Which of the following is correct for $$P_4$$ molecule of white phosphorus?

It has four lone pairs of electrons

It has 6 lone pairs of electrons

It has six $$P-P$$ single bonds

It has three $$P-P$$ single bonds

Single $$N-N$$ bond is weaker than the single $$P-P$$ bond. This is because of:

smaller atomic size of $$N$$ as compared to that of $$P$$

larger $$N-N$$ bond length in comparison to $$P-P$$ bond length

high interelectronic repulsion of the non-bonding electrons, owing to the small $$N-N$$ bond length in comparison to that in $$P-P$$ single bond

higher electronegativity of $$N$$ in comparison to $$P$$

MCQ Questions for Class 12 Engineering Chemistry The P-Block Elements Quiz 7

Which of the following is correct for

P_{4}

$P_4$ molecule of white phosphorus?

0%
It has 6 lone pairs of electrons
0%
It has six $P-P$ single bonds
0%
It has three $P-P$ single bonds
0%
It has four lone pairs of electrons

Maximum number of compounds are known in the case of:

0%
neon
0%
xenon
0%
krypton
0%
argon

Ammonia will be obtained in:

0%
$CaCN_2 +H_2O \to$
0%
$NH_4H_2PO_4 \ \xrightarrow {Heat}$
0%
Both (A) and (B)
0%
None

Which of the following statement are correct?

0%
All the three $N-O$ bond lengths in $HNO_3$ are equal
0%
All $P-Cl$ bond lengths in $PCl_5$ molecule in gaseous state are equal
0%
$P_4$ molecule in white phosphorus have angular strain therefore white phosphorus is very reactive
0%
$PCl$ is ionic in solid state in which cation is tetrahedral and anion is octahedral

For

N H_{2} O H . H C l + N a N O_{2} \to (B) \overset{C u}{\to} (X)_{g a s}

$NH_2OH.HCl +NaNO_2 \to (B) \xrightarrow {Cu} (X)_{gas}$ . Which of the following is correct?

0%
$(B)$ is an amphoteric oxide
0%
$(X)$ is a colourless, diamagnetic gas which combines with $Al$ on heating
0%
$(X)$ can be produced by action of $(Zn +NaOH)$ on $NaNO_2$
0%
none of the above

Electron affinity of noble gas element is:

0%
very high
0%
high
0%
low
0%
zero

Of the following species, one which is non-existent:

0%
$XeF_6$
0%
$XeF_5$
0%
$XeF_4$
0%
$XeF_2$

Which of the following is not correct?

0%
$XeO_3$ has four $\sigma$ and four $\pi$ bonds.
0%
The hybridization of $Xe$ in $XeF_4$ is $sp^3d^2$ .
0%
Among noble gases, the occurrence of argon is highest in air.
0%
Liquid helium is used as cryogenic liquid.

In the Ostwald's process for the manufacture of

H N O_{3}

$HNO_{3}$ , the catalyst used is

0%
Mo
0%
Fe
0%
Ni
0%
Pt

Explanation

4 N H_{3} + 5 O_{2} \to_{1100 K}^{P t - g a u g a e} 4 N O \overset{O_{2}}{⟶} 4 N O_{2} \overset{2 H_{2} O + O_{2}}{⟶} 4 H N O_{3}

$4NH_{3}+5O_{2}\xrightarrow[1100 K]{Pt -gaugae}4NO \ \overset{O_{2}}{\longrightarrow} 4NO_{2} \overset{2H_{2}O+O_{2}}{\longrightarrow} 4HNO_{3}$

So, option D is the correct answer,

The fluoride of xenon with zero dipole moment is:

0%
$XeF_6$
0%
$XeOF_4$
0%
$XeF_4$
0%
$XeO_2F_2$

Explanation

X e F_{4}

$XeF_4$ is symmetrical with a square planar shape, so the resultant dipole moment is zero
1615245_1734258_ans_dbddbdd82659448ab7ccedb68a85ab4a.PNG

1615245_1734258_ans_dbddbdd82659448ab7ccedb68a85ab4a.PNG

Xenon form compounds with fluorine under different conditions. The known fluorides are:

0%
$XeF$
0%
$XeF_{2}$
0%
$XeF_{4}$
0%
$XeF_{3}$

Amongs the following which is not member of apatite family

0%
$Ca_{9}(PO_{4})_{6}.CaCl_{2}$
0%
$Ca_{9}(PO_{4})_{6}.CaF_{2}$
0%
$Ca_{9}(PO_{4})_{2}.Ca(OH)_{2}$
0%
None of these

