Nitrogen family Questions in English

(D) All the given reactions are standard chemical reactions:
$(a)$ Phosphorus reacts with dilute $NaOH$ to form phosphine and sodium hypophosphite.
$(b)$ Bromine reacts with hot and concentrated $NaOH$ to form sodium bromide and sodium bromate.
$(c)$ Nitrogen trichloride undergoes hydrolysis to form ammonia and hypochlorous acid.
$(d)$ Iodine reacts with $NaOH$ to form sodium iodide and sodium hypoiodite.
$(e)$ Aluminum carbide reacts with water to produce aluminum hydroxide and methane.
$(f)$ Dinitrogen tetroxide reacts with water to form a mixture of nitric acid and nitrous acid.
$(g)$ Ammonium nitrate reacts with sodium hydroxide to produce ammonia,sodium nitrate,and water.
Since all reactions $(a)$ through $(g)$ are correctly represented,the correct set includes all of them.

(B) At high temperatures,nitrogen reacts directly with certain metals to form nitrides. Among the given options,aluminum $(Al)$ reacts with nitrogen to form aluminum nitride $(AlN)$.
$Al + \frac{1}{2}N_2 \xrightarrow{\Delta} AlN$

(A) $1$. $P_4O_{10}$ is a powerful dehydrating agent because it has a high affinity for water,reacting to form phosphoric acid $(H_3PO_4)$.
$2$. $P_4O_{10}$ hydrolyses in water to form phosphoric acid,not elemental phosphorus.
$3$. In the structure of $P_4O_{10}$,each $P$ atom is bonded to four oxygen atoms (three bridging oxygen atoms and one terminal double-bonded oxygen atom).
$4$. Red phosphorus does not react with $NaOH$ to form phosphine $(PH_3)$; only white phosphorus reacts with $NaOH$ to produce $PH_3$.

(B) To determine which species contain $X-O-X$ bonds,we analyze the structures of the given compounds:
$1$. $P_4O_{10}$ has a cage structure with $P-O-P$ linkages.
$2$. $P_4O_6$ has a cage structure with $P-O-P$ linkages.
$3$. $H_3P_3O_9$ (trimetaphosphoric acid) is a cyclic structure containing $P-O-P$ linkages.
In option $B$,all three species contain $P-O-P$ bonds.
In option $D$,$H_4P_2O_6$ (hypophosphoric acid) contains a $P-P$ bond,not a $P-O-P$ bond,so it is excluded.

(D) $N_2O_5$ reacts with water to form $HNO_3$ (an oxyacid).
$Cl_2O_7$ reacts with water to form $HClO_4$ (an oxyacid).
$CrO_3$ reacts with water to form $H_2CrO_4$ (an oxyacid).
$N_2O$ is a neutral oxide and does not react with water to form any oxyacid.

(D) $1$. In the $3d$ series,the effective nuclear charge $(Z_{eff})$ increases by $0.15$ for each subsequent element due to poor shielding of $d$-electrons,which is a correct statement.
$2$. $_{90}Th$ (Thorium) is often considered an exception in block classification because it is a $f$-block element but does not contain $f$-electrons in its ground state,which is a correct statement.
$3$. The process $P + e^{-} \to P^{-}$ is exothermic because Phosphorus has a half-filled $p$-orbital $(3p^3)$,making the addition of an electron release energy,which is a correct statement.
$4$. The acidic character of oxoacids of Phosphorus depends on the number of $P-OH$ groups. The order is $H_3PO_4 > H_3PO_3 > H_3PO_2$. The given statement $H_3PO_2 > H_3PO_3 > H_3PO_4$ is false.

(B) The given reaction is $4NH_3 + 5O_2 \xrightarrow{Pt/Rh \text{ catalyst}} 4NO + 6H_2O$.
This reaction represents the catalytic oxidation of ammonia,which is the first step in the industrial production of nitric acid $(HNO_3)$ known as the Ostwald's process.

(A) Allotropy is the property of an element to exist in two or more different physical forms in the same physical state.
In Group $15$ elements,Nitrogen $(N)$ exists as a diatomic molecule $(N_2)$ and does not exhibit allotropy.
Phosphorus,Arsenic,Antimony,and Bismuth all exhibit allotropy due to their ability to form different structural arrangements in the solid state.
Therefore,Nitrogen is the correct answer.

