Carbon family Questions in English

(B) Sulphur exhibits a strong tendency for catenation.
It forms a stable puckered ring structure consisting of eight atoms,represented as $S_8$.

(C) The paramagnetic behavior of a molecule is determined by the presence of unpaired electrons.
$CO_2$ has $16$ valence electrons and all are paired.
$SO_2$ has $18$ valence electrons and all are paired.
$SiO_2$ is a covalent network solid where all electrons are paired.
$ClO_2$ has $19$ valence electrons,meaning it contains an odd number of electrons,which results in at least one unpaired electron.
Therefore,$ClO_2$ exhibits paramagnetic behavior.

(C) The structure of $P_4O_{10}$ consists of a $P_4$ tetrahedron where each phosphorus atom is bonded to three bridging oxygen atoms (forming $P-O-P$ linkages) and one terminal oxygen atom via a double bond (or coordinate bond).
Thus,each phosphorus atom is attached to a total of $4$ oxygen atoms.

(D) $BaO$,$CaO$,and $Na_2O$ are metal oxides which are basic in nature. When they react with water,they form metal hydroxides,resulting in a $pH > 7$.
$P_2O_5$ is a non-metal oxide which is acidic in nature. When it reacts with water,it forms phosphoric acid $(H_3PO_4)$:
$P_2O_5 + 3H_2O \rightarrow 2H_3PO_4$
Since $H_3PO_4$ is an acid,the resulting solution will have a $pH < 7$.

(B) The correct option is $B$.
In phosphorous acid $(H_3PO_3)$,let the oxidation state of $P$ be $x$.
The oxidation state of $H$ is $+1$ and $O$ is $-2$.
Calculation: $3(+1) + x + 3(-2) = 0$.
$3 + x - 6 = 0$.
$x - 3 = 0$,so $x = +3$.

(B) Silicon forms silicides with strongly electropositive metals (like $Na$,$Mg$,$K$,etc.).
In these compounds,silicon acts as the more electronegative element and exhibits an oxidation number of $-4$.

(B) In the periodic table,metallic character increases as we move down a group.
Group $14$ consists of Carbon $(C)$,Silicon $(Si)$,Germanium $(Ge)$,Tin $(Sn)$,and Lead $(Pb)$.
Carbon is a non-metal,Silicon and Germanium are metalloids,while Tin and Lead exhibit metallic character.
Lead $(Pb)$,being the last element of group $14$,is a weak metal.

(A) $CO$ is a neutral oxide.
Neutral oxides do not react with either acids or bases.
Other examples of neutral oxides include $N_2O$ and $NO$.
$SnO_2$ and $ZnO$ are amphoteric oxides,while $SiO_2$ is an acidic oxide.

(D) In group $14$,as we move down the group,the inert pair effect becomes more prominent due to the inertness of the $ns^2$ electron pair.
This leads to an increased tendency to exhibit the $+2$ oxidation state and a decreased tendency to exhibit the $+4$ oxidation state.
Therefore,the stability of the $+2$ oxidation state increases down the group.

(D) $CaO$ is a basic oxide because it reacts with acids to form salt and water.
$CO_2$ is an acidic oxide as it reacts with bases to form carbonates.
$SiO_2$ is a weakly acidic oxide.
$SnO_2$ is an amphoteric oxide,meaning it can react with both acids and bases.

(D) When $SiF_4$ undergoes hydrolysis,it reacts with water to form orthosilicic acid $(H_4SiO_4)$ and hexafluorosilicic acid $(H_2SiF_6)$.
The balanced chemical equation is:
$3SiF_4 + 4H_2O \rightarrow H_4SiO_4 + 2H_2SiF_6$
Among the given options,$H_2SiF_6$ is the acid formed.

(C) Polyphosphates (such as sodium hexametaphosphate or $STPP$) are used as water softening agents because they form soluble complexes with metal cations like $Ca^{2+}$ and $Mg^{2+}$ present in hard water,thereby preventing them from forming precipitates with soap or detergents.

(D) Bleaching powder $(CaOCl_2)$ is produced by the action of chlorine gas on dry slaked lime $(Ca(OH)_2)$.
The chemical reaction is: $Ca(OH)_2 + Cl_2 \rightarrow CaOCl_2 + H_2O$.

