Read the passage given below and answer the following questions from 1 to 5. Hydrogen has the simplest atomic structure among all the elements around us in Nature. In atomic form it consists of only one proton and one electron. However, in elemental form it exists as a diatomic (H_2) molecule and is called dihydrogen. It forms more compounds than any other element. Do you know that the global concern related to energy can be overcome to a great extent by the use of hydrogen as a source of energy? In fact, hydrogen is of great industrial importance as you will learn in this unit. Hydrogen is the first element in the periodic table. However, its placement in the periodic table has been a subject of discussion in the past. As you know by now that the elements in the periodic table are arranged according to their electronic configurations. Hydrogen has electronic configuration 1s^1. On one hand, its electronic configuration is similar to the outer electronic configuration (ns^1) of alkali metals , which belong to the first group of the periodic table. On the other hand, like halogens (with ns^2 np^5 configuration belonging to the seventeenth group of the periodic table), it is short by one electron to the corresponding noble gas configuration, helium (1s^2). Hydrogen, therefore, has resemblance to alkali metals, which lose one electron to form unipositive ions, as well as with halogens, which gain one electron to form uninegative ion. Like alkali metals, hydrogen forms oxides, halides and sulphides. However, unlike alkali metals, it has a very high ionization enthalpy and does not possess metallic characteristics under normal conditions. In fact, in terms of ionization enthalpy, hydrogen resembles more with halogens, _iH of Li is 520 kJ mol^ –1 , F is 1680 kJ mol^ –1 and that of H is 1312 kJ mol^ –1 . Like halogens, it forms a diatomic molecule, combines with elements to form hydrides and a large number of covalent compounds. However, in terms of reactivity, it is very low as compared to halogens. Inspite of the fact that hydrogen, to a certain extent resembles both with alkali metals and halogens, it differs from them as well. Now the pertinent question arises as where should it be placed in the periodic table? Loss of the electron from hydrogen atom results in nucleus (H^+) of ~1.5 10^ –3 pm size. This is extremely small as compared to normal atomic and ionic sizes of 50 to 200pm. As a consequence, H^+ does not exist freely and is always associated with other atoms or molecules. Thus, it is unique in behaviour and is, therefore, best placed separately in the periodic table. Occurrence – Dihydrogen H^2 is the most abundant element in the universe (70 % of the total mass of the universe) and is the principal element in the solar atmosphere. The giant planets Jupiter and Saturn consist mostly of hydrogen. However, due to its light nature, it is much less abundant (0.15% by mass) in the earth’s atmosphere. Of course, in the combined form it constitutes 15.4 % of the earth’s crust and the oceans. In the combined form besides in water, it occurs in plant and animal tissues, carbohydrates, proteins, hydrides including hydrocarbons and many other compounds. Hydrogen has three isotopes: protium, 1H, deuterium, 2H or D and tritium, 3H or T. These isotopes differ from one another in respect of the presence of neutrons. Ordinary hydrogen, protium, has no neutrons, deuterium (also known as heavy hydrogen) has one and tritium has two neutrons in the nucleus. In the year 1934, an American scientist, Harold C. Urey, got Nobel Prize for separating hydrogen isotope of mass number 2 by physical methods. The predominant form is protium. Terrestrial hydrogen contains 0.0156% of deuterium mostly in the form of HD. The tritium concentration is about one atom per 1018 atoms of protium. Of these isotopes, only tritium is radioactive and emits low energy -particles(t1 2 12.33 years) 1. Hydrogen has electronic configuration .. 1. 1s^1 2. 1s^2 3. 2s^1 4. 2s^2 2. Ionisation enthalpy of hydrogen is .. 1. 520 kJ mol^ –1 2. 1312 kJ mol^ –1 3. 1249 kJ mol^ –1 4. 950 kJ mol^ –1 3. Hydrogen has … Isotopes. 1. 1 2. 2 3. 3 4. 4 4. got Nobel Prize for separating hydrogen isotope of mass number 2 by physical methods. 1. Nyholm 2. Gillespie 3. Heitler 4. Harold C. Urey 5. tritium is radioactive and emits low energy…. particles. 1. 2. 3. 4.

1. a) 1s^1 2. b) 1312 kJ mol^ –1 3. c) 3 4. d) Harold C. Urey 5. b)

Read the passage given below and answer the following questions f…

The orbital wave function or $\psi$ for an electronin an atom has no physical meaning. It issimply a mathematical function of thecoordinates of the electron. However, fordifferent orbitals the plots of correspondingwave functions as a function of r (the distancefrom the nucleus) are different. According to the German physicist, MaxBorn, the square of the wave function(i.e.,$\psi^2$) at a point gives the probability densityof the electron at that point. Boundary surface diagrams of constantprobability density for different orbitals give afairly good representation of the shapes of theorbitals. In this representation, a boundarysurface or contour surface is drawn in spacefor an orbital on which the value of probabilitydensity $\mid\psi\mid2$ is constant. In principle manysuch boundary surfaces may be possible.However, for a given orbital, only thatboundary surface diagram of constantprobability density* is taken to be goodrepresentation of the shape of the orbital whichencloses a region or volume in which theprobability of finding the electron is very high,say, 90%.
In hydrogen atom, electron has the same energy when it is in the2s orbital as when it is present in 2p orbital.The orbitals having the same energy are calleddegenerate. The 1s orbital in a hydrogenatom, as said earlier, corresponds to the moststable condition and is called the ground stateand an electron residing in this orbital is moststrongly held by the nucleus.
An electron inthe 2s, 2p or higher orbitals in a hydrogen atomis in excited state.The filling of electrons into the orbitals ofdifferent atoms takes place according to theaufbau principle which is based on the Pauli’sexclusion principle, the Hund’s rule ofmaximum multiplicity and the relativeenergies of the orbitals. Theaufbausprinciple states : In the ground state of theatoms, the orbitals are filled in order oftheir increasing energies. In other words,electrons first occupy the lowest energy orbitalavailable to them and enter into higher energyorbitals only after the lower energy orbitals arefilled.The number of electrons to be filled in variousorbitals is restricted by the exclusion principle,given by the Austrian scientist Wolfgang Pauli(1926). According to this principle : No twoelectrons in an atom can have the sameset of four quantum numbers. Pauliexclusion principle can also be stated as : “Onlytwo electrons may exist in the same orbitaland these electrons must have oppositespin.” This means that the two electrons canhave the same value of three quantum numbersn, l and $m_l$, but must have the opposite spinquantum number.Hund’s Rule of Maximum Multiplicity rule deals with the filling of electrons into the orbitals belonging to the same subshell. It states : pairing ofelectrons in the orbitals belonging to thesame subshell (p, d or f) does not take placeuntil each orbital belonging to thatsubshell has got one electron each i.e., itis singly occupied.
The distribution of electrons into orbitals of anatom is called its electronic configuration.If one keeps in mind the basic rules whichgovern the filling of different atomic orbitals,the electronic configurations of different atomscan be written very easily.The electronic configuration of differentatoms can be represented in two ways. Forexample :

$s^a p^bd^c$…… notation
Orbital diagram

…at a point gives the probability density of the electron at that point.