(D) Neutral oxides are those that do not react with either acids or bases. Among the given options,$N_2O$ (nitrous oxide) is a well-known neutral oxide. $SnO$ and $PbO$ are amphoteric oxides,while $PbO_2$ is an acidic oxide.

(C) The thermal decomposition of ammonium dichromate is given by: $(NH_4)_2Cr_2O_7 \xrightarrow{\Delta} N_2 + Cr_2O_3 + 4H_2O$.
This reaction produces nitrogen gas $(N_2)$,not ammonia $(NH_3)$.
In contrast,the other reactions produce ammonia:
$1$. $NH_4I + KOH \to KI + NH_3 + H_2O$
$2$. $N_2 + 3H_2 \xrightarrow[{Fe/Mo}]{\Delta} 2NH_3$ (Haber process)
$3$. $Mg_3N_2 + 6HCl_{(aq)} \to 3MgCl_2 + 2NH_3$

(B) At room temperature,$N_2$ (nitrogen gas) is considered inert due to the presence of a very strong triple bond $(N \equiv N)$ between the two nitrogen atoms.
The bond dissociation energy of this triple bond is extremely high $(941 \ kJ/mol)$,which makes it chemically unreactive under standard conditions.

(D) $H_4P_2O_7$ is Pyrophosphoric acid,which has an open-chain structure ($HO)_2P(O)-O-P(O)(OH)_2$.
$H_4P_2O_8$ is Peroxodiphosphoric acid,which also has an open-chain structure containing a peroxide linkage ($HO)_2P(O)-O-O-P(O)(OH)_2$.
$(HPO_3)_4$ is Cyclotetrametaphosphoric acid,which has a cyclic structure.
Since both $H_4P_2O_7$ and $H_4P_2O_8$ are not cyclic,the correct option is $(D)$.

(B) The reactions represent the $Ostwald's \ process$ for the manufacture of $HNO_3$.
$1$. $4NH_3 + 5O_2 \xrightarrow[500 \ K]{Pt} 4NO + 6H_2O$. Thus,$A = NO$.
$2$. $2NO + O_2 \to 2NO_2$. Thus,$B = NO_2$.
$3$. $3NO_2 + H_2O \to 2HNO_3 + NO$. Thus,$C = HNO_3$.
Therefore,$A, B$ and $C$ are $NO, NO_2$ and $HNO_3$ respectively.

(D) Phosphine reduces silver nitrate to metallic silver: $PH_3 + 6AgNO_3 + 3H_2O \rightarrow 6Ag + 6HNO_3 + H_3PO_3$.
Organic tissues reduce silver nitrate to metallic silver. In most cases,finely divided silver is obtained,which appears as a black stain.
Silver cyanide dissolves in potassium cyanide to form a complex,potassium dicyanoargentate$(I)$: $AgCN + KCN \rightarrow K[Ag(CN)_2]$.

(B) $(1)$ $Ca_3P_2 + 6H_2O \rightarrow 3Ca(OH)_2 + 2PH_3$ (Forms $PH_3$)
$(2)$ $4H_3PO_3 \xrightarrow{\Delta} 3H_3PO_4 + PH_3$ (Forms $PH_3$)
$(3)$ $P_4 + 3NaOH + 3H_2O \rightarrow 3NaH_2PO_2 + PH_3$ (Forms $PH_3$)
$(4)$ $H_3PO_4 \xrightarrow{\Delta} HPO_3 + H_2O$ (Does not form $PH_3$)
$(5)$ $4HNO_3 + P_4O_{10} \rightarrow 4HPO_3 + 2N_2O_5$ (Does not form $PH_3$)
$(6)$ $P_4 + 20HNO_3 \rightarrow 4H_3PO_4 + 20NO_2 + 4H_2O$ (Does not form $PH_3$)
Thus,reactions $(1)$,$(2)$,and $(3)$ produce $PH_3$ gas. The total count is $3$.

(B) The phenomenon is known as the $inert \ pair \ effect$.
As we move down a group in the $p-$block,the $ns^2$ electrons become increasingly reluctant to participate in bonding due to poor shielding by $d$ and $f$ orbitals.
The energy required to unpair these $ns^2$ electrons is not adequately compensated by the energy released during the formation of two additional bonds.
Therefore,the lower oxidation state becomes more stable than the higher oxidation state.

