Chemistry M.C.Q (1 Marks)

In diamond, each carbon atom undergoes $sp ^3$ hybridisation and linked to four other carbon atoms by using hybridised orbitals in tetrahedral fashion.

Buckminsterfullerene contains six membered and five membered rings and hence is a cage like molecule.

Graphite is very soft and slippery. Hence, it is used as a dry lubricant in machines running at high temperature.

due to inert pair effect $Tl^{+}$is more stable than $Tl^{+3}$.

Non- planar

In graphite each carbon is bonded with three other carbon atoms. So hybridisation of carbon atom is $sp ^{2}$.

Boron belongs to $2^{\text {nd }}$ period and it does not have vacant $d-$orbital.

As a result, $Pb(II)$ is more stable than $Pb(IV)$

$\mathrm{Sn}(\mathrm{IV})$ is more stable than $\mathrm{Sn}(\mathrm{II})$

$\mathrm{Pb}(\mathrm{IV})$ is easily reduced to $\mathrm{Pb}(\mathrm{II})$

$\mathrm{Pb}(\mathrm{IV})$ is oxidizing agent

$\mathrm{Sn}(\mathrm{II})$ is easily oxidized to $\mathrm{Sn}(\mathrm{IV})$

$\mathrm{Sn}(\mathrm{II})$ is reducing agent

$\mathrm{B}(\mathrm{OH})_{3}+\mathrm{H}_{2} \mathrm{O} \longrightarrow\left[\mathrm{B}(\mathrm{OH})_{4}\right]^{(-)}(\mathrm{aq})+\mathrm{H}^{(-)}(\mathrm{aq})$

maximum $C.N.$ of $A l^{+3}$ is six so it forms $AlF_{6}^{3-}$

$\begin{array}{*{20}{c}}
  {Me} \\ 
  | \\ 
  {Me - Si - OH} \\ 
  | \\ 
  {Me} 
\end{array}$ $+$ $\begin{array}{*{20}{c}}
  {Me} \\ 
  | \\ 
  {HO - Si - Me} \\ 
  | \\ 
  {Me} 
\end{array}$ $\xrightarrow[{ - {H_2}O}]{}$ $\mathop {\begin{array}{*{20}{c}}
  {\,\,Me\,\,\,\,\,\,\,\,\,\,\,\,\,Me} \\ 
  {|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|} \\ 
  {Me - Si - O - Si - Me} \\ 
  {|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|} \\ 
  {\,\,Me\,\,\,\,\,\,\,\,\,\,\,\,\,Me} 
\end{array}}\limits_{Final\,product\,(\dim er)} $

These retain metallic conductivity. These resemble the parent metal in chemical properties (reactivity) but differ in physical properties like hardness, melting point, etc.

But diborane, $B _2 H _6$ contains two electrons each, three centred bonds. Each Boron atom is in a link with four hydrogen atoms. This makes tetrahedral geometry.

Hence, each Boron atom is $sp ^3$ - hybridized.

$Al ( OH )_3+ NaOH \rightarrow NaAlO _2+2 H _2 O$

So the $NaAlO _2$ forms $AlO _2^{-}$.

Thus $AlC{l_3}$ can not be obtained by this method

$2NaOH + 2Al + 6{H_2}O \to 2Na[Al{(OH)_4}] + 3{H_2}$

$A{l_2}{O_3}.2{H_2}O + N{a_2}C{O_3} \to 2NaAl{O_2} + C{O_2} + 2{H_2}O$

$2NaAl{O_2} + 3{H_2}O + C{O_2}\xrightarrow{{333\,K}}$ $2Al{(OH)_3} \downarrow  + N{a_2}C{O_3}$

$2Al{(OH)_3}\xrightarrow{{1473\,K}}A{l_2}{O_3} + 3{H_2}O$

$4 Na _3 AlF _6 \leftrightharpoons 12 Na ^{+}+4 Al ^{3+}+12 F ^{-}$

$4 Al ^{3+}+12 e ^{-} \rightarrow 4 Al \text { (at cathode) }$

$12 F ^{-} \rightarrow 6 F _2+12 e ^{-} \text {(at anode) }$

$2 Al _2 O _3+6 F _2 \rightarrow AlF _3+3 O _2$

$B{F_3} < BC{l_3} < BB{r_3} < B{I_3}$.

