Question 14 Marks
Find out the structural formulae of various oxyacids of phosphorus.
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Question 24 Marks
Shravani was performing inorganic qualitative analysis of a salt. To an aqueous solution of that salt, she added silver nitrate. When a white precipitate was formed. On adding ammonium hydroxide to this, she obtained a clear solution. Comment on her observations and write the chemical reactions involved.
Answer
View full question & answer→Shravani was performing inorganic qualitative analysis of a salt. To an aqueous solution of that salt, she added silver nitrate. When a white precipitate was formed. On adding ammonium hydroxide to this, she obtained a clear solution. Comment on her observations and write the chemical reactions involved.
ii. On adding ammonium hydroxide $(NH_4OH)$ to this, the white precipitate of silver chloride gets dissolved and thus, a clear solution is obtained.
ii. On adding ammonium hydroxide $(NH_4OH)$ to this, the white precipitate of silver chloride gets dissolved and thus, a clear solution is obtained.
Question 34 Marks
Explain structure and bonding of diborane.
Answer
View full question & answer→- Electronic configuration of boron is $1s^2 2s^2 2p^1.$ Thus, it has only three valence electrons.
- In diborane, each boron atom is $sp^3$ hybridized. Three of such hybrid orbitals are half filled while the fourth $sp3$ hybrid orbital remains vacant.
- The two half-filled $sp^3$ hybrid orbitals of each $B$ atom overlap with $1s$ orbitals of two terminal H atoms and form four $B – H$ covalent bonds. These bonds are also known as two-centred-two-electron $(2c-2e)$ bonds.
- When $‘1s’$ orbital of each of the remaining two $H$ atoms simultaneously overlap with half-filled hybrid orbital of one B atom and the vacant hybrid orbital of the other $B$ atom, it produces two three-centred-two- electron bonds $(3c-2e)$ or banana bonds.
- Hydrogen atoms involved in $(3c-2e)$ bonds are the bridging $H$ atoms i.e., $H$ atoms in two $B – H – B$ bonds.
- In diborane, two $B$ atoms and four terminal $H$ atoms lie in one plane, while the two bridging H atoms lie symmetrically above and below this plane.
Question 44 Marks
Explain the trend in oxidation state of elements from nitrogen to bismuth.
Answer
View full question & answer→- Group $15$ elements have five valence electrons $(ns^2 np^3).$ Common oxidation states are $-3, +3$ and $+5.$ The range of oxidation state is from $-3$ to $+5.$
- Group $15$ elements exhibit positive oxidation states such as $+3$ and $+5$. Due to inert pair effect, the stability of $+5$ oxidation state decreases and $+3$ oxidation state increases on moving down the group.
- Group 15 elements show tendency to donate electron pairs in $-3$ oxidation state. This tendency is maximum for nitrogen.
- The group 15 elements achieve $+5$ oxidation state only through covalent bonding.
e. g. $NH_3, PH_3, ASH_3, SbH_3,$ and $BiH_3$ contain $3$ covalent bonds. $PCl_5$ and $PF_5$ contain $5$ covalent bonds.
Question 54 Marks
What are silicones? Where are they used?
Answer
View full question & answer→i. a. Silicones are organosilicon polymers having $R_2SiO ($where, $R = CH_3 $ or $C_6H_5 $ group$)$ as a repeating unit held together by
b. Since the empirical formula $R_2SiO ($where $R = CH_3 $ or $C_6H_5 $ group$)$ is similar to that of ketones $(R_2CO)$, these compounds are named as silicones.
ii. Applications: They are used as
b. Since the empirical formula $R_2SiO ($where $R = CH_3 $ or $C_6H_5 $ group$)$ is similar to that of ketones $(R_2CO)$, these compounds are named as silicones.
ii. Applications: They are used as
- insulating material for electrical appliances.
- water proofing of fabrics.
- sealant.
- high temperature lubricants.
- for mixing in paints and enamels to make them resistant to high temperature, sunlight and chemicals.
