Nitrogen family Questions in English

(B) Isoelectronic species are those that have the same number of electrons.
For $CO_2$: Total electrons = $6 + (8 \times 2) = 6 + 16 = 22$.
For $N_2O$: Total electrons = $(7 \times 2) + 8 = 14 + 8 = 22$.
Since both $CO_2$ and $N_2O$ have $22$ electrons,they are isoelectronic.
Therefore,the correct option is $(B)$.

(A) is the correct structure.
$H_3PO_4$ is orthophosphoric acid.
In its structure,the phosphorus atom is bonded to one oxygen atom via a double bond $(P=O)$,and to three hydroxyl groups $(-OH)$ via single bonds.
The structure is represented as $H-O-P(=O)(OH)_2$.

(A) The bond angle in hydrides of group $15$ elements depends on the electronegativity of the central atom and the size of the central atom.
As we move down the group from $N$ to $Sb$,the electronegativity of the central atom decreases,and the size of the central atom increases.
This leads to a decrease in the bond pair-bond pair repulsion and an increase in the bond length,resulting in a decrease in the bond angle.
Therefore,the correct sequence of decreasing bond angle is $NH_3 (107.8^{\circ}) > PH_3 (93.6^{\circ}) > AsH_3 (91.8^{\circ}) > SbH_3 (91.3^{\circ})$.

(A) . Nitrogen does not have vacant $d-$orbitals,so it cannot expand its octet to form $5$ covalent bonds.
$P$ (Phosphorus) has vacant $3d-$orbitals,allowing it to exhibit a $+5$ oxidation state and form $PCl_5$.
Since Nitrogen lacks $d-$orbitals,it is restricted to a maximum covalency of $4$,which is why $NCl_5$ does not exist.

(B) White phosphorus exists as a tetrahedral $P_4$ molecule.
In this structure,each phosphorus atom is bonded to three other phosphorus atoms.
The atomicity $(X)$ of phosphorus is $4$.
Due to the strained tetrahedral geometry,the $P-P-P$ bond angle $(Y)$ is $60^{\circ}$.

(B) The boiling point of hydrides of group $15$ elements generally increases down the group due to an increase in molecular mass and van der Waals forces.
However,$NH_3$ has an anomalously high boiling point due to intermolecular hydrogen bonding.
Among the remaining hydrides ($PH_3$,$AsH_3$,$SbH_3$),$PH_3$ has the lowest molecular mass and thus the weakest van der Waals forces.
Therefore,$PH_3$ has the lowest boiling point.

(B) The basicity of hydrides of Group $15$ elements depends on the availability of the lone pair of electrons on the central atom.
As we move down the group from $N$ to $P$,the atomic size increases.
This causes the lone pair of electrons on the phosphorus atom to be more diffused over a larger volume compared to the nitrogen atom.
Consequently,the electron density on the phosphorus atom is lower,making $PH_3$ a weaker Lewis base than $NH_3$.
Therefore,$PH_3$ is less basic than $NH_3$.

(D) The strength of oxoacids of the same group increases with the increase in the oxidation state of the central atom and decreases as we move down the group due to the increase in atomic size.
Comparing $H_3PO_4$ and $H_3AsO_4$,$P$ is smaller than $As$,making $H_3PO_4$ a stronger acid.
Comparing $H_3PO_4$ and $H_3PO_3$,$H_3PO_4$ has a higher oxidation state $(+5)$ compared to $H_3PO_3$ $(+3)$,making $H_3PO_4$ stronger.
Thus,$H_3PO_4$ is the strongest acid among the given options.

(D) Lewis base is a substance that can donate a lone pair of electrons.
In the group $15$ hydrides $(NH_3, PH_3, AsH_3, SbH_3)$,the central atom has one lone pair of electrons.
The basic strength depends on the availability of this lone pair for donation.
As we move down the group,the size of the central atom increases,and the lone pair occupies a larger orbital,making it less available for donation.
Additionally,the electronegativity of the central atom decreases down the group.
$NH_3$ has the smallest central atom $(N)$,and the lone pair is concentrated in a smaller volume,making it the most available for donation.
Therefore,$NH_3$ is the strongest Lewis base.

