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1. Ionisation enthalpy is the amount of energy required to remove the most loosely bound electron from an isolated gaseous atom in its ground state.

2. It shows the increasing order the magnitude of successive ionization enthalpies. As the positive charge of the ion increases, it becomes more difficult to remove the valence electron due to increase in effective nuclear charge.

The basic theme of organisation of elements in the periodic table is to classify the elements in periods and groups according to their properties. This arrangement makes the study of elements and their compounds simple and systematic. In the periodic table, elements with similar properties are placed in the same group.

Mendeleev’s Periodic Law states that the physical and chemical properties of elements are periodic functions of their atomic weights. On the other hand, the Modern periodic Law states that the physical and chemical properties of elements are periodic functions of their atomic numbers.

1. Mendeleev 

2. Yes, atomic number is the more fundamental property of an element than atomic mass. 

3. The physical and chemical properties of the elements are periodic functions of their atomic numbers.

(a) Merits of Mendeleev’s periodic table:

• Study of elements – Elements are classified into groups with similar properties, thus facilitating the study of properties of elements. 
• Prediction of new elements – Mendeleev left certain vacant places in his table which provided a clue for the discovery of new elements, 

e.g. Eka-AI (for Ga), Eka-Si (for Ge). 

• Determination of correct atomic weights – With the help of this table, doubtful atomic weights of some elements were corrected.

(b) Demerits of Mendeleev’s periodic table:

• Position of hydrogen was not certain. 
• Anomalous pairs of elements – Certain elements of higher atomic weight preceed those with lower atomic weight, 

e.g. I (at.wt.127) was placed after Te (at.wt. 128).

• Lanthanides and Actinides are not given proper places in this periodic table. 
• No proper position for isotopes.

Nuclear mass does not affect the valence shell. Thus, option (c) is the correct answer.

1. Each one of these ions contains 10 electrons and hence these are isoelectronic ions,

2. The ionic radii of isoelectronic ions decrease with the increase in the magnitude of the nuclear charge. Among the isoelectronic ions: N^3- , O^2- , F^- , Na⁺ , Mg²⁺ and Al³⁺ 

nuclear charge increase in the order: 

N^3-  < O²< F^- < Na⁺ < Mg²⁺ < Al³⁺

Therefore, 

the ionic radii decrease in the order: 

N^3-  > O²> F^- > Na⁺ > Mg²⁺ > Al³⁺

Mendeleev Periodic Law states that the properties of the elements are a periodic function of their atomic weights whereas Modern Periodic Law states that the properties of elements are a periodic function of their atomic numbers. Thus, the basic difference in approach between Mendeleev’s Periodic Law and Modern Periodic Law is the change in basis of arrangements of elements from atomic weight to atomic number.

The basic theme of organisation of elements in the periodic table is to facilitate the study of the properties of all the elements and their compounds. On the basis of similarities in chemical properties, the various elements have been divided into different groups. This has made the study of elements simple because their properties are now studied in the form of groups rather than individually.

(a) F > Cl > O > N

Across a period, the oxidising character increases from left to right. Therefore, among F, O and N, oxidising power decreases in the order: F > O > N. However, within a group, oxidising power decreases from top to bottom. Thus, F is a stronger oxidising agent than Cl. Thus, overall decreasing order of oxidising power is F > Cl > O > N and the choice (a) is correct.

(a) N. This is because in N, three 2p-electrons reside in different atomic orbitals in accordance with Hund’s rule (2p\(\frac{1}{x}\)2p\(\frac{1}{y}\)2p\(\frac{1}{z}\) ) whereas in O, two the four 2pelectrons must occupy the same 2p-orbital resulting in an increased electron-electron repulsion (2p\(\frac{2}{x}\) 2p\(\frac{1}{y}\) 2p\(\frac{1}{z}\)). Consequently, it is easier to remove the fourth 2p-electron from O than it is, to remove one of the three 2p-electrons from N.