$\psi\times2$
$-\psi^2$
$\psi$
$\psi^2$

Only …. electrons may exist in the same orbital and these electrons must have opposite spin.

One
Two
Three
Four

…deals with the filling of electrons into the orbitals belonging to the same subshell.

Hund’s Rule of Maximum Multiplicity rule
Pauli’s exclusion principle
Aufbau principle
Werner Heisenberg

Electrons first occupy the …. energy orbital available to them and enter into … energy orbitals.

Lowest, Higher
Higher, Lowest
Middle, Higher
Higher, Middle

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Read the passage given below and answer the following questions from $1$ to $5$.
It is prepared by complete combustion of Carbon and carbon containing fuels in excess Of air.
$\text{C(s)}+\text{O}_2\text{(g)}\xrightarrow{\triangle}\text{CO}_2\text{(g)}$
$\text{CH}_4\text{(g)}+2\text{O}_2\text{(g)}\rightarrow\text{CO}_2\text{(g)}+2\text{H}_2\text{O}\text{(g)}$
In the laboratory it is conveniently Prepared by the action of dilute HCl on calcium Carbonate.
$CaCO_3 (s) + 2HCl (aq) \rightarrow CaCl_2 (aq) + CO_2(g) + H_2O(l)$
$\text{H}_2\text{CO}_3(\text{aq})+\text{H}_2\text{O}\text{(l)}\rightleftharpoons\text{HCO}_3^-\text{(aq)}+\text{H}_3\text{O}^+\text{(aq)}$
$\text{H}\text{CO}_3^-(\text{aq})+\text{H}_2\text{O}\text{(l)}\rightleftharpoons\text{CO}_3^{2-}\text{(aq)}+\text{H}_3\text{O}^+\text{(aq)}$
Buffer system helps to Maintain pH of blood between $7.26$ to $7.42$. Being acidic in nature, it combines with alkalies To form metal carbonates. Carbon dioxide, which is normally present To the extent of $\sim0.03 %$ by volume in the Atmosphere, is removed from it by the process Known as photosynthesis. It is the process By which green plants convert atmospheric $CO_2$ into carbohydrates such as glucose. The Overall chemical change can be expressed as:
$6\text{CO}_2+12\text{H}_2\text{O}\xrightarrow[\text{Chlorophyll}]{\text{hv}}\text{C}_6\text{H}_{12}\text{O}_6+6\text{O}_2$
By this process plants make food for Themselves as well as for animals and human Beings. Unlike $CO$, it is not poisonous. But the Increase in combustion of fossil fuels and Decomposition of limestone for cement Manufacture in recent years seem to increase The $CO_2$ content of the atmosphere. This may Lead to increase in green house effect and Thus, raise the temperature of the atmosphere Which might have serious consequences. Carbon dioxide can be obtained as a solid In the form of dry ice by allowing the liquified $CO_2$ to expand rapidly. Dry ice is used as a Refrigerant for ice-cream and frozen food. Gaseous $CO_2$ is extensively used to carbonate Soft drinks. Being heavy and non-supporter Of combustion it is used as fire extinguisher. A Substantial amount of $CO_2$ is used to Manufacture urea. In $CO_2$ molecule carbon atom undergoes Sp hybridisation. Two sp hybridised orbitals Of carbon atom overlap with two p orbitals of Oxygen atoms to make two sigma bonds while Other two electrons of carbon atom are involved.

In $\text{p}\pi-\text{p}\pi$ bonding with oxyglargeen atom. This Results in its linear shape $[$with both $C–O$ bonds Of equal length $(115 pm)]$ with no dipole Moment. The resonance structures are shown Below: Resonance structures of carbon dioxide.
Silicon Dioxide, $SiO_2 95 \%$ of the earth’s crust is made up of silica And silicates. Silicon dioxide, commonly known As silica, occurs in several crystallographic Forms. Quartz, cristobalite and tridymite are Some of the crystalline forms of silica, and they Are interconvertable at suitable temperature. Silicon dioxide is a covalent, three-dimensional network solid in which each silicon atom is Covalently bonded in a tetrahedral manner to Four oxygen atoms. Each oxygen atom in turn Covalently bonded to another silicon atoms as Shown in diagram. Each corner is Shared with another tetrahedron. The entire Crystal may be considered as giant molecule In which eight membered rings are formed with Alternate silicon and oxygen atoms. Silica in its normal form is almost non- Reactive because of very high $Si—O$ bond Enthalpy. It resists the attack by halogens, Dihydrogen and most of the acids and metals Even at elevated temperatures. However, it is Attacked by HF and NaOH.
$SiO_2 + 2NaOH \rightarrow Na2SiO_3 + H_2O SiO_2 + 4HF \rightarrow SiF_4 + 2H_2O$
Quartz is extensively used as a piezoelectric Material; it has made possible to develop extremely Accurate clocks, modern radio and television Broadcasting and mobile radio communications. Silica gel is used as a drying agent and as a support For chromatographic materials and catalysts. Kieselghur, an amorphous form of silica is used In filtration plants.
Silicones are a group of organosilicon polymers, Which have $(R_2SiO)$ as a repeating unit. The Starting materials for the manufacture of Silicones are alkyl or aryl substituted silicon Chlorides, RnSiCl(4–n), where R is alkyl or aryl Group. When methyl chloride reacts with Silicon in the presence of copper as a catalyst At a temperature $573K$ various types of methyl substituted chlorosilane of formula $MeSiCl_3, Me_2SiCl_2, Me3SiCl$ with small amount of $Me4Si$ Are formed. Hydrolysis of dimethyl- Dichlorosilane, $(CH_3) 2SiCl_2$ followed by Condensation polymerisation yields straight Chain polymers.
A large number of silicates minerals exist in Nature. Some of the examples are feldspar, Zeolites, mica and asbestos. The basic structural unit of silicates is $SiO_4^{4–}$ In which silicon atom is bonded to four Oxygen atoms in tetrahedron fashion. In Silicates either the discrete unit is present or A number of such units are joined together Via corners by sharing $1, 2, 3$ or $4$ oxygen Atoms per silicate units. When silicate units Are linked together, they form chain, ring, Sheet or three-dimensional structures. Negative charge on silicate structure is Neutralised by positively charged metal ions. If all the four corners are shared with other Tetrahedral units, three-dimensional network Is formed. Two important man-made silicates are Glass and cement. Zeolites If aluminium atoms replace few silicon atoms In three-dimensional network of silicon dioxide, Overall structure known as aluminosilicate, Acquires a negative charge. Cations such as $Na+, K+$ Or $Ca_2+$ balance the negative charge. Examples are feldspar and zeolites.
Zeolites are Widely used as a catalyst in petrochemical Industries for cracking of hydrocarbons and Isomerisation, e.g., $ZSM-5$ (A type of zeolite) Used to convert alcohols directly into gasoline. Hydrated zeolites are used as ion exchangers In softening of “hard” water.

