Question 12 Marks
What happens when.
Boric acid is added to water.
AnswerWhen boric acid is added to water, it accepts electrons from ${}^-\text{OH}$ ion.
$\text{B(OH)}_3+2\text{HOH}\rightarrow \big[\text{B(OH)}_4\big]^{-}+\text{H}_3\text{O}^+$
View full question & answer→Question 22 Marks
Explain the following reactions:
Hydrated alumina is treated with aqueous NaOH solution.
AnswerWhen hydrated alumina is added to sodium hydroxide, the former dissolves in the latter because of the formation of sodium meta-aluminate.
$\text{Al}_2\text{O}_3.2\text{H}_2\text{O}+2\text{NaOH}\rightarrow2\text{NaAlO}_2+3\text{H}_2\text{O}$
View full question & answer→Question 32 Marks
How would you explain the lower atomic radius of Ga as compared to Al?
Answer
| Atomic radius (in pm) |
| Aluminium |
143 |
| Gallium |
135 |
Although Ga has one shell more than Al, its size is lesser than Al. This is because of the poor shielding effect of the 3d-electrons. The shielding effect of d-electrons is very poor and the effective nuclear charge experienced by the valence electrons in gallium is much more than it is in the case of Al.
View full question & answer→Question 42 Marks
Is boric acid a protic acid? Explain.
AnswerBoric acid is a Lewis acid; it is not a protonic acid.
Boric acid accepts electrons from hydroxyl ion of $\mathrm{H}_2 \mathrm{O}$ molecule.
$\mathrm{B}(\mathrm{OH})_3+2 \mathrm{HOH} \rightarrow\left[\mathrm{~B}(\mathrm{OH})_4\right]^{-}+\mathrm{H}_3 \mathrm{O}^{+}$
View full question & answer→Question 52 Marks
Give reasons:
Graphite is used as lubricant.
AnswerGraphite has layered structure which are held by weak van der Waals forces. Thus, graphite cleaves easily between the layers, therefore it is very soft and slippery. That’s why it is used as lubricant.
View full question & answer→Question 62 Marks
Rationalise the given statements and give chemical reactions:
Lead is known not to form an iodide, $\mathrm{Pbl}_4$.
AnswerLead is known not to form $\mathrm{Pbl}_4 . \mathrm{Pb}(+4)$ is oxidising in nature and $\mathrm{I}^{-}$is reducing in nature. A combination of $\mathrm{Pb}(\mathrm{IV})$ and iodide ion is not stable. lodide ion is strongly reducing in nature. $\mathrm{Pb}(\mathrm{IV})$ oxidises $\mathrm{I}^{-}$to $\mathrm{I}^2$ and itself gets reduced to $\mathrm{Pb}(\mathrm{II}) . \mathrm{PbI}_4 \rightarrow \mathrm{PbI}_2+\mathrm{I}_2$
View full question & answer→Question 72 Marks
Rationalise the given statements and give chemical reactions:
Lead(IV) chloride is highly unstable towards heat.
AnswerOn moving down group IV, the higher oxidation state becomes unstable because of the inert pair effect. Pb(IV) is highly unstable and when heated, it reduces to Pb(II).
$\text{PbCl}_{4\text{(l)}}+\text{Cl}\xrightarrow{\Delta}\text{PbCl}_{2\text{(s)}}+\text{Cl}_{2(\text{g})}$
View full question & answer→Question 82 Marks
Explain the following reactions:
CO is heated with ZnO.
AnswerWhen CO reacts with ZnO, it reduces ZnO to Zn. CO acts as a reducing agent.
$\text{ZnO}_{\text{(S)}}+\text{CO}_\text{(g)}\xrightarrow{\Delta}\text{Zn}_\text{(s)}+\text{CO}_{2(\text{g})}$
View full question & answer→Question 92 Marks
What happens when.
Aluminium is treated with dilute NaOH.
AnswerAl reacts with dilute NaOH to form sodium tetrahydroxoaluminate(III). Hydrogen gas is liberated in the process.
$2\text{Al}_{\text{(s)}}+2\text{NaOH}_{\text{(eq)}}+6\text{H}_2\text{O}_{\text{(l)}}\rightarrow2\text{Na}^+\big[\text{Al(OH)}_4\big]^-_\text{(eq)}+3\text{H}_{2(\text{g})}$
View full question & answer→Question 102 Marks
What happens when.
Borax is heated strongly.
AnswerWhen heated, borax undergoes various transitions. It first loses water molecules and swells. Then, it turns into a transparent liquid, solidifying to form a glass-like material called borax bead.
$\text{Na}_2\text{B}_4\text{O}_7.10\text{H}_2\text{O} \ \ \ \ \xrightarrow{\Delta} \ \ \ \ \text{Na}_2\text{B}_4\text{O}_7 \ \ \ \ \xrightarrow{\Delta} \ \ \ \ 2\text{NaBO}_2+\text{B}_2\text{O}_3 \\ \ \ \ \ \ \ \ \ \text{Borax}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Sodium} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Boric} \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{metaborate} \ \ \ \ \ \ \ \ \ \ \ \text{anhydride}$
View full question & answer→Question 112 Marks
Explain what happens when boric acid is heated.
AnswerOn heating orthoboric acid $\left(\mathrm{H}_3 \mathrm{BO}_3\right)$ at 370 K or above, it changes to metaboric acid $\left(\mathrm{HBO}_2\right)$. On further heating, this yields boric oxide $\mathrm{B}_2 \mathrm{O}_3$.