Ammonia gas can evolve by heating

0%
$NaNO_{3}$ with zinc dust and sodium hydroxide
0%
dry $(NH_{4})_{2}$ with dil. $HNO_{3}$
0%
dry $(NH_{4})_{2}Cr_{2}O_{7}$
0%
ammonium chloride with lime water

Explanation

(A) N O_{3}^{-} + 9 [H] \overset{Z n / N a O H}{\to} N H_{3} + 3 H_{2} O

$(A)NO_{3}^{-}+9\left [ H \right ]\xrightarrow{Zn/NaOH}NH_{3}+3H_{2}O$

(B) C a C N_{2} + 3 H_{2} O \to C a C O_{3} + 2 N H_{3}

$(B)CaCN_{2}+3H_{2}O\rightarrow CaCO_{3}+2NH_{3}$

(C) (N H_{4})_{2} C r_{2} O_{7} \to N_{2} + C r_{2} O_{3} + 4 H_{2} O; N H_{4} N O_{3} \to N_{2} O + 2 H_{2} O

$(C)(NH_{4})_{2}Cr_{2}O_{7}\rightarrow N_{2}+Cr_{2}O_{3}+4H_{2}O;NH_{4}NO_{3}\rightarrow N_{2}O+2H_{2}O$

(D) N H_{4} C l + C a (O H)_{2} \to 2 N H_{3} + C a C l_{2} + 2 H_{2} O

$(D)NH_{4}Cl+Ca(OH)_{2}\rightarrow 2NH_{3}+CaCl_{2}+2H_{2}O$

A,D

Which of the following names are used for group zero elements?

0%
Rare gases of atmosphere
0%
Noble gases
0%
Inert gases
0%
Rare-earths

Which statement(s) is/are correct?

0%
Radon is obtained from radium
0%
Argon is most abundant noble gas in atmosphere
0%
Monazite is a source of gas
0%
Helium is present in natural gas

The formation of

O_{2}^{+} [P t F_{6}]^{-}

$O_2^+[PtF_6]^-$ is the basis for the formation of xenon fluorides. This is because

0%
$O_2$ and $Xe$ have comparable sizes
0%
Both $O_2$ and $Xe$ are gases
0%
$O_2$ and $Xe$ have comparable ionisation energies
0%
Both (a) and (c)

Calcium cyanide on treatment with steam under pressure gives

N H_{3}

$NH_{3}$ and

0%
$CaHCO_3$
0%
$CaO$
0%
$Ca(OH)_{2}$
0%
$CaCO_{3}$

Explanation

Calcium cyanamide on treatment with steam under pressure gives

N H_{3}

$NH_{3}$ and

C a C O_{3} .

$CaCO_{3}.$

C a C N_{2} + 3 H_{2} O \to_{p r e s s u r e}^{u n d e r} C a C O_{3} + 2 N H_{3}

$CaCN_{2}+3H_{2}O\xrightarrow [ pressure ]{ under } CaCO_{3}+2NH_{3}$

In which of these processes platinum is used as a catalyst?

0%
Oxidation of ammonia to form $HNO_3$
0%
Hardening of oils
0%
Production of synthetic rubber
0%
Synthesis of methanol

Explanation

Platinum acts as a catalyst in the oxidation of ammonia to form nitric oxide. This reaction is used in Ostwald's method of nitric acid preparation.

4 N H_{3} + 5 O_{2} \overset{750^{0} C; P t - c a t a l y s t}{\to} 4 N O + 6 H_{2} O

$4NH_3+ 5O_2 \xrightarrow { 750^0C ;\ Pt-catalyst } 4NO + 6H_2 O$

2 N O + O_{2} \to 2 N O_{2}

$2 NO + O_2 \rightarrow 2NO_2$

4 N O_{2} + O_{2} + 2 H_{2} O \to 4 H N O_{3}

$4NO_2 +O_2 + 2H_2O \rightarrow 4HNO_3$

Which of the following statement is/are correct?

0%
$pK_{a1}$ of maleic acid is less than $pK_{a1}$ of fumaric acid.
0%
$pK_{a2}$ of maleic acid is greater than $pK_{a2}$ of formaric acid.
0%
Phthalic acid is a strong acid than isophthalic acid.
0%
Isopthalic acid is a strong than terephthalic acid.