… is used as a Refrigerant for ice-cream and frozen food.

Dry ice
Wet ice
Crescent Ice
Nugget Ice

$H_2CO_3$ is a …

strong dibasic acid
weak dibasic acid
weak diacidic base
Strong diacidic base

… is extensively used as a piezoelectric Material.

Glass
Ferrite
Quartz
Saphire

… an amorphous form of silica is used In filtration plants.

Ferrite
Quartz
Saphire
Kieselghur

Which of the following is not an example of silicate mineral ?

feldspar
mica
asbestos
hematite

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Read the passage given below and answer the following questions from 1 to 5.
The s-block elements of the Periodic Table are those in which the last electron enters the outermost s-orbital. as the s-orbital can accommodate only two electrons, two Groups (1 & 2) belong to the s-block of the Periodic Table. Group 1 of the Periodic Table consists of the elements: Lithium, sodium, potassium, rubidium, caesium and Francium. They are collectively known as the alkali metals. These are so called because they form hydroxides on Reaction with water which are strongly alkaline in nature. The elements of Group 2 include beryllium, magnesium, Calcium, strontium, barium and radium. These elements With the exception of beryllium are commonly known as The alkaline earth metals. These are so called because their Oxides and hydroxides are alkaline in nature and these Metal oxides are found in the earth’s crust. The general electronic configuration of s-block elements is [noble gas] ns1 for alkali metals and [noble gas] $ns^2$ for Alkaline earth metals.

All the alkali metals have one valence electron, ns1 outside the noble gas core. The loosely held s-electron in the outermost Valence shell of these elements makes them the Most electropositive metals. They readily lose Electron to give monovalent M+ Ions. The monovalent ions (M+) are smaller than the parent atom. Hence they Are never found in free state in nature.
The alkali metal atoms have the largest sizes In a particular period of the periodic table. With increase in atomic number, the atom becomes Larger. the atomic and ionic Radii of alkali metals increase on moving down the group i.e., they increase in size while going From Li to Cs.
The ionization enthalpies of the alkali metals Are considerably low and decrease down the Group from Li to Cs. this is because the effect of increasing size outweighs the increasing Nuclear charge, and the outermost electron is very well screened from the nuclear charge.
The hydration enthalpies of alkali metal ions Decrease with increase in ionic sizes. $Li^+ > Na^+ > K^+ > Rb^+ > Cs^+ > Li^+$ Has maximum degree of hydration and For this reason lithium salts are mostly Hydrated, e.g., $LiCl·2H_2O$.
All the alkali metals are silvery white, soft and Light metals. Because of the large size, these Elements have low density which increases down The group from Li to Cs. However, potassium is Lighter than sodium. The melting and boiling Points of the alkali metals are low indicating Weak metallic bonding due to the presence of Only a single valence electron in them. The alkali Metals and their salts impart characteristic Colour to an oxidizing flame. This is because the Heat from the flame excites the outermost orbital Electron to a higher energy level. When the excited Electron comes back to the ground state . Alkali metals can therefore, be detected by The respective flame tests and can be Determined by flame photometry or atomic Absorption spectroscopy. These elements when Irradiated with light, the light energy absorbed May be sufficient to make an atom lose electron. This property makes caesium and potassium Useful as electrodes in photoelectric cells.
The alkali metals are highly reactive due to Their large size and low ionization enthalpy. The Reactivity of these metals increases down the Group.
Reactivity towards air: The alkali metals Tarnish in dry air due to the formation of Their oxides which in turn react with Moisture to form hydroxides. They burn Vigorously in oxygen forming oxides. Lithium forms monoxide, sodium forms Peroxide, the other metals form Superoxide. The superoxide $O^{2–}$ Ion is Stable only in the presence of large cations Such as K, Rb, Cs.
Reactivity towards water: The alkali Metals react with water to form hydroxide And dihydrogen.
$2\text{M}+2\text{H}_2\text{O}\rightarrow2\text{M}^++2\text{OH}^-+\text{H}_2$
(M = analkali metal)
Reactivity towards dihydrogen: The Alkali metals react with dihydrogen at About 673K (lithium at 1073K) to form Hydrides. All the alkali metal hydrides are Ionic solids with high melting points.
Reactivity towards halogens: The alkali Metals readily react vigorously with Halogens to form ionic halides, $M^+X^–$ .
Reducing nature: The alkali metals are Strong reducing agents, lithium being the Most and sodium the least powerful.
Solutions in liquid ammonia: The alkali Metals dissolve in liquid ammonia giving Deep blue solutions which are conducting in nature.

The general electronic configuration of s-block elements is … for alkali metals.

[noble gas] $ns^1$
[noble gas] $ns^2$
[noble gas] $ns^1np^1$
[noble gas] $ns^1np^2$

The general electronic configuration of s-block elements is … for alkaline earth metals.

noble gas] $ns^1$
[noble gas] $ns^2$
[noble gas] $ns^1np^1$
[noble gas] $ns^1np^2$

The atomic and ionic Radii of alkali metals … on moving down the group.

constant
decrease
increase
All the above

The hydration enthalpies of alkali metal ions … with … in ionic sizes.

increase, decrease
increase, increase
decrease, decrease
decrease, increase

Which of the following element is strong reducing agent?

Lithium
Sodium
Fluorine
Helium

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IUPAC (International Union of Pure and Applied Chemistry) system of nomenclature. Common names are useful and in many cases indispensable, particularly when the alternative systematic names are lengthy and complicated. A systematic name of an organic compound is generally derived by identifying the parent hydrocarbon and the functional group(s) attached to it. By using prefixes and suffixes, the parent name can be modified to obtain the actual name. In a branched-chain compound, small chains of carbon atoms are attached at one or more carbon atoms of the parent chain. The small carbon chains (branches) are called alkyl groups. An alkyl group is derived from a saturated hydrocarbon by removing a hydrogen atom from carbon. Abbreviations are used for some alkyl groups. For example, methyl is abbreviated as Me, ethyl as Et, propyl as Pr and butyl as Bu.

