Nitrogen family Questions in English

(A) The structure of phosphorus trioxide $(P_4O_6)$ consists of four phosphorus atoms at the corners of a tetrahedron,with six oxygen atoms bridging the edges between them. Thus,it has $6$ $P-O-P$ bridges.
In phosphorus pentoxide $(P_4O_{10})$,each phosphorus atom is further bonded to an additional terminal oxygen atom via a coordinate 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 $6$ each.

(B) Sodium pyrophosphate is derived from pyrophosphoric acid,which has the formula $H_4P_2O_7$.
By replacing the four hydrogen atoms with sodium ions,we obtain the salt $Na_4P_2O_7$.

(D) $N_2O_5$ (dinitrogen pentoxide) is a white crystalline solid at room temperature,whereas $NO$,$N_2O$,and $N_2O_3$ are gases or liquids under standard conditions.

(D) The reaction of dilute nitric acid $(HNO_3)$ with copper $(Cu)$ produces nitric oxide $(NO)$.
The balanced chemical equation is:
$3Cu + 8HNO_3 (\text{dilute}) \rightarrow 3Cu(NO_3)_2 + 2NO + 4H_2O$
Therefore,the correct option is $(D)$.

(A) $N_2O$ (nitrous oxide) is used as an anaesthetic in dentistry and surgery because of its mild anaesthetic and analgesic properties,often referred to as laughing gas.

(C) When phosphine $(PH_3)$ is mixed with chlorine gas $(Cl_2)$,it reacts vigorously to form phosphorus pentachloride $(PCl_5)$ and hydrogen chloride $(HCl)$.
The balanced chemical equation is: $PH_3 + 4Cl_2 \to PCl_5 + 3HCl$.
This reaction is highly exothermic,meaning the mixture warms up.

(A) The reaction between ammonia $(NH_3)$ and sodium hypochlorite $(NaOCl)$ is a well-known method for the preparation of hydrazine $(N_2H_4)$.
The chemical equation is: $2NH_3 + NaOCl \rightarrow N_2H_4 + NaCl + H_2O$.
However,in the presence of excess sodium hypochlorite or under certain conditions,hydrazine can be further oxidized to nitrogen gas $(N_2)$.
Given the options provided,$N_2$ is the most stable nitrogen-containing product formed in this reaction pathway.

(B) The nitrogen molecule $(N_2)$ consists of two nitrogen atoms linked by a triple bond $(N \equiv N)$.
The bond dissociation energy of this triple bond is extremely high $(941.4 \ kJ \ mol^{-1})$.
Due to this exceptionally high bond dissociation energy,$N_2$ is chemically inert under standard conditions.
Therefore,the correct option is $B$.

(A) The elements of Group $15$ (Nitrogen family) have a valence shell electronic configuration of $ns^2 np^3$.
They can exhibit a maximum covalency of $5$ due to the presence of vacant $d$-orbitals in elements like Phosphorus $(P)$,Arsenic $(As)$,Antimony $(Sb)$,and Bismuth $(Bi)$.
Among the given options,Phosphorus $(P)$ is the most common example that exhibits a valency of $5$ in compounds like $PCl_5$.

(C) Cyclic metaphosphoric acid is represented by the formula $(HPO_3)_3$ or $H_3P_3O_9$.
In its cyclic structure,three $PO_4$ tetrahedra are linked together by sharing oxygen atoms.
This structure contains a six-membered ring consisting of alternating $P$ and $O$ atoms,resulting in $3$ $P-O-P$ bonds.

(D) The Haber's process is the industrial method for the production of ammonia $(NH_3)$.
In this process,nitrogen gas $(N_2)$ and hydrogen gas $(H_2)$ react in the presence of an iron catalyst to produce ammonia: $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$.

(A) Ammonia is prepared by heating a mixture of calcium hydroxide and ammonium chloride.
$Ca(OH)_{2(s)} + 2NH_4Cl_{(s)} \rightarrow CaCl_{2(s)} + 2H_2O_{(l)} + 2NH_{3(g)}$
Concentrated sulphuric acid $(H_2SO_4)$ and anhydrous calcium chloride $(CaCl_2)$ are not used to dry ammonia because they react with it.
Ammonia is a basic gas,so it reacts with acidic drying agents like $H_2SO_4$ and forms an adduct with $CaCl_2$ $(CaCl_2 \cdot 8NH_3)$.
The gas is passed through fresh quicklime ($CaO$,solid calcium oxide lumps) to effectively dry it.
Ammonia is collected by upward delivery as it is lighter than air.

