Nitrogen family Questions in English

(B) $NO + NO_2 \xrightarrow{253 \ K} N_2O_3 \ (X)$
$N_2O_3 + H_2O \longrightarrow 2HNO_2 \ (Y)$
The anion of $Y$ $(HNO_2)$ is $NO_2^-$.
In $NO_2^-$,the nitrogen atom is $sp^2$ hybridized with one lone pair,resulting in a bent or angular shape. However,among the given options,the geometry of the electron domain is triangular planar. Given the options provided,the intended answer is triangular planar.

(B) When ammonia reacts with excess chlorine,the reaction is as follows:
$NH_3 + 3Cl_2 \longrightarrow NCl_3 + 3HCl$
In this reaction,$NCl_3$ (nitrogen trichloride) is formed as an explosive product along with $HCl$ (hydrogen chloride).

(D) When ammonia reacts with excess chlorine,the reaction is as follows:
$NH_3 + 3Cl_2 \longrightarrow NCl_3 + 3HCl$
In this reaction,nitrogen trichloride $(NCl_3)$ and hydrogen chloride $(HCl)$ are formed as the products.

(C) The reaction of white phosphorus $(P_4)$ with sulphuryl chloride $(SO_2Cl_2)$ produces phosphorus pentachloride $(PCl_5)$ as the compound $X$.
$P_4 + 10SO_2Cl_2 \rightarrow 4PCl_5 + 10SO_2$
When $PCl_5$ undergoes complete hydrolysis,it reacts with water to form phosphoric acid $(H_3PO_4)$ as the compound $Y$.
$PCl_5 + 4H_2O \rightarrow H_3PO_4 + 5HCl$
Therefore,$X$ is $PCl_5$ and $Y$ is $H_3PO_4$.

(B) Hypophosphorous acid $(H_3PO_2)$ acts as a strong reducing agent.
It reduces silver nitrate $(AgNO_3)$ to metallic silver $(Ag)$.
The reaction is: $H_3PO_2 + 4AgNO_3 + 2H_2O \rightarrow 4Ag + 4HNO_3 + H_3PO_4$.
Here,$X$ is $H_3PO_2$ and it gets oxidised to $Y$,which is phosphoric acid $(H_3PO_4)$.

(D) The reaction of white phosphorus $(P_4)$ with thionyl chloride $(SOCl_2)$ is: $P_4 + 8SOCl_2 \rightarrow 4PCl_3 + 4SO_2 + 2S_2Cl_2$. Thus,'$C$' is $PCl_3$.
Hydrolysis of $PCl_3$ gives phosphorous acid $(H_3PO_3)$: $PCl_3 + 3H_2O \rightarrow H_3PO_3 + 3HCl$. Thus,'$O$' is $H_3PO_3$.
Statement $I$: $PCl_3$ has a pyramidal shape due to $sp^3$ hybridization with one lone pair. (Correct)
Statement $II$: $H_3PO_3$ is a dibasic acid because it has two $P-OH$ bonds. (Correct)
Statement $III$: $H_3PO_3$ is a monobasic acid. (Incorrect)
Statement $IV$: $PCl_3$ reacts with acetic acid $(CH_3COOH)$ to give acetyl chloride and phosphorous acid $(H_3PO_3)$: $3CH_3COOH + PCl_3 \rightarrow 3CH_3COCl + H_3PO_3$. (Correct)
Therefore,statements $I$,$II$,and $IV$ are correct.

(A) To determine which oxoacids contain $P-O-P$ bonds,we examine their structures:
$I$. $H_4P_2O_5$ (Pyrophosphorous acid): Contains a $P-O-P$ linkage.
$II$. $H_4P_2O_6$ (Hypophosphoric acid): Contains a $P-P$ bond,not $P-O-P$.
$III$. $H_4P_2O_7$ (Pyrophosphoric acid): Contains a $P-O-P$ linkage.
$IV$. $(HPO_3)_3$ (Cyclotrimetaphosphoric acid): Contains a cyclic structure with $P-O-P$ linkages.
Therefore,$I$,$III$,and $IV$ all contain $P-O-P$ bonds. Given the options provided,both $A$ ($III$ and $IV$) and $C$ ($I$ and $III$) contain correct examples,but $I$,$III$,and $IV$ are the complete set.