(b) F (136 pm) is bigger than F (72 pm). An anion is bigger than its parent atom. Addition of one electron in F results in increased repulsion among the electrons and a decrease in effective nuclear charge. Thus, attraction between nucleus and the electrons decreases and hence size increases.

1. On moving from top to bottom in a given group, size of the atom increases and ionisation enthalpy decreases. Hence, it becomes easier to remove the valence electron.

2. Atoms with octet configuration, half-filled and completely filled configurations have extra stability and hence have higher values of ionization enthalpy.

1. The element is K. 

₁₉K= 1s² 2s² 2p 3s² 3p⁶ 4s¹ 

Group number = 1 

Period number = 4 

Block = s-block

2. (i) Fluorine 

(ii) Chlorine

1. Graph showing the variation of ionisation enthalpy (∆_iH) with atomic number (Z) of the elements of second period.

2. (i) ‘Ne’ is an inert gas and it has closed electron shell with stable octet electronic configuration. Hence it has the maximum ionisation enthalpy in the second period.

(ii) Be has completely filled electronic configuration and a more stable. So ionization enthalpy is high.

Sulphur has greater negative value (-200 kJ mol^-1) for electron gain enthalpy compared to that of oxygen (-141 kJ mol^-1). This is because, due to smaller size of oxygen the added electron experiences much repulsion from the electrons present in the shell. Due to large size, the electron-electron repulsion is much less in sulphur.

1. No. In addition to electronic configuration, ionization enthalpy depends upon other factors like atomic size, nuclear charge and shielding effect/screening effect.

2. In multi-electron atoms, the nuclear charge experienced by the valence electron will be less than the actual charge on the nucleus because it is shielded by inner core of electrons. This is called shielding effect or screening effect.

3. Yes. Ionization enthalpy decreases with increase in shielding effect/screening effect of inner electrons. This is because as a result of shielding of the valence electron by the intervening core of electrons it experiences a net positive charge which is less than the actual charge of the nucleus. In general, shielding is effective when the orbitals in the inner shells are completely filled.

1. Electron gain enthalpy(AegH) is the enthalpy change accompanying the process of addition of an electron to a neutral gaseous atom to convert it into a negative ion. It is a quantitative property of an isolated gaseous atom, which can be measured. Whereas, electronegativity is a qualitative measure of the ability of an atom in a chemical compound to attract shared electrons to itself. It is not a measureable quantity.

2. This is because to remove second electron from a positively charged ion more amount of energy is required due to increase in effective nuclear charge.

1. On moving down a group, electron gain enthalpy becomes less negative because the size of the atom increases and the added electron would be farther from the nucleus.

2. Due to small size of F, the added electron goes to the smaller 2p quantum level and suffers significant replusion from the other electrons present in this level. But due to big size of Cl, the added electron goes to the n = 3p quantum level and occupies a larger region of space and the electron electron repulsion is much less.

1-->>The electron gain enthalpy becomes less negative as we move down a group.

2-->>>The negative electron gain enthalpy of fluorine is less than that of chlorine. It is due to small size of fluorine atom. As a result, there are strong interelectronic repulsions in the relatively small 2p orbitals of fluorine and thus, the incoming electron does not experience much attraction.

1. The statement is correct. Electron gain enthalpies of Be (+240 kJ/mol) and Mg (+230 kJ/mol) are positive because they have stable electronic configurations with fully filled s-orbitals (Be – 2s², Mg – 3s²). Hence, the gain of electron is highly endothermic.

2. Noble gases have stable octet electronic configuration of ns² np⁶ (except He-1 s²) and they have practically no tendency to accept additional electron. They have large positive electron gain enthalpies because the electron has to enter the next higher principal quantum level leading to a very unstable electronic configuration. Hence, energy has to be supplied to add an extra electron.

Correct option is: (D) All of the above

Correct option is: (A) Raw material

1. Electron gain enthalpy(∆_egH) is the enthalpy change accompanying the process of addition of an electron to a neutral gaseous atom to convert it into a negative ion.

