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74 questions · 6 auto-graded MCQ + 68 self-marked written.

Question 11 Mark
Write balanced equations for:
$\mathrm{B}_2 \mathrm{H}_6+\mathrm{NH}_3 \rightarrow$
Answer
$3\text{B}_6\text{H}_6+6\text{NH}_3\rightarrow 3\big[\text{BH}_2(\text{NH}_3)_2\big]^{+}\big[\text{BH}_4\big]^{-}\rightarrow 2\text{B}_3\text{N}_3\text{H}_6+12\text{H}_2 \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Borazene}$
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MCQ 21 Mark
The type of hybridisation of boron in diborane is:
  • A
    $sp$.
  • B
    $\mathrm{sp}^2$.
  • $\mathrm{sp}^3$.
  • D
    $\mathrm{dsp}^2$.
Answer
Correct option: C.
$\mathrm{sp}^3$.
Boron in diborane is $\mathrm{sp}^3$ hybridized.
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Question 31 Mark
Give reasons:
Aluminium wire is used to make transmission cables.
Answer
Aluminium is generally unaffected by air and moisture and it is also good conductor of electricity. That’s why it is used in transmission cables.
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MCQ 41 Mark
Elements of group $14$ :
  • A
    Exhibit oxidation state of $+4 $ only.
  • Exhibit oxidation state of $+2$ and $+4 .$
  • C
    Form $\mathrm{M}^{2-}$ and $\mathrm{M}^{4+}$ ions.
  • D
    Form $\mathrm{M}^{2+}$ and $\mathrm{M}^{4+}$ ions.
Answer
Correct option: B.
Exhibit oxidation state of $+2$ and $+4 .$
Due to inert pair effect, elements of group $14$ exhibit oxidation states of $+2$ and $+4$.
Thus, option $(b)$ is correct.
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MCQ 51 Mark
An aqueous solution of borax is :
  • A
    Neutral.
  • B
    Amphoteric.
  • Basic.
  • D
    Acidic.
Answer
Correct option: C.
Basic.
Borax is a salt of a strong base $(\mathrm{NaOH})$ and a weak acid $\left(\mathrm{H}_3 \mathrm{BO}_3\right)$. It is, therefore, basic in nature.
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Question 61 Mark
Write balanced equations for:
$\mathrm{B}_2 \mathrm{H}_6+\mathrm{H}_2 \mathrm{O} \rightarrow$
Answer
$\ \ \text{B}_2\text{H}_6 \ \ \ \ + \ \ \ \ 6\text{H}_2\text{O} \ \ \ \ \ \ \ \ \rightarrow \ \ \ \ \ \ \ \ \ 2\text{H}_3\text{BO}_3 \ \ \ \ + \ \ \ \ \ 6\text{H}_2 \\ \text{Diborane} \ \ \ \ \ \text{Water} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Orthoboric acid} \ \ \ \ \ \text{Hydrogen}$
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Question 71 Mark
Write balanced equations for:
$\mathrm{NaH}+\mathrm{B}_2 \mathrm{H}_6 \rightarrow$
Answer
$\ \ \ \ \ \text{B}_2\text{H}_6 \ \ \ \ \ + \ \ \ \ \ 2\text{NaH} \ \ \ \ \ \ \xrightarrow{\text{ether}} \ \ \ \ \ 2\text{HaBH}_4 \\ \text{Diborane} \ \ \ \ \ \ \text{Sodium hydride } \ \ \ \ \ \ \ \text{Sodium borohydride}$
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Question 81 Mark
Give reasons:
Diamond is used as an abrasive.
Answer
Diamond is used as an-abrasive because it is an extremely hard substance.