1. Draw the structure of 3-Ethyl-4,4-dimethylheptane. (1)
2. How is the numbering in branched chain hydrocarbon done? (1)
3. Derive the structure of 2-Chlorohexane. (2)
OR
Why $CH _4$ after becoming- $CH _3$ called a methyl group? (2)

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Read the passage given below and answer the following questions from 1 to 5.
The unusual properties of water in the Condensed phase (liquid and solid states) are Due to the presence of extensive hydrogen Bonding between water molecules. This leads To high freezing point, high boiling point, high Heat of vaporisation and high heat of fusion in Comparison to $H_2S$ and $H_2Se$. In comparison To other liquids, water has a higher specific Heat, thermal conductivity, surface tension, Dipole moment and dielectric constant, etc. these properties allow water to play a key role In the biosphere. In the gas phase water is a bent molecule with a bond angle of $104.5^\circ$ , and O–H bond length Of 95.7 pm
It is a highly polar molecule. Its orbital overlap. In the liquid Phase water molecules are associated together By hydrogen bonds. The crystalline form of water is ice. At Atmospheric pressure ice crystallises in the Hexagonal form, but at very low temperatures It condenses to cubic form.
Density of ice is Less than that of water. Therefore, an ice cube Floats on water. In winter season ice formed On the surface of a lake provides thermal Insulation which ensures the survival of the Aquatic life. This fact is of great ecological Significance. Structure of Ice Ice has a highly ordered three dimensional Hydrogen bonded structure. Examination of ice crystals with X-rays shows that each oxygen atom is Surrounded tetrahedrally by four other oxygen Atoms at a distance of 276 pm.
Hydrogen bonding gives ice a rather open Type structure with wide holes. These holes can Hold some other molecules of appropriate size Interstitially.
Water reacts with a large number of Substances. Some of the important reactions Are given below.
Amphoteric Nature: It has the ability to act as an acid as well as a base i.e., it behaves As an amphoteric substance. In the Brönsted Sense it acts as an acid with $NH_3$ and a base with $H_2S.$
$\text{H}_2\text{O}(\text{l})+\text{NH}_3(\text{aq})\rightleftharpoons\text{OH}^-(\text{aq})+\text{NH}^+_4\text{aq}$
$\text{H}_2\text{O}(\text{l})+\text{H}_2\text{S}(\text{aq})\rightleftharpoons\text{H}_3\text{O}^+(\text{aq})+\text{HS}^-\text{(aq)}$
The auto protolysis (self-ionzation) of water takes palace as follow:
$\text{H}_2\text{O}(\text{l})+\text{H}_2\text{O}(\text{l})\rightleftharpoons\text{H}_3\text{O}^+(\text{aq})+\text{OH}^-(\text{aq})$
$\text{acid-1 base-2 (acid-2) base-1}$
$\text{(acid) (base) (conjugate acid) (conjugate base)}$
Redox Reactions Involving Water: Water Can be easily reduced to dihydrogen by highly Electropositive metals.
$2\text{H}_2\text{O}(\text{l})+2\text{Na}\text{(s)}\rightarrow2\text{NaOH}\text{(aq)}+\text{H}_2\text{g}$
Thus. it is a great source of dihydrogen.
water is oxidished to $O_2$ during photosynthesis.
$6\text{CO}_2\text{g}+12\text{H}_2\text{O}(\text{l})\rightarrow\text{C}_6\text{H}_{12}\text{O}_6(\text{aq})+6\text{H}_2\text{O}{\text{l}}+6\text{O}_2\text{(g)}$
With fluorine also it is oxidised to $O_2.$
$2\text{F}_2\text{g}+2\text{H}_2\text{O}(\text{l})\rightarrow4\text{H}^+(\text{aq})+4\text{F}^-(\text{aq})+\text{O}_2\text{(G)}$
Hydrolysis Reaction: Due to high Dielectric constant, it has a very strong Hydrating tendency. It dissolves many ionic Compounds. However, certain covalent and Some ionic compounds are hydrolysed in water.
$\text{P}_4\text{O}_{10}(\text{s})+6\text{H}_2\text{O}(\text{l})\rightarrow4\text{H}_3\text{PO}_4\text{(aq)}$
$\text{SiCl}_4{\text{l}}+2\text{H}_2\text{O}(\text{l})\rightarrow\text{SiO}_2\text{(s)}+4\text{HCl}\text{(aq)}$
Hydrates Formation: From aqueous Solutions many salts can be crystallised as Hydrated salts. Such an association of water Is of different types viz., Coordinated water e.g.,

Hard and Soft Water- Rain water is almost pure (may contain some Dissolved gases from the atmosphere). Being a Good solvent, when it flows on the surface of The earth, it dissolves many salts. Presence of Calcium and magnesium salts in the form of Hydrogencarbonate, chloride and sulphate in Water makes water ‘hard’. Hard water does Not give lather with soap. Water free from Soluble salts of calcium and magnesium is Called Soft water. It gives lather with soap Easily. Temporary hardness is due to the presence of Magnesium and calcium hydrogen- Carbonates. It can be removed by:
Boiling: During boiling, the soluble $Mg(HCO_3)_2$ is converted into insoluble $Mg(OH)_2$ And $Ca(HCO_3)_2$ is changed to insoluble $CaCO_3$. It is because of high solubility product of $Mg(OH)_2$ as compared to that of $MgCO_3$, that $Mg(OH)_2$ is precipitated. These precipitates can Be removed by filtration. Filtrate thus obtained
Will be soft water.
$\text{Mg}(\text{HCO}_3)_2\xrightarrow{\text{Heating}}\text{Mg}(\text{OH})_2\downarrow+2\text{CO}_2\uparrow$
$\text{Ca}(\text{HCO}_3)_2\xrightarrow{\text{Heating}}\text{CaCO}_3\downarrow+\text{H}_2\text{O}+\text{CO}_2\uparrow$
Clark’s method: In this method calculated Amount of lime is added to hard water. It Precipitates out calcium carbonate and Magnesium hydroxide which can be filtered off.
Permanent Hardness is due to the presence of soluble salts of Magnesium and calcium in the form of Chlorides and sulphates in water. Permanent Hardness is not removed by boiling.
$\text{Ca}(\text{Hco}_3)_2+\text{Ca}(\text{OH)}_2\rightarrow2\text{CaCO}_3\downarrow2\text{H}_2\text{O}$
$\text{Mg}(\text{HCO)}_3+2\text{Ca}\text{(Oh)}_2\rightarrow2\text{CaCO}_3\downarrow+\text{Mg}(\text{OH)}_2\downarrow2\text{H}_2\text{O}$
Permanent Hardness is due to the presence of soluble salts of Magnesium and calcium in the form of Chlorides and sulphates in water. Permanent Hardness is not removed by boiling.