$\left[\mathrm{Al}(\mathrm{OH})_{6}\right]^{3-}+3 \mathrm{HCl}$

$\mathrm{Al}_{2} \mathrm{Cl}_{6}+12 \mathrm{H}_{2} \mathrm{O} \rightarrow 2\left[\mathrm{Al}(\mathrm{OH})_{6}\right]^{3-}+12 \mathrm{HCl}$

Thus aluminum chloride is hydrolyzed to form the complex ion containing s hydroxide groups.

$\mathrm{AlCl}_{3}+3 \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{Al}(\mathrm{OH})_{3}+\mathrm{HCl}$

On strongly heating $\mathrm{Al}(\mathrm{OH})_{3}$ is converted into $\mathrm{Al}_{2} \mathrm{O}_{3}$

$\mathrm{Al}(\mathrm{OH})_{3} \rightarrow \mathrm{Al}_{2} \mathrm{O}_{3}+3 \mathrm{H}_{2} \mathrm{O}$

$\mathrm{AlCl}_{3}$ on heating gives $\mathrm{Al}_{2} \mathrm{O}_{3}$

Four hydrogen are connected to $B$ atoms by covalent bonds. So there are four $2$ centered$- 2$ electron bonds

$2$ hydrogen atoms are connected by bridge and it is connected to $2$ boron atoms . So, there are $3$ -centred-$2$- electron bonds

$( H - B ) \rightarrow 2\quad C -2 e$ bonds

$(B-H-B) \rightarrow 3\quad C-2e$ bonds

(B) Correct: Colemanite $= Ca _2 B _6 O _{11} \cdot 5 H _2 O$

(c) Incorrect: Bernarocalcite $= CaB _4 O _7 \cdot NaBO _2 \cdot 8 H _2 O$

(D) Correct: Octahedral form of borax $= Na _2\left[ B _4 O _5( OH )_4\right] \cdot 8 H _2 O$

$\mathrm{AlCl}_{3}+6 \mathrm{H}_{2} \mathrm{O} \quad \rightarrow \quad\left[\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}_{6}\right)\right]^{3+}+3 \mathrm{Cl}^{-}$

Hence, option $\mathrm{B}$ is correct.

$H _3 BO _3 \stackrel{100^{\circ} C }{\longrightarrow} HBO _2+ H _2 O$

metaboric acid form tetraboric acid on heating at $160^{\circ} C$

$4 HBO _2 \stackrel{160^{\circ} C }{\longrightarrow} H _2 B _4 O _7+ H _2 O$

On strong heating, $B _2 O _3$ is produced

$H _2 B _4 O _7 \rightarrow 2 B _2 O _3+ H _2 O$

This order is due to the relative tendency of the halogen atom to back donate its unutilized electrons to the vacant p-orbitals of boron atom.

Due to the back donation of electrons from fluorine to boron, the electron deficiency of boron is reduced and hence, Lewis acidity is decreased.

The tendency for the formation of back bonding is maximum in boron trifluoride and decreases very rapidly from boron trifluoride to boron triodide.

It is electron deficient. The lone pairs of diethyl ether will be donated to the anhydrous AlCI3 due to which it is soluble.

On the other hand, hydrous AlCl3 have the water molecules which make it poor lewis acid.

$\mathrm{So}, \mathrm{S}_{1}>\mathrm{S}_{2}$

$(2)$ $\mathrm{H}_{3} \mathrm{BO}_{3}$ is weak, non-protonic, lewis acid

$(3)$ $\mathrm{H}_{3} \mathrm{BO}_{3}+\mathrm{Na}\left(\mathrm{B}(\mathrm{OH})_{4}\right) \rightarrow$ equimolar amounts are produced and $\mathrm{pH}=\mathrm{pKa}\left(\mathrm{H}_{3} \mathrm{BO}_{3}\right)=9.25,$ lies in basic range.

