31. An atom of carbon has 6 protons. Its mass number is 12. How many neutr

An atom of carbon has 6 protons. Its mass number is 12. How many neutrons are present in an atom of carbon?

12
6
10
14
This question was previously asked in
UPSC NDA-1 – 2016
The correct answer is B) 6.
The mass number (A) of an atom is the total number of protons and neutrons in the nucleus. The number of protons is equal to the atomic number (Z). Therefore, the number of neutrons (N) can be calculated using the formula: N = A – Z.
For the given carbon atom:
Number of protons (Z) = 6
Mass number (A) = 12
Number of neutrons (N) = Mass number (A) – Number of protons (Z) = 12 – 6 = 6.
Thus, there are 6 neutrons in an atom of carbon with mass number 12. This specific isotope is Carbon-12 (¹²C), the most common isotope of carbon.

32. Which one of the following is the shape of BrF₅?

Which one of the following is the shape of BrF₅?

Octahedral
Square planar
Square pyramidal
Trigonal bipyramidal
This question was previously asked in
UPSC Geoscientist – 2023
The shape of BrF₅ is square pyramidal.
To determine the shape of a molecule, VSEPR theory is used. Bromine (Br) is the central atom in BrF₅, having 7 valence electrons. Five Fluorine (F) atoms are bonded to Br, each forming a single bond. This accounts for 5 valence electrons from Br being used in bonding, forming 5 bond pairs. The remaining 7 – 5 = 2 valence electrons form one lone pair on the central Br atom.
The central atom (Br) has 5 bond pairs and 1 lone pair, resulting in a total of 6 electron domains around it. According to VSEPR theory, 6 electron domains arrange themselves in an octahedral electron geometry to minimize repulsion. When one position is occupied by a lone pair and five by bonding pairs, the resulting molecular geometry (shape of the atoms) is square pyramidal. The lone pair occupies an equatorial position to minimize repulsion, pushing the four basal fluorine atoms towards the apex, forming a square base and a pyramid with bromine at the apex.

33. 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 ?

The combining atomic orbitals must have same or nearly the same energy.
The combining atomic orbitals must show lateral overlap with each other.
The combining atomic orbitals must have the same symmetry about the molecular axis.
The combining atomic orbitals must overlap to the maximum extent.
This question was previously asked in
UPSC Geoscientist – 2024
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.

34. 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 ?

Square pyramidal, dsp³
Trigonal bipyramidal, dsp³
Square pyramidal, sp³d
Trigonal bipyramidal, sp³d
This question was previously asked in
UPSC Geoscientist – 2024
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).

35. 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"]

This question was previously asked in
UPSC Geoscientist – 2024
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.

36. Which one among the following is the correct ground state electronic c

Which one among the following is the correct ground state electronic configuration of Cr atom ?

1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁴ 4s²
1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s¹
1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ 4s⁰
1s² 2s² 2p⁶ 3s² 3p³ 4s²
This question was previously asked in
UPSC Geoscientist – 2024
The correct ground state electronic configuration of a Cr (Chromium) atom is 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s¹.
Chromium (atomic number 24) is an exception to the standard Aufbau principle filling order. The configuration 3d⁵ 4s¹ is more stable than 3d⁴ 4s² due to the enhanced stability of a half-filled d subshell.
The electronic configuration follows the Aufbau principle generally, filling orbitals in increasing order of energy (1s, 2s, 2p, 3s, 3p, 4s, 3d…). However, for chromium and copper, promotion of an electron from the 4s orbital to the 3d orbital leads to half-filled (d⁵) or fully filled (d¹⁰) d subshells, which are particularly stable arrangements. The full configuration starts with the filled inner shells (equivalent to Argon, [Ar]), which is 1s² 2s² 2p⁶ 3s² 3p⁶, followed by 3d⁵ 4s¹.

37. Which one of the following is the correct electronic configuration of

Which one of the following is the correct electronic configuration of copper?

