Nitrogen family Questions in English

(A) The electronegativity $(EN)$ values for Group $15$ elements are as follows:

Element	$EN$
$N$	$3.0$
$P$	$2.1$
$As$	$2.0$
$Sb$	$1.9$
$Bi$	$1.9$

Given that the difference between $EN(X)$ and $EN(P)$ is greater than the difference between $EN(P)$ and $EN(Y)$.
Let $X = N$ $(EN = 3.0)$ and $Y = As$ $(EN = 2.0)$.
$|EN(N) - EN(P)| = |3.0 - 2.1| = 0.9$
$|EN(P) - EN(As)| = |2.1 - 2.0| = 0.1$
Since $0.9 > 0.1$,the condition is satisfied. Thus,$X$ is $N$ and $Y$ is $As$.

(A) Stability: The bond dissociation enthalpy decreases down the group as the size of the central atom increases,so stability decreases: $NH_{3} > PH_{3} > AsH_{3} > SbH_{3} > BiH_{3}$. Thus,statement $A$ is correct.
Basicity: The electron density on the central atom decreases as the size increases,so basicity decreases: $NH_{3} > PH_{3} > AsH_{3} > SbH_{3} > BiH_{3}$. Thus,statement $B$ is correct.
Reducing character: As stability decreases,the tendency to release $H_{2}$ increases,so reducing character increases: $NH_{3} < PH_{3} < AsH_{3} < SbH_{3} < BiH_{3}$. Thus,statement $C$ is correct.
Boiling point: $NH_{3}$ has a higher boiling point than $PH_{3}$ due to hydrogen bonding. The trend is $PH_{3} < AsH_{3} < NH_{3} < SbH_{3} < BiH_{3}$. Thus,statement $D$ is incorrect.

(B) Statement-$I$: The electronegativity order for the given elements is $N > P > As > Sb$. Thus,$X = N$ (most electronegative) and $Y = Sb$ (least electronegative). The oxide $N_{2}O_{3}$ is acidic,and $Sb_{2}O_{3}$ is amphoteric. Therefore,Statement-$I$ is true.
Statement-$II$: $BCl_{3}$ is covalent and undergoes hydrolysis in water to form boric acid,$B(OH)_{3}$ (or $H_{3}BO_{3}$),and $HCl$. The reaction is $BCl_{3} + 3H_{2}O \rightarrow B(OH)_{3} + 3HCl$. The statement claims it produces $[B(OH)_{4}]^{-}$ and $[B(H_{2}O)_{6}]^{3+}$,which is incorrect. Therefore,Statement-$II$ is false.

(C) Statement $I$: The bond length in $HX$ increases as the size of the halogen increases $(HF < HCl < HBr < HI)$. The halogen with the longest bond is $I$,which has the largest covalent radius,not the smallest. Thus,Statement $I$ is false.
Statement $II$: The boiling point order for group $15$ hydrides is $PH_3 < AsH_3 < NH_3 < SbH_3 < BiH_3$. The hydride with the lowest boiling point is $PH_3$. The element $E$ is $P$ (phosphorus). The maximum covalency of phosphorus is $6$ (e.g.,in $[PF_6]^-$),not $4$. Thus,Statement $II$ is false.

(C) The reaction of $EH_{3}$ with $K_{2}HgI_{4}$ (Nessler's reagent) in an alkaline medium to form a brown precipitate of basic mercury$(II)$amido-iodine is a characteristic test for ammonia $(NH_{3})$.
Therefore,the element $E$ is nitrogen $(N)$.
The first ionization enthalpy values for the first elements of groups $13, 14, 15$,and $16$ are $B$ $(801 \ kJ \ mol^{-1})$,$C$ $(1086 \ kJ \ mol^{-1})$,$N$ $(1402 \ kJ \ mol^{-1})$,and $O$ $(1314 \ kJ \ mol^{-1})$.
Since $E$ is $N$,its first ionization enthalpy is $1402 \ kJ \ mol^{-1}$.