2. Effective nuclear charge, atomic size, electronic configuration

3. Electron gain enthalpy becomes more negative with increase in the atomic number across a period. The effective nuclear charge increases from left to right across a period and consequently, it will be easier to add an electron to a smaller atom since the added electron on an average would be closer to the nucleus. Thus more energy is released.

1. The periodical repetition of elements with similar properties after certain regular intervals when the elements are arranged in the order of increasing atomic numbers is called periodicity,

 2. Electronic configuration.

Correct Answer - Option 3 : EURO - 6

The correct answer is EURO - 6

​ INDIA BHARAT STAGE VI EMISSION STANDARDS

•  On January 6, 2016 the Indian Ministry of Road Transport and Highways (MoRTH) announced its decision to leapfrog from BS-IV to BS-VI emission standards in an accelerated fashion, with full implementation of BS-VI level emission standards beginning in 2020.
  • This announcement was supported by corresponding actions taken by the Ministry of Petroleum and Natural Gas (MoPNG) to ensure a nationwide supply of BS-VI fuel along with the proposed BS-VI emission standard implementation date of April 1, 2020.
  • The announcement by the MoRTH effectively withdrew a draft regulatory roadmap released in November 2015 that proposed the adoption of BS V and BS VI emission standards in 2019 and 2021, respectively.
  • The proposed BS-VI regulation will be the first global instance of leapfrogging from the Euro 4/IV level directly to Euro 6/VI level motor vehicle emission standards.
  • This precedent-setting step marks a new path forward for all developing country markets to follow to accelerate the adoption of clean vehicle technologies and fuels.
  • The notification came at a time when many Indian cities are struggling with severe air-quality problems and is an important step in addressing these issues. 

1. Wireless communication technologies using radio waves Bluetooth: This technology uses radio waves in the frequency range of 2.402 GHz to 2.480 GHz. And transmit data in short distance. Mobile phones, Laptops, tablets etc use Bluetooth technology to transmit data. 

2. Wi Fi(Wireless Fidelity): It uses radio waves to transmit information across a network in a range 2.4 GHz to 5 GHz in short distance. Nowadays this technology is used to access internet in Laptops, Desktops, tablets. Mobile phones etc.

AM and FM radio broadcast and mobile phones make use of Radio waves medium for transmission.

Answer is (d) (2/3) x^3/2 + k

order is 2 degree is not defined

Let sin^-1 x = θ then. 

x = sin θ ⇒ x = cos(π/2 - θ) 

⇒ cos^-1x = π/2 - θ ⇒ θ + cos^-1 x = π/2

∴ sin^-1 x + cos^-1 x = π/2

f(x) = 8x³, g(x)= x^1/3

fog (x) = f(g(x))

= f(x^1/3) =8(x³)^1/3 =8x

gof (x) = g(f(x)) = g(8x³)

= (8x³)^1/3 = 2x

U = {1, 2, 3 … 10), A = {2, 3, 5, 7} 

B = {2, 4, 6, 8, 10} B¹ = {3, 5, 7,9} 

[A ∩ B¹ = {3,5,7)]

A = {a, b}, B = {1, 2, 3} 

B × A= {(1, a) (1, b) (2, a) (2, b) (3, a) (3, b)}

Let x = sec q ⇒ q = sec^-1 x.

Now cot^-1(1/√x² - 1) cot^-1(1/√sec² θ - 1)

cot^-1(1/tanθ) = cot^-1(cot θ) = θ = sec^-1 x.