Question 64 Marks
Explain the trend of the following in group 13 elements :
A. atomic radii
B. ionization enthalpy
C. electron affinity
Answer
View full question & answer→A. Atomic radii:
- In group 13, on moving down the group, the atomic radii increases from B to Al.
- However, there is an anomaly observed in the atomic radius of gallium due to the presence of 3d electrons. These inner 3d electrons offer poor shielding effect and thus, valence shell electrons of Ga experience greater nuclear attraction. As a result, atomic radius of gallium is less than that of aluminium.
- However, the atomic radii again increases from Ga to Tl.
- Therefore, the atomic radii of the group 13 elements varies in the following order:
B < Al > Ga < In < Tl
B. Ionization enthalpy:
- Ionization enthalpies show irregular trend in the group 13 elements.
- As we move down the group, effective nuclear charge decreases due to addition of new shells in the atom of the elements which leads to increased screening effect. Thus, it becomes easier to remove valence shell electrons and hence, ionization enthalpy decreases from B to Al as expected.
- However, there is a marginal difference in the ionization enthalpy from Al to Tl.
- The ionization enthalpy increases slightly for Ga but decreases from Ga to In.
In case of Ga, there are 10 d-electrons in its inner electronic configuration which shield the nuclear charge less effectively than the s and p-electrons and therefore, the outer electron is held fairly strongly by the nucleus. As a result, the ionization enthalpy increases slightly. - Number of d electrons and extent of screening effect in indium is same as that in gallium. However, the atomic size increases from Ga to In. Due to this, the first ionization enthalpy of In decreases.
- The last element Tl has 10 d-electrons and 14 f-electrons in its inner electronic configuration which exert still smaller shielding effect on the outer electrons. Consequently, its first ionization enthalpy increases considerably.
C. Electron affinity:
a. Electron affinity shows irregular trend. It first increases from B to A1 and then decreases. The less electron affinity of boron is due to its smaller size. Adding an electron to the 2p orbital in boron leads to a greater repulsion than adding an electron to the larger 3p orbital of aluminium.
b. From Al to Tl, electron affinity decreases. This is because, nuclear charge increases but simultaneously the number of shells in the atoms also increases. As a result, the effective nuclear charge decreases down the group resulting in increased atomic size and thus, it becomes difficult to add an electron to a larger atom. The electron affinity of Ga and In is same.
Note: Electron affinity of group 13 elements:
Question 74 Marks
Write a note on the oxidation state of p-block elements with respect to groups 13, 14 and 15 elements.
Answer
View full question & answer→- Oxidation state is the primary chemical property of all elements.
- The highest oxidation state exhibited by the p-block elements is equal to the total number of valence electrons i.e., the sum of s- and p-electrons present in the valence shell. This is sometimes called the group oxidation state.
- In boron, carbon and nitrogen families, the group oxidation state is the most stable oxidation state for the lighter elements.
- Besides, the elements of groups 13, 14 and 15 exhibit other oxidation states which are lower than the group oxidation state by two units.
- The lower oxidation states become increasingly stable as we move down to heavier elements in the groups.