(D) Bleaching powder $(CaOCl_2)$ is a yellowish-white powder that is soluble in water,but a small amount of insoluble lime is always left behind due to the presence of impurities. It reacts with dilute acids to release chlorine gas $(CaOCl_2 + H_2SO_4 \rightarrow CaSO_4 + H_2O + Cl_2)$. It acts as an oxidising agent. The statement that it is 'highly soluble in water' is incorrect because it is only sparingly soluble,leaving behind a residue of calcium hydroxide.

(A) The major constituents of Portland cement are Lime ($CaO$,$50-60\%$),Silica ($SiO_2$,$20-25\%$),Alumina ($Al_2O_3$,$5-10\%$),and small amounts of other oxides. Among the given options,Silica is the most significant constituent after lime.

(C) Carbon $(C)$ has the electronic configuration $2s^2 2p^2$ and lacks $d$-orbitals in its valence shell,limiting its covalency to $4$.
Silicon $(Si)$ has the electronic configuration $3s^2 3p^2$ and possesses vacant $3d$-orbitals.
Due to the availability of these low-lying vacant $d$-orbitals,silicon can expand its octet and accommodate more ligands,allowing it to exhibit a coordination number of $6$ (e.g.,in $[SiF_6]^{2-}$).

(A) Silicon tetrachloride $(SiCl_4)$ undergoes hydrolysis in the presence of water to form silicon dioxide $(SiO_2)$ and hydrogen chloride $(HCl)$.
The reaction is: $SiCl_4 + 2H_2O \rightarrow SiO_2 + 4HCl$.

(C) In the laboratory,silicon is prepared by the reduction of silica $(SiO_2)$ with magnesium $(Mg)$.
The chemical reaction is:
$SiO_2 + 2Mg \rightarrow Si + 2MgO$

(D) Red lead is chemically known as trilead tetroxide with the formula $Pb_3O_4$.
Upon heating,it decomposes into lead$(II)$ oxide $(PbO)$ and oxygen gas $(O_2)$:
$2Pb_3O_4(s) \rightarrow 6PbO(s) + O_2(g)$.

(A, D) In the carbon family $(Group \ 14)$:
$1$. $C$ (Carbon) is a non-metal.
$2$. $Si$ (Silicon) is a metalloid.
$3$. $Ge$ (Germanium) is a metalloid.
$4$. $Sn$ (Tin) and $Pb$ (Lead) are metals.
Since both $Si$ and $Ge$ are metalloids,the question as stated has two correct options. However,in many contexts,$Ge$ is specifically highlighted as a metalloid. Given the options,both $Si$ and $Ge$ are metalloids.

(B) The term 'lead pencil' is a misnomer. It does not contain lead $(Pb)$.
Instead,it contains a mixture of $Graphite$ and clay,which is used as the core of the pencil.

(C) At high temperatures,$CaC_2$ (calcium carbide) reacts with $N_2$ gas to form calcium cyanamide $(CaCN_2)$ and carbon. The reaction is: $CaC_2 + N_2 \rightarrow CaCN_2 + C$. This mixture of $CaCN_2$ and $C$ is known as nitrolim,which is used as a fertilizer.

(A) The correct method for preparing silicon in the laboratory is by the reduction of silica $(SiO_2)$ with magnesium metal.
The chemical reaction is: $SiO_2 + 2Mg \to Si + 2MgO$.

(D) The inert pair effect is the reluctance of the $ns^2$ electron pair to participate in bonding due to poor shielding by intervening $d$ and $f$ orbitals.
It becomes more prominent as we move down the group in the $p$-block elements.
Among the given elements of group $14$ $(C, Si, Ge, Sn, Pb)$,$Pb$ is at the bottom of the group.
Therefore,the inert pair effect is most prominent in $Pb$.

(A) Glass is primarily composed of silica $(SiO_2)$. When it reacts with hydrofluoric acid $(HF)$,it forms silicon tetrafluoride $(SiF_4)$ and water.
The reaction is: $SiO_2 + 4HF \rightarrow SiF_4 + 2H_2O$.
Further,$SiF_4$ can react with excess $HF$ to form hydrofluorosilicic acid $(H_2SiF_6)$,but the primary product of the reaction with glass is $SiF_4$.

(A) smoke screen is a mass of dense artificial smoke used to conceal military areas or operations from an enemy. It is produced by calcium phosphide $(Ca_3P_2)$.
When $Ca_3P_2$ reacts with water,it produces phosphine gas $(PH_3)$:
$Ca_3P_2 + 6H_2O \rightarrow 3Ca(OH)_2 + 2PH_3$
The phosphine $(PH_3)$ formed spontaneously catches fire in the air to produce phosphorus pentoxide $(P_2O_5)$:
$2PH_3 + 4O_2 \rightarrow P_2O_5 + 3H_2O$
$P_2O_5$ forms a white,dense smoke screen.

