric acid $(H_4P_2O_7)$ has $4$ $P-OH$ groups,making it a tetrabasic acid.

(B) Metaphosphoric acid is represented by the formula $(HPO_3)_n$.
Among the given options,$HPO_3$ represents the empirical formula of metaphosphoric acid.
$H_3PO_4$ is orthophosphoric acid,$H_3PO_3$ is orthophosphorous acid,and $H_3PO_2$ is hypophosphorous acid.

(A) The anhydride of an acid is obtained by removing water molecules from the acid until no hydrogen remains.
For orthophosphoric acid $(H_3PO_4)$:
$4H_3PO_4 \rightarrow 2P_2O_5 + 6H_2O$
Since $P_2O_5$ exists as a dimer,the formula is $P_4O_{10}$.
Thus,$P_4O_{10}$ is the anhydride of orthophosphoric acid.

(D) Orthophosphoric acid $(H_3PO_4)$ undergoes dehydration upon heating.
At $600\,^oC$,it loses water molecules to form metaphosphoric acid $(HPO_3)$.
The reaction is: $H_3PO_4 \xrightarrow{\Delta} HPO_3 + H_2O$.

(D) In the nitrogen family (Group $15$),as we move down the group,the metallic character increases due to the increase in atomic size and decrease in ionization energy.
Nitrogen and phosphorus are non-metals,arsenic and antimony are metalloids,and bismuth is a metal.
Ionic character is observed when a metal reacts with a non-metal.
$NI_3$,$NF_3$,and $NCl_3$ are covalent compounds because nitrogen is a non-metal.
$BiF_3$ is an ionic compound because bismuth is a metal and fluorine is a highly electronegative non-metal.

(D) The molecule $N_2O_4$ (dinitrogen tetroxide) is formed by the dimerization of two $NO_2$ molecules.
In $NO_2$,the nitrogen atom has one unpaired electron,making it paramagnetic.
When two $NO_2$ molecules combine to form $N_2O_4$,the two unpaired electrons on the nitrogen atoms form a $N-N$ single bond.
As a result,$N_2O_4$ has no unpaired electrons,which makes it diamagnetic.

(B) Pure nitric acid $(HNO_3)$ is a colorless liquid. When exposed to light,it undergoes photochemical decomposition to form nitrogen dioxide $(NO_2)$,water $(H_2O)$,and oxygen $(O_2)$.
The chemical equation for this reaction is: $4HNO_3 \rightarrow 4NO_2 + 2H_2O + O_2$.
The brown color is due to the presence of $NO_2$ gas.

(A) The $HNO_3$ (nitric acid) is commercially manufactured by the $Ostwald$ process.
In this process,ammonia $(NH_3)$ is catalytically oxidized to nitric oxide $(NO)$,which is then oxidized to nitrogen dioxide $(NO_2)$.
Finally,$NO_2$ is absorbed in water to produce $HNO_3$ with high yield.

(C) The stability of nitrogen trihalides towards hydrolysis depends on the presence of vacant $d$-orbitals on the central atom and the electronegativity of the halogen atom.
$NF_3$ is the only nitrogen trihalide that is stable and does not undergo hydrolysis by water.
This is because nitrogen does not have vacant $d$-orbitals to accept the lone pair of electrons from water,and the $N-F$ bond is very strong due to high electronegativity of fluorine,making the attack of water molecules difficult.
In contrast,other nitrogen trihalides like $NCl_3$,$NBr_3$,and $NI_3$ undergo hydrolysis because the larger halogen atoms can facilitate the reaction or have lower bond energies.

(C) Black phosphorus is the most stable allotrope of phosphorus.
It exists in two forms: $\alpha$-black phosphorus and $\beta$-black phosphorus.
$\alpha$-black phosphorus is formed when red phosphorus is heated in a sealed tube at $803 \ K$.
$\beta$-black phosphorus is prepared by heating white phosphorus at $473 \ K$ under high pressure.
It has a layered,graphite-like structure and is a good conductor of electricity due to the presence of delocalized electrons in its layered structure.