Which of the following is a hard acid?

0%
$Br_2$
0%
$Cd^{2+}$
0%
$CO_2$
0%
$Fe^{3+}$

Fill in the blanks :
Potassium nitrate crystals on heating strongly yield potassium nitrite and ________ gas.

0%
oxygen
0%
phosphoric
0%
nitric acid
0%
sulphurous

Point out in which of the following properties oxygen differs from the rest of the members of its family (Group -

V I A

$VIA$ ).

0%
High value of ionisation energies
0%
Oxidation states $(2,4,6)$
0%
Polymorphism
0%
Formation of hydrides

Explanation

Oxygen does not show

+ 4

$+4$ and

+ 6

$+6$ oxidation state. Oxygen shows

- 2, - 1, - \frac{1}{2}, 0, + 1

$-2,-1, -\dfrac12, 0, +1$ and

+ 2

$+2$ oxidation state.

Calcium cyanamide on treatment with steam produce

0%
$CaCO_{3}+NH_{3}$
0%
$CaHCO_{3}+NH_{3}$
0%
$CaO+NH_{3}$
0%
$Ca(OH)_{2}+NH_{3}$

Explanation

Calcium cyanamide on treatment with steam produce

C a C O_{3} + N H_{3} .

$CaCO_{3}+NH_{3}.$

C a C N_{2} + 3 H_{2} O \to C a C O_{3} + 2 N H_{3}

$CaCN_2+3H_2O\to CaCO_3+2NH_3$

Which is the most abundant element on the earth's crust among group

15

$15$ elements?

0%
$N$
0%
$As$
0%
$P$
0%
$Sb$

Ammonia can be obtained by adding water to?

0%
Ammonium chloride
0%
Ammonium nitrite
0%
Magnesium nitride
0%
Magnesium nitrate

Nitrogen atom of amino group is ______hybridised.

0%
$sp$
0%
$sp^2$
0%
$sp^3$
0%
$sp^3d$

Which of the following compound does not form

p π - p π

$p \pi - p \pi$ bond?

0%
$SO_{3}$
0%
$NO_{3}^{-}$
0%
$SO_{4}^{2-}$
0%
$CO_{3}^{2-}$

Consider the following reactions:

A_{x} + y B_{2} \to_{s u p p l y o f a i r}^{l i m i t e d} c o m p o u n d^{'} P^{'} \to_{e x c e s s a i r}^{+ z B_{2}} C o m p o u n d^{'} Q^{'}

$A_{x} + yB_{2} \xrightarrow [supply \,of \,air]{limited} compound \,'P' \xrightarrow [excess \,air]{+zB_{2}} Compound \,'Q'$
If atomic number of elements A and B are

15

$15$ and

8

$8$ respectively, then according to the given information the correct statement(s) is/are:

0%
$(B-A-B)$ bond angle in compound $'Q' > (B-A-B)$ bond angle in compound 'P'
0%
$(A-B)$ bond length in compound $'Q' < (A-B)$ bond length in compound 'P'
0%
Value of $x + y + z$ is $9$
0%
Compound 'P' is $P_{2}O_{5}$ and compound 'Q' is $P_{4}O_{10}.$

Explanation

\underset{(A_{x})}{P_{4}} + \underset{(y B_{2})}{3 O_{2}} \to_{s u p p l y o f a i r}^{L i m i t e d} \underset{(P)}{P_{4} O_{6}} \to_{e x c e s s a i r}^{2 O_{2}} \underset{(Q)}{P_{4} O_{10}}

$\underset{(A_x)}{P_4} + \underset{(yB_2)}{3O_2} \xrightarrow[supply \,of \,air]{Limited} \underset{(P)}{P_4O_6} \xrightarrow[excess \,air]{2O_2} \underset{(Q)}{P_4O_{10}}$

According to Bent's rule :

x_{1} > x_{2}, α < β

$x_{1} > x_{2}, \alpha < \beta$

x + y + z = 4 + 3 + 2 = 9

$x + y + z = 4 + 3 + 2 = 9$

Hence, Options "A", "B" & "C" are correct answers.

