UPSC Geoscientist Archives

1. Which one of the following types of isomerisms is shown by [CoCl₂(en)₂

Which one of the following types of isomerisms is shown by [CoCl₂(en)₂]Cl ?

[amp_mcq option1=”Optical isomerism only” option2=”Geometrical isomerism only” option3=”Both ionisation and geometrical isomerisms” option4=”Both geometrical and optical isomerisms” correct=”option4″]

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The complex ion is [CoCl₂(en)₂]⁺, which is an octahedral complex with coordination number 6. It exhibits both geometrical and optical isomerism.

The complex [CoCl₂(en)₂]⁺ is of the type [Ma₂b₂] (where ‘a’ is Cl and ‘b’ is the bidentate ligand ‘en’ represented by two coordination points). Such octahedral complexes can exist as cis and trans geometrical isomers. The trans isomer (where the two Cl ligands are opposite to each other) has a plane of symmetry and is optically inactive. The cis isomer (where the two Cl ligands are adjacent) lacks a plane of symmetry and is chiral, existing as a pair of enantiomers (non-superimposable mirror images), which are optically active.

Geometrical isomerism is shown by the existence of cis and trans forms. Optical isomerism is shown by the cis isomer due to its chirality. Ionisation isomerism involves the exchange of an ion inside the coordination sphere with a counter ion outside, which is not applicable here as the counter ion (Cl⁻) does not correspond to a ligand inside the sphere that can be exchanged. Linkage isomerism requires ambidentate ligands, which are not present here (en and Cl are not ambidentate).

2. Which one of the following is the IUPAC name of [Ag(NH₃)₂] [Ag(CN)₂] ?

Which one of the following is the IUPAC name of [Ag(NH₃)₂] [Ag(CN)₂] ?

[amp_mcq option1=”Diammine dicyanidodisilver(I)” option2=”Diammine dicyanidoargentate(I)” option3=”Diamminesilver(I) dicyanidoargentate(I)” option4=”Dicyanidoargentate(I) diamminesilver(I)” correct=”option3″]

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The compound is a salt consisting of a complex cation and a complex anion: [Ag(NH₃)₂]⁺ [Ag(CN)₂]⁻. The IUPAC name is derived by naming the cation complex followed by the anion complex.

For the cation [Ag(NH₃)₂]⁺: The metal is Silver (Ag). Ammonia (NH₃) is an neutral ligand named ‘ammine’. There are two ammine ligands (diammine). The charge of the complex is +1. Since NH₃ is neutral, the oxidation state of Ag is +1. As it is a cation, the metal name is used as is, followed by the oxidation state in Roman numerals. Name: Diamminesilver(I).
For the anion [Ag(CN)₂]⁻: The metal is Silver (Ag). Cyanide (CN⁻) is an anionic ligand named ‘cyanido’ or ‘cyano’. There are two cyanido ligands (dicyanido). The charge of the complex is -1. Since CN⁻ has a charge of -1, Ag + 2(-1) = -1, so the oxidation state of Ag is +1. As it is an anion, the suffix ‘-ate’ is added to the Latin name of the metal (Argentum for Silver), followed by the oxidation state. Name: Dicyanidoargentate(I).

Combining the cation and anion names gives the full IUPAC name: Diamminesilver(I) dicyanidoargentate(I). This matches option C. Option A incorrectly uses ‘disilver’ and doesn’t separate the cation and anion naming properly. Option B combines parts of both names but doesn’t represent the ionic structure accurately. Option D reverses the order of cation and anion names.

3. What is the colour of anhydrous CuSO₄ ?

What is the colour of anhydrous CuSO₄ ?

[amp_mcq option1=”White” option2=”Blue” option3=”Green” option4=”Yellow” correct=”option1″]

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The colour of anhydrous CuSO₄ is white.

Copper(II) sulfate exists in several hydration states. The most common form is copper(II) sulfate pentahydrate (CuSO₄·5H₂O), which is a bright blue crystalline solid. The blue colour is due to the presence of hydrated copper(II) ions, [Cu(H₂O)₄]²⁺, where water molecules act as ligands coordinated to the Cu²⁺ ion. The d-d electronic transitions within the Cu²⁺ ion, influenced by the surrounding water ligands, cause the absorption of certain wavelengths of visible light (specifically red-orange light), resulting in the complementary blue colour being observed.
When copper(II) sulfate pentahydrate is heated, it loses its water of crystallization. The blue pentahydrate first turns into pale blue CuSO₄·3H₂O, then blue-green CuSO₄·H₂O, and finally becomes anhydrous copper(II) sulfate (CuSO₄), which is a white powder. In the anhydrous state, the crystal structure changes, and the Cu²⁺ ion is no longer surrounded by water ligands in the same way. This alters the energy levels of the d orbitals, and the absorption characteristics change such that no visible light is strongly absorbed, causing the compound to appear white.