Adduct form $^{\mathrm{n}}$ order $\mathrm{BI}_{3}>\mathrm{BBr}_{3}>\mathrm{BCl}_{3}>\mathrm{BF}_{3}$

As in $\mathrm{BF}_{3}$ strong back bonding occur. So adduct form $^{\mathrm{n}}$ decreases. Back bonding $\alpha \frac{1}{\text { adduct }}$ form $^{\text {n }}$

$\mathrm{BF}_{3}+3 \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{B}(\mathrm{OH})_{3}+3 \mathrm{HF}$

$3 \mathrm{HF}+3 \mathrm{BF}_{3} \rightarrow 3 \mathrm{H}\left[\mathrm{BF}_{4}\right]$

$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_$

$4 \mathrm{BF}_{3}+3 \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{B}(\mathrm{OH})_{3}+3 \mathrm{H}\left[\mathrm{BF}_{4}\right]$

(ii) Size of halides increase down the group, hence the ability of back bonding decreases making the molecule more acidic.

Hence, correct order of Lewis acid for boron-halides is $\mathrm{Bl}_{3}>\mathrm{BBr}_{3}>\mathrm{BCl}_{3}>\mathrm{BF}_{3}$

(ii) Size of halides increase down the group, hence the ability of back bonding decreases making the molecule more acidic.

Hence, correct order of Lewis acid for boron-halides is $\mathrm{Bl}_{3}>\mathrm{BBr}_{3}>\mathrm{BCl}_{3}>\mathrm{BF}_{3}$

Thus alkane is formed instead of acetylene.

$\mathrm{Na}_{2} \mathrm{B}_{4} \mathrm{O}_{7}+2 \mathrm{HCl}+5 \mathrm{H}_{2} \mathrm{O} \rightarrow 4 \mathrm{H}_{3} \mathrm{BO}_{3}+2 \mathrm{NaCl}$

Hence, the correct answer is option B.

The structure of $\mathrm{B}_{2} \mathrm{H}_{6}$ is represented as follows:

In it, two electrons of $\mathrm{B}$ -H bond are involved in the formation of three centre bond, these bonds are represented by a dotted line.

Hence, option A is correct.

Anhydrous aluminium chloride is hydrolysed partially with the moisture in the atmosphere to give HCl gas.

This HCI combines with the moisture in the air and appears white in colour.

$\mathrm{AlCl}_{3}+3 \mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{Al}(\mathrm{OH})_{3}+3 \mathrm{HCl}(\text { white fumes })$

Hence, the correct answer is option D.

Explanation:

Borax on strong heating loses its water of crystallization and then shrinks forming a transparent glassy bead of sodium metaborate and boric anhydride. Boric anhydride, being non-volatile displaces more volatile acidic oxides and combine with basic oxides present to form metaborates which are identified through their characteristic colours

$NH_3$ can react with $B_2H_6,$

${B_2}{H_6} + 2N{H_3}\xrightarrow{{warm}}{[B{H_2}.2N{H_3}]^ + }.{[B{H_4}]^ - }$ $\xrightarrow{{200{\,^o}C}}{B_3}{N_3}{H_6}$ (Inorganic benzene)

$X$ is $HBO _2$ which is metaboric acid and $Y$ is $H _2 B _4 O _7$ which is tetraboric acid.

Perchloric acid has no peroxy linkage $(HClO_4)$

Solid $SO_3$ have cyclic structure.

$B_3N_3H_6$ $\left( {\mu  = 0} \right)$ Borazine

$Sn{O_2} + 4HCl \to SnC{l_4} + 2{H_2}O$

$\mathop {3Si{F_4}}\limits_{{\rm{Silicon\, \,tetrafluoride}}} + \mathop {4{H_2}O}\limits_{{\rm{Water}}} \to 2{H_2}Si{F_6} + \mathop {{H_4}Si{O_4}}\limits_{{\rm{White\, silicic\, acid}}} $

Conclusion: hence the answer option (C) is correct.