$\text{H}_3\text{BO}_3 \ \ \ \ \ \xrightarrow[370\text{k}]{\Delta}\ \ \ \ \ \ \text{HBO}_2 \ \ \ \ \ \ \xrightarrow[\text{red hot}]{\Delta} \ \ \ \ \ \ \text{B}_2\text{O}_3 \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Metaboric acid} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Boric oxide}$
View full question & answer→Question 122 Marks
Give reasons:
A mixture of dilute NaOH and aluminium pieces is used to open drain.
AnswerNaOH reacts with Al to evolve $\mathrm{H}_2$ gas. Thus the pressure of the gas produced can be used for clogged drains.
$2 \mathrm{Al}(\mathrm{s})+2 \mathrm{NaOH}(\mathrm{aq})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow 2 \mathrm{NaAlO}_2(\mathrm{aq})+3 \mathrm{H}_2(\mathrm{g})$
View full question & answer→Question 132 Marks
Give reasons:
Aluminium utensils should not be kept in water overnight.
AnswerGenerally, aluminium metal does not react with water quickly but, when it is kept overnight, it reacts slowly with water in presence of air.
$2 \mathrm{Al}(\mathrm{s})+\mathrm{O}_2(\mathrm{~g})+\mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{Al}_2 \mathrm{O}_3(\mathrm{~S})+\mathrm{H}_2(\mathrm{~g})$
a very small amount of (in ppm) $\mathrm{Al}^{3+}$ produced in the solution is injurious to health if the water is used for drinking purposes.
View full question & answer→Question 142 Marks
Give reasons:
Aluminium alloys are used to make aircraft body.
AnswerAlloys of aluminium, like duralumin, is used to make aircraft body due to some of its property like toughness, lightness and resistant to corrosion.
View full question & answer→Question 152 Marks
Explain why is there a phenomenal decrease in ionization enthalpy from carbon to silicon?
AnswerIonisation enthalpy of carbon (the first element of group 14) is very high (1086kJ/ mol). This is expected owing to its small size. However, on moving down the group to silicon, there is a sharp decrease in the enthalpy (786kJ). This is because of an appreciable increase in the atomic sizes of elements on moving down the group.
View full question & answer→Question 162 Marks
What happens when.
$\mathrm{BF}_3$ is reacted with ammonia?
Answer$\mathrm{BF}_3$ (a Lewis acid) reacts with $\mathrm{NH}_3$ (a Lewis base) to form an adduct. This results in a complete octet around B in $\mathrm{BF}_3$.
$\text{F}_3\text{B}+:\text{NH}_3\rightarrow \text{F}_3\text{B}\leftarrow:\text{NH}_3$
View full question & answer→Question 172 Marks
Rationalise the given statements and give chemical reactions:
Lead(II) chloride reacts with $\mathrm{Cl}_2$ to give $\mathrm{PbCl}_4$.
AnswerLead belongs to group 14 of the periodic table. The two oxidation states displayed by this group is +2 and +4 . On moving down the group, the +2 oxidation state becomes more stable and the +4 oxidation state becomes less stable. This is because of the inert pair effect. Hence, $\mathrm{PbCl}_4$ is much less stable than $\mathrm{PbCl}_2$. However, the formation of $\mathrm{PbCl}_4$ takes place when chlorine gas is bubbled through a saturated solution of $\mathrm{PlCl}_2$.
$\text{PbCl}_{2\text{(s)}}+\text{Cl}_{2\text{(s)}}\rightarrow\text{PbCl}_{4\text{(l)}}$
View full question & answer→Question 182 Marks
Give reasons:
Conc. $\mathrm{HNO}^3$ can be transported in aluminium container.
AnswerAl reacts with cone. $\mathrm{HNO}_3$ to form a very thin film of aluminium oxide on its surface which protects it from further reaction.
$2 \mathrm{Al}(\mathrm{~s})+6 \mathrm{HNO}_3 \text { (conc.) } \rightarrow \mathrm{Al}_2 \mathrm{O}_3(\mathrm{~s})+6 \mathrm{NO}_2(\mathrm{~g})+3 \mathrm{H}_2 \mathrm{O}(\mathrm{l})$
View full question & answer→Question 192 Marks
- What is litharge chemically, Is it basic or acidic?
- At what temperature, lead of lead pencil melts?
Answer
- PbO is litharge chemically. It is basic oxide.
- Lead of lead pencils is made up of graphite which has melting point 3170°.
View full question & answer→Question 202 Marks
Element A reacts with aqueous NaOH solution to form B and H . Aqueous solution of B is hetated to $323-33 \mathrm{~K}$ and on passing $\mathrm{CO}_2$ into it, $\mathrm{Na}_2 \mathrm{CO}_3$ and C were formed. When C is heated to $1200^{\circ} \mathrm{C}, \mathrm{Al}_2 \mathrm{O}_3$ is formed. identify A,B, and C and also write the reaction involved.
AnswerThe reaction involved are as follow.
$2 \mathrm{Al}+2 \mathrm{NaOH}+2 \mathrm{H}_2 \mathrm{O} \longrightarrow 2 \mathrm{NaAlO}_2+3 \mathrm{H}_2 \uparrow$
$2 \mathrm{NaAlO}_2+\mathrm{CO}_2+3 \mathrm{H}_2 \mathrm{O} \xrightarrow[\Delta]{\Delta}+\mathrm{Na}_2 \mathrm{O}_3+2 \mathrm{Al}(\mathrm{OH})_3$
$2 \mathrm{Al}(\mathrm{OH})_3 \xrightarrow{1200^{\circ} \mathrm{C}} \mathrm{Al}_2 \mathrm{O}_3+3 \mathrm{H}_2 \mathrm{O}$
Thus, $\mathrm{A}, \mathrm{B}$, and C are $\mathrm{Al}, \mathrm{NaAlO}_2$ and $\mathrm{Al}(\mathrm{OH})_3$ respectively.