(D) The basicity of nitrogen halides 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 of the lone pair on the nitrogen atom towards itself.
This significantly reduces the availability of the lone pair for donation,making $NF_3$ the least basic among the nitrogen halides.
As the electronegativity of the halogen decreases from $F$ to $I$,the $-I$ effect decreases,and the basicity increases.

(B) The electronic configuration of nitrogen is $1s^2 2s^2 2p^3$.
It has $5$ valence electrons in its outermost shell.
It can lose up to $5$ electrons to reach an oxidation state of $+5$ when bonded to more electronegative elements like oxygen or fluorine.
It can also gain $3$ electrons to complete its octet,resulting in an oxidation state of $-3$.
Therefore,nitrogen exhibits oxidation states ranging from $-3$ to $+5$.

(A) $H_3PO_3$ has the structure $(HO)_2P(O)H$. It contains two $P-OH$ bonds,making it dibasic.
It also contains one $P-H$ bond,which makes it a reducing agent.
In contrast,$H_3PO_4$ has the structure $(HO)_3P(O)$,which is tribasic and non-reducing because it lacks $P-H$ bonds.

(B) metalloid is an element that exhibits properties of both metals and non-metals. Among the given options,$Sb$ (Antimony) is a well-known metalloid. $Pb$ (Lead) and $Bi$ (Bismuth) are post-transition metals,and $Zn$ (Zinc) is a transition metal.

(A) The metallic character of elements in group $15$ ($VA$ group) increases as we move down the group.
Since acidic character is inversely proportional to metallic character,the acidity of their pentoxides decreases down the group.

(D) Although nitrogen and phosphorus belong to the same group $(15)$,nitrogen cannot exhibit pentavalency due to the absence of $d$-orbitals in its valence shell.
Phosphorus,being larger in size,has available $d$-orbitals and a greater ability to accommodate electrons,making its pentavalent state more stable.

(B) Thallium is a $p$-block element belonging to group $13$ and period $6$.
The electronic configuration of Thallium is $[Xe] \, 4f^{14} \, 5d^{10} \, 6s^2 \, 6p^1$.
The $d$ and $f$ electrons of inner shells provide poor shielding,which increases the effective nuclear charge on the $6s^2$ electrons,making them reluctant to participate in bonding. This phenomenon is known as the inert pair effect.
Due to this effect,the $6s^2$ electrons remain paired,while the $6p^1$ electron is easily lost,leading to the $+1$ oxidation state. The $+3$ state is also possible but less stable than the $+1$ state for Thallium.

(B) $Zn$ (Zinc) is an amphoteric metal that reacts with a boiling solution of $NaOH$ (caustic soda) to produce sodium zincate and hydrogen gas.
The chemical equation is:
$Zn + 2NaOH + 2H_2O \to Na_2[Zn(OH)_4] + H_2 \uparrow$

(B) When $Sn$ (tin) reacts with an excess of hot concentrated $NaOH$ solution,it forms sodium stannate $(Na_2SnO_3)$ and releases hydrogen gas.
The balanced chemical equation is:
$Sn + 2NaOH + H_2O \to Na_2SnO_3 + 2H_2$.

(D) Phosphine $(PH_3)$ is prepared by the hydrolysis of metal phosphides.
$1$. Calcium phosphide $(Ca_3P_2)$ reacts with water to produce phosphine:
$Ca_3P_2 + 6H_2O \to 3Ca(OH)_2 + 2PH_3$
$2$. Potassium phosphide $(K_3P)$ also reacts with water to produce phosphine:
$K_3P + 3H_2O \to 3KOH + PH_3$
Therefore,both $(b)$ and $(c)$ are correct.

(A) Gallium is a silvery,soft metal.
It has a very low melting point of $29.76\ ^oC$,which is close to room temperature.
Because it remains liquid over a wide temperature range,it is used in high-temperature thermometers.
Therefore,the correct metal is Gallium.