In the gas phase water is a bent molecule with a bond angle of:

$104.5^\circ$
$94.5^\circ$
$110.5^\circ$
$95.5^\circ$

At Atmospheric pressure ice crystallises in the … form.

Cubic
Hexagonal
Octagonal
Pentagonal

Water free from Soluble salts of calcium and magnesium is called …

hard water
dry water
soft water
None of above

Water has…. Nature.

acidic
basic
neutral
amphoteric

Water is…. Molecule.

Polar
Non- Polar
Ionic
All the above

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Read the passage given below and answer the following questions from 1 to 5.
Alkaline earth elements have two electrons in the s - orbital of the valence shell. Their general electronic configuration may be represented as [noble gas] $ns^2$. Like alkali metals, the compounds of these elements are also predominantly ionic.

The atomic and ionic radii of the alkaline earth metals are smaller than those of the corresponding alkali metals in the same periods. This is due to the increased nuclear charge in these elements. Within the group, the atomic and ionic radii increase with increase in atomic number.
The alkaline earth metals have low ionization enthalpies due to fairly large size of the atoms. Since the atomic size increases down the group, their ionization enthalpy decreases. The first ionisation enthalpies of the alkaline earth metals are higher than those of the corresponding Group 1 metals. This is due to their small size as compared to the corresponding alkali metals. It is interesting to note that the second ionisation enthalpies of the alkaline earth metals are smaller than those of the corresponding alkali metals.
Like alkali metal ions, the hydration enthalpies of alkaline earth metal ions decrease with increase in ionic size down the group. $Be^{2+}> Mg^2+ > Ca^{2+} > Sr^{2+} > Ba^{2+}$ The hydration enthalpies of alkaline earth metal ions are larger than those of alkali metal ions. Thus, compounds of alkaline earth metals are more extensively hydrated than those of alkali metals, e.g., $MgCl_2$ and $CaCl_2$ exist as $MgCl_{2.}6H_2O$ and $CaCl_2· 6H_2O$ while NaCl and KCl do not form such hydrates.
The alkaline earth metals, in general, are silvery white, lustrous and relatively soft but harder than the alkali metals. Beryllium and magnesium appear to be somewhat greyish. The melting and boiling points of these metals are higher than the corresponding alkali metals due to smaller sizes. The trend is, however, not systematic. Because of the low ionisation enthalpies, they are strongly electropositive in nature. The electropositive character increases down the group from Be to Ba. Calcium, strontium and barium impart characteristic brick red, crimson and apple green colours respectively to the flame. In flame the electrons are excited to higher energy levels and when they drop back to the ground state, energy is emitted in the form of visible light. The electrons in beryllium and magnesium are too strongly bound to get excited by flame. Hence, these elements do not impart any colour to the flame. The flame test for Ca, Sr and Ba is helpful in their detection in qualitative analysis and estimation by flame photometry. The alkaline earth metals like those of alkali metals have high electrical and thermal conductivities which are typical characteristics of metals.
Chemical Properties- The alkaline earth metals are less reactive than the alkali metals. The reactivity of these elements increases on going down the group.
i) Reactivity towards air and water: Beryllium and magnesium are kinetically inert to oxygen and water because of the formation of an oxide film on their surface. Magnesium is more electropositive and burns with dazzling brilliance in air to give $MgO$ and $Mg_3N_2$. Calcium, strontium and barium are readily attacked by air to form the oxide and nitride.
ii) Reactivity towards the halogens: All the alkaline earth metals combine with halogen at elevated temperatures forming their halides.
$M+X_2\rightarrow Mx_2(x=F, Cl, Br, l)$
iii) Reactivity towards hydrogen: All the elements except beryllium combine with hydrogen upon heating to form their hydrides, $MH_2. BeH_2$, however, can be prepared by the reaction of $BeCl_2$ with $LiAlH_4$
2 $BeCl_2+ LiAlH_4\rightarrow 2BeH_2+ LiCl+AlCl_3$
iv) Reactivity towards acids: The alkaline earth metals readily react with acids liberating dihydrogen. $M + 2HCl \rightarrow MCl_2 + H_2$
v) Reducing nature: Like alkali metals, the alkaline earth metals are strong reducing agents. This is indicated by large negative values of their reduction potentials. However their reducing power is less than those of their corresponding alkali metals. Beryllium has less negative value compared to other alkaline earth metals
vi) Solutions in liquid ammonia: Like alkali metals, the alkaline earth metals dissolve in liquid ammonia to give deep blue black solutions forming ammoniated ions.
$\text{M}+(\text{x+y})\text{NH}_3\rightarrow \big[\text{M}(\text{NH}_3)\text{X}\big]^{2+}+2\big[\text{e}(\text{NH}_3)\text{y}\big]^-$
From these solutions, the ammoniates, $\big[\text{M}(\text{NH}_3)6\big]^{2+}$can be recovered.
Beryllium is used in the manufacture of alloys. Copper - beryllium alloys are used in the preparation of high strength springs. Metallic beryllium is used for making windows of X-ray tubes. Magnesium forms alloys with aluminium, zinc, manganese and tin. Magnesium-aluminium alloys being light in mass are used in air-craft construction. Magnesium (powder and ribbon) is used in flash powders and bulbs, incendiary bombs and signals. A suspension of magnesium hydroxide in water (called milk of magnesia) is used as antacid in medicine. Magnesium carbonate is an ingredient of toothpaste. Calcium is used in the extraction of metals from oxides which are difficult to reduce with carbon. Calcium and barium metals, owing to their reactivity with oxygen and nitrogen at elevated temperatures, have often been used to remove air from vacuum tubes. Radium salts are used in radiotherapy, for example, in the treatment of cancer.

The atomic and ionic radii of the alkaline earth metals are … than those of the corresponding alkali metals in the same periods.

smaller
bigger
different
None of above

Within the group, the atomic and ionic radii of alkaline earth metals … with … in atomic number.

increase, decrease
increase, increase
decrease, increase
decrease, decrease

Alkaline earth elements have … electrons in the s -orbital of the valence shell.