The reaction of anhydrous copper sulfate with water is exothermic and results in the formation of the blue hydrated salt. This reaction is often used as a test for the presence of water.

4. Which one of the following conditions is not correct for the combinati

Which one of the following conditions is not correct for the combination of atomic orbitals to form molecular orbitals ?

[amp_mcq option1=”The combining atomic orbitals must have same or nearly the same energy.” option2=”The combining atomic orbitals must show lateral overlap with each other.” option3=”The combining atomic orbitals must have the same symmetry about the molecular axis.” option4=”The combining atomic orbitals must overlap to the maximum extent.” correct=”option2″]

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The condition that the combining atomic orbitals must show lateral overlap with each other is not correct for the combination of atomic orbitals to form molecular orbitals.

The essential conditions for the linear combination of atomic orbitals (LCAO) to form molecular orbitals are:
1. **The combining atomic orbitals must have comparable energies.** This allows for effective interaction and mixing of orbitals. (Statement A is correct)
2. **The combining atomic orbitals must have the same symmetry about the molecular axis.** Orbitals with different symmetries (e.g., s orbital and p$_x$ orbital if the molecular axis is z) cannot effectively overlap. (Statement C is correct)
3. **The combining atomic orbitals must overlap to the maximum possible extent.** Significant overlap leads to strong bonding (and antibonding) molecular orbitals. (Statement D is correct)
Statement B claims that combining atomic orbitals *must* show lateral overlap. This is incorrect. Atomic orbitals can overlap axially (end-to-end), which leads to the formation of sigma ($\sigma$) molecular orbitals, or laterally (side-by-side), which leads to the formation of pi ($\pi$) molecular orbitals. Both types of overlap are valid and common ways for atomic orbitals to combine. It is not a requirement that *only* lateral overlap must occur. For instance, the formation of H₂ molecule involves axial overlap of two 1s orbitals.

Sigma bonds are formed by axial overlap (s-s, s-p$_z$, p$_z$-p$_z$ along the internuclear axis). Pi bonds are formed by lateral overlap (p$_x$-p$_x$, p$_y$-p$_y$). A double bond consists of one sigma and one pi bond, while a triple bond consists of one sigma and two pi bonds.

5. Which one of the following represents the correct combination of shape

Which one of the following represents the correct combination of shape and hybridisation of PF₅ ?

[amp_mcq option1=”Square pyramidal, dsp³” option2=”Trigonal bipyramidal, dsp³” option3=”Square pyramidal, sp³d” option4=”Trigonal bipyramidal, sp³d” correct=”option4″]

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The correct combination of shape and hybridisation of PF₅ is Trigonal bipyramidal, sp³d.

In PF₅, Phosphorus (P) is the central atom. P has 5 valence electrons (Group 15). It forms 5 single bonds with 5 Fluorine atoms.
Number of sigma bonds around P = 5.
Number of lone pairs on P = (Valence electrons on P – number of bonding electrons) / 2 = (5 – 5*1) / 2 = 0 lone pairs.
The steric number (number of sigma bonds + number of lone pairs) = 5 + 0 = 5.
A steric number of 5 corresponds to sp³d (or dsp³) hybridization. This hybridization involves one s orbital, three p orbitals, and one d orbital combining to form five hybrid orbitals.
According to VSEPR theory, when there are 5 electron groups and no lone pairs around the central atom, the electron group geometry is trigonal bipyramidal, and the molecular shape is also trigonal bipyramidal.
The five hybrid orbitals in sp³d hybridization are directed towards the vertices of a trigonal bipyramid.

In PF₅, the three equatorial P-F bonds are slightly shorter and stronger than the two axial P-F bonds. The F-P-F bond angles are 90° (axial-equatorial), 120° (equatorial-equatorial), and 180° (axial-axial).

6. Which one among the following is the correct arrangement of molecules

Which one among the following is the correct arrangement of molecules in increasing order of their dipole moment ?

[amp_mcq option1=”NF₃ < NH₃ < BF₃" option2="BF₃ < NH₃ < NF₃" option3="BF₃ < NF₃ < NH₃" option4="NF₃ < BF₃ < NH₃" correct="option3"]

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The correct arrangement of molecules in increasing order of their dipole moment is BF₃ < NF₃ < NH₃.