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Question 91 Mark
Write balanced equations for:
$\mathrm{BF}_3+\mathrm{LiH} \rightarrow$
Answer
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ 2\text{BF}_3 \ \ \ \ \ + \ \ \ \ \ 6\text{LiH} \ \ \ \ \ \ \ \ \ \rightarrow \ \ \ \ \ \ \ \ \ \text{B}_2\text{H}_6 \ \ \ \ \ + \ \ \ \ \ 6\text{LiF} \\ \ \ \text{Boron trifluride} \ \ \ \text{Lithium hydride} \ \ \ \ \ \ \ \text{Diiborane} \ \ \ \text{Lithium fluoride}$
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Question 101 Mark
Write balanced equations for:
$\text{H}_3\text{BO}_3\xrightarrow{\Delta}$
Answer
$4\text{H}_3\text{BO}_3 \ \ \ \ \ \xrightarrow[-4\text{H}_2\text{O}] \ \ \ \ \ {}4\text{HBO}_2 \ \ \ \ \ \xrightarrow[-\text{H}_2\text{O}]{411\text{k}} \ \ \ \ \ \text{H}_2\text{B}_4\text{O}_7 \ \ \ \ \ \xrightarrow[-\text{H}_2\text{O}]{\text{Red hot}} \ \ \ \ \ 2\text{B}_2\text{O}_3 \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Metaboric acid} \ \ \ \ \ \ \ \ \ \ \ \text{Tetraboric acid} \ \ \ \ \ \ \ \ \ \ \ \text{Boron trioxide}$
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MCQ 111 Mark
Boric acid is polymeric due to :
  • A
    Its acidic nature.
  • The presence of hydrogen bonds.
  • C
    Its monobasic nature.
  • D
    Its geometry.
Answer
Correct option: B.
The presence of hydrogen bonds.
Boric acid is polymeric because of the presence of hydrogen bonds. In the given figure, the dotted lines represent hydrogen bonds.
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MCQ 121 Mark
Thermodynamically the most stable form of carbon is :
  • A
    Diamond.
  • Graphite.
  • C
    Fullerenes.
  • D
    Coal.
Answer
Correct option: B.
Graphite.
Graphite is thermodynamically the most stable form of carbon.
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Question 131 Mark
Write balanced equations for:
Al + NaOH →
Answer
$2\text{Al} \ \ \ \ \ + \ \ \ \ \ \text{NaOH} \ \ \ \ \ + \ \ \ \ \ 6\text{H}_2\text{O} \ \ \ \ \ \rightarrow \ \ \ \ \ 2\text{Na}^+\big[\text{Al(OH)_4}\big]^-_{\text{(eq)}} \ \ \ \ \ + \ \ \ \ \ 3\text{H}_2 \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Sodium tetrahydroxoaluminate (III)}$
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Question 141 Mark
What name is given to the compounds formed by more electropositive elements with carbon?
Answer
Ionic compounds e.g. calcium carbide, $\text{Ca}^{2+}(\text{C}\equiv\text{C})^{2-}.$
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Question 161 Mark
Are all the B-H bonds in diborane equivalent?
Answer
No, there are two types of bonds in diborane two electron normal bonds and three centred two electron bonds.
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Question 171 Mark
Among tri halides of nitrogen which one is least basic and why?
Answer
$\text{NF}_3$ is least basic because 'F' is most electronegative therefore lone pair of electron will be less available.
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Question 181 Mark
Atomic radius of gallium (135 pm) is less than that of aluminium (143 pm). Give reason.
OR
How would you justify the lower atomic radius of Ga as compared to Al?
Answer
It is due to poor shielding effect of 3d-electrons due to which effective nuclear charge increases in Ga, therefore, it is smaller than Al.
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MCQ 201 Mark
Choose the wrong statement :
  • $\mathrm{CO}_2$ and $\mathrm{SiO}_2$ both are linear in geometry.
  • B
    $\mathrm{SnCl}_2$ is solid, $\mathrm{SnCl}_4$ is liquid.
  • C
    Carborundum is $\ce{SiC}.$
  • D
    Zeotites can act as sieves.
Answer
Correct option: A.
$\mathrm{CO}_2$ and $\mathrm{SiO}_2$ both are linear in geometry.
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Question 211 Mark
Why does C differ from rest of elements?
Answer
Carbon has smallest size, highest ionisation enthalpy and high electronegativity and does not have d-orbitals, therefore, it differs from rest of the elements.
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Question 221 Mark
What is inert pair effect?
Answer
In the elements of 4th, 5th and 6th period of the p-block the electrons present in the intervening dand f-orbitals do not shield the s-electrons of the valence shell effectively. As a result, $\text{ns}^2$-electrons remain more tightly held by the nucleus and hence, do not participate in bonding. This is called inert pair effect.