Question 84 Marks
Complete and write the balanced chemical equations for$:$
$i. \ \mathrm{Ca}_2 \mathrm{~B}_6 \mathrm{O}_{11}+\mathrm{Na}_2 \mathrm{CO}_3 \rightarrow$
$ii. \ \mathrm{CoO}+\mathrm{B}_2 \mathrm{O}_3 \rightarrow$
$iii. \ \mathrm{AgCl}+\mathrm{NH}_3 \rightarrow$
$iv.\ \mathrm{ZnSOS}_4+2 \mathrm{NH}_4 \mathrm{OH} \rightarrow$
$v. \ a \cdot 2 \mathrm{KI}+\mathrm{HgCl}_2 \rightarrow$
$b. \ 2 \mathrm{KI}+\mathrm{Hgl}_2 \rightarrow$
View full question & answer→$i. \ \mathrm{Ca}_2 \mathrm{~B}_6 \mathrm{O}_{11}+\mathrm{Na}_2 \mathrm{CO}_3 \rightarrow$
$ii. \ \mathrm{CoO}+\mathrm{B}_2 \mathrm{O}_3 \rightarrow$
$iii. \ \mathrm{AgCl}+\mathrm{NH}_3 \rightarrow$
$iv.\ \mathrm{ZnSOS}_4+2 \mathrm{NH}_4 \mathrm{OH} \rightarrow$
$v. \ a \cdot 2 \mathrm{KI}+\mathrm{HgCl}_2 \rightarrow$
$b. \ 2 \mathrm{KI}+\mathrm{Hgl}_2 \rightarrow$
Question 94 Marks
How is ammonia manufactured by Haber process?
Answer
View full question & answer→On the large scale, ammonia is prepared by direct combination of dinitrogen and dihydrogen by Haber process.
In this process, dinitrogen reacts with dihydrogen under high pressure of $200 \times 10^5 Pa (200$ atm$)$ and temperature around $700 K$ to produce ammonia.
$N_{2(g)} + 2H_{2(g)} ⇌ 2NH_{3(g)}; \triangle _fH^\circ = -46.1 \ kJ \ mol^{-1}$
Iron oxide with trace amounts of $K_2O$ and $Al_2O_3$ is used as catalyst in Haber process.
High pressure favours the formation of ammonia as equilibrium is attained rapidly under these conditions.
In this process, dinitrogen reacts with dihydrogen under high pressure of $200 \times 10^5 Pa (200$ atm$)$ and temperature around $700 K$ to produce ammonia.
$N_{2(g)} + 2H_{2(g)} ⇌ 2NH_{3(g)}; \triangle _fH^\circ = -46.1 \ kJ \ mol^{-1}$
Iron oxide with trace amounts of $K_2O$ and $Al_2O_3$ is used as catalyst in Haber process.
High pressure favours the formation of ammonia as equilibrium is attained rapidly under these conditions.
Question 104 Marks
Explain borax bead test.
Answer
View full question & answer→$i.$ Borax bead test is used to detect coloured transition metal ions.
$ii.$ On heating, borax loses water molecules and swells up. On further heating, it melts and forms a transparent liquid, which solidifies into a glass like material known as borax bead.
$iii.$ The borax bead consists of sodium metaborate and boric anhydride, which reacts with metals salts to form coloured bead.
e.g. When borax is heated in a Bunsen burner flame with $\ce{CoO}$ on a loop of platinum wire, a blue coloured $\ce{Co(BO_2)_2}$ bead is formed.
$\underset{\text { Cobalt oxide }}{ \ce{CoO} }+\underset{\begin{array}{c}\text { Boric } \\ \text { anhydride }\end{array}}{ \ce{B_2O_3}} \longrightarrow \underset{\begin{array}{c}\text { Cobalt } \\ \text { metaborate } \\ \text { (Blue bead) }\end{array}}{ \ce{Co} \left( \ce{BO} _2\right)_2}$
$ii.$ On heating, borax loses water molecules and swells up. On further heating, it melts and forms a transparent liquid, which solidifies into a glass like material known as borax bead.
$iii.$ The borax bead consists of sodium metaborate and boric anhydride, which reacts with metals salts to form coloured bead.
e.g. When borax is heated in a Bunsen burner flame with $\ce{CoO}$ on a loop of platinum wire, a blue coloured $\ce{Co(BO_2)_2}$ bead is formed.
$\underset{\text { Cobalt oxide }}{ \ce{CoO} }+\underset{\begin{array}{c}\text { Boric } \\ \text { anhydride }\end{array}}{ \ce{B_2O_3}} \longrightarrow \underset{\begin{array}{c}\text { Cobalt } \\ \text { metaborate } \\ \text { (Blue bead) }\end{array}}{ \ce{Co} \left( \ce{BO} _2\right)_2}$
Question 114 Marks
Explain the structure of silicon dioxide.