1832113_1861367_ans_df660c049ab74d6fb403d0847af5d3ed.png

Oxygen gas is not produced from the following decomposition reaction :

0%
$K_2Cr_2O_7 \xrightarrow{\Delta}$
0%
$Ag_2C_2O_4 \xrightarrow{\Delta}$
0%
$Pb(NO_3)_2 \xrightarrow{\Delta}$
0%
$Ag_2CO_3 \xrightarrow{\Delta}$

Explanation

Thermal decomposition of silver oxalate gives metallic silver and

C O_{2}

$CO_2$ gas.

The reaction is as follows:

A g_{2} C_{2} O_{4} \overset{Δ}{\to} 2 A g (s) + 2 C O_{2} (g)

$Ag_2C_2O_4\xrightarrow{\Delta} 2Ag(s)+2CO_2(g)$

Which of the following catalyst is used in the Haber's process for the manufacture of ammonia?

0%
Platinized asbestos
0%
Iron with a suitable promoter
0%
Copper oxide
0%
Alumina

Explanation

Hint: The method of preparation of ammonia from nitrogen and hydrogen is called Haber’s process.

Explanation:

In Haber’s process ammonia is produced by the reaction of $1$ molecule of $N_2$ and $3$ molecules of $H_2$

The reaction is- $N_{2(g)}+3H_{2(g)} \rightarrow 2NH_{3(g)}$ .

This reaction is spontaneous at low temperatures and high pressure.
At STP, the reaction is quite slow that is why there is a need for a catalyst that speeds up the reaction.
In this reaction, iron is used as a catalyst. Here elemental $Mo$ is used as a promoter.
Promoters are the ones that improve the performance of the catalyst so the reaction will drive faster.

Final Answer: Correct option (B).

A solution containing

B r^{-}

$Br^{-}$ ions is treated with each of the following. Which of these will not liberate bromine gas (in acidic medium) ?

0%
$Cl_{2}$
0%
$HI$
0%
$I_{2}$
0%
$SO_{3}$

Explanation

C l_{2}

$Cl_{2}$ and

S O_{3}

$SO_{3}$ being stronger oxidising agent, oxidise

B r^{-}

$Br^{-}$ to

B r_{2}

$Br_{2}$ gas

2 B r^{-} + C l_{2} \to B r_{2} + 2 C l^{-}

$2Br^{-}+Cl_{2}\rightarrow Br_{2}+2Cl^{-}$

S O_{3} \to S O_{2} + [O]

$SO_{3}\rightarrow SO_{2}+[O]$

2 B r^{-} + 2 H^{+} + [O] \to B r_{2} + H_{2} O

$2Br^{-}+2H^{+}+[O]\rightarrow Br_{2}+H_{2}O$

2 B r^{-} + 2 H^{+} + 2 H^{+} \to B r_{2} + H_{2} O + S O_{2}

$2Br^{-}+2H^{+}+2H^{+}\rightarrow Br_{2}+H_{2}O+SO_{2}$
HI being reducing agent and

I_{2}

$I_{2}$ being weak oxidising agent (comes after bromine) don't liberate bromine from

B r_{-}

$Br_{-}$ .

Which is soluble in water :

0%
AgCl
0%
AgBr
0%
AgI
0%
AgF

Explanation

Ionic compounds are soluble in water
In AgF

F^{-}

$F^-$ is small onion with high charge density so less polarisation and less covalent character in the whole group of Ag holdies. AgF is most Ionic in nature.

An element with variable (more than one) valency stored under water is:

0%
silicon
0%
iron
0%
phosphorus
0%
gold

Explanation

Phosphorous can accept either 3 electrons or donate 5 electrons depending on the neighborhood element. Hence, it has variable valency. The allotropic form of phosphorus (white phosphorus) is less stable and therefore highly reactive because of angular strain in

P_{4}

$P_4$ molecule where the angles are only

60^{o}

$60^o$ .

It readily catches fire in air to give dense white fumes of

P_{4} O_{10}

$P_{4}O_{10}$ . Therefore, it is kept in water to avoid oxidation by the oxygen present in the air. It is kept in water because it is insoluble in water.

