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Chemistry 1 Marks Question

The p-Block Elements · Haryana Board (HBSE) STD 11 Science (Haryana - English Medium). 74 questions.

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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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