(A) Hypophosphorus acid is $H_3PO_2$.
In its structure,the central phosphorus atom is bonded to one oxygen atom by a double bond $(P=O)$,one hydroxyl group $(-OH)$,and two hydrogen atoms directly attached to the phosphorus atom $(P-H)$.
This structure is represented by option $A$.

(B) The thermal decomposition of ammonium nitrate $(NH_4NO_3)$ at $250 \, ^oC$ yields nitrous oxide $(N_2O)$ and water vapor.
The balanced chemical equation is:
$NH_4NO_3(s) \xrightarrow{\Delta} N_2O(g) + 2H_2O(g)$
Thus,the correct option is $B$.

(C) The oxidation state of phosphorus $(P)$ in its compounds ranges from $-3$ to $+5$.
For example,in $PH_3$,the oxidation state of $P$ is $-3$.
In $H_3PO_4$,the oxidation state of $P$ is $+5$.

(B) The correct answer is $(B)$.
$BiCl_5$ does not exist because the $+3$ oxidation state of $Bi$ is significantly more stable than the $+5$ oxidation state due to the inert pair effect,which makes the formation of $Bi(V)$ compounds thermodynamically unfavorable.

(B) The thermal stability of the hydrides of Group $15$ elements decreases down the group as the size of the central atom increases,which leads to a decrease in the bond dissociation energy.
The order of stability is: $NH_3 > PH_3 > AsH_3 > SbH_3 > BiH_3$.
Therefore,$SbH_3$ is the least stable among the given options.

(A) The basicity of hydrides of group $15$ elements $(NH_3, PH_3, AsH_3, SbH_3, BiH_3)$ depends on the availability of the lone pair of electrons on the central atom.
As we move down the group,the atomic size increases and the electron density on the central atom decreases,making the lone pair less available for donation.
Therefore,the basicity decreases down the group.
$NH_3$ has the highest electron density on the nitrogen atom,making it the most basic hydride among them.

(C) The nitrogen family (Group $15$ elements) consists of Nitrogen $(N)$,Phosphorus $(P)$,Arsenic $(As)$,Antimony $(Sb)$,and Bismuth $(Bi)$.

$I$. Element	$II$. Atomic Number
$1$. $N$	$7$
$2$. $P$	$15$
$3$. $As$	$33$
$4$. $Sb$	$51$
$5$. $Bi$	$83$

The $3$rd member of the nitrogen family is Arsenic $(As)$,which has an atomic number of $33$.

(C) $H_3PO_4$ (orthophosphoric acid) is a tribasic acid because it contains three replaceable hydrogen atoms attached to oxygen atoms as $P-OH$ groups.
The structure is $O=P(OH)_3$.

(D) The thermal decomposition of ammonium dichromate is given by the following reaction:
$(NH_4)_2Cr_2O_7 \to N_2 + Cr_2O_3 + 4H_2O$
As shown in the equation,the gas evolved is nitrogen $(N_2)$.

(D) The reaction of ammonia $(NH_3)$ with hypochlorite anion $(OCl^-)$ is a method for the preparation of hydrazine $(N_2H_4)$.
The balanced chemical equation is:
$2NH_3 + OCl^- \to N_2H_4 + Cl^- + H_2O$
However,in the presence of excess ammonia and specific conditions,the reaction can also produce ammonium chloride $(NH_4Cl)$ as a byproduct.
Therefore,both $NH_4Cl$ and $N_2H_4$ can be formed.

(A) The acidic character of oxides in Group $15$ decreases down the group as the metallic character increases.
$P$ (Phosphorus) is a non-metal and forms a strongly acidic oxide $(P_4O_{10})$.
$As$ (Arsenic) and $Sb$ (Antimony) form amphoteric oxides.
$Bi$ (Bismuth) forms a basic oxide.
Therefore,the correct option is $(A)$.

(D) The electronic configuration of nitrogen $(Z=7)$ is $1s^2, 2s^2, 2p^3$.
Nitrogen belongs to the second period and lacks $d$-orbitals in its valence shell.
Because of the absence of vacant $d$-orbitals,nitrogen cannot expand its octet to form pentahalides (like $NF_5$).
Other elements of group $15$ (like $P, As, Sb$) possess vacant $d$-orbitals,allowing them to exhibit a higher covalency of $5$ and form pentahalides.