Answer
View full question & answer→Silicon dioxide $(SiO_2),$ is also known as silica.
It is a covalent three$-$dimensional network solid.
In $SiO_2,$ each silicon atom is covalently bound in tetrahedral manner to four oxygen atoms.
The crystal contains eight membered rings having alternate silicon and oxygen atoms.
It is a covalent three$-$dimensional network solid.
In $SiO_2,$ each silicon atom is covalently bound in tetrahedral manner to four oxygen atoms.
The crystal contains eight membered rings having alternate silicon and oxygen atoms.
Question 124 Marks
Enlist properties of
i. white phosphorus.
ii. red phosphorus.
i. white phosphorus.
ii. red phosphorus.
Answer
View full question & answer→i. Properties of white phosphorus:
- It is translucent white waxy solid.
- It glows in the dark (chemiluminescence).
- It is insoluble in water but dissolves in boiling NaOH solution.
- It is poisonous.
- It is stable and less reactive.
- It is odourless and possess iron grey lustre.
- It does not glow in the dark.
- It is insoluble in water.
- It is nonpoisonous.
Question 134 Marks
Describe the properties of carbon nanotubes.
Question 144 Marks
Explain the structure of diamond.
Question 154 Marks
Explain catenation of group $14$ elements.
Answer
View full question & answer→$i.$ The property of self$-$linking of atoms of an element by covalent bonds to form chains and rings is called catenation.
$ii.$ The strength of the element$-$element bond determines the tendency of an element to form a chain.
$iii.$ Among the elements of group $14,$ the bond strength is maximum for $C-C$ bond $(348 \ kJ mol^{-1}).$ Hence, carbon has maximum tendency for catenation.
$ii.$ The strength of the element$-$element bond determines the tendency of an element to form a chain.
$iii.$ Among the elements of group $14,$ the bond strength is maximum for $C-C$ bond $(348 \ kJ mol^{-1}).$ Hence, carbon has maximum tendency for catenation.
| Bond | Bond strength $($Bond enthalpy $kJ mol^{-1})$ |
| $C-C$ | $348$ |
| $Si-Si$ | $297$ |
| $Ge-Ge$ | $260$ |
| $Sn-Sn$ | $240$ |
Question 164 Marks
The values of the first ionization enthalpy of $Al, Si$ and $P$ are $577, 786$ and $1012 \ kJ mol^{-1}$ respectively. Explain the observed trend.
Answer
View full question & answer→The trend shows increasing first ionization enthalpy from $Al$ to $Si$ to$ P.$
$Al, Si$ and $P$ belong to the third period in the periodic table and hence, they have same valence shell.
As we move across a period from left to right, the nuclear charge increases. Due to this, electrons in the valence shell are held more tightly by the nucleus as we go from $Al$ to $Si$ to $P.$
Therefore, more energy is required to remove an electron from its outermost shell.
Hence, the value of first ionization enthalpy increases from $Al$ to $Si$ to $P.$
$Al, Si$ and $P$ belong to the third period in the periodic table and hence, they have same valence shell.
As we move across a period from left to right, the nuclear charge increases. Due to this, electrons in the valence shell are held more tightly by the nucleus as we go from $Al$ to $Si$ to $P.$
Therefore, more energy is required to remove an electron from its outermost shell.
Hence, the value of first ionization enthalpy increases from $Al$ to $Si$ to $P.$
Question 174 Marks
Why the atomic radius of Gallium is less than that of aluminium?
Answer
View full question & answer→Atomic radius of the elements increases down the group due to addition of new shells.
Electronic configuration of $Al$ is $[Ne]3s^23p^1$ while that of $Ga$ is $[Ar]3d^{10}4s^24p^1.$
As $Al$ does not have $d-$electrons, it offers an exception to this trend.