View full question & answer→Question 212 Marks
Explain the following:
Electron gain enthalpy of chlorine is more negative as compared to fluorine.
AnswerDue to small size, the electron-electron repulsions in the relatively compact 2p-subshell of F are quite strong and hence the incoming electron is not accepted with the same ease as in case of bigger Cl atom where repulsions are comparatively weak. Thus, electron gain enthalpy of chlorine is more negative as compared to that of fluorine.
View full question & answer→Question 222 Marks
Which of the following is not hydrolysed by water and why?
$\text{BF}_3,\text{BCl}$ and $\text{BBr}_3$.
Answer$\mathrm{BF}_3$ is not easily hydrolysed by water.It forms an adduct $\mathrm{BF}_3 \cdot \mathrm{OH}_2$ whereas $\mathrm{BCl}_3$ and $\mathrm{BBr}_3$ are hydrolysed to given boric acid and HCl or HBr respectively. This is because the $\mathrm{B}-\mathrm{F}$ bond in $\mathrm{BF}_3$ is very strong due to extensive $\mathrm{p} \pi$ - $\mathrm{p} \pi$ back bonding. As a result, it is not hydrolysed by water. The B-F bond energy is far larger than B-OH bond energy and cannot be compensated.
However, in $\mathrm{BCl}_3$ and $\mathrm{BBr}_3$, the corresponding $\mathrm{B}-\mathrm{Cl}$ and $\mathrm{B}-\mathrm{Br}$ bond energy is relatively less than $\mathrm{B}-\mathrm{F}$ because of inefficient $р \pi-p \pi$ back bonding. Therefore, these get hydrolysed. View full question & answer→Question 232 Marks
- Elemental silicon does not form graphite like structure as carbon does. Give reason.
- Arrange hyrides of group 14 elements in increasing order of reducing power.
Answer
- Silicon cannot form $\text{p}\pi-\text{p}\pi$ bond like 'C' forms in graphite due to bigger atomic size, therefore, it can not form graphite like structure.
- CH4 < SiH4 < GeH4 < SnH4 < PbH3 is increasing order of reducing power because bond dissociation enthalpy decreases with increases in bond length due to increase in atomic size of group 14 element.
View full question & answer→Question 242 Marks
What property of anhydrous $\mathrm{AlCl}_3$ makes it a very good preparative reagent in organic chemistry?
AnswerIt acts as Lewis acid. It generates electrophile by accepting negativity charged ion, therefore, helps in carrying out electrophilic substitution reactions such as Friedel crafts Alkylation and Acylation.
View full question & answer→Question 252 Marks
- What are fullerenes? How are they prepared?
- Classify the following compounds into acidic, basic and amphoteric oxides:
$\mathrm{Al}_2 \mathrm{O}_3, \mathrm{Cl}_2 \mathrm{O}_7$Answeri. Fullerenes are allotropes of carbon having structure similar to soccer ball. They are made by heating graphite in electric arc in presence of inert gases such as helium or argon.
ii. $\mathrm{Al}_2 \mathrm{O}_3$ is amphoteric whereas $\mathrm{Cl}_2 \mathrm{O}_7$ is acidic oxide.
View full question & answer→Question 262 Marks
Mention the chief reason for the anomalous behavior of boron in Group 13 of the periodic table.
AnswerIt is due to small size and higher ionisation enthalpy and high charge/ radius ratio of boron i.e. high polarizing power, high electronegativity and absence of d-orbitals.
View full question & answer→Question 272 Marks
Explain the following:
$\mathrm{PbX}_2$ is more stable than $\mathrm{PbX}_4$.
AnswerDue to inert pair effect, +2 oxidation state of Pb is more stable than its +4 oxidation state. Consequently, PbX 2 in which the oxidation state of Pb is +2 is more stable than $\mathrm{PbX}_4$ in which the oxidation state of Pb is +4 .
View full question & answer→Question 282 Marks
Why does boron form stable electron deficient compounds?
AnswerBoron has three valence electrons, it will share three electrons with other elements to form electron deficient compounds which are stable due to back bonding i.e. lone pair of electrons of halogen is denoted to vacant p-orbital of 'B' forming $\text{p}\pi-\text{p}\pi$ bond making it stable.
View full question & answer→Question 292 Marks
- Of Bi(V) and Sb(V) which may be a stronger oxidising agent and why?
- Complete the following chemical equation:
$\text{Ca}_3\text{P}_2+\text{H}_2\text{O}\xrightarrow{\ \ \ \ \ \ \ \ }$Answer
- Bi(V) is stronger oxidising agent due to inert pair effect as Bi(III) is more stable as compared to Sb(III).
- $\text{Ca}_3\text{P}_3+6\text{H}_2\text{O}\xrightarrow{\ \ \ \ \ \ \ }3\text{Ca(OH})_2+2\text{PH}_3$
View full question & answer→Question 302 Marks
Identify the compounds A, X and Z in the following reactions:
$\text{A}+2\text{HCl}+5\text{H}_2\text{O}\rightarrow2\text{NaCl}+\text{X}$
$\text{X}\frac{\triangle}{370\text{K}}\text{HBO}_2\frac{\triangle}{>370\text{K}}\text{Z}$
Answer$\text{Na}_2\text{B}_4\text{O}_7+2\text{HCl}+5\text{H}_2\text{O}\rightarrow2\text{NaCl}+4\text{H}_3\text{BO}_3\\ \ \ \ \ \text{(A)}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{(X)}\\ \ \ \ ^ \text{(Borax)} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ^{\text{(Boric acid)}}$
$\text{H}_3\text{BO}_3\xrightarrow{\triangle,370\text{K}}\text{HBO}_2\ \ +\text{H}_2\text{O}\\\ \ \text{(X)}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ^\text{Metaboric acid}$
$4\text{HBO}_2\xrightarrow[-\text{H}_2\text{O}]{\triangle,370\text{K}}[\text{H}_2\text{B}_4\text{O}_7]\xrightarrow{\text{Red heat}}2\text{B}_2\text{O}_3\ \ \ +\text{H}_2\text{O}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ^{\text{Tetraboric acid}}\ \ \ \ \ \ \ \ \ ^\text{Boric trioxide(Z)}$
View full question & answer→Question 312 Marks
What happens when:
- Diborane burns in presence of oxygen.