(B) As we move from $NH_3$ to $SbH_3$,the electronegativity of the central atom $M$ decreases.
Due to the decrease in electronegativity,the bond pair of electrons shifts away from the central atom towards the hydrogen atom.
According to Drago's rule,as the electronegativity of the central atom decreases,the hybridization of the central atom becomes less significant,and the $M-H$ bonds are formed using almost pure $p$-orbitals of the central atom.
Since $p$-orbitals are oriented at $90^o$ to each other,the bond angle approaches $90^o$.

(A) The highest oxidation state for elements in Group-$15$ is $+5$,which remains unchanged as we move down the group.

(A) . Hypophosphorus acid $(H_3PO_2)$ is a monobasic acid which acts as a strong reducing agent.
In this molecule,two $P-H$ bonds are responsible for its reducing character,and one $O-H$ bond is responsible for its monobasic acid character.

(B) $N_2O$ (Nitrous oxide) is commonly known as Laughing gas because it induces euphoria and laughter when inhaled.

(B) The balanced chemical equation for the reaction between nitric acid $(HNO_3)$ and phosphorus pentoxide $(P_4O_{10})$ is:
$4HNO_3 + P_4O_{10} \to 4HPO_3 + 2N_2O_5$
Comparing this with the given reaction $4HNO_3 + P_4O_{10} \to 4HPO_3 + x$,we find that $x = 2N_2O_5$.
Therefore,the product $x$ is dinitrogen pentoxide $(N_2O_5)$.

(B) The chemical formula of hypophosphorous acid is $H_3PO_2$.
In its structure,the phosphorus atom is bonded to one oxygen atom via a double bond $(P=O)$,one hydroxyl group $(-OH)$,and two hydrogen atoms directly ($P-H$ bonds).
Therefore,there are $2$ hydrogen atoms directly attached to the phosphorus atom.

(B) The reaction between equimolar amounts of nitric oxide $(NO)$ and nitrogen dioxide $(NO_2)$ at low temperatures $(-30\,^oC)$ produces dinitrogen trioxide $(N_2O_3)$,which exists as a blue liquid.
The chemical equation is: $NO_{(g)} + NO_{2_{(g)}} \xrightarrow{-30\,^oC} N_2O_{3_{(l)}} \text{ (Blue liquid)}$

(C) Black phosphorus is the most thermodynamically stable allotropic form of phosphorus.
This is because it has a highly ordered polymeric structure and the highest ignition temperature among all phosphorus allotropes.

(A) Nitrogen is unique among the group $15$ elements because it can form a wide variety of oxides in all oxidation states from $+I$ to $+V$.
These include $N_2O$ $(+I)$,$NO$ $(+II)$,$N_2O_3$ $(+III)$,$NO_2$ $(+IV)$,and $N_2O_5$ $(+V)$.

(C) The boiling points of the hydrides of Group $15$ elements are influenced by both molecular mass and hydrogen bonding.
$SbH_3$ has the highest boiling point due to its large molecular mass and strong van der Waals forces.
$NH_3$ has a higher boiling point than $AsH_3$ and $PH_3$ due to the presence of intermolecular hydrogen bonding.
$AsH_3$ and $PH_3$ follow the trend based on molecular mass.
The boiling points are: $SbH_3 (254 \ K) > NH_3 (238 \ K) > AsH_3 (211 \ K) > PH_3 (185 \ K)$.
Thus,the correct order is $SbH_3 > NH_3 > AsH_3 > PH_3$.

(D) In the electrothermal process,calcium phosphate $(Ca_3(PO_4)_2)$ is heated with silica $(SiO_2)$ and coke $(C)$.
The silica displaces phosphorus pentoxide $(P_2O_5)$ from calcium phosphate.
The reaction is: $Ca_3(PO_4)_2 + 3SiO_2 \to 3CaSiO_3 + P_2O_5$.
Subsequently,$P_2O_5$ is reduced by coke to form phosphorus $(P_4)$.

(B) The structure of $P_4O_{10}$ consists of a central $P_4$ tetrahedron where each $P$ atom is bonded to three oxygen atoms forming $P-O-P$ bridges,and each $P$ atom is also double-bonded to a terminal oxygen atom $(P=O)$.
There are $6$ $P-O-P$ bridges (involving $12$ $P-O$ single bonds) and $4$ $P=O$ bonds.
Total number of $P-O$ bonds (including both single and double bonds) is $12 + 4 = 16$.

