Chemistry

Lifetimes of the molecules in the excited states are often measured by using pulsed radiation source of duration nearly in the nanosecond range. If the radiation source has the duration of 2 \mathrm{~ns} and the number of photons emitted during the pulse source is 2.5 \times 10^{15} J, calculate the energy of the source.

Frequency of radiation $(\nu)$, $\nu=\frac{1}{2.0 \times 10^{-9} s}$ $\nu=5.0 \times 10^{8} s^{-1}$ Energy $(E)$ of source $=$ Nhv Where, $N$ is the no. photons emitted $\mathrm{h}$ is Planck's...

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What is the energy in joules, required to shift the electron of the hydrogen atom from the first Bohr orbit to the fifth Bohr orbit and what is the wavelength of the light emitted when the electron returns to the ground state? The ground state electron energy is -2.18\times10^{-11} ergs. The ground-state electron energy is -2.18\times10^{-11}ergs.

$ E_{5}=\frac{-\left(2.18 \times 10^{-18}\right) Z^{2}}{(n)^{2}} $ Where, $Z$ denotes the atom's atomic number Ground state energy $=-2.18 \times 10^{-11}$ ergs $=-2.18 \times 10^{-11} \times...

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The quantum numbers of six electrons are given below. Arrange them in order of increasing energies. If any of these combination(s) has/have the same energy lists: n = 4, l = 2, ml = –2 , ms = –1/2 n = 3, l = 2, ml= 1 , ms = +1/2 n = 4, l = 1, ml = 0 , ms = +1/2 n = 3, l = 2, ml = –2 , ms = –1/2 n = 3, l = 1, ml = –1 , ms= +1/2 n = 4, l = 1, ml = 0 , ms = +1/2

The 4d, 3d, 4p, 3d, 3p, and 4p orbitals are home to electrons 1, 2, 3, 4, 5, and 6. (respectively). Ranking these orbitals in the increasing order of energies: (3p) < (3d) < (4p) < (4d).

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A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives 3.38 g carbon dioxide, 0.690 g of water and no other products. A volume of 10.0 L (measured at STP) of this welding gas is found to weigh 11.6 g. Find: (i) Empirical formula (ii) Molar mass of the gas, and (iii) Molecular formula

(i) Empirical formula 1 mole of $CO_{ 2 }$ contains 12 g of carbon Therefore, 3.38 g of $CO_{ 2 }$ will contain carbon = $\frac{ 12 \; g }{ 44 \; g } \; \times 3.38 \; g$ = 0.9217 g   18 g of...

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Pressure versus volume graph for a real gas and an ideal gas is shown in Fig. 5.4. Answer the following questions based on this graph. (i) Interpret the behaviour of real gas with respect to an ideal gas at low pressure. (ii) Interpret the behaviour of real gas with respect to an ideal gas at high pressure. (iii)Mark the pressure and volume by drawing a line at the point where real gas behaves as an ideal gas.

(i) At low pressure as the dark blue curve and the sky blue curve are approaching each other, it shows that the real gas is behaving as an ideal gas at a low pressure. (ii) At high pressure as the...

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The variation of pressure with the volume of the gas at different temperatures can be graphically represented as shown in Fig. 5.3. Based on this graph answer the following questions. (i) How will the volume of a gas change if its pressure is increased at constant temperature? (ii) At constant pressure, how will the volume of a gas change if the temperature is increased from 200K to 400K?

(i) As the temperature is constant, and the pressure is increasing and the change in the volume is seen as exponentially decreasing. (ii) At constant pressure, by increasing the temperature there is...

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The relation between the pressure exerted by an ideal gas (Pideal) and observed pressure (Pearl) is given by the equation: Pideal = Preal+ an2/V2 If the pressure is taken in Nm-2, the number of moles in mol and volume in m3, Calculate the unit of ‘a’. What will be the unit of ‘a’ when pressure is in atmosphere and volume in dm3?

We know that: Pideal = Preal + an2/V2 Pideal – Preal= an2/V2 Nm-2 = a*mol2/m6 A = Nm4mol-2 The unit of ‘a’ when the pressure is taken in Nm-2, number of moles in “mol” and volume in m3 is Nm4mol-2...