- Diborane is hydrolysed.
Answer
- When diborane burns in presence of oxygen $\text{B}_2\text{O}_3$ and $\text{H}_2\text{O}$ is formed.
$\text{B}_2\text{H}_6+3\text{O}_2\xrightarrow{ \ \ \ \ \ \ \ \ }\text{B}_2\text{O}_3+3\text{H}_2\text{O}$
- When diborane is hydrolysed, it forms boric acid and $\text{H}_2$ gas is formed.
$\text{B}_2\text{H}_6+3\text{H}_2\text{O}\xrightarrow{\ \ \ \ \ \ }2\text{H}_3\text{BO}_3+3\text{H}_2$ View full question & answer→Question 322 Marks
Gallium has higher I.E. than Al. Explain.
AnswerIt is due to poor shielding effect of d-electrons in Ga, effective nuclear charge increases as compared to Al, therefore, I.E. of Ga is higher than Al. Secondly, Ga is smaller than Al.
View full question & answer→Question 332 Marks
Explain the following:
Silicon forms $\text{SiF}^{2-}_6$ ion whereas corresponding fluoro compound of carbon is not known.
AnswerSilicon has vacant d orbital in its valence shall due to which it can accommodate 6 electrons from fluorine atoms whereas carbone does not have d orbital and cannot expand its covalence beyond four.
View full question & answer→Question 342 Marks
Explain why $\mathrm{BCl}_3$ is monomer but $\mathrm{BH}_3$ exists as $\mathrm{B}_2 \mathrm{H}_6$.
AnswerIn $\mathrm{BCl}_3$, there is back bonding, i.e., lone pair of electrons is donated from chlorine to boron thus reducing its electron deficiency.
In $\mathrm{BH}_3$, back bonding is not possible but H -bridges are possible, therefore, it exists as dimer.
View full question & answer→Question 352 Marks
- Which out of $TICl_3$, GaCl, InCl, TICl undergo disproportionation reaction and why?
- Why is boric acid used in carom boards for smooth gliding of pawns.
Answer
- GaCl and InCl will undergo disproportionation reaction because they can get oxidised to +3 oxidation state and get reduced to oxidation state equal to zero.
$3\text{GaCl}\xrightarrow{\ \ \ \ \ \ \ \ }2\text{Ga}+\text{GaCl}_3$
$3\text{InCl}\xrightarrow{\ \ \ \ \ \ \ \ }2\text{In}+\text{InCl}_3$
- H-bonding in $H_3BO_3$ gives it a layered structure which has weak forces of attraction due to which it is soft and makes carom boards for smooth gliding of pawn.
View full question & answer→Question 362 Marks
- Name a mineral of ‘Al' which does not contain oxygen. Give its use.
- In which light Germanium is transparent?
Answeri. Cryolite, $\mathrm{Na}_3 \mathrm{AlF}_6$ is mineral of ' $\mathrm{Al}$' which does not contain oxygen. It dissolves bauxite and increases its conductance before electrolytic reduction.
ii. In infrared region, germanium is transparent.
View full question & answer→Question 372 Marks
Explain the following:
$\mathrm{TI}\left(\mathrm{NO}_3\right)_3$ acts as an oxidising agent.
AnswerDue ta strong inert pair effect, the +3 oxidation state of TI is less stable than its +1 oxidation state. Since in $\mathrm{TI}\left(\mathrm{NO}_3\right)_3$, oxidation state of TI is +3 , therefore, it can easily gain two electrons to form $\mathrm{TINO}_3$ in which the oxidation state of TI is +1 . Consequently, $\mathrm{TI}\left(\mathrm{NO}_3\right)_3$ acts as an oxidising agent.
View full question & answer→Question 382 Marks
Account for the following:
Why $\mathrm{BF}_3$ is less acidic than $\mathrm{BCl}_3$ though fluorine is more electronegative than chlorine?
Answer$\mathrm{BF}_3$ is less acidic because back bonding (donation of lone pair of halogen to vacant p -orbital of ' B ') is most effective due to $2 p-2 p$ overlapping whereas it is less effective in $\mathrm{BCl}_3$ due to $2 p-3 p$ overlapping.
View full question & answer→Question 392 Marks
How does electron deficient compound $\mathrm{BF}_3$ achieve electronic saturation, i.e. fully occupied outer electron shells?
Answer$\mathrm{BF}_3$ achieve it by the following ways.
i. Multiple bonding or $p \pi-p \pi$ back bonding, e.g. $\mathrm{BF}_3$ in which a lone pair of electrons present in 2 p -orbital of one of the fluorine atoms may be transferred to the vacant p-orbital on the boron atom.
ii. Formation of complexes in which electrons are received from a donor molecule, e.g. $\mathrm{F}_3 \mathrm{B} \leftarrow \mathrm{NH}_3$. Boron compounds, thus, behave as Lewis acids.
View full question & answer→Question 402 Marks
Carbon and silicon are mainly tetravalent but Ge, Sn and Pb show divalency. Give reason.
AnswerGe, Sn, Pb are divalent due to inert pair effect due to poor shielding effect of d and f-electrons which is not there in carbon and silicon as they do not have electrons in d-orbitals.