(C) When aluminium oxide $(Al_2O_3)$ and carbon $(C)$ are strongly heated in a stream of dry chlorine gas $(Cl_2)$,the reaction occurs as follows:
$Al_2O_3 + 3C + 3Cl_2 \xrightarrow{\Delta} 2AlCl_3 + 3CO$
The product formed is anhydrous aluminium chloride $(AlCl_3)$.

(C) Aluminium metal is not attacked by nitric acid of any concentration because of the thin and unreactive protective layer of aluminium oxide $(Al_2O_3)$ formed on the metallic surface due to the reaction of aluminium metal with oxygen of air.

(D) $PbO_2$ is an amphoteric oxide because it reacts with both acids and bases.
For example,it reacts with $HCl$ to form $PbCl_2$ and with $NaOH$ to form sodium plumbate $(Na_2PbO_3)$.

(B) The inert pair effect becomes significant for the $6^{th}$ and $7^{th}$ period $p-$block elements. Due to the poor shielding of $d$ and $f$ orbitals,the $ns^2$ electrons are held tightly by the nucleus and do not participate in bonding,making the $+1$ oxidation state more stable than the $+3$ state for $Tl$.

(C) Lapis lazuli is a deep-blue metamorphic rock used as a semi-precious stone.
It is primarily composed of the mineral lazurite,which is a sulphur-containing sodium alumino silicate.
The chemical composition is approximately $3Na_2O \cdot 3Al_2O_3 \cdot 6SiO_2 \cdot 2Na_2S$.

(C) When tin $(Sn)$ reacts with concentrated nitric acid $(HNO_3)$,it gets oxidized to metastannic acid $(H_2SnO_3)$.
The chemical equation is:
$Sn + 4HNO_3 \to H_2SnO_3 + 4NO_2 + H_2O$

(A) Litharge is one of the natural mineral forms of lead$(II)$ oxide,$PbO$.
Litharge is a secondary mineral which forms from the oxidation of galena ores.

(D) Lead $(II, IV)$ oxide,also known as red lead or triplumbic tetroxide,is a bright red or orange pigment.
Chemically,it is represented as $Pb_3O_4$,which can also be written as $2PbO \cdot PbO_2$.
It is widely used in the manufacture of lead glass,rust-proof paints,and in the preparation of $Sindhur$.

(B) Elements like $I_2$,$N_2$,and $O_2$ exist as stable diatomic molecules.
Phosphorus,due to its smaller size and high repulsion between lone pairs,prefers to form a tetrahedral $P_4$ molecule rather than a diatomic $P_2$ molecule.

(B) The brown ring test is used to detect the presence of nitrate ions $(NO_3^-)$ in a solution.
In this test,$Fe(II)$ ions react with $NO$ (nitric oxide) to form a brown-colored coordination complex.
The reaction is: $[Fe(H_2O)_6]^{2+} + NO \to [Fe(H_2O)_5(NO)]^{2+} + H_2O$.
This complex,$[Fe(H_2O)_5(NO)]^{2+}$,is responsible for the brown ring observed at the junction of the two layers.

(B) The chemical formula for metaphosphoric acid is $HPO_3$.
It is a polymeric acid often existing as cyclic trimers or higher polymers,represented by the empirical formula $(HPO_3)_n$.

(A) Ammonia $(NH_3)$ is a basic gas.
To dry a basic gas,we must use a basic drying agent.
If an acidic drying agent is used,it will react with the ammonia to form a salt.
$CaCl_2$ forms an adduct with $NH_3$ $(CaCl_2 \cdot 8NH_3)$.
$P_2O_5$ and $H_2SO_4$ are acidic and will react with $NH_3$ to form ammonium salts.
Therefore,$CaO$ (quicklime) is the only suitable basic drying agent among the given options.

(A) The correct answer is $A$. White phosphorus is soluble in $CS_2$,whereas red phosphorus is insoluble in $CS_2$. Both forms consist of $P_4$ units (same kind of atoms),both can be oxidized by heating in air,and they can be interconverted under specific conditions.