Let \(\frac12\)cos^-1\(\frac34\) = θ 

⇒ cos 2θ = 3/4

⇒ 2 cos²θ - 1 = 3/4

⇒ 2 cos²θ = 1 + 3/4 = 7/4

⇒ cos²θ = 7/8

⇒ cos θ = \(\frac{\sqrt7}{2\sqrt2}\) 

\(\therefore\) sin θ = \(\frac1{2\sqrt2}\)

\(\therefore\) tan θ = \(\frac{sin\theta}{cos\theta}\) = \(\frac{1/2\sqrt2}{\sqrt 7/2\sqrt2}\)  = \(\frac1{\sqrt7}\) 

\(\therefore\) θ = tan^-1\(\frac1{\sqrt7}\)

\(\therefore\) θ = tan^-1\(\frac1{\sqrt7}\)

⇒ \(\frac12\) cos^-1\(\frac34\) = tan^-1\(\frac1{\sqrt7}\) 

\(\therefore\) tan²(\(\frac12\)cos^-1\(\frac34\)) = tan²(tan^-1\(\frac1{\sqrt7}\)) = [tan(tan^-1\(\frac1{\sqrt7}\))]²

 = (\(\frac1{\sqrt7}\))² = \(\frac17\)

Correct option is D. 100%

Study of Population is meaning of demography

Study of population is meaning of demography.

Study of population is related with Demography.

The correct answer is ​Syed Muhammad Nizamuddin Auliya.

• Amir Khosrow was a royal poet and a disciple of Syed Muhammad Nizamuddin Auliya.

• Syed Muhammad Nizamuddin Auliya was one of the most famous Sufi saints from the Indian subcontinent region.
• He is also known as Hazrat Nizamuddin and Mehboob-e-Illahi (Beloved of God).
• He was a Sufi saint of the Chisti order.
• His shrine is located in New Delhi.
• Nizamuddin Auliya was born in 1238 AD in Badaun, Uttar Pradesh.

​

The correct answer is Both 'A' and 'R' are correct and 'R' is not the correct explanation for 'A'.

• The Bhakti movement emerged during the medieval period across the nation and religion.
  • The Bhakti Saints mainly used the regional languages for spreading their teaching among the masses.
  • Some of the famous Bhakti saints are Bhagat Namdev, Saint Kabir Das, Chaitanya, Meera Bai and many others.
• The Delhi Sultans as well as the Mughals made Persian their official language. 
  • Indo-Persian culture flourished during the invaders from foreign lands.
  • The emperor Humayun made Persian as his court language.

The correct answer is 2015.

• ASTROSAT was launched on September 28, 2015.
• It is India's first space observatory class satellite dedicated to Astronomy.
• launched onboard PSLV from Satish Dhawan Space Centre SHAR, Sriharikota.

• The Current chairman of ISRO is K Sivan.
• He is also the Secretary of the Department of Space.
• Recently the Indian government increased his tenure by 1 year.

• India’s communication satellite CMS-01 was successfully launched by PSLV-C50 on  December 17, 2020, from the Satish Dhawan Space Centre (SDSC) SHAR, Sriharikota.
• It replaced the former communication satellite GSAT-12.

The correct answer is 2 only.

• Polar Satellite Launch Vehicle (PSLV) is the third generation launch vehicle of India. Hence, Statement 1 is incorrect.
• It is the first Indian launch vehicle to be equipped with liquid stages. Hence, Statement 2 is correct.
• After its first successful launch in October 1994, PSLV emerged as the reliable and versatile workhorse launch vehicle of India with 39 consecutively successful missions by June 2017.
• During the 1994-2017 period, the vehicle has launched 48 Indian satellites and 209 satellites for customers from abroad.
• Besides, the vehicle successfully launched two spacecraft – Chandrayaan-1 in 2008 and Mars Orbiter Spacecraft in 2013 – that later travelled to Moon and Mars respectively.
• PSLV earned its title 'the Workhorse of ISRO' through consistently delivering various satellites to Low Earth Orbits, particularly the IRS series of satellites.
• It can take up to 1,750 kg of payload to Sun-Synchronous Polar Orbits of 600 km altitude.
• Due to its unmatched reliability, PSLV has also been used to launch various satellites into Geosynchronous and Geostationary orbits, like satellites from the IRNSS constellation.
• PSLV uses 6 solid rocket strap-on motors to augment the thrust provided by the first stage in its PSLV-G and PSLV-XL variants.
• However, strap-ons are not used in the core alone version (PSLV-CA).