Zero
One
Two
Three

Ionization enthalpy …. down the group of alkaline earth metals.

first increases then decreases
first decreases then increases
increase
decreases

The hydration enthalpies of alkaline earth metal ions … with … in ionic size down the group.

increase, decrease
increase, increase
decrease, increase
decrease, decrease

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The ionic character of metallic halides tends toward covalent nature as per Fajan's rule. Such covalent halides behave as non-metal in their higher oxidation states. The property to hydrolyse to give oxy-acids of the element and corresponding hydro halogen acid for most non-metallic elements proceeds exceptionally in the way, keeping oxidation number of element and halide sam in oxo-acids.
Non-polar halides are immiscible in water, as they do not show hydrolysis, but halides of some elements with empty d-orbital undergo hydrolysis. Stability of halides of the higher state is governed by the inert-pair effect.

1. How does halide undergo hydrolysis to give oxy-acids of underlined element $PCl _3$ ? (1)
2. Out of $NCl _3$ and $BCl _3$ undergoes hydrolysis to form oxy-acids? Write the chemical reaction for the correct answer. (1)
3. Out of $PbCl _4, PbF _4, PbI _4$ and $PbBr _4$ which one doesn't exist? (2)
OR
Non-Polar halides are immiscible in water. Why? (2)

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Read the passage given below and answer the following questions from 1 to 5. Uses of Dihydrogen:

The largest single use of dihydrogen is in the synthesis of ammonia which is used in the manufacture of nitric acid and nitrogenous
Dihydrogen is used in the manufacture of vanaspati fat by the hydrogenation of polyunsaturated vegetable oils like soyabean, cotton seeds
It is used in the manufacture of bulk organic chemicals, particularly
It is widely used for the manufacture of metal
It is used for the preparation of hydrogen chloride, a highly useful
In metallurgical processes, it is used to reduce heavy metal oxides to
Atomic hydrogen and oxy-hydrogen torches find use for cutting and welding purposes. Atomic hydrogen atoms (produced by dissociation of dihydrogen with the help of an electric arc) are allowed to recombine on the surface to be welded to generate the temperature of 4000
It is used as a rocket fuel in space
Dihydrogen is used in fuel cells for generating electrical energy. It has many advantages over the conventional fossil fuels and electric It does not produce any pollution and releases greater energy per unit mass of fuel in comparison to gasoline and other fuels.

Dihydrogen, under certain reaction conditions, combines with almost all elements, except noble gases, to form binary compounds, called hydrides. If ‘E’ is the symbol of an element then hydride can be expressed as EHx $($e.g., Mg $H_2)$ or EmHn $($e.g.,$B_2H_6)$. The hydrides are classified into three categories:

Ionic or saline or saltlike hydrides
Covalent or molecular hydrides
Metallic or non-stoichiometric hydrides

Ionic or Saline Hydrides are stoichiometric compounds of dihydrogen formed with most of the s-block elements which are highly electropositive in character. However, significant covalent character is found in the lighter metal hydrides such as LiH, $BeH_2$ and $MgH_2$. In fact Be $H_2$ and Mg $H_2$ are polymeric in structure. The ionic hydrides are crystalline, non-volatile and non- conducting in solid state. However, their melts conduct electricity and on electrolysis liberate dihydrogen gas at anode, which confirms the existence of $H^{–} ion$. Covalent or Molecular Hydride Dihydrogen forms molecular compounds with most of the p-block elements. Most familiar examples are $CH_4, NH_3, H_2O$ and $HF.$ For convenience hydrogen compounds of non- metals have also been considered as hydrides. Being covalent, they are volatile compounds. Molecular hydrides are further classified according to the relative numbers of electrons and bonds in their Lewis structure into:

electron-deficient,
electron-precise, and
electron-rich

An electron-deficient hydride, as the name suggests, has too few electrons for writing its conventional Lewis structure. Diborane $(B_2H_6)$ is an example. In fact all elements of group 13 will form electron-deficient compounds. They act as Lewis acids i.e., electron acceptors. Electron-precise compounds have the required number of electrons to write their conventional Lewis structures. All elements of group 14 form such compounds (e.g., $CH_4$) which are tetrahedral in geometry. Electron-rich hydrides have excess electrons which are present as lone pairs. Elements of group 15-17 form such compounds. (NH3 has 1 - lone pair, $H_2O – 2$ and $HF –3$ lone pairs). What do you expect from the behaviour of such compounds ? They will behave as Lewis bases i.e., electron donors. The presence of lone pairs on highly electronegative atoms like N, O and F in hydrides results in hydrogen bond formation between the molecules. This leads to the association of molecules. Metallic or Non-stoichiometric (or Interstitial ) Hydrides are formed by many d- block and f-block elements. However, the metals of group 7, 8 and 9 do not form hydride. Even from group 6, only chromium forms CrH. These hydrides conduct heat and electricity though not as efficiently as their parent metals do. Unlike saline hydrides, they are almost always non- stoichiometric, being deficient in hydrogen. For example, $La H_{2.87}, Yb H_{2.55}, TiH1_{.5–1.8}, ZrH_{1.3–1.75}$, etc. In such hydrides, the law of constant composition does not hold good. Earlier it was thought that in these hydrides, hydrogen occupies interstices in the metal lattice producing distortion without any change in its type. Consequently, they were termed as interstitial hydrides. However, recent studies have shown that except for hydrides of Ni, Pd, Ce and Ac, other hydrides of this class have lattice different from that of the parent metal. The property of absorption of hydrogen on transition metals is widely used in catalytic reduction / hydrogenation reactions for the preparation of large number of compounds. Some of the metals (e.g., Pd, Pt) can accommodate a very large volume of hydrogen and, therefore, can be used as its storage media. This property has high potential for hydrogen storage and as a source of energy. A major part of all living organisms is made up of water. Human body has about 65% and some plants have as much as 95% water. It is a crucial compound for the survival of all life forms. It is a solvent of great importance. The distribution of water over the earth’s surface is not uniform.

Dihydrogen, under certain reaction conditions, combines with almost all elements, except …

Noble gases
Halogens
Alkali metals
Alkaline earth metal

Covalent or Molecular Hydride Dihydrogen forms molecular compounds with most of the p-block elements. Most familiar example is:

$CH_4$
$NH_3$
$H_2O$
All the above

All elements of group 14 form such compounds have … geometry.

pyramidal
tetrahedral
bilateral
spherical

From group 6, only … forms hydride.

molybdenum
tungsten
chromium
seaborgium

Which of the following hydride is/ are deficient in hydrogen.

