2019

121. Which one of the following statements is correct?

Which one of the following statements is correct?

[amp_mcq option1=”Euplectella and Spongilla belong to different phyla of Animalia.” option2=”Sea anemone belongs to phylum Echinodermata.” option3=”Holothuria is a member of phylum Mollusca.” option4=”Scolopendra (Centipede) is a member of phylum Arthropoda.” correct=”option4″]

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Scolopendra (Centipede) is indeed a member of the phylum Arthropoda.
Arthropoda is the largest phylum in the animal kingdom and includes insects, spiders, crustaceans, and myriapods (like centipedes and millipedes). Centipedes belong to the class Chilopoda within Arthropoda.
Let’s analyze the other options: A) Euplectella and Spongilla are both types of sponges and belong to the phylum Porifera, so they are in the same phylum. B) Sea anemones belong to the phylum Cnidaria (or Coelenterata), not Echinodermata. C) Holothuria (Sea cucumber) belongs to the phylum Echinodermata, not Mollusca.

122. Catalytic converter transforms waste gases from the engines of many ca

Catalytic converter transforms waste gases from the engines of many cars into carbon dioxide, nitrogen and water. The catalyst is made of

[amp_mcq option1=”platinum and copper” option2=”molybdenum and copper” option3=”platinum and rhodium” option4=”rhodium and molybdenum” correct=”option3″]

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Catalytic converters use precious metals like platinum, palladium, and rhodium as catalysts to convert harmful exhaust gases into less harmful substances.
Platinum and rhodium are commonly used together in catalytic converters, particularly in three-way converters, which target carbon monoxide (CO), unburnt hydrocarbons (HC), and nitrogen oxides (NOx).
Platinum is effective for oxidizing CO and HCs. Rhodium is effective for reducing NOx. Palladium is also used, often in conjunction with platinum, and is effective for oxidizing CO and HCs. Copper and molybdenum are not typically used as the primary catalysts in automotive catalytic converters.

123. The number of elements in the lanthanoids of the periodic table is

The number of elements in the lanthanoids of the periodic table is

[amp_mcq option1=”8″ option2=”18″ option3=”14″ option4=”32″ correct=”option3″]

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The number of elements in the lanthanoids of the periodic table is 14.
The lanthanoids (or lanthanides) are a series of chemical elements comprising the 15 elements from lanthanum (atomic number 57) through lutetium (atomic number 71). However, the term “lanthanoid series” usually refers to the 14 elements immediately following Lanthanum, from Cerium (atomic number 58) to Lutetium (atomic number 71), as these are the elements where the 4f subshell is progressively filled. Lanthanum itself has the electron configuration [Xe] 5d¹ 6s², not filling the 4f subshell, but it is chemically similar to the other lanthanoids and is often included in the series or considered the first element.

In the context of the f-block of the periodic table, which is typically displayed as two rows below the main body, the lanthanoids constitute the upper row. This row consists of 14 elements, starting after Barium (Z=56) and before Hafnium (Z=72) in the main table. These 14 elements are Cerium (58), Praseodymium (59), Neodymium (60), Promethium (61), Samarium (62), Europium (63), Gadolinium (64), Terbium (65), Dysprosium (66), Holmium (67), Erbium (68), Thulium (69), Ytterbium (70), and Lutetium (71).

Given the options, 14 is present and is the standard number of elements represented in the f-block lanthanoid series, corresponding to the filling of the 4f orbitals. While the definition can sometimes include Lanthanum (making it 15), 14 is the number that fits the typical arrangement and f-orbital filling.

The actinoids are the second series in the f-block, starting from Actinium (Z=89) or Thorium (Z=90) up to Lawrencium (Z=103), also containing 14 elements where the 5f subshell is filled. The lanthanoids and actinoids are sometimes collectively called the inner transition metals.