View full question & answer→Question 412 Marks
- Name an element of group 13 which is best reducing agent and why?
- Why is $\mathrm{TI}^{3+}$ good oxidising agent?
Answer
- 'Al’ is best reducing agent because it has lowest standard reduction potential.
$\text{E}^\text{o}_{\frac{\text{Al}^{3+}}{\text{Al}}}=-1.66\text{V}.$
- $\mathrm{TI}^{3+}$ gain 2 electrons to form $\mathrm{Ti}^{+}$which is more stable due to inert pair effect.
View full question & answer→Question 422 Marks
- Why is $\mathrm{B}-\mathrm{F}$ bond dissociation energy in $\mathrm{BF}_3$ is higher than $\mathrm{C}-\mathrm{F}$ in $\mathrm{CF}_4$?
- What is formula of borax?
Answeri. It is because there is significant $\mathrm{p} \pi-\mathrm{p} \pi$ interaction between B and F in BF , whereas there is no possibility of such interaction between C and F in $\mathrm{CF}_4$.
ii. $\mathrm{Na}_2 \mathrm{~B}_4 0_7 \cdot 10 \mathrm{H}_2 \mathrm{O}$ is formula of borax.
View full question & answer→Question 432 Marks
Explain the following:
Aluminium forms $\left[\mathrm{AlF}_6\right]^{3-}$ ion but boron does not form $\left[\mathrm{BF}_6\right]^{3-}$ ion.
AnswerAl has vacant d-orbitals and hence can expand its coordination number from 4 to 6 and hence forms octahedral $\left[\mathrm{A}_1 \mathrm{~F}_6\right]^{3-}$ ion in which Al undergoes $\mathrm{sp}^3 \mathrm{~d}^2$ hybridisation. In contrast, B does not have d -orbitals. Therefore, it can have a maximum coordination number of 4 . Therefore, B forms $\left[\mathrm{BF}_4\right]^{-}$(in which B is $\mathrm{sp}^3$-hybridised) but not $\left[\mathrm{BF}_6\right]^{3-}$.
View full question & answer→Question 442 Marks
Arrange the hydrides of group 14 in increasing order of thermal stability.
Answer$\mathrm{PbH}_4<\mathrm{SnH}_4<\mathrm{GeH}_4<\mathrm{SiH}_4<\mathrm{CH}_4$ because bond dissociation enthalpy increases due to small bond length as atomic size of element decreases from Pb to C.
View full question & answer→Question 452 Marks
Account for the following.
- Graphite is used as lubricant.
- Diamond is used as an abrasive.
Answer
- Graphite has layered structure. Layers are held together by weak van der Waals' forces and hence can be made to slip over one another. Therefore, graphite acts as a dry.lubricant.
- In diamond, each $sp^3$ hybridised carbon atom is linked to four other carbon atoms. It has three-dimensional network of carbon atoms. It is very difficult to break extended covalent bonding and therefore diamond is a hardest substance on the earth. That's why it is used as an abrasive.
View full question & answer→Question 462 Marks
Complete the following chemical equations:
$\text{Z}+3\text{LiAlH}_4\rightarrow\text{X}+3\text{LiF}+3\text{AlF}_3$
$\text{X}+6\text{H}_2\text{O}\rightarrow\text{Y}+6\text{H}_2$
$3\text{X}+3\text{O}_2\xrightarrow{\triangle}\text{B}_2\text{O}_3+3\text{H}_2\text{O}$
Answer$4\text{BF}_3+3\text{LiAlH}_4\rightarrow2\text{B}_2\text{H}_6+3\text{LiF}+3\text{A1F}_3$
$\text{B}_2\text{H}_6(\text{g})+6\text{H}_20(1)\rightarrow2\text{B}(\text{OH})_3\text{(aq)}+6\text{H}_2\text{(g)}$
$\text{B}_2\text{H}_6+3\text{O}_2\xrightarrow{\triangle}\text{B}_2\text{O}_3+3\text{H}_2\text{O}$
$\text{Z}=\text{BF}_3,\text{X}=\text{B}_2\text{H}_6\text{ and Y=B(OH)}_3\text{ or H}_3\text{BO}_3$
View full question & answer→Question 472 Marks
i. $\mathrm{Ca}_2$
ii. $\mathrm{CaC}_2$
iii. $\mathrm{H}_2 \mathrm{CO}_3$
iv. $HCN$
v. $CO$
Answer
|
S. No.
|
Corbon compousds
|
Oxidation state
|
|
i.
|
$CO_2$
|
+4
|
|
ii.
|
$CaCO_3$
|
-1
|
|
iii.
|
$H_2CO_3$
|
+4
|
|
iv.
|
$HCN$
|
+2
|
|
v.
|
$CO$
|
+2
|
View full question & answer→Question 482 Marks
- Carbon dioxide is non-polar while water is polar. What conclusion do you draw about their structures from these.
- What is dry ice? Why is it so called?
Answer
-
is linear, bond moments are equal and opposite, net dipole moment is zero.
Water is bent molecule, it has net dipole moment.
- Solid $\mathrm{CO}_2$ is called dry ice because it directly changes into vapours.
View full question & answer→Question 492 Marks
Why are pentahalides more covalent than trihalides?
AnswerIt is because pentavalent cation has more polarising power than trivalent cation due to smaller size and higher charge. Therefore, pentahalides have more tendency to form covalent bond.
View full question & answer→Question 502 Marks
Explain the following:
Gallium has higher ionisation enthalpy than aluminium.
AnswerDue to ineffective shielding of valence electrons by the intervening 3d electrons, the effective nuclear charge on Ga is slightly higher than that on Al and hence the $\Delta \mathrm{H}_{\mathrm{i}}$ of gallium is slightly higher than that of Al. Boron has three electrons in the valence shell. Because of its small size and high sum of the first three ionisation enthalpies $(\text{i.e.,}\triangle).$
View full question & answer→Question 512 Marks
How would you explain the following observations?