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Question 231 Mark
Out of carbon and silicon which can form multiple bond and why?
Answer
Carbon can form multiple bonds because it is smaller in size.
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Question 241 Mark
The colour of bead in borax bead test is mainly due to formation of which compound?
Answer
Metal metaborate e.g. $\mathrm{Co}\left(\mathrm{BO}_2\right)_2$ is blue, $\mathrm{Cr}\left(\mathrm{BO}_2\right)_3$ is green.
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Question 251 Mark
Why does $\text{BF}^{3-}_6$ not exist but $\text{AlF}^{3-}_6$ exist?
Answer
It is because 'B' does not have vacant d-orbitals whereas as Al has d-orbitals of form $\text{AlF}^{3-}_6.$
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Question 261 Mark
Give two uses of silicones.
Answer
  1. They are used as sealants and electrical insulators.
  2. They are used in surgical and cosmetic implants because they biocompatible.
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Question 271 Mark
Draw the structure of the following:
  1. $\text{PCl}_3$
  2. $\text{NO}_3^-$
  3. $\text{P}_4$
  4. $\text{PCl}_5$
Answer
  1.  
  1.  
  1.  
  1.  
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Question 281 Mark
Mention the type of hybrid orbitals of silicon in $\text{SiF}^{2-}_6$ ion.
Answer
$\text{sp}^3\text{d}^2$.
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Question 291 Mark
Why does boron resemble Si?
Answer
Both have similar charge over radius ratio, i.e., similar polarising power.
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Question 311 Mark
Why is boron metalloid?
Answer
Boron resembles both with metals and non-metals, therefore, it is metalloid.
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Question 321 Mark
Name the member of group 14 that forms the most acidic oxide.
Answer
Among monoxides, CO is neutral and GeO is acidic while among dioxides, $\mathrm{CO}_2, \mathrm{SiO}_2$ are acidic, $\mathrm{GeO}_2$ is also acidic but less acidic than $\mathrm{SiO}_2$.
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Question 331 Mark
C and Si are always tetravalent but Ge, Sn, Pb show divalency.
Answer
$\mathrm{Ge}, \mathrm{Sn}, \mathrm{Pb}$ show divalency due to inert pair effect, $\mathrm{Pb}^{2+}$ is more stable than $\mathrm{Pb}^{4+}$.
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Question 341 Mark
Write the formula of inorganic benzene.
Answer
$\mathrm{B}_3 \mathrm{N}_3 \mathrm{H}_6$ is formula of inorganic benzene.
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Question 351 Mark
A gaseous substance dissolves in water giving a pale blue solution which decolourises $\mathrm{KMnO}_4$ and oxides KI to $\mathrm{I}_2$. Identify the gaseous substance.
Answer
$\mathrm{N}_2 \mathrm{O}_3$ is gaseous substance.
$\mathrm{N}_2 \mathrm{O}_3+\mathrm{H}_2 \mathrm{O} \longrightarrow 2 \mathrm{HNO}_2$ (pale blue solution)
It decolouries $\mathrm{KM}_{\mathrm{n}} \mathrm{O}_4$ oxidises $\mathrm{HNO}_2$ to $\mathrm{HNO}_3$ and gets reduced to $\mathrm{M}_{\mathrm{n}^{2+}}$ which is colourless.
$\mathrm{HNO}_2$ oxidises $\mathrm{KI}$ to $\mathrm{I}_2$ because $\mathrm{I}^{-}$is good reducing agent.
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Question 361 Mark
  1. Name the salt of phosphorus which is used as water softener in detergents.
  2. What happens when silver reacts with $\mathrm{PCl}_5$? Write chemical equation.
Answer
a. Sodium tri poly phosphate $\left(\mathrm{Na}_5 \mathrm{P}_3 \mathrm{O}_{10}\right)$ is used as water softener in detergents.
b. $\ce{AgCl}$ and $\mathrm{PCl}_3$ is formed,
$\text{PCl}_5+2\text{Ag}\xrightarrow{\ \ \ \ \ \ \\ }2\text{AgCl}+\text{PCl}_3.$
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Question 371 Mark
How will you prepare an aluminosilicate?