$\mathrm{Ex} .$ Talc $\mathrm{Mg}\left(\mathrm{Si}_{2} \mathrm{O}_{5}\right)_{2} \mathrm{Mg}(\mathrm{OH})_{2}$

The structure of silicates has been found with the help of X-ray diffraction techniques.

Hence,option B is correct.

Here $A$ is $Hg$.

Compound A is silica (SiO $_{2}$ ) made of Si and $\mathrm{O}_{2}$, two most occuring elements into the earth crust, having a polymeric tetraheadral network structure. Compound B is C O, a poisonous gas which is the most stable diatomic molecule.

CO is neutral and $\mathrm{CO}_{2}$ is acidic in nature.

$2 \mathrm{KOH}+\mathrm{CO}_{2} \rightarrow \mathrm{K}_{2} \mathrm{CO}_{3}+\mathrm{H}_{2} \mathrm{O}$

$\mathrm{PbO}_{2}+\mathrm{HCl} \rightarrow \mathrm{PbCl}_{2}+\mathrm{H}_{2} \mathrm{O}+\mathrm{Cl}_{2}$

$\mathrm{P} \mathrm{b}_{3} \mathrm{O}_{4}+8 \mathrm{HCl} \rightarrow 4 \mathrm{H}_{2} \mathrm{O}+3 \mathrm{P} \mathrm{bCl}_{2}+\mathrm{Cl}_{2}$

$\mathrm{X}, \mathrm{Y}$ and $\mathrm{Z}$ are $\mathrm{Pb}_{3} \mathrm{O}_{4}, \mathrm{P} \mathrm{bO}_{2}, \mathrm{P} \mathrm{bCl}_{2}$ respectively.

$\mathrm{R}_{2} \mathrm{SiCl}_{2}=$ Chain polymer

$\mathrm{R}_{3} \mathrm{SiCl}_{3}=$ Dimer

$\mathrm{PbO}_{2}$ contains $\mathrm{O}^{2-}$ ions and Option $(\mathrm{A}) \mathrm{Pb}$ is $+4,$ whereas $(\mathrm{B}),(\mathrm{C})$ and (D) contains $[\mathrm{O}-\mathrm{O}]^{2-}$ ions (peroxide ions).

Thus Option A is correct.

$\mathrm{C}+2 \mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{CO}_{2} \uparrow+2 \mathrm{SO}_{2} \uparrow+2 \mathrm{H}_{2} \mathrm{O}$

$\mathrm{H}_{2} \mathrm{SO}_{4}$ is reduced to $\mathrm{SO}_{2}$ and $\mathrm{C}$ is oxidised to $\mathrm{CO}_{2}$

The chain length of Silicone polymer can be controlled by adding $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{SiCl}$, which blocks the ends as shown below:

$\left(\mathrm{CH}_{3}\right)_{3} \mathrm{SiCl} \frac{\mathrm{H}_{2} \mathrm{O}}{-\mathrm{HCl}}\left(\mathrm{CH}_{3}\right)_{3} \mathrm{SiOH}$

$B{e_2}C + 4H - OH \to 2Be{(OH)_2} + C{H_4}$

However, Me $_{3} \mathrm{SiCl}$ is not a monomer for a high molecular mass silicone polymer as it contains only one Cl atom and at best, it can form a dimer.

Hence, the correct answer is option D.

Non existance of $PbI _4$ can be explained on the basis of strong oxidising nature of $Pb_ 4^{+}$

$Pb ^{4+}+2 I ^{-} \rightarrow Pb ^{2+}+ I _2$

C atoms are also present in one half of the tetrahedral voids. There are 8 tetrahedral voids in fcc structure.

This accounts for remaining $8 \times \frac{1}{2}=4 \mathrm{C}$ atoms

Thus, total $4+4=8 \mathrm{C}$ atoms are present per unit cell of diamond.

The starting materials for the manufacture of silicones are alkyl or aryl substituted silicon chlorides, $\mathrm{R}_{\mathrm{n}} \mathrm{SiCl}_{(4-\mathrm{n})}$, where $\mathrm{R}$ is alkyl or aryl group.

On hydrolysis of this alkyl or aryl substituted silicon chlorides the silicones are formed.