(D) The given reaction is: $P_4 + 3 NaOH + 3 H_2 O \rightarrow PH_3 + 3 NaH_2 PO_2$.
Here,$Q$ is phosphine $(PH_3)$.
Let us analyze the products of the given reactions:
$A$: $Ca_3 P_2 + 6 H_2 O \rightarrow 3 Ca(OH)_2 + 2 PH_3$ (Produces $PH_3$)
$B$: $4 H_3 PO_3 \stackrel{\Delta}{\longrightarrow} 3 H_3 PO_4 + PH_3$ (Produces $PH_3$)
$C$: $PH_4 I + KOH \rightarrow KI + H_2 O + PH_3$ (Produces $PH_3$)
$D$: $PCl_3 + 3 H_2 O \rightarrow H_3 PO_3 + 3 HCl$ (Does not produce $PH_3$)
Therefore,$Q$ is not the product in reaction $D$.

(A) $P_4 + 3 NaOH + 3 H_2 O \xrightarrow{CO_2} 3 NaH_2 PO_2 + PH_3$
$(X = NaH_2 PO_2)$
$NaH_2 PO_2 + HCl_{(aq)} \rightarrow H_3 PO_2 + NaCl$
$(Y = H_3 PO_2)$
$H_3 PO_2$ is hypophosphorous acid or phosphinic acid.
It is a monobasic acid because it has only one $P-OH$ bond.
It acts as a strong reducing agent.
The oxidation state of $P$ in $H_3 PO_2$ is $+1$.
Therefore,the statement that it is called phosphonic acid is incorrect,as phosphonic acid is $H_3 PO_3$.

(A) Phosphorous acid $(H_3PO_3)$ has two $P-OH$ bonds,making it a dibasic acid.
Phosphoric acid $(H_3PO_4)$ has three $P-OH$ bonds,making it a tribasic acid.
Therefore,they are respectively dibasic and tribasic acids.

(C) The structure of pyrophosphoric acid $(H_4P_2O_7)$ is shown below:
$HO-P(=O)(OH)-O-P(=O)(OH)-OH$
From the structure:
- Number of $P-O-H$ bonds = $4$
- Number of $P=O$ bonds = $2$
- Number of $P-O-P$ bonds = $1$
Thus,the correct option is $C$.

(C) The incorrect statement is that white phosphorus $(P_4)$ molecules contain four $P-P$ covalent bonds. In a white phosphorus molecule,there are four lone pairs of electrons on the phosphorus atoms and six $P-P$ covalent bonds. The structure of white phosphorus $(P_4)$ is a tetrahedron where each phosphorus atom is bonded to three other phosphorus atoms,resulting in a total of six $P-P$ bonds.

(B) The molecular formula of metaphosphoric acid is $HPO_3$.
It has the general formula $(HPO_3)_n$,where $n$ denotes the number of phosphoric acid units present in the ring,with $n = 3$ or more.
In metaphosphoric acid,the oxidation state of phosphorus is $+5$ with a smaller number of $H$-atoms compared to orthophosphoric acid.

(C) The structure of $P_4O_{10}$ consists of a tetrahedral arrangement of four phosphorus atoms.
Each phosphorus atom is bonded to one terminal oxygen atom via a double bond $(P=O)$.
There are six oxygen atoms that act as bridges between the phosphorus atoms,forming $P-O-P$ linkages.
Therefore,the total number of bridging oxygen atoms is $6$.