Answer
Aluminosilicate is prepared by substituting some of the Si atoms in the three dimensional network of Sio, by Al atoms.
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Question 381 Mark
Arrange C-C, Ge-Ge, Si-Si, Sn-Sn in decreasing order of bond energy.
Answer
C-C > Si-Si > Ge, Si-Si, Sn-Sn.
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Question 391 Mark
What is the formula of kernite, an ore of boron?
Answer
Formula of kernite, $\text{Na}_2[\text{B}_4\text{O}_5(\text{OH})_4]$ or $\text{Na}_2\text{B}_4\text{O}_7.2\text{H}_2\text{O}$
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Question 401 Mark
Name the compound of boron which have the similar structure to graphite.
Answer
Boron nitride, (BN) resembles with graphite in structure as:
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Question 411 Mark
Which one is more soluble in diethyl ether, anhydrous $\mathrm{AlCl}_3$ or hydrated $\mathrm{AlCl}_3$ ? Explain in terms of bonding.
 
Answer
Anhydrous $\mathrm{AlCl}_3$ is an electron-deficient compound while hydrated $\mathrm{AlCl}_3$ is not. Therefore, anhydrous $\mathrm{AlCl}_3$ is more soluble in diethyl ether because the oxygen atom of ether donates a pair of electrons to the vacant p-orbital on the Al atom in $\mathrm{AlCl}_3$ forming a coordinate bond.

In case of hydrated $\mathrm{AlCl}_3, \mathrm{Al}$ is not electron deficient since $\mathrm{H}_2 \mathrm{O}$ has already donated a pair of electrons to it.
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Question 421 Mark
Why the elements of the second row $($first short period$)$ show a number of differences in properties from other members of their respective families?
Answer
The differences in the properties of the first member of a group from those of the other members are due to.
  1. The smaller size of the atom,
  2. Presence of one inner shell of only two electrons and.
  3. Absence of $d-$orbitals.
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Question 431 Mark
Which of the following does not produce $\mathrm{N}_2$ gas by the thermal decomposition.
$\text{Ba}(\text{N}_3)_2,(\text{NH}_4)_2\text{Cr}_2\text{O}_7,\text{NH}_4\text{NO}_2,(\text{NH}_4)_2\text{SO}_4.$
Answer
$\left(\mathrm{NH}_3\right)_2 \mathrm{SO}_4$ does not give nitrogen on thermal decompositions. Others give $\mathrm{N}_2$.
$\mathrm{Ba}\left(\mathrm{~N}_3\right)_2 \xrightarrow{\text { Heat }} \mathrm{Ba}+3 \mathrm{~N}_2$
$\left(\mathrm{NH}_4\right)_2 \mathrm{Cr}_2 \mathrm{O}_7 \xrightarrow{\text { Heat }} \mathrm{Na}+\mathrm{Cr}_2 \mathrm{O}_3+4 \mathrm{H}_2 \mathrm{O}$
$\mathrm{NH}_4 \mathrm{NO}_2 \xrightarrow{\Delta} \mathrm{~N}_2+2 \mathrm{H}_2 \mathrm{O}$
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Question 451 Mark
Why does silicon show a higher covalency than carbon?
Answer
Si because of the presence of vacant d-orbitals can show a covalency upto six while C because of the absence of d-orbitals cannot have a covalency of more than four.
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Question 461 Mark
Mention two important ores of boron.
Answer
i. Kermite, $\mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7 .2 \mathrm{H}_2 \mathrm{O}$
ii. Colemanite $\mathrm{Ca}_3 \mathrm{~B}_6 \mathrm{O}_{11} .5 \mathrm{H}_2 \mathrm{O}$
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Question 481 Mark
What is bond angles $\mathrm{Cl}-\mathrm{P}-\mathrm{Cl}$ in $\mathrm{PCI}_5$ ?
Answer
$120^\circ$ and $90^\circ$
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Question 501 Mark
Arrange the hydrides of group 14 in increasing order of reducing power.
Answer
$\text{CH}_4<\text{SiH}_4<\text{GeH}_4<\text{SnH}_4<\text{Pb}$
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Question 511 Mark
Why does gallium undergo disproportionation reactions?