Dipole moment is a measure of the net polarity of a molecule, resulting from the vector sum of individual bond dipoles and contributions from lone pairs.
– **BF₃**: Boron trifluoride has a central Boron atom bonded to three Fluorine atoms. Boron is sp² hybridized, and the molecule is trigonal planar. The B-F bonds are polar, but the three bond dipoles are symmetrically arranged at 120° to each other in a plane. Their vector sum is zero, so BF₃ has a zero net dipole moment.
– **NF₃**: Nitrogen trifluoride has a central Nitrogen atom bonded to three Fluorine atoms and one lone pair. Nitrogen is sp³ hybridized, and the molecule is trigonal pyramidal. N-F bonds are polar (F is more electronegative). The bond dipoles point towards the F atoms, away from N. The lone pair contributes a dipole moment pointing away from N. The N-F bond dipoles and the lone pair dipole are in opposite directions along the molecular axis, partially canceling each other. This results in a small net dipole moment for NF₃.
– **NH₃**: Ammonia has a central Nitrogen atom bonded to three Hydrogen atoms and one lone pair. Nitrogen is sp³ hybridized, and the molecule is trigonal pyramidal. N-H bonds are polar (N is more electronegative). The bond dipoles point towards the N atom. The lone pair contributes a dipole moment pointing away from N. Both the N-H bond dipoles and the lone pair dipole point in the same general direction (upwards along the molecular axis), reinforcing each other. This results in a significant net dipole moment for NH₃.
Comparing NF₃ and NH₃, the dipole moment of NH₃ is significantly larger than that of NF₃.
Therefore, the order of increasing dipole moment is BF₃ (0) < NF₃ < NH₃.

Typical dipole moment values (in Debye): BF₃ ≈ 0 D, NF₃ ≈ 0.23 D, NH₃ ≈ 1.47 D. The difference between NH₃ and NF₃ highlights the importance of the direction of bond dipoles relative to the lone pair dipole in determining the net molecular dipole moment.

7. Which one among the following elements has the most negative electron

Which one among the following elements has the most negative electron gain enthalpy ?

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

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Among the given elements, Chlorine (Cl) has the most negative electron gain enthalpy.

Electron gain enthalpy is the energy change when an electron is added to a neutral gaseous atom to form a uninegative anion. A more negative value indicates a greater tendency to accept an electron.
Fluorine (F) and Chlorine (Cl) are halogens (Group 17), which generally have high (very negative) electron gain enthalpies because adding an electron gives them a stable noble gas configuration. Phosphorus (P) and Sulfur (S) are in periods 3. P is in Group 15, S is in Group 16.
Electron gain enthalpy generally becomes more negative across a period and less negative down a group.
Halogens (Group 17) have the most negative electron gain enthalpies. Comparing F and Cl, due to the small size of F, the added electron experiences significant electron-electron repulsion in the compact 2p subshell. In Cl, the added electron enters the larger 3p subshell, where repulsions are less significant. Consequently, Cl has a more negative electron gain enthalpy than F.
Comparing P and S: S (Group 16) has a more negative electron gain enthalpy than P (Group 15). Adding an electron to S (3p⁴) gives a stable 3p⁵ configuration. Adding an electron to P (3p³) disrupts the stable half-filled configuration, making it less favorable, leading to a less negative or even positive electron gain enthalpy for P.
Generally, electron gain enthalpy order for these elements is approximately: P < S < F < Cl (from least negative to most negative). Therefore, Cl has the most negative electron gain enthalpy among the given options.

While halogens generally have the most negative electron gain enthalpies, the electron gain enthalpy of chlorine is more negative than that of fluorine. Values (kJ/mol): F = -328, Cl = -349, S = -200, P = -72.

8. Which one among the following is the correct arrangement in increasing

Which one among the following is the correct arrangement in increasing order of ionic radii of O²⁻, F⁻, Na⁺, Mg²⁺ ?

[amp_mcq option1=”Mg²⁺ < Na⁺ < F⁻ < O²⁻" option2="O²⁻ < F⁻ < Na⁺ < Mg²⁺" option3="Mg²⁺ < Na⁺ < O²⁻ < F⁻" option4="O²⁻ < F⁻ < Mg²⁺ < Na⁺" correct="option1"]

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The correct arrangement in increasing order of ionic radii is Mg²⁺ < Na⁺ < F⁻ < O²⁻.