$La H_2._{87}$
$Yb H_2._{55}$
TiH5–1.8
All of above

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The idea of oxidation number has been invariably applied to define oxidation, reduction, oxidising agent (oxidant), reducing agent (reductant) and the redox reaction. To summarise, we may say that:
Oxidation: An increase in the oxidation number of the element in the given substance.
Reduction: A decrease in the oxidation number of the element in the given substance.
Oxidising agent: A reagent which can increase the oxidation number of an element in a given substance. These reagents are called as oxidants also.
Reducing agent: A reagent which lowers the oxidation number of an element in a given substance. These reagents are also called as reductants.
Redox reactions: Reactions which involve change in oxidation number of the interacting species.
Types of Redox Reactions
1.) Combination reactions -A combination reaction may be denoted in the manner:
$A + B → C$
Either A and B or both A and B must be in the elemental form for such a reaction to be a redox reaction. All combustion reactions, which make use of elemental dioxygen, as well as other reactions involving elements other than dioxygen, are redox reactions. Some important examples of this category are:

2.) Decomposition reactions- Decomposition reactions are the opposite of combination reactions. Precisely, a decomposition reaction leads to the breakdown of a compound into two or more components at least one of which must be in the elemental state.
Examples of this class of reactions are:

It may carefully be noted that there is no change in the oxidation number of hydrogen in methane under combination reactions and that of potassium in potassium chlorate in reaction. This may also be noted here that all decomposition reactions are not redox reactions. For example, decomposition of calcium carbonate is not a redox reaction.
3.) Displacement reactions- In a displacement reaction, an ion (or an atom) in a compound is replaced by an ion (or an atom) of another element. It may be denoted as:
$X + YZ → XZ + Y$
Displacement reactions fit into two categories: metal displacement and non-metal displacement.
(a) Metal displacement: A metal in a compound can be displaced by another metal in the uncombined state. Metal displacement reactions find many applications in metallurgical processes in which pure metals are obtained from their compounds in ores.
(b) Non-metal displacement: The non-metal displacement redox reactions include hydrogen displacement and a rarely occurring reaction involving oxygen displacement. All alkali metals and some alkaline earth metals (Ca, Sr, and Ba) which are very good reductants, will displace hydrogen from cold water. Many metals, including those which do not react with cold water, are capable of displacing hydrogen from acids. Dihydrogen from acids may even be produced by such metals which do not react with steam. Cadmium and tin are the examples of such metals.
4.) Disproportionation reactions – Disproportionation reactions are a special type of redox reactions. In a disproportionation reaction an element in one oxidation state is simultaneously oxidised and reduced. One of the reacting substances in a disproportionation reaction always contains an element that can exist in at least three oxidation states. The element in the form of reacting substance is in the intermediate oxidation state; and both higher and lower oxidation states of that element are formed in the reaction. The decomposition of hydrogen peroxide is a familiar example of the reaction, where oxygen experiences disproportionation.

Here the oxygen of peroxide, which is present in –1 state, is converted to zero oxidation state in $O2$ and decreases to –2 oxidation state in $H_2O$.

In … an ion (or an atom) in a compound is replaced by an ion (or an atom) of another element.

displacement reaction
decomposition reaction
disproportionation reaction
combination reaction

leads to the breakdown of a compound into two or more components at least one of which must be in the elemental state.

displacement reaction
decomposition reaction
disproportionation reaction
combination reaction

In …. an element in one oxidation state is simultaneously oxidised and reduced.

displacement reaction
decomposition reaction
disproportionation reaction
combination reaction

Reactions which involve change in oxidation number of the interacting species…

Exothermic reaction
Endothermic reaction
Neutralization reaction
Redox reaction

One of the reacting substances in a disproportionation reaction always contains an element that can exist in at least … oxidation states.