Among given options, option A is the only alkyl substituted silicone chloride and hence it is correct answer.

Among given options, option $A$ is the only alkyl substituted silicone chloride and hence it is correct answer.

These silicates can be cleaved easily just like graphite.

Hence, the correct option is $\mathrm{A}$

$\mathrm{Si}+4 \mathrm{HF} \rightarrow \mathrm{SiF}_{4}+4 \mathrm{H}^{+}$

$\begin{array}{l}\downarrow \mathrm{HF} \\\mathrm{H}_{2} \mathrm{SiF}_{6}\end{array}$

The reaction is as follows:

$\mathrm{SiCl}_{4}+4 \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{Si}(\mathrm{OH})_{4}+4 \mathrm{HCl}$

Hence, silicic acid is formed when $\mathrm{SiCl}_{4}$ undergoes hydrolysis.

Hence, option $\mathrm{C}$ is the correct answer.

$PbF _2\,<\, PbCl _2\,>\, PbBr _2\,<\, PbI _2$

$4B{{F}_{3}}+3{{H}_{2}}O\xrightarrow{R.T}{{H}_{3}}B{{O}_{3}}+\underbrace{3H[B{{F}_{4}}]}_{\begin{smallmatrix} 
 does\,not\,reaect \\ 
 with\,\,silica\,\,(Si{{O}_{2}})\, 
\end{smallmatrix}}$

$(b) \rightarrow$ fabe

for linear silicones $R_2 SiCl _2$ is used

$(c) \rightarrow$ true

$(d) \rightarrow SiO _2+ HF \rightarrow H_2 SiF _6$

$a, c$ and $d$

The reactions are as follows:

$\mathrm{SiO}_{2}+4 \mathrm{HF} \rightarrow \mathrm{SiF}_{4}+2 \mathrm{H}_{2} \mathrm{O}$

$3 \mathrm{SiF}_{4}+4 \mathrm{H}_{2} \mathrm{O} \rightarrow 2 \mathrm{H}_{2} \mathrm{SiF}_{6}+\mathrm{H}_{4} \mathrm{SiO}_{4}$

Among given options, option $A$ is the only alkyl substituted silicone chloride and hence it is correct answer.

Statement $II$ : The atomic size of aluminum is greater than that of gallium.

${[\text { Number of neutrons }=6]}$

$\mathrm{x}=6$

$\mathrm{~B}+\mathrm{O}_2 \rightarrow \mathrm{B}_2 \mathrm{O}_3$

Oxidation state of $\mathrm{B}$ in $\mathrm{B}_2 \mathrm{O}_3=+3$

So, $\mathrm{y}=3$

Hence $x+y=9$

$B$. Electronegativity order $\mathrm{B}>\mathrm{Al}<\mathrm{Ga}<\operatorname{In}<\mathrm{T} \ell$

$D$. $\mathrm{B}, \mathrm{Al}$ are more stable in $+3$ oxidation state

So, only $C, E$ statements are correct.

Statement - $II$ $\Rightarrow$ False

$\mathrm{Ga}$ is used to measure high temperature

$\therefore$ Stability of $A \ell^{+1}<\mathrm{Ga}^{+1}<\mathrm{In}^{+1}<\mathrm{T} \ell^{+1}$

Ionic radius $\left(\mathrm{M}^{+3} / \mathrm{pm}\right): \mathrm{T} \ell>\mathrm{In}>\mathrm{Ga}>\mathrm{A} \ell>\mathrm{B}$

$\left(\Delta_{\mathrm{IE}} \mathrm{H}\right)_1\left[\frac{\mathrm{kJ}}{\mathrm{mol}}\right]: \mathrm{B}>\mathrm{T} \ell>\mathrm{A} \ell \approx \mathrm{Ga}>\mathrm{In}$

Atomic radius (in $\mathrm{pm}$ ) : $\mathrm{T} \ell>\mathrm{In}>\mathrm{A} \ell>\mathrm{Ga}>\mathrm{B}$

$(B)$ EN does not decrease gradually from $\mathrm{C}$ to $\mathrm{Pb}$.

$(C)$ Correct.