As we go from $Al$ down to $Ga$ the nuclear charge increases by $18$ units. Out of the $18$ electrons added, $10$ electrons are in the inner $3d$ subshell of $Ga.$ These $d-$electrons offer poor shielding effect.
Therefore, the effect of attraction due to increased nuclear charge is experienced prominently by the outer electrons of $Ga$ and thus, its atomic radius becomes smaller than that of $Al.$
Hence, the atomic radius of gallium is less than that of aluminium.
Electronic configuration of $Al$ is $[Ne]3s^23p^1$ while that of $Ga$ is $[Ar]3d^{10}4s^24p^1.$
As $Al$ does not have $d-$electrons, it offers an exception to this trend.
As we go from $Al$ down to $Ga$ the nuclear charge increases by $18$ units. Out of the $18$ electrons added, $10$ electrons are in the inner $3d$ subshell of $Ga.$ These $d-$electrons offer poor shielding effect.
Therefore, the effect of attraction due to increased nuclear charge is experienced prominently by the outer electrons of $Ga$ and thus, its atomic radius becomes smaller than that of $Al.$
Hence, the atomic radius of gallium is less than that of aluminium.
Question 184 Marks
Write condensed electronic configurations of the following elements.
$
{ }_{13} Al ,{ }_{49} In ,{ }_{14} Si ,{ }_{50} Sn ,{ }_{15} P ,{ }_{33} As
$
Answer
View full question & answer→Condensed electronic configurations of :
i. ${ }_{13} Al :[ Ne ] 3 s ^2 3 p^1$
ii. $49 In$ : $[ Kr ] 4 d ^{10} 5 s^2 5 p^1$
iii. ${ }_{14} Si :[ Ne ] 3 s^2 3 p^2$
iv. ${ }_{50} S n:[K r] 4 d^{10} 5 s^2 5 p^2$
v. ${ }_{15} P :[ Ne ] 3 s ^2 3 p ^3$
vi. ${ }_{33}$ As: $[A r] 3 d^{10} 4 s^2 4 p^3$
i. ${ }_{13} Al :[ Ne ] 3 s ^2 3 p^1$
ii. $49 In$ : $[ Kr ] 4 d ^{10} 5 s^2 5 p^1$
iii. ${ }_{14} Si :[ Ne ] 3 s^2 3 p^2$
iv. ${ }_{50} S n:[K r] 4 d^{10} 5 s^2 5 p^2$
v. ${ }_{15} P :[ Ne ] 3 s ^2 3 p ^3$
vi. ${ }_{33}$ As: $[A r] 3 d^{10} 4 s^2 4 p^3$
| Group 13 (Boron family) | Group 14 (Carbon family) | Group 15 (Nitrogen family) | |||||
| Element | Condensed electronic configuration | Element | Condensed electronic configuration | Element | Condensed electronic configuration | ||
| _(sB) | [He]2s^(2)2p^(1) | 6 | [Hc]2s^(2)2p^(2) | _(7)N | [He]2s^(2)2p^(3) | ||
| _(13)Al | [Ne]3s^(2)3p^(-) | _(14)Si | [Ne]3s^(2)3p^(2) | _(15)P | [Ne]3s^(2)3p^(3) | ||
| _(31)Ga | [Ar]3d^(10)4s^(2)4p^(1) | _(32)Ge | [Ar]3d^(10)4s^(2)4p^(2) | _(33)As | [Ar]3d^(10)4s^(2)4p^(3) | ||
| _(49) In | [Kr]4d^(10)5s^(2)5p^(1) | _(50)Sn | [Kr]4d^(10)5s^(2)5p^(2) | _(51)Sb | [Kr]4d^(10)5s^(2)5p^(3) | ||
| _(81)Tl | [Xe]4f^(14)5d^(16)6s^(2)6p^(1) | _(82)Pb | [Xe]4f^(44)5d^(10)6s^(2)6p^(2) | _(83)Bi | [Xe]4f^(4)5d^(10)6s^(2)6p^(3) | ||