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For real gases the relation between p, V and T are given by van der Waals equation: [(P + an2) / V2](V – nb) = nRT Where‘a’ and ‘b’ are van der Waals constants, ‘nb’ is approximately equal to the total volume of the molecules of a gas. ‘a’ is the measure of the magnitude of intermolecular attraction. (i) Arrange the following gases in the increasing order of ‘b’. Give reason. O2, CO2, H2, He (ii) Arrange the following gases in the decreasing order of magnitude of ‘a’. Give reason. CH4, O2, H2

(i) The increasing order of ‘b’ is as follows: He < H2< O2< CO2. As the Vander Waals constant ‘b’ is approximately equal to the total volume of the molecules of a gas. (ii)The decreasing...

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Gases possess characteristic critical temperature which depends upon the magnitude of intermolecular forces between the particles. Following are the critical temperatures of some gases. Gases H2 He O2 N2 Critical temperature in Kelvin 33.2 5.3 154.3 126 From the above data what would be the order of liquefaction of these gases? Start writing the order from the gas liquefying first (i) H2, He, O2, N2 (ii) He, O2, H2, N2 (iii) N2, O2, He, H2 (iv) O2, N2, H2, He

The correct option is (iv) O2, N2, H2, He.

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If a liquid compound decomposes at its boiling point, which method(s) can you choose for its purification. It is known that the compound is stable at low pressure, steam volatile and insoluble in water.

Because the liquid component decomposes near its boiling point, indicating that it is heat-sensitive, we purify it using "Steam distillation." For temperature-sensitive materials, this is done.

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During the hearing of a court case, the judge suspected that some changes in the documents had been carried out. He asked the forensic department to check the ink used at two different places. According to you which technique can give the best results? (i) Column chromatography (ii) Solvent extraction (iii) Distillation (iv) Thin-layer chromatography

Option IV is the correct answer.

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Assertion (A): If aluminium atoms replace a few silicon atoms in three the dimensional network of silicon dioxide, the overall structure acquires a negative charge. Reason (R): Aluminium is trivalent while silicon is tetravalent. (i) Both A and R are correct and R is the correct explanation of A. (ii) Both A and R are correct but R is not the correct explanation of A. (iii) Both A and R are not correct (iv) A is not correct but R is correct.

Correct Option is (i)

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Assertion (A): The black body is an ideal body that emits and absorbs radiations of all frequencies. Reason (R): The frequency of radiation emitted by a body goes from a lower frequency to higher frequency with an increase in temperature.

(i) Both A and R are true and R is the correct explanation of A. (ii) Both A and R are true but R is not the explanation of A. (iii) A is true and R is false. (iv) Both A and R are false.  ...

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According to de Broglie, the matter should exhibit dual behavior that is both particle and wave-like properties. However, a cricket ball of mass 100 g does not move like a wave when it is thrown by a bowler at a speed of 100 km/h. Calculate the wavelength of the ball and explain why it does not show wave nature.

Calculation: Given, Mass, m = 100g / 0.1kg Velocity = 100km/h Velocity =100×1000 / 60×60 Velocity = 1000/36m/s λ =h/mν λ = 2.387 × 10-34 m

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The Balmer series in the hydrogen spectrum corresponds to the transition from n1 = 2 to n2 = 3,4,………. This series lies in the visible region. Calculate the wavenumber of the line associated with the transition in Balmer series when the electron moves to n = 4 orbit. (RH= 109677 cm-1)

Calculation: According to Bohr’s model for the hydrogen atom; ν = RH(1/n12-1/ n22)cm-1 Given, n1 = 2 n2 = 4 H (Rydberg’s constant) = 109677 Wave number = 109677 ( ¼-1/16) Hence, Wave number =...

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A vessel contains 1.6 g of dioxygen at STP (273.15K, 1 atm pressure). The gas is now transferred to another vessel at a constant temperature, where the pressure becomes half of the original pressure. Calculate: (i) the volume of the new vessel. (ii) a number of molecules of dioxygen.