An element X of group 15 exists as diatomic molecule and combines with hydrogen at 773 K in presence of the catalyst to form a compound, ammonia which has a characteristic pungent smell. The element X is :

0%
nitrogen
0%
oxygen
0%
phosphorous
0%
sulphur

Phosphide and phosphate ions are:

0%
$PO_{4}^{3-}$ and $P^{3-}$
0%
$P^{3-}$ and $PO_{4}^{3-}$
0%
$PO_{3}^{4-}$ and $P^{4-}$
0%
$P^{4-}$ and $PO_{3}^{4-}$

Explanation

Phosphide and phosphate ions are

P^{3 -}

$P^{3-}$ and

P O_{4}^{3 -}

$PO_{4}^{3-}$ respectively. Thus, sodium phosphide can be represented as

N a_{3} P

$Na_3P$ and sodium phosphate can be represented as

N a_{3} P O_{4}

$Na_3PO_4$ .

The steps for the manufacture of nitric acid by Ostwald process are given. Arrange them in sequence.
(a) Recirculation of nitric oxide
(b) Addition of

H_{2} O

$H_2O$ to nitrogen dioxide
(c) Oxidation of nitric oxide.
(d) Catalytic oxidation of ammonia

0%
d c b a
0%
d c a b
0%
c d b a
0%
c d a b

Explanation

In the first step of Ostwald process, ammonia is oxidised to form nitric oxide.

4 N H_{3} + 5 O_{2} \leftrightarrow 4 N O + 6 H_{2} O

$4NH_3 + 5O_2 \leftrightarrow 4NO + 6H_2O$

In the second step, nitric oxide is oxidised to form nitrogen dioxide.

2 N O + O_{2} \leftrightarrow 2 N O_{2}

$2NO + O_2 \leftrightarrow 2NO_2$

Then, nitrogen dioxide is mixed with water in an absorption tower to get nitric acid.

3 N O_{2} + H_{2} O \to 2 H N O_{3} + N O

$3NO_2 + H_2O \rightarrow 2HNO_3 + NO$

Lastly, nitrogen dioxide is recirculated in the absorption tower to ensure that more and more of it is absorbed with water to increase the concentration of nitric acid. The vapours of nitric acid are then passed over concentrated sulphuric acid that absorbs water from it to increase its concentration further.

Nitrogen can exhibit which of the following oxidation states:

0%
$+1$
0%
$+2$
0%
$+3$
0%
all of the above

Explanation

Nitrogen forms oxides in which nitrogen exhibits each of its positive oxidation numbers from +1 to +5.

N_{2} O

$N_2O$ has nitrogen in the +1 oxidation state,

N O

$NO$ has +2 and

N O_{2}

$NO_2$ has +4 oxidation state.

The ions

M g^{+ 2}

$Mg^{+2}$ and

P O_{4}^{- 3}

$PO_{4}^{-3}$ combine to form compounds:

0%
$Mg_3(PO_4)_2$
0%
$Mg_2(PO_4)_2$
0%
$Mg_2(PO_4)_3$
0%
$Mg_3(PO_4)_3$

Explanation

The ions

M g^{+ 2}

$Mg^{+2}$ and

P O_{4}^{- 3}

$PO_{4}^{-3}$ combine to form compounds

M g_{3} (P O_{4})_{2}

$Mg_3(PO_4)_2$ . Total charge on cation

= 3 (2 +) = + 6

$=3(2+)=+6$ . Total charge on anion

= 2 (- 3) = - 6

$=2(-3)=-6$ . Hence, total charge on cation

=

$=$ total charge on anion.

Magnesium on heating to redness in an atmosphere of

N_{2}

$N_{2}$ and then on treating with

H_{2} O

$H_{2}O$ gives?

0%
$NH_{3}$
0%
$H_{2}$
0%
$N_{2}$
0%
$O_{2}$

Explanation

Megnesium when heated with

N_{2}

$N_2$ in atmosphere it forms magnesium nitride(

M g_{3} N_{2}

$Mg_3N_2$ ) which on hydrolysis(reaction with water) gives ammonia.

3 M g + N_{2} \to M g_{3} N_{2}

$3Mg+N_2\rightarrow Mg_3N_2$

M g_{3} N_{2} + 6 H_{2} O \to 3 M g (O H)_{2} + 2 N H_{3}

$Mg_3N_2+6H_2O\rightarrow 3Mg(OH)_2+2NH_3$ .

Hence option A is correct.

The elements with atomic numbers 9, 17, 35, 53 and 85 belong to noble gases columnz__________.

0%
True
0%
False

N H_{3}

$NH_{3}$ (Ammonia) has __________shape.