1
2
3
4

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Read the passage given below and answer the following questions from $(i)$ to $(v).$
Predicting the Direction of the Reaction- The equilibrium constant helps in predicting the direction in which a given reaction will proceed at any stage. For this purpose, we calculate the reaction quotient $Q$. The reaction quotient, $Q (Qc$ with molar concentrations and $QP$ with partial pressures$)$ is defined in the same way as the equilibrium constant Kc except that the concentrations in $Qc$ are not necessarily equilibrium values. For a general reaction:
$\text{a}\text{A}+\text{b}\text{B}\rightleftharpoons\text{c}\text{C}+\text{d}\text{D}$
$\text{Q}\text{c}=\frac{[\text{C}]^\text{c}[\text{D}]^\text{d}}{[\text{A}]^\text{a}[\text{B}]^\text{b}}$
Then,
If $Qc > Kc,$ the reaction will proceed in the direction of reactants (reverse reaction).
If $Qc < Kc,$ the reaction will proceed in the direction of the products (forward reaction).
If $Qc = Kc,$ the reaction mixture is already at equilibrium. Consider the gaseous reaction of $H_2$ with $I_2$ ,
$\text{H}_{2(\text{g})}+\text{l}_{2(\text{g})}\rightleftharpoons2\text{Hl}_{(\text{g})};\text{kc}=57.0\text{at}700\text{k}.$
Suppose we have molar concentrations $[H_2 ]t =0.10M, [I_2 ]t = 0.20 M$ and $[HI]t = 0.40 M.$ (the subscript t on the concentration symbols means that the concentrations were measured at some arbitrary time t, not necessarily at equilibrium). Thus, the reaction quotient, Qc at this stage of the reaction is given by,
$\text{Qc}=\frac{[\text{Hl}]\text{t}^2}{[\text{H}]^2]_\text{t}[\text{l}_2]_\text{t}}=\frac{(0.40)_2}{(0.10)\times(0.20)}=8.0$
Now, in this case, $Qc (8.0)$ does not equal Kc $(57.0)$, so the mixture of $H2 _{(g)}, I2 _{(g)} $ and $HI_{(g)} $ is not at equilibrium; that is, more $H2 _{(g)} $ and $I 2 _{(g)} $ will react to form more $HI_{(g)} $ and their concentrations will decrease till $Qc = Kc.$ The reaction quotient, Qc is useful in predicting the direction of reaction by comparing the values of Qc and Kc.Thus, we can make the following generalisations concerning the direction of the reaction
If $Qc < Kc,$ net reaction goes from left to right
If $Qc > Kc,$ net reaction goes from right to left.
If $Qc = Kc,$ no net reaction occurs.
Calculating Equilibrium Concentrations In case of a problem in which we know the initial concentrations but do not know any of the equilibrium concentrations, the following three steps shall be followed:
Step 1) Write the balanced equation for the reaction.
Step 2) Under the balanced equation, make a table that lists for each substance involved in the reaction: $(a)$ the initial concentration, $(b)$ the change in concentration on going to equilibrium, and $(c)$ the equilibrium concentration. In constructing the table, define $x$ as the concentration (mol/L) of one of the substances that reacts on going to equilibrium, then use the stoichiometry of the reaction to determine the concentrations of the other substances in terms of $x.$
Step 3) Substitute the equilibrium concentrations into the equilibrium equation for the reaction and solve for x. If you are to solve a quadratic equation choose the mathematical solution that makes chemical sense.
Step 4) Calculate the equilibrium concentrations from the calculated value of $x.$
Step 5) Check your results by substituting them into the equilibrium equation.
Relationship between equilibrium constant K, reaction quotient Q and gibbs energy G The value of Kc for a reaction does not depend on the rate of the reaction. However, it is directly related to the thermodynamics of the reaction and in particular, to the change in Gibbs energy, $\triangle\text{G}.$ If,
$\triangle\text{G}$ is negative, then the reaction is spontaneous and proceeds in the forward direction.
$\triangle\text{G}$ is positive, then reaction is considered non-spontaneous. Instead, as reverse reaction would have a negative $\triangle\text{G},$ the products of the forward reaction shall be converted to the reactants.
$\triangle\text{G}$ is 0, reaction has achieved equilibrium; at this point, there is no longer any free energy left to drive the reaction. A mathematical expression of this thermodynamic view of equilibrium can be described by the following equation:
$\triangle\text{G}=\triangle\text{G}^\phi+\text{RT}\text{lnQ}$
where, $\triangle\text{G}^\phi$ is standard Gibbs energy. At equilibrium, when $\triangle\text{G}=0$ and $Q = Kc,$ the equation becomes,
$\triangle\text{G}=\text{G}^\phi+\text{RT}\text{lnk}=0$
$\triangle\text{G}^\phi=-\text{RT}\text{lnk}$
$\text{Ink}=\frac{-\triangle\text{G}^\phi}{\text{RT}}$
Taking antilog of both sides, we get,
$\text{K}=\text{e}-\frac{\triangle\text{G}0}{\text{RT}}$
Hence, using the equation, the reaction spontaneity can be interpreted in terms of the value of $\triangle\text{G}^\phi.$
If $\triangle\text{G}^\phi>0$ then $\frac{-\triangle\text{G}^\phi}{\text{RT}}$ is positive, and $>1$, making $K > 1$, which implies a spontaneous reaction or the reaction which proceeds in the forward direction to such an extent that the products are present predominantly.
If $\triangle\text{G}^\phi>0,$ then $\frac{-\triangle\text{G}^\phi}{\text{RT}}$ is negative, and $< 1$, that is, $K < 1$, which implies a non-spontaneous reaction or a reaction which proceeds in the forward direction to such a small degree that only a very minute quantity of product is formed.
Factors affecting equilibria One of the principal goals of chemical synthesis is to maximise the conversion of the reactants to products while minimizing the expenditure of energy. This implies maximum yield of products at mild temperature and pressure conditions. If it does not happen, then the experimental conditions need to be adjusted. For example, in the Haber process for the synthesis of ammonia from $N_2$ and $H_2,$ the choice of experimental conditions is of real economic importance. Annual world production of ammonia is about hundred million tones, primarily for use as fertilizers. Equilibrium constant, Kc is independent of initial concentrations. But if a system at equilibrium is subjected to a change in the concentration of one or more of the reacting substances, then the system is no longer at equilibrium; and net reaction takes place in some direction until the system returns to equilibrium once again. Similarly, a change in temperature or pressure of the system may also alter the equilibrium. In order to decide what course the reaction adopts and make a qualitative prediction about the effect of a change in conditions on equilibrium we use Le Chatelier’s principle. It states that a change in any of the factors that determine the equilibrium conditions of a system will cause the system to change in such a manner so as to reduce or to counteract the effect of the change. This is applicable to all physical and chemical equilibria.
Effect of Concentration Change In general, when equilibrium is disturbed by the addition/removal of any reactant/ products, Le Chatelier’s principle predicts that:
The concentration stress of an added reactant/product is relieved by net reaction in the direction that consumes the added substance.
The concentration stress of a removed reactant/product is relieved by net reaction in the direction that replenishes the removed substance. or in other words, “When the concentration of any of the reactants or products in a reaction at equilibrium is changed, the composition of the equilibrium mixture changes so as to minimize the effect of concentration changes”. Let us take the reaction,
$\text{H}_{2(\text{g})}+\text{l}_{2(\text{g})}\rightleftharpoons2\text{Hl}_{(\text{g})}$
If $H_2$ is added to the reaction mixture at equilibrium, then the equilibrium of the reaction is disturbed. In order to restore it, the reaction proceeds in a direction wherein $H_2$ is consumed, i.e., more of $H_2$ and $I_2$ react to form HI and finally the equilibrium shifts in right (forward) direction. This is in accordance with the Le Chatelier’s principle which implies that in case of addition of a reactant/product, a new equilibrium will be set up in which the concentration of the reactant/product should be less than what it was after the addition but more than what it was in the original mixture. The same point can be explained in terms of the reaction quotient, $Qc,$

Case Study Questions Class 11 Chemistry – Equilibrium

$\text{Qc}=\frac{[\text{HI}]^2}{[\text{H}]_2[\text{I}]_2}$
Addition of hydrogen at equilibrium results in value of $Qc$ being less than $Kc$. Thus, in order to attain equilibrium again reaction moves in the forward direction. Similarly, we can say that removal of a product also boosts the forward reaction and increases the concentration of the products and this has great commercial application in cases of reactions, where the product is a gas or a volatile substance. In case of manufacture of ammonia, ammonia is liquified and removed from the reaction mixture so that reaction keeps moving in forward direction. Similarly, in the large scale production of $CaO$ (used as important building material) from $CaCO_3$, constant removal of $CO_2$ from the kiln drives the reaction to completion. It should be remembered that continuous removal of a product maintains Qc at a value less than Kc and reaction continues to move in the forward direction.

If … the reaction will proceed in the direction of reactants (reverse reaction).

$Qc > Kc$
$Qc < Kc$
$Qc = Kc$
None of above

If … the reaction will proceed in the direction of the products (forward reaction).

$Qc > Kc$
$Qc < Kc$
$Qc = Kc$
None of above

If … the reaction mixture is already at equilibrium. Consider the gaseous reaction.

$Qc > Kc$
$Qc < Kc$
$Qc = Kc$
All of above

If $\triangle\text{G}$ is …. then the reaction is spontaneous and proceeds in the forward direction.

Zero
Positive
Negative
None of above

$\triangle\text{G}$ is … reaction has achieved equilibrium; at this point, there is no longer any free energy left to drive the reaction.

Zero
Positive
Negative
None of above

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