(C) The number of ionisable hydrogen atoms in an oxoacid of phosphorus is equal to the number of $P-OH$ bonds present in the molecule.
$1$. $H_3PO_4$ (Orthophosphoric acid): Contains $3$ $P-OH$ bonds,so it has $3$ ionisable hydrogen atoms.
$2$. $H_3PO_3$ (Phosphorous acid): Contains $2$ $P-OH$ bonds,so it has $2$ ionisable hydrogen atoms.
$3$. $H_3PO_2$ (Hypophosphorous acid): Contains $1$ $P-OH$ bond,so it has $1$ ionisable hydrogen atom.
$4$. $H_4P_2O_6$ (Hypophosphoric acid): Contains $4$ $P-OH$ bonds,so it has $4$ ionisable hydrogen atoms.
Comparing the number of ionisable hydrogen atoms:
$IV (4) > I (3) > II (2) > III (1)$
Therefore,the decreasing order is $IV > I > II > III$.

(B) When white phosphorus $(P_4)$ is boiled with concentrated $NaOH$ solution,phosphine $(PH_3)$ gas is produced in the laboratory.
$P_4 + 3NaOH + 3H_2O \longrightarrow PH_3 + 3NaH_2PO_2$

(A) The correct matches are:
$A. H_3PO_2$: Prepared by the reaction of white $P_4$ with alkali. $(A-II)$
$B. H_4P_2O_5$: Prepared by the reaction of $PCl_3$ with $H_3PO_3$. $(B-III)$
$C. H_3PO_4$: Prepared by the reaction of $P_2O_5$ with $H_2O$. $(C-IV)$
$D. H_4P_2O_7$: Prepared by the thermal dehydration $(\Delta)$ of $H_3PO_4$. $(D-V)$
Thus,the correct matching is $A-II, B-III, C-IV, D-V$.

(C) White phosphorus $(P_4)$ undergoes disproportionation reaction when dissolved in boiling $NaOH$ solution in an inert atmosphere (e.g.,$CO_2$ or $N_2$ atmosphere) to produce phosphine $(PH_3)$ and sodium hypophosphite $(NaH_2PO_2)$.
The chemical equation for the reaction is:
$P_4 + 3NaOH + 3H_2O \longrightarrow PH_3 + 3NaH_2PO_2$
Therefore,the products formed are phosphine and sodium hypophosphite.
Hence,option $C$ is correct.

(B) The chemical formula of hypophosphorus acid (also known as phosphinic acid) is $H_3PO_2$.
It has a tetrahedral geometry with $sp^3$ hybridization of the phosphorus atom.

(C) The chemical formula of pyrophosphoric acid is $H_4P_2O_7$.
It has a structure with one $P-O-P$ linkage,four $P-OH$ bonds,and two $P=O$ bonds.
Since it has four $P-OH$ groups,it is a tetrabasic acid.
It does not contain any $P-H$ bonds.
Orthophosphoric acid $(H_3PO_4)$ is prepared by dissolving $P_4O_{10}$ in water: $P_4O_{10} + 6H_2O \longrightarrow 4H_3PO_4$.
Therefore,the statement that pyrophosphoric acid contains two $P-H$ bonds is incorrect.

(C) The presence of $P-H$ bonds in phosphorus oxyacids is determined by their structures:
$1$. $H_3PO_4$ (Orthophosphoric acid): Contains three $P-OH$ bonds and one $P=O$ bond. No $P-H$ bond.
$2$. $H_3PO_3$ (Phosphorous acid): Contains two $P-OH$ bonds,one $P=O$ bond,and one $P-H$ bond.
$3$. $H_3PO_2$ (Hypophosphorous acid): Contains one $P-OH$ bond,one $P=O$ bond,and two $P-H$ bonds.
Therefore,the pair $H_3PO_3$ and $H_3PO_2$ contains $P-H$ bonds.

(B) The structure of $P_4O_6$ consists of four phosphorus atoms arranged at the corners of a tetrahedron. Each phosphorus atom is bonded to three oxygen atoms,which act as bridges between the phosphorus atoms. Thus,each phosphorus atom is linked to $3$ oxygen atoms.