BeO is almost insoluble but $\mathrm{BeSO}_4$ is soluble in water
AnswerBeO is almost insoluble in water and $\mathrm{BeSO}_4$ is soluble in water. $\mathrm{Be}^{2+}$ is a small cation with a high polarizing power and $\mathrm{O}^{2-}$ is a small anion. The size compatibility of $\mathrm{Be}^{2+}$ and $\mathrm{O}^{2-}$ is high. Therefore, the lattice energy released during their formation is also very high. When BeO is dissolved in water, the hydration energy of its ions is not sufficient to overcome the high lattice energy. Therefore, BeO is insoluble in water. On the other hand, $\mathrm{SO}_4^{2-}$ ion is a large anion. Hence, $\mathrm{Be}^{2+}$ can easily polarize $\mathrm{SO}_4^{2-}$ ions, making $\mathrm{BeSO}_4$ unstable. Thus, the lattice energy of $\mathrm{BeSO}_4$ is not very high and so it is soluble in water.
View full question & answer→Question 522 Marks
Why is $\mathrm{PbCl}_4$ good oxidizing agent?
OR
Boron forms only covalent compounds.
Answer$\mathrm{PbCl}_4$ is good oxidising agent because $\mathrm{Pb}^{2+}$ is more stable than $\mathrm{Pb}^{4+}$ due to inert pair effect, therefore, $\mathrm{Pb}^{4+}$ readily changes into $\mathrm{Pb}^{2+}$ by gaining two electrons.
View full question & answer→Question 532 Marks
Arrange the following in increasing order of Lewis acid character:
$\mathrm{BF}_3, \mathrm{BFI}_3, \mathrm{BBr}_3 \text { and } \mathrm{BI}_3$
Answer$\mathrm{Bl}_3>\mathrm{BBr}_3>\mathrm{BCl}_3>\mathrm{BF}_3$
$\because$ Back bonding (donation of lone pair of halogen to vacant ' p ' orbital of B ) is most effective in $\mathrm{BF}_3$ due to $2 \mathrm{p}-2 \mathrm{p}$ overlapping and least effective in $\mathrm{Bl}_3$ due to $2 p-5$ p overlapping.
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Between $\mathrm{AlF}_3$ and $\mathrm{AlCl}_3$, which one will have a higher melting point?
Answer$\mathrm{AlF}_3$ is more ionic, therefore, has higher melting point as compared to $\mathrm{AlCl}_3$ because ' F ' is more electronegative than Cl . Higher the difference in electro-negativity more will be ionic character.
View full question & answer→Question 552 Marks
How is that silicon atoms can have a coordination number more than four but carbon atoms cannot?
AnswerSilicon has vacant d-orbitals, therefore, it can have coordination number more than four but carbon cannot have because it does not have d-orbitals. It can not expand its valency beyond four.
View full question & answer→Question 562 Marks
Like CO, why its analog of SiO is not stable?
AnswerCO is a resonance hybrid of the following two structures,
Thus, CO contains pπ-pπ multiple bonds. This is due to the reason that carbon has a strong tendency to form pπ-pπ multiple bonds due to its small size and high electronegativity. Silicon, on the other hand, because of its bigger size and lower electronegativity has no tendency to form pπ-pπ multiple bonds and hence, Si does not form SiO molecule analogous to CO molecule.
View full question & answer→Question 572 Marks
- What is drikold and for what purpose it is used?
- Why does lime water turn milky when $\text{CO}_2$ is passed through it?
Answer
- Solid carbon dioxide is known as dry ice. It is soft, white snow like substances and looks like ice. However, it does not wet a piece of cloth or paper because it sublimes without melting. Solid carbon dioxide is used as a refrigerant under the commercial name of drikold.
- When carbon dioxide is passed through lime water it turn milky due to the formation of $\text{CaCO}_3$.
But when carbon dioxide is passed continuously, the milkiness disappears due to the formation of soluble,$\text{Ca(HCO}_3)_2.$
$\text{Ca(OH})_32+\text{CO}_2\xrightarrow{\ \ \ \ \ \ \ }\text{CaCO}_3+\text{H}_2\text{O}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Milkiness}\\\xrightarrow{\frac{\text{CO}_2}{\text{H}_2\text{O}}}\text{Ca(HCO}_3)_2\\\ \ \ \ \ \ \ \ \ \text{Souble in water}$ View full question & answer→Question 582 Marks
- How is diborane prepared in the laboratory? Draw its structure.
- Explain why $\mathrm{CO}_2$ is a gas whereas $\mathrm{SiO}_2$ is a solid.
Answer
- $2\text{NaBH}_4+\text{I}_2\xrightarrow{\ \ \ \ \ \ \ }\text{B}_2\text{H}_6+2\text{NaI}+\text{H}_2$
- $\mathrm{CO}_2$ exists as descrete molecules which are held together by weak van der Waals' forces of attraction whereas $\mathrm{SiO}_2$ is three dimensional covalent solid.
View full question & answer→Question 592 Marks
Explain the following:
Boron does not exist as $B^{3+}$ ion.
AnswerBoron has three electrons in the valence shell. Because of its small size and high sum of the first three ionisation enthalpies $(\text{i.e.,}\triangle_\text{i}\text{H}_1+\triangle_\text{i}\text{H}_2+\triangle_\text{i}\text{H}_3),$ boron does not lose all its valence electrons to form $B^{3+}$ ions.