(i) Calculation: Moles of oxygen = 1.6/32 Moles of oxygen = 0.05mol 1 mol of oxygen= 22.4L (at STP) Volume of Oxygen (V1) = 22.4 × 0.05 Volume of Oxygen (V1) = 1.12L V2 =? P1 = 1atm P2 = ½ P2 =...

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Assertion (A): One atomic mass unit is defined as one-twelfth of the mass of one carbon-12 atom. Reason (R): Carbon-12 isotope is the most abundant isotope of carbon and has been chosen as the standard. (i) Both A and R are true and R is the correct explanation of A. (ii) Both A and R are true but R is not the correct explanation of A. (iii) A is true but R is false. (iv) Both A and R are false.

Correct Answer: (ii) Both A and R are true but R is not the correct explanation of A Explanation: The carbon 12 isotope defines the mass of atoms and molecules.

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Assertion (A): The empirical mass of ethene is half of its molecular mass. Reason (R): The empirical formula represents the simplest whole-number the ratio of various atoms present in a compound. (i) Both A and R are true and R is the correct explanation of A. (ii) A is true but R is false. (iii) A is false but R is true. (iv) Both A and R are false.

Correct Answer: (i) Both A and R are true and R is the correct explanation of A Explanation: The empirical formula represents the simplest whole-number the ratio of various atoms present in a...

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If two elements can combine to form more than one compound, the masses of one element that combine with a fixed mass of the other element, are in whole-number ratio. (a) Is this statement true? (b) If yes, according to which law? (c) Give one example related to this law.

(a) If two elements can combine to form more than one compound, the masses of one element that combine with a fixed mass of the other element, are in whole-number ratio and this statement is true....

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One of the statements of Dalton’s atomic theory is given below: “Compounds are formed when atoms of different elements combine in a fixed ratio” Which of the following laws is not related to this statement? (i) Law of conservation of mass (ii) Law of definite proportions (iii) Law of multiple proportions (iv) Avogadro’s law

Correct Answers: (i) Law of conservation of mass; (iv) Avogadro's law Explanation: According to the Dalton's atomic theory, The Chemical compounds are formed when atoms of various elements join in a...

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Which of the following statements is correct about the reaction given below: 4Fe(s) + 3O2(g) → 2Fe2O3(g) (i) The total mass of iron and oxygen in reactants = total mass of iron and oxygen in product therefore it follows the law of conservation of mass. (ii) The total mass of reactants = total mass of product; therefore, the law of multiple proportions is followed. (iii) Amount of Fe2O3 can be increased by taking any one of the reactants (iron or oxygen) in excess. (iv) Amount of Fe2O3 produced will decrease if the amount of any one of the reactants (iron or oxygen) is taken in excess.

Correct Answer: (i) The total mass of iron and oxygen in reactants = total mass of iron and oxygen in product therefore it follows the law of conservation of mass. Explanation: From the reaction,...

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Which of the following statements about a compound is incorrect? (i) A molecule of a compound has atoms of different elements. (ii) A compound cannot be separated into its constituent elements by physical methods of separation. (iii) A compound retains the physical properties of its constituent elements. (iv) The ratio of atoms of different elements in a compound is fixed.

Correct Answer: (iii) A compound retains the physical properties of its constituent elements Explanation: Molecule of a compound is made up of atoms of various elements which cannot be separated...

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One mole of any substance contains 6.022 × 1023 atoms/molecules. Number of molecules of H2SO4 present in 100 mL of 0.02M H2SO4 solution is ______. (i) 12.044 × 1020 molecules (ii) 6.022 × 1023 molecules (iii) 1 × 1023 molecules (iv) 12.044 × 1023molecules

Correct Answer: (i) 12.044 × 1020 molecules Explanation: Moles of H2SO4​= Molarity of H2SO4​×Volume of solution (L) Hence, the number of molecules of H2SO4 present in 100 mL of 0.02M H2SO4 solution...

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Hydrogen bonds are formed in many compounds e.g., H2O, HF, NH3. The boiling point of such compounds depends to a large extent on the strength of hydrogen bond and the number of hydrogen bonds. The correct decreasing order of the boiling points of the above compounds is : (i) HF > H2O > NH3 (ii) H2O > HF > NH3 (iii) NH3 > HF > H2O (iv) NH3 > H2O > HF

Solution: Option (ii) is the answer.