(A) The basicity of an oxoacid is determined by the number of $OH$ groups directly attached to the central atom.
$H_3PO_3$ (Phosphorous acid) has two $OH$ groups and one $P-H$ bond,so its basicity is $2$.
$H_2SO_4$ (Sulfuric acid) has two $OH$ groups attached to the sulfur atom,so its basicity is $2$.
$H_3PO_2$ (Hypophosphorous acid) has one $OH$ group and two $P-H$ bonds,so its basicity is $1$.
$H_2SO_3$ (Sulfurous acid) has two $OH$ groups,so its basicity is $2$.
$H_3PO_4$ (Phosphoric acid) has three $OH$ groups,so its basicity is $3$.
$H_2S_2O_8$ (Peroxodisulfuric acid) has two $OH$ groups,so its basicity is $2$.
Therefore,the pair $H_3PO_3$ and $H_2SO_4$ both have a basicity of $2$.

(B) $1$. For option $A$: The boiling point order of group $16$ hydrides is $H_2S < H_2Se < H_2Te < H_2O$. $H_2O$ has the highest boiling point due to hydrogen bonding. Thus,$A$ is incorrect.
$2$. For option $B$: The acidic nature of nitrogen oxides increases with the oxidation state of nitrogen. The oxidation states are: $N_2O (+1), NO (+2), N_2O_3 (+3), N_2O_4 (+4), N_2O_5 (+5)$. The order $N_2O < NO < N_2O_3 < N_2O_4 < N_2O_5$ is correct.
$3$. For option $C$: The acidic strength of hydrohalic acids increases down the group as bond dissociation energy decreases: $HF < HCl < HBr < HI$. Thus,$C$ is incorrect.
$4$. For option $D$: The bond angle decreases down the group as the electronegativity of the central atom decreases: $H_2O (104.5^{\circ}) > H_2S (92^{\circ}) > H_2Se (91^{\circ}) > H_2Te (90^{\circ})$. Thus,$D$ is incorrect.

(B) The reaction of white phosphorus with sodium hydroxide is a disproportionation reaction:
$P_4 + 3 NaOH + 3 H_2O \rightarrow PH_3 + 3 NaH_2PO_2$
Here,$X$ is phosphine $(PH_3)$.
Next,phosphine reacts with copper$(II)$ sulfate:
$2 PH_3 + 3 CuSO_4 \rightarrow Cu_3P_2 + 3 H_2SO_4$
Thus,$Y$ is copper phosphide $(Cu_3P_2)$.

(D) The bond energy depends on the bond length. As the size of the central atom increases,the bond length increases,and the bond energy decreases.
The order of atomic size is $N < P < As$.
Therefore,the order of bond energy is $N-H > P-H > As-H$.
The bond energy values are $N-H = 389 \ kJ \ mol^{-1}$,$P-H = 318 \ kJ \ mol^{-1}$,and $As-H = 247 \ kJ \ mol^{-1}$.
Thus,the bond energies for $P-H$,$As-H$,and $N-H$ are $318, 247, 389$ respectively.

(B) The nitrogen dioxide molecule $(NO_2)$ has a total of $17$ valence electrons (odd number of electrons).
Due to the presence of an unpaired electron on the nitrogen atom,$NO_2$ exhibits paramagnetic behavior.
Other oxides like $SO_2$,$SiO_2$,and $CO_2$ have all electrons paired,making them diamagnetic.

(B) $_{82}Pb = [Xe] 4f^{14} 5d^{10} 6s^{2} 6p^{2}$
$_{83}Bi = [Xe] 4f^{14} 5d^{10} 6s^{2} 6p^{3}$
Due to the inert pair effect,the $6s^{2}$ electrons are held tightly by the nucleus due to relativistic contraction and poor shielding by $4f$ and $5d$ electrons.
Consequently,$Pb$ shows a stable bivalency ($+2$ oxidation state) and $Bi$ shows a stable trivalency ($+3$ oxidation state) instead of their group oxidation states of $+4$ and $+5$ respectively.

(B) The inert pair effect is the tendency of the two electrons in the outermost $s$-orbital to remain unshared in compounds of post-transition elements,particularly in the $p$-block elements of the $6^{th}$ period and some $5^{th}$ period elements.
$Sn$ ($5^{th}$ period),$Pb$ ($6^{th}$ period),and $In$ ($5^{th}$ period) are $p$-block elements that exhibit the inert pair effect.
$Fe$ is a $d$-block element (transition metal) and does not exhibit the inert pair effect.