View full question & answer→Question 602 Marks
Starting from $\mathrm{SiCl}_4$ prepare the following in steps not exceeding the number given in parenthness (give reactions only).
i. Silicon.
ii. Linear silicon containing methyl groups only.
iii. $\mathrm{Na}_2 \mathrm{SiO}_3$.
Answer
- $3\text{SiCl}_4+4\text{Al}\xrightarrow{\ \ \ \ \ \ \ }4\text{AlCl}_3+3\text{Si}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Molten}$
- $\text{Si}+2\text{CH}_3\text{Cl}\xrightarrow[570\text{ K}]{\text{Cu powder}}(\text{CH}_3)_2\text{SiCl}_2$
View full question & answer→Question 612 Marks
Explain the nature of boric acid as a Lewis acid in water.
AnswerBoric acid is a weak monobasic acid and acts as a Lewis acid by accepting electrons from a hydroxyl ion.
$\text{B(OH)}_3+2\text{H}_2\text{O}\rightarrow[\text{B(OH)}_4]^-+\text{H}_3\text{O}^+$
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$\mathrm{BCl}_3$ exists but $\mathrm{BH}_3$ does not. Explain.
AnswerIn $\mathrm{BCl}_3, \mathrm{Cl}$ donates lone pair of electrons to vacant p-orbital of boron (back-bonding) making it more stable whereas in $\mathrm{BH}_3$ back-bonding is not possible, therefore, it exists as dimer. Secondly, in $\mathrm{BCl}_3$, chlorine being bigger in size cannot form bridged bonds like hydrogen in $\mathrm{B}_2 \mathrm{H}_6$.
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Why does C differ from rest of elements?
AnswerCarbon has smallest size, highest ionisation enthalpy and high electronegativity and does not have d-orbitals, therefore, it differs from rest of the elements.
View full question & answer→Question 642 Marks
$\mathrm{AlCl}_3$ exist a as dimer while $\mathrm{BCl}_3$ exist as monomer, why?
Answer'Al' has d-orbital which can accept a lone pair from ' Cl ' forming coordinate bond, $\mathrm{AlCl}_3$ exist as dimer.
In $\mathrm{BCl}_3$, 'Cl' donates a pair of electron to vacant p -orbital of ' B ' forming $\mathrm{p} \pi-\mathrm{p} \pi$ bond making it stable. It exist as $\mathrm{BCl}_3$
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Arrange the following compounds in decreasing order of property indicated against each. Give reason for your answer: $\mathrm{BCl}_3, \mathrm{AlCl}_3, \mathrm{AgCl}_3, \mathrm{InCl}_3, \mathrm{TICl}_3$
(Stability of +3 oxidation state.)
Answer$\mathrm{BCl}_3>\mathrm{AlCl}_3>\mathrm{GaCl}_3>\mathrm{InCl}_3>\mathrm{TICl}_3$ because of inert pair effect stability of +3 oxidation state decreases.
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Carbon monoxide is readily absorbed by ammoniacal cuprous chloride solution but carbon dioxide is not. Explain.
AnswerDue to the presence of a lone pair of electrons on carbon in Co, it acts as a Lewis base (or ligand) and thus forms a soluble complex with ammoniacal cuprous chloride solution.
$\text{CaCl}+\text{NH}_3+:\text{CO}\xrightarrow{\\ \ \ \ \ \ }[\text{Cu(CONH}_3]^+\text{Cl}^-\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Soluble complex}$
On the other hand, $\mathrm{CO}_2$ does not act as a Lewis base since it does not have a lone pair of electrons on the carbon atom and hence, does not dissolve in ammoniacal cuprous chloride solution.
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Complete the following chemical reaction equations:
- $\text{I}_2+\text{HNO}_3\xrightarrow{\ \ \ \ \ \ \ \ \ }\\ \ \ \ \ \ \ \ \ \ ^\text{(conc.)}$
- $\text{HgCl}_2+\text{PH}_3\xrightarrow{\ \ \ \ \ \ \ \ \ \ }$
Answer
- $\text{I}_2+10\text{HNO}_3\xrightarrow{\ \ \ \ \ \ }2\text{HIO}_3+10\text{NO}_2+4\text{H}_2\text{O}\\ \ \ \ \ \ \ \ \ \ \ \ ^\text{(conc.)}$
- $3\text{HgCl}_2+2\text{PH}_3\xrightarrow{\ \ \ \ \ \ \ }\text{Hg}_3\text{P}_2+6\text{HCl}$
View full question & answer→Question 682 Marks
Give suitable reasons for the following:
i. $\left[\mathrm{SiF}_6\right]^{2-}$ is known whereas $\left[\mathrm{SiCl}_6\right]^{2-}$ not.
ii. Diamond is covalent, yet it has high melting point.
Answer
- It is because 'F' is smaller in size, more electronegative and stronger oxidising agent than chlorine.
- It is due to three dimensional compact structure, it has strong covalent bonds, which requires high amount of energy, therefore, high melting point.
View full question & answer→Question 692 Marks
- Complete the following chemical equations:
- $\text{Fe}_2\text{O}_3+3\text{CO}\xrightarrow{\Delta\ \ }$
- $\text{CaCO}_3+2\text{HCl}\xrightarrow{\ \ \ \ \ \ \ }$
- Write a brief account on the following:
-
Diamond is covalent, yet it has high melting point.
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Atomic radius of gallium (135 pm) is less than that of aluminium (143 pm).
-
Graphite is a good conductor of electricity but diamond is insulator.
Answer
-
- $\text{Fe}_2\text{O}_3+3\text{CO}\xrightarrow{\Delta\ \ }2\text{Fe}+3\text{CO}_2$
- $\text{CaCO}_3+2\text{HCl}\xrightarrow{\ \ \ \ \ \ \ }\text{CaCl}_2+\text{H}_2\text{O}+\text{CO}_2$
-
- It is due to strong $\text{C}-\text{C}$ bonding and compact structure.