(C) $1$. Reducing nature: The bond dissociation energy decreases down the group $(NH_3 > PH_3 > AsH_3 > SbH_3 > BiH_3)$,so the reducing nature increases from $NH_3$ to $BiH_3$. ($A$ is correct).
$2$. Tendency to donate lone pair: Electronegativity of the central atom decreases and the size of the central atom increases down the group,making the lone pair more diffuse. Thus,the basicity (tendency to donate a lone pair) decreases from $NH_3$ to $BiH_3$. ($B$ is correct).
$3$. Stability: Bond dissociation energy decreases down the group as the $E-H$ bond length increases,so thermal stability decreases from $NH_3$ to $BiH_3$. ($C$ is correct).
$4$. Bond angle: The bond angle depends on the electronegativity of the central atom. As electronegativity decreases down the group,the bond pairs are further away from the central atom,which decreases the bond angle. The order is $NH_3 (107.8^{\circ}) > PH_3 (93.6^{\circ}) > AsH_3 (91.8^{\circ}) > SbH_3 (91.3^{\circ})$. ($D$ is correct).
Therefore,all statements $A, B, C$,and $D$ are correct.

(A) The brown ring test is a standard laboratory test used to detect the presence of nitrate ions $(NO_3^-)$ in a solution.
When a freshly prepared ferrous sulphate $(FeSO_4)$ solution is added to a solution containing nitrate ions,followed by the careful addition of concentrated sulphuric acid $(H_2SO_4)$ along the sides of the test tube,a brown ring is formed at the interface.
The chemical reaction involves the reduction of nitrate ions to nitric oxide $(NO)$ by $Fe^{2+}$ ions.
$NO_3^- + 3Fe^{2+} + 4H^+ \rightarrow NO + 3Fe^{3+} + 2H_2O$
The nitric oxide $(NO)$ then reacts with the hydrated ferrous sulphate complex to form the brown-coloured nitroso-ferrous sulphate complex,$[Fe(H_2O)_5(NO)]SO_4$.
Therefore,the gas '$X$' is $NO$ and the compound '$Y$' is $[Fe(H_2O)_5(NO)]SO_4$.

(D) Statement $A$ is correct: Nitrogen compounds in intermediate oxidation states like $+1, +2, +3, +4$ are generally unstable and undergo disproportionation in acidic media.
Statement $B$ is incorrect: Nitrogen forms $ppi-ppi$ multiple bonds,not $dpi-ppi$ multiple bonds,because it lacks $d$-orbitals.
Statement $C$ is incorrect: The $N-N$ single bond is weaker than the $P-P$ single bond due to high inter-electronic repulsion between the lone pairs of nitrogen atoms.
Statement $D$ is incorrect: Density increases down the group as atomic mass increases significantly compared to volume.
Statement $E$ is correct: Nitrogen lacks $d$-orbitals in its valence shell,so it can only use one $s$ and three $p$ orbitals for bonding,limiting its maximum covalency to $4$ (e.g.,in $NH_{4}^{+}$).
Therefore,statements $A$ and $E$ are correct.

(C) The electronegativity of $Ge$ on the Pauling scale is $2.0$. Among group $13$ elements,Boron $(B)$ has an electronegativity of $2.0$.
However,in the context of group $13$ chemistry,the properties described (oxidizing nature of $M^{3+}$ and positive reduction potential) are characteristic of heavier elements like Thallium $(Tl)$.
$Tl^{3+}$ is a strong oxidizing agent because of the inert pair effect,which makes $Tl^+$ more stable than $Tl^{3+}$.
Thus,$Tl^{3+}$ acts as an oxidizing agent ($B$ is correct) and has a positive standard electrode potential ($D$ is correct).
Therefore,statements $B$ and $D$ are correct.

(C) Nitrogen has a small atomic size and high electronegativity,which allows it to form stable $p\pi$-$p\pi$ multiple bonds with itself (as in $N_{2}$).
Nitrogen lacks vacant $d$-orbitals in its valence shell,hence it cannot form $d\pi$-$p\pi$ bonds.
Therefore,the statement that nitrogen can form $d\pi$-$p\pi$ bonds with oxygen is incorrect.