The given ions O²⁻, F⁻, Na⁺, and Mg²⁺ are isoelectronic species, meaning they all have the same number of electrons (10 electrons, with the electronic configuration $1s^2 2s^2 2p^6$).
For isoelectronic species, the ionic radius decreases as the nuclear charge (atomic number) increases. A higher nuclear charge exerts a stronger attraction on the same number of electrons, pulling them closer to the nucleus and reducing the ionic size.
The atomic numbers are: Oxygen (O) = 8, Fluorine (F) = 9, Sodium (Na) = 11, Magnesium (Mg) = 12.
The effective nuclear charges are approximately +8 for O²⁻, +9 for F⁻, +11 for Na⁺, and +12 for Mg²⁺.
As the nuclear charge increases (from +8 to +12), the attraction on the 10 electrons increases, leading to a decrease in ionic size.
Order of increasing nuclear charge: O²⁻ (+8) < F⁻ (+9) < Na⁺ (+11) < Mg²⁺ (+12). Order of decreasing ionic radii: O²⁻ > F⁻ > Na⁺ > Mg²⁺.
Order of increasing ionic radii: Mg²⁺ < Na⁺ < F⁻ < O²⁻.

Cations are generally smaller than their parent atoms because they lose valence electrons and the remaining electrons are held more tightly by the nucleus. Anions are generally larger than their parent atoms because they gain electrons, increasing electron-electron repulsion and expanding the electron cloud. Among isoelectronic species, negative ions are larger than positive ions, and within negative ions, the size increases with increasing negative charge. Within positive ions, the size decreases with increasing positive charge.

9. Consider the following statements: 1. A special provision was made

Consider the following statements:

1. A special provision was made in the Constitution of India by the Constitution (Eighty-ninth Amendment) Act, 2003 to establish the National Commission for Scheduled Tribes.
2. National Commission for Backward Classes, National Commission for Minorities and National Commission for Scheduled Castes are constitutional bodies.
3. National Commission for Women is not a constitutional body.

Which of the statements given above is/are not correct ?

[amp_mcq option1=”1, 2 and 3″ option2=”1 and 3 only” option3=”2 only” option4=”3 only” correct=”option3″]

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Only statement 2 is not correct.

– Statement 1 is correct. The National Commission for Scheduled Tribes was established as a separate constitutional body under Article 338A by the Constitution (Eighty-ninth Amendment) Act, 2003, splitting the original National Commission for Scheduled Castes and Scheduled Tribes.
– Statement 3 is correct. The National Commission for Women was constituted as a statutory body under the National Commission for Women Act, 1990. It is not a constitutional body.
– Statement 2 is incorrect. The National Commission for Backward Classes (NCBC) was given constitutional status by the 102nd Amendment Act, 2018 (adding Article 338B). The National Commission for Scheduled Castes (NCSC) is a constitutional body (Article 338). However, the National Commission for Minorities is a statutory body, established under the National Commission for Minorities Act, 1992. Since the statement claims all three (NCBC, NCM, and NCSC) are constitutional bodies, it is incorrect.

Constitutional bodies are those explicitly mentioned in the Constitution of India. Statutory bodies are created by an Act of Parliament.

10. Which of the following statements is/are correct as per the Constituti

Which of the following statements is/are correct as per the Constitution of India ?

1. The Parliament of India has the power to make provisions with respect to delimitation of constituencies.
2. The validity of a law made by the Parliament of India with respect to delimitation of constituencies shall not be called in question in a court.

Select the answer using the code given below :

[amp_mcq option1=”1 only” option2=”2 only” option3=”Both 1 and 2″ option4=”Neither 1 nor 2″ correct=”option3″]

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Both statements regarding delimitation of constituencies as per the Constitution of India are correct.

– Statement 1 is correct. Article 82 of the Constitution empowers the Parliament to enact a law to provide for the readjustment of seats in the Lok Sabha and the division of each State into territorial constituencies after every census. Similarly, Article 170 deals with the composition and delimitation of constituencies for State Legislative Assemblies, with provisions for delimitation laws made by or under the authority of Parliament. Parliament establishes a Delimitation Commission for this purpose.
– Statement 2 is correct. Article 329(a) of the Constitution specifically prohibits courts from inquiring into the validity of any law relating to the delimitation of constituencies or the allotment of seats to such constituencies made under Article 82 or Article 170. This provides judicial immunity to delimitation laws.

Delimitation is the process of fixing limits or boundaries of territorial constituencies in a country or a province having legislative body. The process is carried out by a Delimitation Commission, whose orders have the force of law and cannot be called in question before any court.