(B) Disproportionation is a reaction where an element in an intermediate oxidation state is simultaneously oxidized and reduced.
In the case of $N_2O_4$,the oxidation state of nitrogen is $+4$.
When $N_2O_4$ reacts with $NaOH$,it forms a mixture of nitrite $(NO_2^-)$ and nitrate $(NO_3^-)$ salts:
$N_2O_4 + 2NaOH \rightarrow NaNO_2 + NaNO_3 + H_2O$.
Here,nitrogen is reduced from $+4$ to $+3$ in $NaNO_2$ and oxidized from $+4$ to $+5$ in $NaNO_3$.

(A) $N_2O_5$ is a colorless solid at room temperature.
$N_2O$ is a colorless gas.
$NO$ is a colorless gas.
Therefore,only $N_2O_5$ fits the description of a colorless solid.

(D) . Thermal stability of hydrides decreases down the group as the $M-H$ bond dissociation energy decreases. Thus,$NH_3 > PH_3 > AsH_3$ is correct.
$b$. Reducing power increases down the group as thermal stability decreases. Thus,$NH_3 < PH_3 < AsH_3$ is correct.
$c$. Lewis basic character of nitrogen trihalides is $NCl_3 > NBr_3 > NI_3$ because the electronegativity of the halogen decreases,making the lone pair on nitrogen more available. The given order $NI_3 > NBr_3 > NCl_3$ is incorrect.
$d$. Acidic character of oxides decreases down the group as metallic character increases. Thus,$N_2O_5 > P_2O_5 > As_2O_5$ is correct.
Therefore,statements $a, b,$ and $d$ are correct.

(B) The reaction between equimolar amounts of $NO$ (nitric oxide) and $NO_2$ (nitrogen dioxide) at $-30\,^{\circ}C$ results in the formation of dinitrogen trioxide $(N_2O_3)$.
$NO(g) + NO_2(g) \xrightarrow{-30\,^{\circ}C} N_2O_3(l)$
$N_2O_3$ exists as a blue-colored liquid at low temperatures.

(C) $Mn_3O_4$ is a mixed oxide of $MnO$ and $MnO_2$,which can be represented as $MnO \cdot MnO_2$.
$BF_3$ does not undergo hydrolysis,while $SiF_4$ and $BiCl_3$ undergo hydrolysis.
Therefore,statement $(C)$ is the only correct statement.

(C) $(a) \ NH_4NO_2 \xrightarrow{\Delta } N_2 + 2H_2O$ is correct.
$(b) \ NH_4NO_3 \xrightarrow{\Delta } N_2O + 2H_2O$ is correct.
$(c) \ Na_2CO_3$ is thermally stable and does not decompose into $Na_2O$ and $CO_2$ under normal heating conditions.
$(d) \ NH_4Cl \xrightarrow{\Delta } NH_3 + HCl$ is correct.
Therefore,the correct reactions are $(a), (b),$ and $(d)$.

(C) The strong reducing behaviour of $H_3PO_2$ (hypophosphorous acid) is due to the presence of $P-H$ bonds.
In the structure of $H_3PO_2$,there is one $P=O$ bond,one $-OH$ group,and two $P-H$ bonds.
Any oxyacid of phosphorus that contains at least one $P-H$ bond acts as a strong reducing agent because the hydrogen atom directly bonded to phosphorus is easily oxidized.

(A) Heating $(NH_4)_2Cr_2O_7$ results in the evolution of nitrogen gas: $(NH_4)_2Cr_2O_7 \xrightarrow{\Delta} N_2 + Cr_2O_3 + 4H_2O$.
$KClO_3$ decomposes to give $KCl$ and $O_2$.
$NaNO_3$ decomposes to give $NaNO_2$ and $O_2$.
$K_2Cr_2O_7$ decomposes to give $K_2CrO_4$,$Cr_2O_3$,and $O_2$.
Therefore,$(NH_4)_2Cr_2O_7$ is the compound that does not release oxygen upon heating.

(B) $NO_2$ is a brown coloured gas and is paramagnetic due to the presence of an odd electron in the $sp^2$ hybrid orbital.
Upon dimerisation,$2NO_2$ (brown,paramagnetic) converts to $N_2O_4$ (colourless,diamagnetic).
Therefore,the statement that it is coloured in its dimeric form is incorrect.