0%
rectangular
0%
pyramidal
0%
trigonal-bi-pyramidal
0%
planar

Explanation

Ammonia has triogonal pyramidal shape and the nitrogen in ammonia undergoes

S P^{3}

$SP^{3}$ hybridization.

Hence option B is correct.

Nitrogen is liberated by the thermal decomposition of only :

0%
$NH_4NO_2$
0%
$NaN_3$
0%
$(NH_4)_2Cr_2O_7$
0%
All of these

Explanation

Nitrogen is liberated by the thermal decomposition of ammonium nitrite

(N H_{4} N O_{2}), N a N_{3}, (N H_{4})_{2} C r_{2} O_{7}

$(NH_4NO_2), NaN_3, (NH_4)_2Cr_2O_7$

Which of the following belongs to the halogen family?

0%
Francium
0%
Polonium
0%
Radium
0%
Astatine

At low temperatures, nitrogen:

0%
is poisonous
0%
is highly soluble in water
0%
can be liquefied on compression
0%
dissociates to form nascent nitrogen

Single

N - N

$N-N$ bond is weaker than the single

P - P

$P-P$ bond. This is because of:

0%
larger $N-N$ bond length in comparison to $P-P$ bond length
0%
high interelectronic repulsion of the non-bonding electrons, owing to the small $N-N$ bond length in comparison to that in $P-P$ single bond
0%
higher electronegativity of $N$ in comparison to $P$
0%
smaller atomic size of $N$ as compared to that of $P$

Explanation

N - N

$N-N$ single bond is weaker than

P - P

$P-P$ bond due to smaller size of

N

$N$ as compared to

P

$P$ . Smaller size of

N

$N$ leads to smaller

N - N

$N-N$ bond length. As a result, the lone pair of electrons on the both the

N

$N$ atoms repel each other leading to unstability or weakening of

N - N

$N-N$ bond. Because of larger size of

P

$P$ atom,

P - P

$P-P$ bond length is more and lone pair-lone pair repulsion between

P

$P$ atoms is less which makes the

P - P

$P-P$ bond stronger than

N - N

$N-N$ bond.

P C l_{5}

$PCl_5$ molecule has:

0%
three fold axis of symmetry
0%
two fold axis of symmetry
0%
both $A$ and $B$
0%
none of these

Explanation

Considering the structure as shown in above image and hybridization of

P C l_{5}

$PCl_5$ , we can say that it has both three fold axis of symmetry and two fold axis of symmetry.
1062891_114756_ans_399fe63f71094b8aad8dc0397490fd25.PNG

1062891_114756_ans_399fe63f71094b8aad8dc0397490fd25.PNG

________ bacteria are microbes that work in the absence of oxygen.

0%
Anaerobic
0%
Aerobic
0%
Gram-positive
0%
Gram-negative

The compound

I F_{7}

$IF_7$ is named as Iodine heptafluoride.

0%
True
0%
False

Explanation

The compound

I F_{7}

$IF_7$ is named as Iodine heptafluoride. Iodine is named first as it is less electronegative. Fluorine is named second as it is more electronegative. The prefix 'hepta' indicates seven. It has pentagonal bipyramidal geometry with

s p^{3} d^{3}

$sp^3d^3$ hybridisation of iodine atom.

Hydride of nitrogen is called :

0%
Hydrogen sulphide
0%
Water
0%
Ammonia
0%
Nitrous oxide

Explanation

Hydride of nitrogen is the binary compound formed by the combination of hydrogen and nitrogen .

The hydride formed with nitrogen is ammonia (

N H_{3}

$NH_3$ ).

Hence option C is correct.

Nitrogen reacts with a metal

M

$M$ to produce a compound

X

$X$ . This compound

X

$X$ , when dissolved in water forms pungent gas

Y

$Y$ which can be identified easily by the formation of white fumes in the presence of

H C l

$HCl$ .

Identify the compound

X

$X$ .

0%
Iron nitride
0%
Calcium nitride
0%
Nitrolim
0%
Sodium nitride

Explanation

Calcium nitride dissolves in water to give ammonia gas.

Compound

X

$X$ is

C a_{3} N_{2}

$Ca_3N_2$ .

C a_{3} N_{2} + 6 H_{2} O \to 3 C a (O H)_{2} + 2 N H_{3}

${Ca_3}N_2 + 6H_2O \rightarrow 3Ca (OH)_2+2{NH_3}$

Hence, (B) is the correct answer.

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