- 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.
- It is due to the presence of free electrons in graphite but not in diamond.
View full question & answer→Question 702 Marks
Why do boron halides form addition compounds with ammonia and amines?
AnswerIt is because they are electron deficient i.e. need a pair of electron to complete their octet. $\mathrm{NH}_3$ and $\mathrm{RNH}_2$ have lone pair of electron which can be donated to boron halides.
View full question & answer→Question 712 Marks
What type of glass is obtained when borax is added to sodium carbonate, silica and calcium carbonate?
AnswerPyrex glass is obtained which is heat resistant. It can withstand high temperature. It is also called Borosil glass.
View full question & answer→Question 722 Marks
Why is $\mathrm{BF}_3$ planar molecule but $\mathrm{NH}_3$ is pyramidal?
AnswerIn $\mathrm{BF}_3$ ' B ' is $\mathrm{sp}^2$ hybridised whereas in $\mathrm{NH}_3{ }^{\prime} \mathrm{N}^{\prime}$ is $\mathrm{sp}^3$ hybridised with one lone pair of electrons, therefore, it is pyramidal while $\mathrm{BF}_3$ is planar.
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What happens when:
- Silicon dioxide is treated with excess of HF?
- $\mathrm{N}\left(\mathrm{CH}_3\right)_2 \mathrm{SI}(\mathrm{OH})_3$ is treated with $\left(\mathrm{CH}_3\right)_3 \mathrm{SiOH}$ ?
Answer
- Hydrofluorosilicic acid is obtained.
$\text{SiO}_2+4\text{HF}\xrightarrow{\ \ \ \ \ \ \ \ }\text{SiF}_4+2\text{H}_2\text{O}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \big\downarrow^{_\text{2HF}}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{H}_2\text{SiF}_6$
- Length of the silicone chain is controlled if $\left(\mathrm{CH}_3\right)_3 \mathrm{SiOH}$ is used.
View full question & answer→Question 742 Marks
Account for the following observations:
$\mathrm{AlCl}_3$ is a Lewis acid.
AnswerElectronic configuration of $\mathrm{Al} 2,8,3$. In $\mathrm{AlCl}_3$ it forms three bonds and hence outer shell has 6 electrons. Now Al needs two more electrons to complete its octet. By definition those which accepts electrons are called lewis acids. So $\mathrm{AlCl}_3$ is a Lewis acid.
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Give one chemical reaction to show that:
-
Tin (II) is a reducing agent whereas lead (II) is not.
-
Tin (II) chloride is a reducing agent.
Answer
-
Both tin and lead show two oxidation states of +2 and +4 due to inert pair effect. But the inert pair effect is more prominent in case of Pb than in Sn . In other words, +2 oxidation state of Sn is less stable than its +4 oxidation state. Therefore, Sn (II) acts as a reducing agent and gets converted into more stable Sn (IV) by losing two electrons. e.g. it reduces $Fe ^{3+}$ to $Fe ^{2+}$ ions.
$2\text{Fe}^{3+}+\text{Sn}^{2+}\xrightarrow{\ \ \ \ \ \ \ \ \ \ }2\text{Fe}^{2+}+\text{Sn}^{4+}$
In contrast, +2 oxidation state of Pb is more stable than its +4 oxidation state. In other words, Pb (II) does not lose electrons easily and hence does not act as a reducing agent.
-
Being a reducing agent, tin (II) chloride reduces mercuric chloride $\left( HgCl _2\right)$ to mercurous chloride $\left( Hg _2 Cl _2\right)$ and ferric salts to ferrous salts.
$\text{SnCl}_2+2\text{HgCl}_ 2\xrightarrow{\ \ \ \ \ \ \ \ }\text{SnCl}_4+2\text{FeCl}_2$
$\text{SnCl}_2+2\text{FeCl}_3\xrightarrow{\ \ \ \ \ \ }\text{SnCl}_4+2\text{FeCl}_2$ View full question & answer→Question 762 Marks
- $\text{p}\pi-\text{d}\pi$ bonding is present in
- $\text{N(SiH}_3)_3$
- $\text{N(CH}_3)_3$
- $\text{N}_2\text{O}_4$
- $\text{NO}^-_3$
- Which of the following has ionic bonding?
$\text{PbCl}_2,\text{PbCl}_4,\text{CCl}_4,\text{SiCl}_4$Answeri. $\mathrm{N}(\mathrm{SiH})_3$ because ' Si ' has d -orbitals and accepts lone pair of electron from nitrogen forming $\mathrm{p} \pi-\pi \mathrm{p}$ bond.
ii. $\mathrm{PbCl}_2$ has ionic bonding because $\mathrm{Pb}^{2+}$ has less polarizing power.
View full question & answer→Question 772 Marks
Why is boric acid $\left(\mathrm{H}_2 \mathrm{BO}_3\right)$ monobasic acid?
AnswerIt accepts a pair of electrons from $\mathrm{OH}^{-}$ion of $\mathrm{H}_2 \mathrm{O}$ therefore, it is monobasic acid. $\mathrm{B}(\mathrm{OH}) \_3+\mathrm{H}_2 \mathrm{O} \longrightarrow\left[\mathrm{B}(\mathrm{OH})_4\right]^{-}+\mathrm{H}^{+}$
View full question & answer→Question 782 Marks
Which one of the following elements exhibits +1 oxidation state as well?
AI, B, Ca, Ti, Be
AnswerTI shows +1 oxidation state due to inert pair effect i.e. $\mathrm{ns}^2$ electron does not take part in bond formation due to poor shielding effect of d and f -electrons is stronger attracted towards nucleus.
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