(D) Both $H_4P_4O_{12}$ (cyclotetrametaphosphoric acid) and $H_3P_3O_9$ (cyclotrimetaphosphoric acid) are cyclic metaphosphoric acids with the general formula $(HPO_3)_n$.
In these cyclic structures,each phosphorus atom is bonded to one $OH$ group,one $P=O$ group,and two oxygen atoms that form the $P-O-P$ linkages in the ring.
For $H_3P_3O_9$ $(n=3)$: It has $3$ $P=O$ bonds,$3$ $P-O-H$ bonds,and $3$ $P-O-P$ bonds.
For $H_4P_4O_{12}$ $(n=4)$: It has $4$ $P=O$ bonds,$4$ $P-O-H$ bonds,and $4$ $P-O-P$ bonds.
Since the number of bonds depends on the value of $n$,none of the options provided are identical for both molecules.

(A) Phosphorus exists as a tetra-atomic molecule $P_4$.
When $P_4$ reacts with excess oxygen,it forms phosphorus pentoxide $(P_4O_{10})$,which is a strong dehydrating agent.
The chemical reaction is: $P_4 + 5O_2 \rightarrow P_4O_{10}$.

(A) The chemical formula of pyrophosphorous acid is $H_4P_2O_5$.
It is formed by the condensation of two molecules of phosphorous acid $(H_3PO_3)$ with the loss of one water molecule: $2H_3PO_3 \rightarrow H_4P_2O_5 + H_2O$.
The structure is $(HO)(H)P(=O)-O-P(H)(=O)(OH)$.
In this structure:
- There are $2$ $P-OH$ bonds.
- There are $2$ $P-H$ bonds.
- There are $2$ $P=O$ bonds.
Therefore,the number of $P-OH$,$P-H$,and $P=O$ linkages are $2, 2, 2$ respectively.

(B) Thermal decomposition of ammonium nitrite $(NH_4NO_2)$ yields nitrogen gas and water:
$NH_4NO_2 \xrightarrow{\Delta} N_2 + 2H_2O$
Other ammonium salts decompose differently: $NH_4NO_3$ gives $N_2O$,$NH_4Cl$ sublimes or dissociates into $NH_3$ and $HCl$,and $Mg(NH_4)PO_4$ gives $Mg_2P_2O_7$.

(C) The chemical formula for cyclotrimetaphosphoric acid is $(HPO_3)_3$.
Its structure consists of a six-membered ring containing alternating phosphorus and oxygen atoms $(P-O-P-O-P-O)$.
Each phosphorus atom is bonded to one hydroxyl group $(-OH)$ and one terminal oxygen atom via a double bond $(P=O)$.
Counting the bonds:
$1$. There are $3$ $P=O$ double bonds.
$2$. The single bonds are: $3$ $P-O$ bonds in the ring,$3$ $P-OH$ bonds,and $3$ $O-H$ bonds,totaling $3 + 3 + 3 = 9$ single bonds.
Wait,let us re-examine the structure:
- $3$ $P=O$ double bonds.
- Ring bonds: $3$ $P-O$ and $3$ $P-O$ (total $6$ $P-O$ single bonds in the ring).
- External bonds: $3$ $P-OH$ and $3$ $O-H$ (total $6$ single bonds).
- Total single bonds = $6 + 6 = 12$.
Thus,there are $3$ double bonds and $12$ single bonds.

(C) An acid that is tribasic can form three series of salts by replacing its three replaceable hydrogen atoms one by one.
$H_3PO_4$ is a tribasic acid because it contains three $P-OH$ bonds.
It reacts with bases in three steps:
$1. H_3PO_4 + NaOH \rightarrow NaH_2PO_4 + H_2O$
$2. NaH_2PO_4 + NaOH \rightarrow Na_2HPO_4 + H_2O$
$3. Na_2HPO_4 + NaOH \rightarrow Na_3PO_4 + H_2O$
Thus,$H_3PO_4$ forms three series of salts.

(A) The thermal decomposition of ammonium dichromate is given by: $(NH_4)_2Cr_2O_7 \rightarrow Cr_2O_3 + 4H_2O + N_2$.
The thermal decomposition of barium azide is given by: $Ba(N_3)_2 \rightarrow Ba + 3N_2$.
Thus,$N_2$ gas is produced in both cases.