(A) Aerated water is water that has been artificially impregnated with a large amount of gas,specifically $carbon \ dioxide$ $(CO_2)$.
This gas is either injected under pressure or occurs due to natural geological processes.

(A) Silicon is an important constituent of rocks in the form of silica $(SiO_2)$ and silicates.
$A$. Rocks: Silicon is a major component of the Earth's crust in the form of silica and silicates.
$B$. Amalgams: These are alloys of mercury $(Hg)$ with other metals.
$C$. Chlorophyll: This pigment contains magnesium $(Mg)$ at its center.
$D$. Haemoglobin: This protein contains iron $(Fe)$ as its central metal ion.

(D) The hydrolysis of $SiF_4$ occurs in two steps.
First,$SiF_4$ reacts with water to form silicic acid $(Si(OH)_4)$ and hydrofluoric acid $(HF)$.
$SiF_4 + 4H_2O \rightarrow Si(OH)_4 + 4HF$.
The $HF$ produced then reacts with excess $SiF_4$ to form fluorosilicic acid: $2HF + SiF_4 \rightarrow H_2SiF_6$.
However,the primary product of the hydrolysis of $SiF_4$ is $Si(OH)_4$ (orthosilicic acid).

(B) Zeolites are aluminosilicates with a three-dimensional network structure.
In this structure,some $SiO_4$ units are replaced by $AlO_4$ tetrahedra.
They do not contain $AlO_6$ units.
Therefore,the statement that $SiO_4^{4-}$ units are replaced by $AlO_4^{5-}$ and $AlO_6^{9-}$ ions is false.
Zeolites possess an open,cage-like structure that allows them to act as cation exchangers and adsorb small molecules.

(C) The correct option is $(C)$.
$Pb_3O_4$ (red lead) is a mixed oxide.
It is composed of two oxides of lead,$PbO$ and $PbO_2$,in a $2:1$ molar ratio.
It can be represented as $2PbO \cdot PbO_2$.

(D) The correct answer is $(d)$.
In the carbon family (Group $14$),as we move down the group,the stability of the $+2$ oxidation state increases due to the inert pair effect,which is a consequence of poor shielding by $d$ and $f$ electrons.

(D) In sheet silicates,three out of four oxygen atoms of the $[SiO_4]^{4-}$ unit are shared with adjacent tetrahedra.
In pyrosilicates,only one oxygen atom is shared.
In linear chain silicates,two oxygen atoms per tetrahedron are shared.
In three-dimensional silicates,all four oxygen atoms are shared.

(A) $Pb_3O_4$ is known as red lead or minium.
It is a mixed oxide of lead,which can be represented as $2PbO \cdot PbO_2$.

(C) The chemical formula for white lead is $2PbCO_3 \cdot Pb(OH)_2$.
It is a basic lead carbonate.

(C) The ionic character of a compound depends on the electronegativity difference between the bonded atoms and the oxidation state of the metal.
As we move down group $14$,the metallic character of the elements increases,which enhances the ionic nature of their halides.
Additionally,due to the inert pair effect,the lower oxidation state $(+2)$ becomes more stable for heavier elements like $Pb$.
$PbCl_2$ involves $Pb$ in the $+2$ oxidation state,which is more ionic compared to the covalent $PbCl_4$ (where $Pb$ is in the $+4$ state) and the covalent halides of lighter elements like $C$ and $Si$.
Therefore,$PbCl_2$ is the most ionic compound among the given options.

(B) Lead dioxide,$PbO_2$,is a strong oxidizing agent. It is widely used in the manufacture of matches,dyes,and rubber substitutes.

(C) $Pb_3O_4$ (red lead) is a mixed oxide,which is formed by the combination of $2PbO$ (lead$(II)$ oxide) and $PbO_2$ (lead$(IV)$ oxide).
$2PbO + PbO_2 \rightarrow Pb_3O_4$

(B) Among the given elements,tin $(Sn)$ exhibits the phenomenon of allotropy.
$\beta$-tin (the metallic form,or white tin),which is stable at and above room temperature,is malleable.
In contrast,$\alpha$-tin (the non-metallic form,or gray tin),which is stable below $13.2^{\circ}C$,is brittle.