(B) To determine the number of $P-H$ bonds,we examine the structures of the given oxoacids of phosphorus:
$1$. $H_3PO_2$ (Hypophosphorous acid): It has two $P-H$ bonds and one $P=O$ bond.
$2$. $H_3PO_3$ (Phosphorous acid): It has one $P-H$ bond and one $P=O$ bond.
$3$. $H_3PO_4$ (Orthophosphoric acid): It has zero $P-H$ bonds and one $P=O$ bond.
$4$. $H_4P_2O_7$ (Pyrophosphoric acid): It has zero $P-H$ bonds.
Comparing these,$H_3PO_2$ contains the highest number of $P-H$ bonds (two).

(A) Active nitrogen is produced by passing an electric discharge through $N_2$ gas at very low pressure.
Under these conditions,the $N_2$ molecules dissociate into highly reactive nitrogen atoms,which are referred to as active nitrogen.

(C) The oxide $N_2O_4$ (dinitrogen tetroxide) is a mixed anhydride of two acids,namely nitrous acid $(HNO_2)$ and nitric acid $(HNO_3)$.
The reaction is given by: $N_2O_4 + H_2O \rightarrow HNO_2 + HNO_3$.

(D) The basicity of nitrogen trihalides $(NX_3)$ depends on the availability of the lone pair of electrons on the nitrogen atom.
In $NF_3$,the fluorine atom is highly electronegative.
It exerts a strong $-I$ effect,which pulls the electron density away from the nitrogen atom towards itself.
This significantly reduces the availability of the lone pair on the nitrogen atom,making $NF_3$ the least basic among the nitrogen trihalides.
Therefore,the correct option is $D$.

(A) The dehydration of nitric acid $(HNO_3)$ by phosphorus pentoxide $(P_4O_{10})$ is a standard laboratory method for the preparation of dinitrogen pentoxide $(N_2O_5)$.
The chemical equation for the reaction is:
$4HNO_3 + P_4O_{10} \rightarrow 2N_2O_5 + 4HPO_3$.

(D) Heating $NH_4Cl$ gives $NH_3$ and $HCl$.
Heating $(NH_4)_2SO_4$ gives $NH_3$ and $H_2SO_4$.
Heating $NH_4H_2PO_4$ gives $NH_3$ and $H_3PO_4$.
Heating $NH_4NO_2$ (Ammonium nitrite) undergoes thermal decomposition to produce nitrogen gas $(N_2)$ and water $(H_2O)$,not ammonia $(NH_3)$.
The reaction is: $NH_4NO_2 \rightarrow N_2 + 2H_2O$.

(B) Ammonia $(NH_3)$ is a basic gas. To dry it,a basic drying agent must be used.
$P_4O_{10}$ is acidic,$H_2SO_4$ is acidic,and $CaCl_2$ forms an adduct with $NH_3$ $(CaCl_2 \cdot 8NH_3)$.
Therefore,quicklime $(CaO)$ is used as a drying agent for ammonia because it is basic and does not react with $NH_3$.

(C) Cyclic metaphosphoric acid is represented by the formula $(HPO_3)_3$,which is also known as trimetaphosphoric acid.
In its cyclic structure,three $PO_4$ tetrahedra are linked together by sharing oxygen atoms.
The structure consists of a six-membered ring containing alternating phosphorus and oxygen atoms $(P-O-P-O-P-O)$.
In this cyclic structure,there are $3$ $P-O-P$ linkages.

(C) The brown ring test is used to detect the presence of nitrate ions $(NO_3^-)$ in a solution.
When $FeSO_4$ is added to a solution containing nitrate ions followed by the addition of concentrated $H_2SO_4$ along the sides of the test tube,a brown ring is formed at the junction of the two layers.
The brown ring is due to the formation of the complex $[Fe(H_2O)_5(NO)]SO_4$.
In this reaction,the nitrate ion is reduced to nitric oxide $(NO)$,which then reacts with the hydrated ferrous ion to form the brown-colored complex.

(B) Pure $N_2$ is prepared by the thermal decomposition of sodium azide or barium azide. However,in the laboratory,it is commonly prepared by the reaction of aqueous solutions of ammonium chloride $(NH_4Cl)$ and sodium nitrite $(NaNO_2)$:
$NH_4Cl(aq) + NaNO_2(aq) \rightarrow N_2(g) + 2H_2O(l) + NaCl(aq)$.
This reaction produces $N_2$ gas.