(B) Nitrogen forms a stable $N \equiv N$ triple bond due to effective $2p\pi - 2p\pi$ overlapping.
In phosphorus,the atomic size is larger,which makes $3p\pi - 3p\pi$ bonding ineffective and weak.
Therefore,phosphorus prefers to form single bonds and exists as a $P_4$ tetrahedral molecule to satisfy its valency.

(C) The hydrolysis of $PCl_5$ occurs in two steps:
$1$. Partial hydrolysis: $PCl_5 + H_2O \rightarrow POCl_3 + 2HCl$
$2$. Complete hydrolysis: $POCl_3 + 3H_2O \rightarrow H_3PO_4 + 3HCl$
Overall reaction: $PCl_5 + 4H_2O \rightarrow H_3PO_4 + 5HCl$
In this reaction,the oxidation state of phosphorus remains $+5$ in both $PCl_5$ and $H_3PO_4$.

(A) Alums are double salts with the general formula $M_2SO_4 \cdot M'_2(SO_4)_3 \cdot 24H_2O$,where $M$ is a monovalent cation (such as $K^+$,$Na^+$,or $NH_4^+$) and $M'$ is a trivalent cation (such as $Al^{3+}$,$Cr^{3+}$,or $Fe^{3+}$).
Therefore,the correct general formula is $M_2SO_4 \cdot M'_2(SO_4)_3 \cdot 24H_2O$.

(A) Pure phosphine $(PH_3)$ is non-inflammable.
However,it often contains impurities like diphosphine $(P_2H_4)$ which is spontaneously inflammable in air.
Therefore,the presence of $P_2H_4$ makes the $PH_3$ gas inflammable.

(C) The chemical formula of pyrophosphoric acid is $H_4P_2O_7$.
Its structure consists of two $P(=O)(OH)_2$ units linked by an oxygen atom,represented as $(HO)_2P(=O)-O-P(=O)(OH)_2$.
By observing the structure,we can count the number of hydroxyl $(-OH)$ groups attached to the phosphorus atoms.
There are $4$ hydroxyl groups present in the molecule.

(B) The molecular formula of hypophosphorous acid is $H_3PO_2$.
In its structure,the phosphorus atom is bonded to one oxygen atom via a double bond,one $-OH$ group via a single bond,and two hydrogen atoms directly via single bonds.
Therefore,there are $2$ hydrogen atoms directly attached to the phosphorus atom.
Hence,option $B$ is correct.

(NONE) All the given reactions are standard thermal decomposition reactions.
$A$: $4HNO_3 \xrightarrow{\Delta} 2H_2O + 4NO_2 + O_2$ is correct.
$B$: $2Pb(NO_3)_2 \xrightarrow{\Delta} 2PbO + 4NO_2 + O_2$ is correct.
$C$: $MgCl_2 \cdot 6H_2O \xrightarrow{\Delta} MgO + 2HCl + 5H_2O$ is correct.
$D$: $Fe_2(SO_4)_3 \xrightarrow{\Delta} Fe_2O_3 + 3SO_3$ is correct.
Since all reactions are chemically correct,there is no incorrect reaction among the options provided.

(D) $NH_3$ is a Lewis base.
$H_2SO_4$ is an acid,$P_2O_5$ is an acidic oxide,and $CaCl_2$ forms an adduct with $NH_3$.
$NH_3$ reacts with $H_2SO_4$ to form $(NH_4)_2SO_4$.
$NH_3$ reacts with $P_2O_5$ to form ammonium phosphates.
$NH_3$ reacts with $CaCl_2$ to form an addition compound $CaCl_2 \cdot 8NH_3$.
Therefore,these substances cannot be used to dry $NH_3$.

(B) $PCl_3$ reacts with moisture and undergoes hydrolysis to produce $HCl$ gas,which appears as white fumes.
The chemical reaction is: $PCl_3 + 3H_2O \rightarrow H_3PO_3 + 3HCl$
Since $HCl$ is a gas that forms white fumes in moist air,$PCl_3$ is observed to fume in the presence of moisture.