$(D)$ Correct.

$(E)$ Range of oxidation state of carbon ; $-4$ to $+4$

In case of aluminium, acidified aqueous solution forms octahedral $\left[\mathrm{Al}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}$ ion.

$C$               $2.5$

$Si$              $1.8$

$Ge$            $1.8$

$Sn$            $1.8$

$Pb$            $1.9$

The electronegativity values for elements from $\mathrm{Si}$ to $\mathrm{Pb}$ are almost same. So Statement $\mathrm{I}$ is false.

Carbon does not have d-orbitals so can not form $\mathrm{p} \pi-\mathrm{d} \pi$ Bond with itself. Due to properties of catenation and $\mathrm{p} \pi-\mathrm{p} \pi$ bond formation. carbon is able to show allotropic forms.

On treatment with metal salt, boric anhydride forms metaborate of the metal which gives different colours in oxidising and reducing flame.

For example, in the case of copper sulphate, following reactions occur.

$CuSO _4+ B _2 O _3 \longrightarrow Cu \left( BO _2\right)_2+ SO _3$

Two reactions may take place in reducing flame (Luminous flame)

$(i)$ The blue-green $Cu \left( BO _2\right)_2$ is reduced to colourless cuprous metaborate as :

$2 Cu ( BO _2 )_2+2 NaBO _2+ C \underset{\text { flame }}{\stackrel{\text { Luminous }}{\longrightarrow}} 2 CuBO _2+$ $ Na _2 B _4 O _7+ CO$

$(ii)$ Cupric metaborate may be reduced to metallic copper and bead appears red opaque.

$2 Cu \left( BO _2\right)_2+4 NaBO _2+2 C \underset{\text { fuminous }}{\stackrel{\text { flame }}{\longrightarrow}} 2 Cu +$ $2 Na _2 B _4 O _7+2 CO$

$BF _3 < BCl _3 < BBr _3 < BI _3$ (lewis acidic nature)

$2 Ga + Ga ^{+} GaCl _4^{-}+2 Al _2 Cl _6 \stackrel{190^{\circ}}{\longrightarrow} 4 Ga ^{+} AlCl _4^{-}$

Ga is cationic part of salt $GaAlCl _4$.

$B _8 N _3 H _6+9 H _2 O \rightarrow 3 NH _3+3 H _8 BO _3+3 H _2$

$Sr ( OH )_2$ is basic in nature.

These two undergo acid - base reaction to form a salt.

$Be ( OH )_2+ Sr ( OH )_2 \rightarrow Sr \left[ Be ( OH )_4\right]$

(salt)

$Ni$ is used in Hydrogenation of unsaturated fat to make edible fats.

Statements$-II$ is false as hydride of Silicon is electron precise and neither electron deficient nor electron rich.

$CuSO _4 \stackrel{\Delta}{\longrightarrow} CuO + SO _3$

$CuO + B _2 O _3 \rightarrow Cu \left( BO _2\right)_2$

$B$ : $801\,kJ / mol$

$Al : 577\,kJ / mol$

Ga : $579\,kJ / mol$

$Ga :[ Ar ] 3 d ^{10} 4 s ^2 4 p ^1$

Syn gas $\rightarrow$ Coal gasification

$\left( C _{\text {(Redhot coke) }}+ H _2 O ( g ) \rightarrow CO + H _2\right)$

Nickel catalyst $\rightarrow$ Water gas production

Brine solution $\rightarrow$ Production

(aq. $NaCl ) \quad\left(\begin{array}{l} H _2 \rightarrow \text { Cathode } \\ Cl _2 \rightarrow \text { anode }\end{array}\right)$

$Na _{2} B _{4} O _{7} \stackrel{\Delta}{\longrightarrow} 2 NaBO _{2} \text { (sodium meta borate) }+ B _{2} O _{3}$

$B _{2} O _{3}+ CoO \rightarrow Co \left( BO _{2}\right)_{2} \text { (cobalt (II) meta borate) }$

$\quad\quad\quad\quad\quad\quad\quad$Blue Bead

This is the method to prepare $BeH _{2}$