(B) $NO_2$ is a mixed anhydride because it reacts with water to form two different acids,nitrous acid $(HNO_2)$ and nitric acid $(HNO_3)$.
The reaction is: $2NO_2 + H_2O \to HNO_2 + HNO_3$.

(D) White phosphorus $(P_4)$ is highly reactive and spontaneously ignites in air to form phosphorus pentoxide $(P_4O_{10})$,which is a new compound. The reaction is: $P_4 + 5O_2 \rightarrow P_4O_{10}$.

(A) Superphosphate is formed by the reaction of calcium phosphate,$Ca_3(PO_4)_2$,with dilute sulphuric acid,$H_2SO_4$.
The chemical reaction is: $Ca_3(PO_4)_2 + 2H_2SO_4 \rightarrow Ca(H_2PO_4)_2 + 2CaSO_4$.
The resulting mixture of calcium dihydrogen phosphate and calcium sulphate is known as superphosphate of lime.

(B) $PbO_2$ acts as a strong oxidizing agent. When it reacts with concentrated $HNO_3$,it undergoes reduction while oxidizing the acid or water present,leading to the evolution of oxygen gas. The reaction is: $2PbO_2 + 4HNO_3 \rightarrow 2Pb(NO_3)_2 + 2H_2O + O_2 \uparrow$. Therefore,the correct option is $B$.

(A) The acidic character of oxides of elements in the same group decreases as we move down the group.
This is because the metallic character increases down the group,which leads to a decrease in the non-metallic character and electronegativity.
Since $N$,$P$,$As$,and $Sb$ belong to Group $15$,the order of acidic strength is $N_2O_5 > P_2O_5 > As_2O_5 > Sb_2O_5$.
Therefore,$N_2O_5$ is the most acidic oxide.

(B) The magnetic property of a molecule depends on the presence of unpaired electrons.
$NO$ has $11$ valence electrons (odd number),making it paramagnetic.
$N_2O_4$ has an even number of electrons and all electrons are paired,making it diamagnetic.
$N_2O_5$ is diamagnetic,but $N_2O_4$ is the most commonly cited example of a diamagnetic nitrogen oxide in this context.
$N_2O$ is also diamagnetic,but $N_2O_4$ is the standard answer for this specific question type in textbooks.
Therefore,the correct option is $B$.

(D) The thermal decomposition reactions for the given compounds are as follows:
$1$. $NH_4NO_2 \xrightarrow{\Delta} N_2 + 2H_2O$
$2$. $2NaN_3 \xrightarrow{\Delta} 2Na + 3N_2$
$3$. $(NH_4)_2Cr_2O_7 \xrightarrow{\Delta} N_2 + Cr_2O_3 + 4H_2O$
Since all three compounds liberate $N_2$ gas upon thermal decomposition,the correct option is $D$.

(B) Yellow phosphorus consists of discrete $P_4$ tetrahedral molecules with high angular strain,making it highly reactive.
In contrast,red phosphorus exists as a polymeric chain structure where $P_4$ tetrahedra are linked together.
This polymeric structure reduces the strain and makes it significantly less reactive than yellow phosphorus.

(A) The acidity of halides of group $15$ elements depends on the tendency of the central atom to accept a lone pair of electrons (Lewis acidity).
As we move down the group,the size of the central atom increases and its electronegativity decreases.
$PCl_3$ has the smallest central atom with the highest electronegativity among the given group $15$ halides,making it the most effective at accepting electron pairs.
Therefore,$PCl_3$ is the most acidic.

(C) The chemical formula $H_3PO_2$ corresponds to hypophosphorus acid (also known as phosphinic acid).
In its structure,there is only one $P-OH$ group,which means only one hydrogen atom is ionizable.
Therefore,the basicity of $H_3PO_2$ is $1$.

(C) The electronic configuration of $NO_2$ involves a total of $17$ valence electrons.
Since the total number of valence electrons is odd,there is an unpaired electron present on the nitrogen atom.
Due to the presence of this unpaired electron,$NO_2$ is paramagnetic in nature.
Other oxides like $N_2O_3$,$N_2O$,and $N_2O_5$ have an even number of electrons and are diamagnetic.

(B) The reaction between nitric oxide $(NO)$ and oxygen $(O_2)$ produces nitrogen dioxide $(NO_2)$,which is a deep brown coloured gas.
The chemical equation is: $2NO(g) + O_2(g) \to 2NO_2(g)$ (Deep brown gas).