(B) The orange solid is ammonium dichromate,$(NH_{4})_{2}Cr_{2}O_{7}$.
Upon heating,it undergoes thermal decomposition to produce nitrogen gas (colourless),chromium$(III)$ oxide (green solid),and water vapor:
$(NH_{4})_{2}Cr_{2}O_{7} \xrightarrow{\Delta} N_{2} \uparrow + Cr_{2}O_{3} + 4H_{2}O$
The green solid,$Cr_{2}O_{3}$,can be reduced to metallic chromium by aluminium powder in a thermite reaction:
$Cr_{2}O_{3} + 2Al \longrightarrow 2Cr + Al_{2}O_{3}$

(A) The industrial manufacture of nitric acid is carried out by the $Ostwald$ process. The steps are as follows:
$1$. Catalytic oxidation of ammonia: $4 NH_{3} + 5 O_{2} \xrightarrow[800-900^{\circ} C]{Pt \text{ gauge}} 4 NO + 6 H_{2} O$
$2$. Oxidation of nitric oxide to nitrogen dioxide: $2 NO + O_{2} \longrightarrow 2 NO_{2}$
$3$. Absorption of nitrogen dioxide in water: $4 NO_{2} + 2 H_{2} O + O_{2} \longrightarrow 4 HNO_{3}$
Thus,the intermediate compounds formed are nitric oxide $(NO)$ and nitrogen dioxide $(NO_{2})$.

(A) The structure of $P_4O_{10}$ consists of a central $P_4$ tetrahedron where each edge is bridged by an oxygen atom,forming $P-O-P$ linkages.
There are $6$ edges in a tetrahedron,and each edge is bridged by one oxygen atom.
Therefore,there are $6$ $P-O-P$ linkages in the $P_4O_{10}$ molecule.

(C) The structure of $H_3PO_3$ (Phosphorous acid) contains two $-OH$ groups and one $P-H$ bond.
Thus,it has $2$ $-OH$ groups.
The structure of $H_3PO_4$ (Orthophosphoric acid) contains three $-OH$ groups.
Thus,it has $3$ $-OH$ groups.
Therefore,the number of $-OH$ groups in $H_3PO_3$ and $H_3PO_4$ are $2$ and $3$ respectively.

(A, C, D) White phosphorus $(P_{4})$ consists of a discrete tetrahedral molecule.
It contains $6$ $P-P$ single bonds and $4$ lone pairs of electrons (one on each phosphorus atom).
The $P-P-P$ bond angle in the tetrahedral structure is $60^{\circ}$.
Therefore,options $A$,$C$,and $D$ are correct characteristics of white phosphorus.

(C) The structure of Pyrophosphoric acid $(H_4P_2O_7)$ consists of two phosphate units linked by an oxygen atom,forming a $P-O-P$ bond.
Its structure is $(HO)_2P(O)-O-P(O)(OH)_2$.
Other acids like Hypophosphorus acid $(H_3PO_2)$,Phosphorus acid $(H_3PO_3)$,and Orthophosphoric acid $(H_3PO_4)$ do not contain a $P-O-P$ linkage.

(C) An anhydride is a compound formed by the removal of water from an acid.
For phosphoric acid $(H_3 PO_4)$,the dehydration reaction is:
$4 H_3 PO_4 \longrightarrow P_4 O_{10} + 6 H_2 O$
Thus,$P_4 O_{10}$ is the anhydride of orthophosphoric acid $(H_3 PO_4)$.

(B) In a group,as we move downwards,the metallic character increases and the non-metallic character decreases.
Consequently,the acidic nature of oxides decreases down the group.
Among the given oxides of Group $15$ elements,$P$ is at the top of the group,followed by $As$,$Sb$,and $Bi$.
Therefore,$P_2O_5$ is the most acidic oxide.