124. Out of the elements phosphorus (P), sulphur (S), chlorine (Cl) and flu

Out of the elements phosphorus (P), sulphur (S), chlorine (Cl) and fluorine (F), the elements having the most negative and least negative electron gain enthalpy, respectively are

[amp_mcq option1=”Cl and P” option2=”F and S” option3=”F and Cl” option4=”P and S” correct=”option1″]

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The elements having the most negative and least negative electron gain enthalpy among P, S, Cl, and F, respectively, are Cl and P.
Electron gain enthalpy is the energy change when an electron is added to a neutral gaseous atom to form a negative ion. A negative value means energy is released, indicating the atom has an affinity for electrons. A more negative value signifies a stronger affinity.
We are comparing P (Group 15), S (Group 16), Cl (Group 17), and F (Group 17).
General trends:
– Across a period (left to right), electron gain enthalpy generally becomes more negative (atoms become smaller and nuclear charge increases, attracting the added electron more strongly). Group 18 (noble gases) have positive electron gain enthalpies as they have a stable electron configuration. Group 15 elements have relatively less negative (or slightly positive) values due to the stability of the half-filled p-subshell.
– Down a group, electron gain enthalpy generally becomes less negative (the added electron is further from the nucleus in a larger shell and experiences more shielding).

Let’s consider the given elements:
– F (Period 2, Group 17)
– Cl (Period 3, Group 17)
– S (Period 3, Group 16)
– P (Period 3, Group 15)

Comparing F and Cl (Group 17): Although the general trend is less negative down a group, Cl has a more negative electron gain enthalpy (-349 kJ/mol) than F (-328 kJ/mol). This is an anomaly due to the very small size of F, where the added electron experiences significant repulsion from existing electrons in the compact 2p subshell.
Comparing Cl and S (same period, different groups): Electron gain enthalpy becomes more negative across the period. Cl is in Group 17, S is in Group 16. So, Cl should have a more negative value than S. (Cl: -349 kJ/mol, S: -200 kJ/mol). This is consistent.
Comparing S and P (same period, different groups): S is in Group 16, P is in Group 15. Group 15 elements have unusually low (less negative) electron gain enthalpies due to the stable half-filled configuration. So, P should have a less negative value than S. (S: -200 kJ/mol, P: -74 kJ/mol). This is consistent.

Ranking the elements by electron gain enthalpy from most negative to least negative:
Cl (-349) > F (-328) > S (-200) > P (-74)

The element with the most negative electron gain enthalpy is Cl.
The element with the least negative electron gain enthalpy is P.

The pair is (Cl, P).

Electron gain enthalpy can be positive for some elements, meaning energy must be supplied to add an electron. Noble gases typically have positive electron gain enthalpies. Elements in Group 2 (alkaline earth metals) and Group 12 also tend to have values close to zero or positive for the first electron gain enthalpy.

125. Which one of the following elements forms compounds with pronounced co

Which one of the following elements forms compounds with pronounced covalent character?

[amp_mcq option1=”Lithium (Li)” option2=”Sodium (Na)” option3=”Potassium (K)” option4=”Rubidium (Rb)” correct=”option1″]

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Lithium (Li) forms compounds with the most pronounced covalent character among the given options.
The formation of covalent character in predominantly ionic compounds is explained by Fajans’ Rules. These rules state that a compound is more likely to have covalent character if:
1. The cation is small.
2. The cation has a high charge (not applicable here as all form +1 ions).
3. The anion is large (not directly comparable here as the anion is not specified, but assume it’s a common one like a halide).
4. The cation has a pseudo noble gas configuration (not applicable here as alkali metals form noble gas configuration ions).

The given elements are alkali metals: Lithium (Li), Sodium (Na), Potassium (K), and Rubidium (Rb). When they form compounds, they form +1 ions: Li⁺, Na⁺, K⁺, Rb⁺.
Their ionic radii increase down the group: Li⁺ < Na⁺ < K⁺ < Rb⁺. According to Fajans' Rules, smaller cations have higher polarizing power (the ability to distort the electron cloud of the anion), which leads to increased covalent character in the bond. Li⁺ is the smallest cation among Li⁺, Na⁺, K⁺, and Rb⁺. Therefore, Li⁺ has the highest polarizing power and forms compounds with anions that exhibit the most pronounced covalent character compared to the compounds of Na, K, and Rb with the same anion. For example, LiCl has significantly more covalent character than NaCl, KCl, or RbCl.