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Assertion (A): Electron gain enthalpy becomes less negative as we go down a group.
Reason (R): Size of the atom increases on going down the group and the added electron would be farther from the nucleus.
(a) Assertion and reason both are correct statements but reason is not correct explanation of assertion.
(b) Assertion and reason both are correct statements and reason is correct explanation of assertion.
(c) Assertion and reason both are wrong statement.
(d) Assertion is wrong statement but reason is correct statement.

(b) As one moves down the group, the electron gain enthalpy decreases because the atomic size grows and the new electron is further away from the nucleus.

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Assertion (A): Boron has a smaller first ionization enthalpy than beryllium. Reason (R): The penetration of a 2s electron to the nucleus is more than the 2p electron, hence, 2p electron is more shielded by the inner core of electrons that the 2s electrons.
(a) Assertion and reason both are correct statements but reason is not correct explanation of assertion.
(b) Assertion is correct statement but reason is wrong statement.
(c) Assertion and reason both are correct statements and reason is correct explanation of assertion.
(d) Assertion and reason both are wrong statement.

(c) Because beryllium (1s2 2s2) has a fully filled, boron (1s2 2s2 2p1) has a lower initial ionisation enthalpy than beryllium (1s2 2s2). s-subshell. When compared to 2s-electrons, 2s-electrons are...

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Assertion and Reason Type Questions
In the following questions a statement of Assertion (A) followed by a statement of Reason (R) is given.
Choose the correct option out of the choices given below each question.
Assertion (A): Generally, ionization enthalpy increases from left to right in a period.
Reason (R): When successive electrons are added to the orbitals in the same principal quantum level, the shielding effect of inner core of electrons does not increase very much to compensate for the increased attraction of the electron to the nucleus.
(a) Assertion is correct statement and reason is wrong statement.
(b) Assertion and reason both are correct statements and reason is correct explanation of assertion.
(c) Assertion and reason both are wrong statements.
(d) Assertion is wrong statement and reason is correct statement.

(b) As atomic size decreases, the ionization enthalpy increases from left to right over time. The effective nuclear charge of the electrons in the subshell is about the same.

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Polarity in a molecule and hence the dipole moment depends primarily on electronegativity of the constituent atoms and shape of a molecule. Which of Does the following have the highest dipole moment? (i) CO2 (ii) HI (iii) H2O (iv) SO2

H2O has a high dipole moment because Oxygen is highly electronegative. This attracts the electron from Hydrogen towards it with greater charge. Therefore, H2O...

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The statement that is not correct for periodic classification of elements is:
(i) The properties of elements are periodic function of their atomic numbers.
(ii) Non-metallic elements are less in number than metallic elements.
(iii) For transition elements, the 3d-orbitals are filled with electrons after3p-orbitals and before 4s-orbitals.
(iv) The first ionisation enthalpies of elements generally increase withincrease in atomic number as we go along a period.

Option (iii) is the answer. The Aufbau principle describes how electrons first fill low-energy orbitals (near to the nucleus) before moving on to higher-energy orbitals. They fill the orbitals...

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Ionisation enthalpies of elements of the second period are given below: Ionisation enthalpy/ kcal mol–1: 520, 899, 801, 1086, 1402, 1314, 1681, 2080. Match the correct enthalpy with the elements and complete the graph given in Fig. 3.1. Also, write symbols of elements with their atomic number.

Solution: N has a higher first ionisation enthalpy than O, despite the fact that O has a higher nuclear charge. This is because the electron in N must be removed from a more stable, exactly...

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An element belongs to the 3rd period and group-13 of the periodic table. Which of the following properties will be shown by the element?
(i) Good conductor of electricity
(ii) Liquid, metallic
(iii) Solid, metallic
(iv) Solid, non-metallic

Option (i) and (iii) are the answers. The element belonging to 3rd period and 13th group is aluminium which is a metal. Hence, it is solid, metallic and good conductor of electricity.