Answer
Gallium shows both +1 and +3 oxidation states due to inert pair effect. However, its +3 oxidation state is more stable than +1 oxidation state, hence undergoes disproportionation to form gallium and more stable +3 gallium ions in aqueous solution as shown below.

or $3\text{Ga}^+(\text{aq})\xrightarrow{\ \ \ \ \ \ \ \ \ }2\text{Ga}(\text{s})+\text{Ga}^{3+}(\text{aq})$
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Question 521 Mark
Silicones are used for making waterproof fabrics. Give reason.
Answer
Silicones are synthetic polymers containing repeated units of $\mathrm{R}_2 \mathrm{SiO}$ where R is alkyl group.Therefore, these are water repellants i.e. do not absorb water and are used for making waterproof fabrics.
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Question 531 Mark
  1. Why is $\mathrm{PCl}_5$ known but $\mathrm{Pl}_5$ is not?
  2. Red phosphorus is denser than white phosphorus, why?
Answer
  1. $'Cl'$ is small in size as compared to $I$, therefore fire $CI$ atoms can be easily accommodated around $‘P\ '$ atom but five bigger $I$ atoms can not be accommodate due to stearic henderace. $P-I$ bond is weaker than $P-Cl$ bond.
  2. Red phosphorus has polymeric structure and more inter molecular force of attraction as compared to white phosphorus $($monomeric$)$, therefore, it is denser.
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Question 541 Mark
Give one example of zeolite.
Answer
Feldspar $\left(\mathrm{KAISi}_3 \mathrm{O}_8\right)$, Zeolite $\left(\mathrm{NaAlSi}_2 \mathrm{O}_6 \cdot \mathrm{H}_2 \mathrm{O}\right)$
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Question 551 Mark
Why is $\mathrm{B}-\mathrm{X}$ bond distance in $\mathrm{BX}_3$ shorter than the theoretically expected value?
Answer
This is due to pπ-pπ back bonding of the completely filled p-orbital of halogen X into the empty p-orbital of boron.
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Question 561 Mark
Why the tri halide of group 13 elements fume in moist air?
Answer
Boron forms only covalent compounds because it does not form $\text{B}^{3+}$ ion. High energy is required to remove 3 electrons.
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Question 571 Mark
$\mathrm{AlF}_3$ is high melting solid but $\mathrm{AlCl}_3$ is low melting. Explain.
Answer
$\mathrm{AlF}_3$ is high melting solid because it is ionic in nature. On the other hand, $\mathrm{AICI}_3$ is covalent in nature and hence is a low melting solid.
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Question 581 Mark
  1. Why is electron gain enthalpy of group $15$ less than group $16$ ?
  2. Nitric oxide becomes brown when released in air, why?
Answer
  1. It is because group $15$ have outer electronic configuration $\ce{ns^2 np^3}$ i.e. half filled p-orbitals which are more stable, therefore, have least tendency to gain electron as compared to group $\ce{16 (ns^2 - np^4)}$
  2. Nitric oxide reacts with air to form nitrogen dioxide which is brown in colour.
$2\text{NO}(\text{g})+\text{O}_2(\text{g})\xrightarrow{\ \ \ \ \ \ \ \ \ }2\text{NO}_2(\text{g})$
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Question 591 Mark
a. $\mathrm{Bi}_2 \mathrm{O}_3$ is treated with $\mathrm{HNO}_3$. Write balanced equation for the reaction.
b. Why does Bismuth not form pentahalide?
Answer
  1. $\text{BI}_2\text{O}_3+6\text{HNO}_3\xrightarrow{\ \ \ \ \ \ \ \ \ \ }2\text{Bi}(\text{NO}_3)_3+3\text{H}_2\text{O}$
  2. $\mathrm{Bi}^{3+}$ is more stable than $\mathrm{Bi}^{5+}$, therefore Bismuth does not penta$-$halide.
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Question 601 Mark
Although pure silicon is an insulator, then how does it behave as semiconductor on heating?
Answer
Silicon becomes semiconductor on heating because electrons become free to move.