Lithium exhibits several properties that are anomalous compared to the other alkali metals, often showing similarities to magnesium (diagonal relationship). This pronounced covalent character is one such anomaly, explaining why compounds like LiCl are soluble in organic solvents and why Li₂CO₃ is less stable to heat than other alkali metal carbonates.

126. Two reactants in a flask produce bubbles of gas that turn limewater mi

Two reactants in a flask produce bubbles of gas that turn limewater milky. The gas is

[amp_mcq option1=”SO₂” option2=”NO₂” option3=”CO” option4=”CO₂” correct=”option4″]

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The gas that turns limewater milky is CO₂.
Limewater is a dilute aqueous solution of calcium hydroxide, Ca(OH)₂. It is a common test reagent for carbon dioxide (CO₂).
When carbon dioxide gas is bubbled through limewater, it reacts with calcium hydroxide to form a precipitate of calcium carbonate (CaCO₃), which is insoluble in water and makes the solution appear milky or cloudy.
The chemical reaction is:
Ca(OH)₂(aq) + CO₂(g) → CaCO₃(s) + H₂O(l)

If excess carbon dioxide is bubbled through the milky solution, the calcium carbonate precipitate reacts further to form soluble calcium bicarbonate, Ca(HCO₃)₂, and the milky appearance disappears.
CaCO₃(s) + CO₂(g) + H₂O(l) → Ca(HCO₃)₂(aq)

Other gases listed:
– SO₂ (Sulfur dioxide) also reacts with limewater, but forms calcium sulfite (CaSO₃), which is also insoluble and causes turbidity. However, the reaction with CO₂ is the most common and characteristic test for CO₂.
– NO₂ (Nitrogen dioxide) is an acidic gas and would react with Ca(OH)₂, but it typically doesn’t produce a milky precipitate with limewater as characteristically as CO₂.
– CO (Carbon monoxide) is a neutral gas and does not react with limewater.

Given the standard chemical tests, turning limewater milky is the definitive test for CO₂.

This reaction is commonly used in experiments to detect the production of carbon dioxide, such as in respiration or combustion.

127. The horizontal component of the earth’s magnetic field is zero at

The horizontal component of the earth’s magnetic field is zero at

[amp_mcq option1=”magnetic equator” option2=”magnetic poles” option3=”South and North Poles” option4=”nowhere” correct=”option2″]

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The horizontal component of the earth’s magnetic field is zero at magnetic poles.
The Earth’s magnetic field lines emerge from near the geographic South Pole (which is the North magnetic pole) and enter near the geographic North Pole (which is the South magnetic pole).
– At the magnetic poles (where a compass needle points vertically downwards or upwards), the magnetic field lines are essentially perpendicular to the Earth’s surface. Therefore, the magnetic field vector has only a vertical component, and the horizontal component is zero.
– At the magnetic equator, the magnetic field lines are approximately parallel to the Earth’s surface. Therefore, the magnetic field vector is primarily horizontal, and the vertical component is zero.
– The geographic poles (South and North Poles) are points on the Earth’s rotational axis and do not necessarily coincide with the magnetic poles. The horizontal component is generally not zero at the geographic poles unless they happen to coincide perfectly with the magnetic poles (which they do not).
The Earth’s magnetic poles are not fixed and drift over time. The angle between the magnetic north and geographic north is called the magnetic declination. The angle between the horizontal plane and the Earth’s magnetic field line is called the magnetic dip or inclination; the dip is 90 degrees at the magnetic poles and 0 degrees at the magnetic equator.

128. Stars twinkle in the sky at night because

Stars twinkle in the sky at night because

[amp_mcq option1=”refractive index of the atmosphere changes due to the change of temperature” option2=”stars emit light in the form of pulses” option3=”of interference of light coming from different stars” option4=”of diffraction of light” correct=”option1″]

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Stars twinkle in the sky at night because refractive index of the atmosphere changes due to the change of temperature.
Twinkling of stars (scintillation) is caused by atmospheric refraction.
– Light from distant stars travels through the Earth’s atmosphere before reaching our eyes.
– The atmosphere is not uniform; it consists of layers with varying temperatures and densities.
– Variations in temperature and density cause variations in the refractive index of the air.
– As light from a star passes through these turbulent layers with changing refractive index, it undergoes continuous refraction in random directions.
– This causes fluctuations in the apparent position and brightness of the star as seen from Earth. These rapid fluctuations are perceived as twinkling.
– Planets, being much closer, appear as extended sources of light rather than point sources. The light from different parts of a planet’s disc undergoes similar but independent variations, which average out, so planets do not twinkle noticeably.