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Ionic radii vary in
(i) inverse proportion to the effective nuclear charge.
(ii) inverse proportion to the square of effective nuclear charge.
(iii) direct proportion to the screening effect.
(iv) direct proportion to the square of screening effect.

Option (i) and (iii) are the answers. Ionic radii decreases as the effective nuclear charge increases due to inverse proportional relation. Also, ionic radii increases as the screening effect...

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In which of the following options order of arrangement does not agree with
the variation of property indicated against it?
(i) Al3+ < Mg2+ < Na+ < F– (increasing ionic size)
(ii) B < C < N < O (increasing first ionisation enthalpy)
(iii) I < Br < Cl < F (increasing electron gain enthalpy)
(iv) Li < Na < K < Rb (increasing metallic radius)

Option (ii) and (iiii) are the answers. For increasing first ionization enthalpy, the order should be: B < C < O < N For increasing electron gain enthalpy, the order should be: I < Br...

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Which of the following statements are correct?
(i) Helium has the highest first ionisation enthalpy in the periodic table.
(ii) Chlorine has less negative electron gain enthalpy than fluorine.
(iii) Mercury and bromine are liquids at room temperature.
(iv) In any period, the atomic radius of alkali metal is the highest.

Option (i), (iii) and (iv) are the answers. Because of its larger size and lower electronic repulsion, chlorine has a higher negative electron gain enthalpy than fluorine.  

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Those elements impart colour to the flame on heating in it, the atoms of which require low energy for the ionisation (i.e., absorb energy in the visible region of the spectrum). The elements of which of the following groups will impart colour to the flame?
(i) 2
(ii) 13
(iii) 1
(iv) 17

Option (i) and (iii) are the answers. Ionization enthalpies are low in group 1 (alkali metals) and group 2 (alkaline earth metals). As a result, they give flame colour.

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Electronic configurations of four elements A, B, C and D are given below :

    \[\left( \mathbf{A} \right)\text{ }\mathbf{1}{{\mathbf{s}}^{\mathbf{2}}}~\mathbf{2}{{\mathbf{s}}^{\mathbf{2}}}~\mathbf{2}{{\mathbf{p}}^{\mathbf{6}}}~\]


    \[\left( \mathbf{B} \right)\text{ }\mathbf{1}{{\mathbf{s}}^{\mathbf{2}}}~\mathbf{2}{{\mathbf{s}}^{\mathbf{2}}}~\mathbf{2}{{\mathbf{p}}^{\mathbf{4}}}\]


    \[\left( \mathbf{C} \right)\text{ }\mathbf{1}{{\mathbf{s}}^{\mathbf{2}}}~\mathbf{2}{{\mathbf{s}}^{\mathbf{2}}}~\mathbf{2}{{\mathbf{p}}^{\mathbf{6}}}~\mathbf{3}{{\mathbf{s}}^{\mathbf{1}}}~\]


    \[\left( \mathbf{D} \right)\text{ }\mathbf{1}{{\mathbf{s}}^{\mathbf{2}}}~\mathbf{2}{{\mathbf{s}}^{\mathbf{2}}}~\mathbf{2}{{\mathbf{p}}^{\mathbf{5}}}\]


Which of the following is the correct order of increasing tendency to gain electron :
(i) A < C < B < D
(ii) A < B < C < D
(iii) D < B < C < A
(iv) D < A < B < C

Option (i) is the answer. (a) A – Is2 2s2 2p6 – Noble gas configuration B -1s2 2s2 2p4 – 2 electrons short of stable configuration C – 1s2 2s2 2p6 3.?1 – Requires one electron to complete 5-orbital...