(B) The decomposition reactions are as follows:
$1. \text{Ammonium nitrate: } NH_4NO_{3(s)} \xrightarrow{\Delta} 2H_2O_{(g)} + N_2O_{(g)}$
$2. \text{Sodium nitrate: } 2NaNO_{3(s)} \xrightarrow{\Delta} 2NaNO_{2(s)} + O_{2(g)}$
$3. \text{Silver nitrate: } 2AgNO_{3(s)} \xrightarrow{\Delta} 2Ag_{(s)} + 2NO_{2(g)} + O_{2(g)}$
$4. \text{Lead nitrate: } 2Pb(NO_3)_{2(s)} \xrightarrow{\Delta} 2PbO_{(s)} + 4NO_{2(g)} + O_{2(g)}$
Among these,$NH_4NO_3$ decomposes completely into gaseous products ($H_2O$ vapor and $N_2O$ gas),leaving no solid residue. Therefore,the correct option is $(B)$.

(C) Non-metals form acidic oxides. Phosphorus is a non-metal that forms a solid oxide $P_4O_{10}$ (or $P_2O_5$).
$P_4 + 5O_2 \to P_4O_{10}$
When $P_4O_{10}$ dissolves in water,it forms phosphoric acid,which is an acidic solution:
$P_4O_{10} + 6H_2O \to 4H_3PO_4$
Thus,the correct element is Phosphorus.

(A) White phosphorus is kept in water to prevent it from coming in contact with atmospheric oxygen,as it has a very low ignition temperature of approximately $30\,^{\circ}C$ and catches fire spontaneously in air.

(C) The thermal decomposition of lead nitrate $(Pb(NO_3)_2)$ upon strong heating is represented by the following chemical equation:
$2Pb(NO_3)_2(s) \rightarrow 2PbO(s) + 4NO_2(g) + O_2(g)$
As shown in the reaction,the gases evolved are nitrogen dioxide $(NO_2)$ and oxygen $(O_2)$.

(B) When calcium carbide $(CaC_2)$ is heated with nitrogen at a temperature of $500 - 600 ^\circ C$,it forms calcium cyanamide $(CaCN_2)$ and carbon $(C)$.
This mixture is known as nitrolim.
The chemical reaction is:
$CaC_2 + N_2 \xrightarrow{500 - 600 ^\circ C} CaCN_2 + C$

(D) Superphosphate of lime is a mixture of calcium dihydrogen phosphate,$Ca(H_{2}PO_{4})_{2}$,and calcium sulfate,$CaSO_{4}$.
It is prepared by the action of sulfuric acid on phosphate rock (fluorapatite or phosphorite).
The reaction is: $Ca_{3}(PO_{4})_{2} + 2H_{2}SO_{4} \rightarrow Ca(H_{2}PO_{4})_{2} + 2CaSO_{4}$.

(D) Superphosphate of lime is produced by the action of concentrated sulfuric acid on phosphate rock (calcium phosphate).
The chemical reaction is: $Ca_3(PO_4)_2 + 2H_2SO_4 + 4H_2O \rightarrow Ca(H_2PO_4)_2 \cdot H_2O + 2CaSO_4 \cdot 2H_2O$.
Thus,it is a mixture of monocalcium phosphate monohydrate and calcium sulfate dihydrate.

(B) $P_4O_{10}$ is an acidic oxide,while $NH_3$ is a basic gas.
When $NH_3$ is passed over $P_4O_{10}$ for drying,they react with each other to form ammonium phosphate.
Therefore,$P_4O_{10}$ cannot be used as a drying agent for $NH_3$.

(C) $P_2O_5$ (phosphorus pentoxide) has a very high affinity for water.
It is widely used in laboratories and industries as a powerful dehydrating agent to remove water from various organic and inorganic compounds.

(C) The rubbing surface on the sides of a safety matchbox is coated with a mixture of powdered glass and red phosphorus $(P_4)$. The match head contains potassium chlorate $(KClO_3)$,antimony sulphide $(Sb_2S_3)$,and glue. When the match is struck against the side,the heat generated converts some red phosphorus into white phosphorus,which reacts with potassium chlorate to produce enough heat to ignite the antimony sulphide.

(C) Bone black is an impure form of $Carbon$. It is obtained by the destructive distillation of bones. It contains about $10-12\%$ of finely divided $Carbon$ deposited on a porous framework of calcium phosphate $(Ca_3(PO_4)_2)$.