(D) The correct answer is $D$.
Pyrophosphoric acid $(H_4P_2O_7)$ is a tetrabasic acid because it contains four $P-OH$ groups in its structure.
The structure is: $HO-P(=O)(OH)-O-P(=O)(OH)-OH$.
Since it has four replaceable hydrogen atoms attached to oxygen atoms,it is tetrabasic.

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

(A) The correct answer is $A$.
$NCl_5$ is not known because nitrogen lacks $d$-orbitals in its valence shell,which prevents it from expanding its covalency beyond $4$.

(B) Phosphorus exists in the form of discrete tetrahedral molecules where four phosphorus atoms are linked together by single covalent bonds.
Therefore,the chemical formula for the phosphorus molecule is $P_4$.

(D) The $P_4$ molecule has a tetrahedral structure.
It contains six $P-P$ single bonds.
Each phosphorus atom has one lone pair,so there are four lone pairs in total.
The bond angle $P-P-P$ is $60^o$.
Therefore,both $(a)$ and $(c)$ are correct statements.

(A) The correct answer is $(A)$.
The Birkeland-Eyde process is an industrial method used for the fixation of atmospheric nitrogen to produce nitric acid $(HNO_3)$.
In this process,atmospheric air is passed through an electric arc furnace at very high temperatures (about $3000 \ ^\circ C$),where nitrogen and oxygen react to form nitric oxide $(NO)$: $N_2 + O_2 \rightleftharpoons 2NO$.
Therefore,the primary raw material used in this process is air.

(D) Among the different allotropes of phosphorus,$White \ phosphorus$ is the most reactive.
This is due to the high angular strain in the $P_4$ molecules,where the bond angle is only $60^\circ$,making it highly unstable and reactive compared to other allotropes like $Red$ or $Black \ phosphorus$.

(B) In the laboratory,phosphine $(PH_3)$ is prepared by heating white phosphorus with a concentrated aqueous solution of caustic potash $(KOH)$ or caustic soda $(NaOH)$ in an inert atmosphere of $CO_2$.
The chemical reaction is:
$P_4 + 3KOH + 3H_2O \to PH_3 + 3KH_2PO_2$
Therefore,the correct option is $(B)$.

(D) The metallic character increases down the group in the $p$-block elements.
In Group $15$,the elements are $N, P, As, Sb, Bi$.
$N$ and $P$ are non-metals.
$As$ and $Sb$ are metalloids.
$Bi$ is a metal.
Therefore,$Bismuth$ $(Bi)$ is the most metallic element among the given options.

(B) The chemical formula of orthophosphoric acid is $H_3PO_4$.
In its structure,the phosphorus atom is bonded to one oxygen atom via a double bond and to three hydroxyl $(-OH)$ groups via single bonds.
The basicity of an oxoacid is defined by the number of ionizable hydrogen atoms attached to oxygen atoms.
Since there are $3$ $-OH$ groups present in $H_3PO_4$,it can release $3$ protons ($H^+$ ions) in an aqueous solution.
Therefore,the basicity of orthophosphoric acid is $3$.

(A) Heating $KNO_3$ produces potassium nitrite and oxygen gas: $2KNO_{3(s)} \xrightarrow{\Delta} 2KNO_{2(s)} + O_{2(g)}$.
Heating heavy metal nitrates like $Pb(NO_3)_2$,$Cu(NO_3)_2$,and $AgNO_3$ produces nitrogen dioxide $(NO_2)$ along with oxygen and the metal oxide.

(A) When ammonia is heated and passed over solid copper$(II)$ oxide $(CuO)$ at high temperatures,it acts as a reducing agent and reduces $CuO$ to metallic copper,while ammonia itself is oxidized to nitrogen gas.
The balanced chemical equation is:
$2 NH_{3(g)} + 3 CuO_{(s)} \rightarrow N_{2(g)} + 3 Cu_{(s)} + 3 H_2O_{(g)}$
Thus,the gas evolved is $N_2$.