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Comprehension given below is followed by some multiple-choice questions.
Each question has one correct option. Choose the correct option.
In the modern periodic table, elements are arranged in order of increasingatomic numbers which are related to the electronic configuration. Depending upon the type of orbitals receiving the last electron, the elements in the periodic the table has been divided into four blocks, viz, s, p, d and f. The modern periodic table consists of 7 periods and 18 groups. Each period begins with the filling of a new energy shell. In accordance with the Aufbau principle, the seven periods (1 to 7) have 2, 8, 8, 18, 18, 32 and 32 elements respectively. The seventh period is still incomplete. To avoid the periodic table being too long, the two series of f-block elements, called lanthanoids and actinoids are placed at the bottom of the main body of the periodic table.
(i) The element with atomic number 57 belongs to
(a) s-block
(b) p-block
(c) d-block
(d) f-block
(ii) The last element of the p-block in 6th period is represented by the outermost electronic configuration.
(a) 7s2 7p6
(b) 5f 14 6d10 7s 2 7p 0
(c) 4f 14 5d10 6s2 6p6
(d) 4f 14 5d10 6s 2 6p 4
(iii) Which of the elements whose atomic numbers are given below, cannot be accommodated in the present set up of the long form of the periodic table?
(a) 107
(b) 118
(c) 126
(d) 102
(e) The electronic configuration of the element which is just above the element with atomic number 43 in the same group is ________.
(a) 1s2 2s2 2p6 3s2 3p6 3d5 4s2
(b) 1s2 2s2 2p6 3s2 3p6 3d5 4s3 4p6
(c) 1s2 2s2 2p6 3s2 3p6 3d6 4s2
(d) 1s2 2s2 2p6 3s2 3p6 3d7 4s2
(v) The elements with atomic numbers 35, 53 and 85 are all ________.

(a) noble gases
(b) halogens
(c) heavy metals
(d) light metals

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The formation of the oxide ion, O2- (g), from oxygen atom requires first an exothermic and then an endothermic step as shown below:

    \[\mathbf{O}\text{ }\left( \mathbf{g} \right)\text{ }+\text{ }\mathbf{e}\text{ }\to \text{ }{{\mathbf{O}}^{-}}\left( \mathbf{g} \right)\text{ };\text{ };\text{ }\text{ }\mathbf{HV}\text{ }=\text{ }\text{ }\mathbf{141}\text{ }\mathbf{kJ}\text{ }\mathbf{mol}\mathbf{1}\]


    \[{{\mathbf{O}}^{-}}\left( \mathbf{g} \right)\text{ }+\text{ }\mathbf{e}\text{ }\to \text{ }{{\mathbf{O}}^{\mathbf{2}-}}\left( \mathbf{g} \right)\text{ };\text{ }\text{ }\mathbf{HV}\text{ }=\text{ }+\text{ }\mathbf{780}\text{ }\mathbf{kJ}\text{ }\mathbf{mol}\mathbf{1}\]


Thus the process of formation of O2– in the gas phase is unfavourable even though O2- is isoelectronic with neon. It is due to the fact that
(i) oxygen is more electronegative.
(ii) addition of electron in oxygen results in larger size of the ion.
(iii) electron repulsion outweighs the stability gained by achieving a noble gas configuration.
(iv) O- ion has a comparatively smaller size than an oxygen atom.

Option (iii) is the answer. This is due to the fact that when an electron is introduced to a negatively charged ion, it is repelled rather than attracted. As a result, the addition of the second...

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Isostructural species are those which have the same shape and hybridisation. Among the given species identify the isostructural pairs. (i) [NF3 and BF3] (ii) [BF4- and NH4+] (iii) [BCl3 and BrCl3] (iv) [NH3 and NO3-]

From a structural standpoint, we can see that, NF3 is pyramidal whereas BF3 is planar triangular. BF4- and NH4+ ions are tetrahedral in structure. BCl3 is triangular planar and BrCl3 is...

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The period number in the long form of the periodic table is equal to
(i) magnetic quantum number of any element of the period.
(ii) an atomic number of any element of the period.
(iii) maximum Principal quantum number of any element of the period.
(iv) maximum Azimuthal quantum number of any element of the period.

Option (iii) is the answer. Period number = maximum n of any element where 'n' stands for the principle quantum number. It determines the element's period number. Mg, for example, has a maximum main...

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The order of screening effect of electrons of s, p, d and f orbitals of a given shell of an atom on its outer shell electrons is:
(i) s > p > d > f
(ii) f > d > p > s
(iii) p < d < s > f
(iv) f > p > s > d

Option (i) is the answer. In every atom with more than one electron shell, this effect, known as the screening effect, describes the decrease in attraction between an electron and the nucleus. The...

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