Option A correctly identifies the cause: changes in atmospheric refractive index due to temperature variations (and hence density variations) lead to varying refraction of starlight.
Option B is incorrect; stars emit light continuously.
Option C is incorrect; twinkling is an effect on light from a single star due to atmospheric effects, not interference from different stars.
Option D is incorrect; while diffraction occurs, twinkling is primarily an effect of refraction due to atmospheric turbulence.

Atmospheric refraction is also responsible for phenomena like the apparent flattening of the sun at sunrise/sunset and the fact that we can see the sun just before it rises and just after it sets. The degree of twinkling is affected by atmospheric conditions (turbulence).

129. The minimum length of a plane mirror to see your full length image is

The minimum length of a plane mirror to see your full length image is

[amp_mcq option1=”one quarter of your height” option2=”one-third of your height” option3=”half of your height” option4=”equal to your height” correct=”option3″]

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The minimum length of a plane mirror to see your full length image is half of your height.
This is a fundamental principle of reflection from a plane mirror. To see the full image of an object in a plane mirror, the minimum vertical extent of the mirror required is half the vertical extent of the object.

Consider a person standing in front of a mirror. Light rays from the top of their head reflect off the top part of the mirror and enter their eyes. Light rays from their feet reflect off the bottom part of the mirror and enter their eyes.
– The angle of incidence equals the angle of reflection.
– The law of reflection dictates that to see the top of your head, the top edge of the mirror must be halfway between the top of your head and your eye level.
– Similarly, to see your feet, the bottom edge of the mirror must be halfway between your feet and your eye level.
– The distance between these two points on the mirror is the minimum required length. This distance is (1/2 * distance from head to eye) + (1/2 * distance from eye to feet). Since (distance from head to eye) + (distance from eye to feet) equals the total height of the person, the required mirror length is half the person’s height.

The distance of the person from the mirror does not affect the *minimum length* required, although it does affect the *field of view*.

This principle is utilized in everyday life, for instance, when installing full-length mirrors. The mirror doesn’t need to be as tall as the person.

130. A Kelvin thermometer and a Fahrenheit thermometer both give the same r

A Kelvin thermometer and a Fahrenheit thermometer both give the same reading for a certain sample. The corresponding Celsius temperature is about

[amp_mcq option1=”301 °C” option2=”614 °C” option3=”276 °C” option4=”273 °C” correct=”option1″]

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The corresponding Celsius temperature is about 301 °C.
We are given that a Kelvin thermometer and a Fahrenheit thermometer give the same reading for a certain sample. Let this reading be x.
So, the temperature in Kelvin (K) is x, and the temperature in Fahrenheit (F) is x.
We need to find the corresponding temperature in Celsius (C).

The conversion formulas between these scales are:
1. Fahrenheit to Celsius: C = (F – 32) * 5/9
2. Kelvin to Celsius: C = K – 273.15 (or often approximated as C = K – 273)

Let’s use the exact conversion K = C + 273.15 and F = (9/5)C + 32.
Since K = x and F = x, we have:
x = C + 273.15 (Equation 1)
x = (9/5)C + 32 (Equation 2)

Equating the right sides of Equation 1 and Equation 2:
C + 273.15 = (9/5)C + 32

Rearrange the terms to solve for C:
273.15 – 32 = (9/5)C – C
241.15 = (9/5 – 5/5)C
241.15 = (4/5)C

C = (241.15 * 5) / 4
C = 1205.75 / 4
C = 301.4375

Rounding to the nearest whole number or considering the options, the corresponding Celsius temperature is about 301 °C.

The question asks for “about” the Celsius temperature, indicating an approximation is acceptable. Using the approximation C = K – 273 would yield C = 301.4375 – 0.15 ≈ 301.2875, still close to 301 °C. The unusual point where Kelvin and Fahrenheit scales read the same is 301.4375 in both K and °F.
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