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Question 611 Mark
Name the elements of group 13 which forms only covalent compounds.
Answer
Boron forms only covalent compounds because it does not form $\text{B}^{3+}$ ion. High energy is required to remove 3 electrons.
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Question 621 Mark
Why is $\mathrm{BF}_3$ weaker Lewis acid than $\mathrm{BCI}_3$ ?
Answer
$\mathrm{BF}_3$ is weaker Lewis acid than $\mathrm{BCl}_3$ because of more effective back bonding in case of F due to smaller size than Cl.
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Question 631 Mark
Why $\mathrm{SnCl}_4$ is more covalent than $\mathrm{SnCl}_2$ ?
Answer
It is because $\mathrm{Sn}^{4+}$ has more polarizing power, therefore, more tendency to form covalent bond than $\mathrm{Sn}^{2+}$.
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Question 641 Mark
$\mathrm{NaCl}_3$ gets hydrolysed while $\mathrm{NF}_3$ does not, why?
Answer
In $\mathrm{NCl}_3$, 'Cl' has vacan d-orbitals to accept a lone pair of electrons Donated by O of $\mathrm{H}_2 \mathrm{O}$, molecule but in $\mathrm{NF}_3$, 'F' does not d-orbitals.
$\text{NCl}_3+3\text{H}_2\text{O}\xrightarrow{\ \ \ \ \ }\text{NH}_3+3\text{HCl}$
$\text{NF}_3+\text{H}_2\text{O}\xrightarrow{\ \ \ \ \ \ \ \ }\text{No reaction}$.
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Question 651 Mark
  1. Why does tendency to show catenation decreases down the group?
  2. Why is $\mathrm{N}_2 \mathrm{O}$, more acidic than $\mathrm{N}_2 \mathrm{O}_3$ ?
Answer
  1. It is because bond dissociation enthalpy decreases due to increase in bond length with the increase in atomic size of element.
  2. Higher the oxidation state, more will be acidic character, that is why $\mathrm{N}^2 \mathrm{O}^5$ is more acidic than $\mathrm{N}_2 \mathrm{O}_3$.
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Question 661 Mark
Write the formula of dry ice.
Answer
Dry ice is solid $\mathrm{CO}_2$ in which there are van der Wall's forces of attraction.
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Question 671 Mark
Why does boron trifluoride behave as Lewis acid?
Answer
It is because 'B' shares three electrons with fluorine and its octet is not complete, therefore, it behaves as Lewis acid.
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Question 681 Mark
Which of the following is acidic and why?
$\mathrm{SiO}_2, \mathrm{Al}_2 \mathrm{O}_3, \mathrm{PbO}_2, \mathrm{SnO}_2$
Answer
$\mathrm{SiO}_2$ is acidic oxide because Si shows non-metallic properties. $\mathrm{Al}_2 \mathrm{O}_3, \mathrm{PbO}_2$ and $\mathrm{SnO}_2$ are amphoteric.
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Question 701 Mark
Why does boron not form $\text{B}^{3+}$ ion?
Answer
Boron does not form $\text{B}^{3+}$ ion due to smallest atomic size and highest ionisation enthalpy.
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Question 711 Mark
What are germanes and plumbares?
Answer
Boron resembles both with metals and non-metals, therefore, it is metalloid.
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Question 721 Mark
Why is diamond a bad conductor of electricity but a good conductor of heat?
Answer
Conduction of electricity essentially requires movement of electrons while conduction of heat does not necessarily require movement of electrons.
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Question 731 Mark
$\mathrm{TI}\left(\mathrm{NO}_3\right)_3$ acts as an oxidising agent Explain.
Answer
Due to inert pair effect, Tl in +1 oxidation state is more stable than that of +3 oxidation state. Therefore, $\mathrm{TI}\left(\mathrm{NO}_3\right)_3$ acts as an oxidising agent and readily reduced to $\mathrm{TINO}_3$.
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Question 741 Mark
Which of the following are Lewis acids:
$\text{PH}_3,\text{BCl}_3,\text{AlCl}_3,\text{SiCl}_4.$
Answer
$\mathrm{BCl}_3$ and $\mathrm{AlCl}_3$ are Lewis acid because their octet is not complete.
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