(D) Let us analyze each option:
$A$. Borax $(Na_2[B_4O_5(OH)_4] \cdot 8H_2O)$ contains the $[B_4O_5(OH)_4]^{2-}$ ion,which consists of two $BO_4$ tetrahedra and two $BO_3$ triangles,forming two six-membered rings. This is correct.
$B$. $P_4O_{10}$ has an adamantane-like structure containing $4$ six-membered rings (each formed by $P-O-P-O-P-O$ linkages). This is correct.
$C$. Beryl $(Be_3Al_2Si_6O_{18})$ is a cyclic silicate containing $[Si_6O_{18}]^{12-}$ rings. In this structure,each $Si$ atom is bonded to $4$ oxygen atoms,where $2$ oxygen atoms are shared and $2$ are unshared (monovalent). Since there are $6$ $Si$ atoms,there are $6 \times 2 = 12$ monovalent oxygen atoms. This is correct.
$D$. $(HPO_3)_3$ (trimetaphosphoric acid) is a cyclic compound with a $P_3O_9$ ring structure where $P$ atoms are linked via oxygen atoms ($P-O-P$ linkages). There are no $P-P$ linkages present in this structure. Thus,this statement is incorrect.

(D) $1$. The reaction of $NH_3$ with red hot $CuO$ is: $3CuO + 2NH_3 \xrightarrow{\Delta} 3Cu + N_2 + 3H_2O$. This reaction produces $N_2$ gas.
$2$. The thermal decomposition of sodium azide $(NaN_3)$ is: $2NaN_3 \xrightarrow{\Delta} 2Na + 3N_2$. This is a standard method for the preparation of very pure $N_2$.
$3$. The thermal decomposition of barium azide $(Ba(N_3)_2)$ is: $Ba(N_3)_2 \xrightarrow{\Delta} Ba + 3N_2$. This is also a standard method for the preparation of very pure $N_2$.
$4$. Since all the given reactions produce $N_2$,the correct answer is that none of these options fail to produce $N_2$.

(C) In the structure of $P_4O_6$,there are $4$ phosphorus atoms arranged at the corners of a tetrahedron.
Each phosphorus atom is bonded to $3$ oxygen atoms,and each oxygen atom acts as a bridge between two phosphorus atoms.
There are $6$ edges in a tetrahedron,and each edge contains one oxygen atom bridging two phosphorus atoms,resulting in $6$ $P-O-P$ linkages.
Since each $P-O-P$ linkage contains $2$ $P-O$ bonds,the total number of $P-O$ bonds is $6 \times 2 = 12$.

(A) The structures of the given nitrogen oxides are as follows:
$1$. $N_2O_3$: It has a direct $N-N$ bond $(O=N-NO_2)$.
$2$. $N_2O_4$: It has a direct $N-N$ bond $(O_2N-NO_2)$.
$3$. $N_2O_5$: It has an $N-O-N$ linkage $(O_2N-O-NO_2)$ and does not contain a direct $N-N$ bond.
Therefore,$N_2O_3$ and $N_2O_4$ contain a nitrogen-nitrogen bond.

(D) The structure of pyrophosphoric acid $(H_4P_2O_7)$ consists of two $PO_4$ tetrahedra linked by an oxygen atom $(P-O-P)$.
Each phosphorus atom is bonded to two $OH$ groups,one terminal oxygen atom via a double bond,and one bridging oxygen atom.
Therefore,the total number of $P-OH$ bonds is $4$ (two per phosphorus atom).
To calculate the oxidation state of $P$ in $H_4P_2O_7$:
Let the oxidation state of $P$ be $x$.
$4(+1) + 2(x) + 7(-2) = 0$
$4 + 2x - 14 = 0$
$2x = 10$
$x = +5$
Thus,the number of $P-OH$ bonds is $4$ and the oxidation state of phosphorus is $+5$.

(B) Group $15$ elements have a stable half-filled $ns^2 np^3$ electronic configuration.
Due to this extra stability,they have higher ionization enthalpy values compared to Group $16$ elements in the corresponding periods.
Therefore,Group $16$ elements have lower ionization enthalpy values compared to Group $15$ elements.

(C) $H_4P_2O_6$ (Hypophosphoric acid) contains a $P-P$ bond.
Its structural formula is $(HO)_2P(O)-P(O)(OH)_2$.

(B) The correct order for basic strength of group $15$ hydrides is $NH_3 > PH_3 > AsH_3 > SbH_3 > BiH_3$.
As the size of the central atom increases,the electron density on the central atom decreases,making it less available for donation,thus decreasing basic strength.
Therefore,the sequence $NH_3 < PH_3 < AsH_3 < SbH_3$ is incorrect as it represents increasing basic strength,whereas it should be decreasing.