1s²2s²2p⁶3s²3p⁶3d¹⁰4s¹
1s²2s²2p⁶3s²3p⁶3d⁹4s²
1s²2s²2p⁶3s²3p⁶3d¹⁰4s²
1s²2s²2p⁶3s²3p⁶3d⁸4s²
This question was previously asked in
UPSC Geoscientist – 2023
Copper (Cu) has an atomic number of 29. The Aufbau principle and Hund’s rule predict the filling of orbitals in increasing order of energy. The expected electronic configuration based on the strict Aufbau principle would be 1s²2s²2p⁶3s²3p⁶4s²3d⁹.
However, elements like Copper (and Chromium) are exceptions to this rule. There is a slight energy difference between the 4s and 3d orbitals. The configuration with a completely filled or half-filled d subshell is more stable than one that is nearly filled.
In the case of Copper, promoting one electron from the 4s orbital to the 3d orbital results in the configuration 1s²2s²2p⁶3s²3p⁶3d¹⁰4s¹. This configuration has a stable, completely filled 3d subshell and a half-filled 4s subshell, which is energetically more favorable than the 3d⁹4s² configuration.
Therefore, the correct electronic configuration of copper is 1s²2s²2p⁶3s²3p⁶3d¹⁰4s¹.
Copper is an exception to the standard Aufbau principle; its ground state electronic configuration involves the transfer of an electron from the 4s to the 3d orbital to achieve a more stable completely filled 3d subshell.
Other elements exhibiting similar exceptions include Chromium (Cr, Z=24) with configuration [Ar] 3d⁵ 4s¹ instead of [Ar] 3d⁴ 4s². These exceptions highlight the complex interplay of electron-electron repulsion and orbital energies.

38. What is the total number of orbitals associated with the principal qua

What is the total number of orbitals associated with the principal quantum number 3?

3
6
9
12
This question was previously asked in
UPSC Geoscientist – 2023
The principal quantum number (n) defines the energy level or shell. For a given principal quantum number n, the number of possible subshells is equal to n. These subshells are characterized by the azimuthal quantum number (l), which can take integer values from 0 to n-1.
For n = 3, the possible values of l are 0, 1, and 2.
l = 0 corresponds to the s subshell. The number of orbitals in an s subshell is 1 (m_l = 0).
l = 1 corresponds to the p subshell. The number of orbitals in a p subshell is 3 (m_l = -1, 0, +1).
l = 2 corresponds to the d subshell. The number of orbitals in a d subshell is 5 (m_l = -2, -1, 0, +1, +2).
The total number of orbitals associated with a principal quantum number n is the sum of the number of orbitals in each subshell:
Total orbitals for n=3 = (number of orbitals for l=0) + (number of orbitals for l=1) + (number of orbitals for l=2)
Total orbitals for n=3 = 1 (3s) + 3 (3p) + 5 (3d) = 9.
Alternatively, the total number of orbitals in a shell with principal quantum number n is given by n².
For n = 3, total orbitals = 3² = 9.
For a given principal quantum number n, the total number of atomic orbitals in that shell is n².
Each atomic orbital can hold a maximum of 2 electrons with opposite spins (Pauli Exclusion Principle). Thus, the maximum number of electrons in a shell with principal quantum number n is 2n². For n=3, the maximum number of electrons is 2 * 3² = 18.

39. Which one of the following is the correct order of increase of the ato

Which one of the following is the correct order of increase of the atomic radius of the elements?

[amp_mcq option1=”C < B < Si < Al" option2="C < B < Al < Si" option3="C < Si < B < Al" option4="Si < Al < C < B" correct="option1"]

This question was previously asked in
UPSC Geoscientist – 2023
The elements are C (Carbon), B (Boron), Si (Silicon), and Al (Aluminum).
C and B are in Period 2, Group 14 and 13 respectively.
Si and Al are in Period 3, Group 14 and 13 respectively.
Atomic radius generally decreases across a period from left to right due to increasing nuclear charge pulling the electron cloud closer. So, B > C.
Atomic radius generally increases down a group due to the addition of electron shells. So, Si > C and Al > B.
Within Period 3, Al is to the left of Si, so Al > Si.
Comparing elements across periods and groups: Period 3 elements (Si, Al) are generally larger than Period 2 elements (C, B).
Combining these trends:
Within Period 2: C < B Within Period 3: Si < Al Comparing across periods: C and B are smaller than Si and Al. Approximate covalent radii (in picometers, pm): C: 70 pm B: 85 pm Si: 110 pm Al: 125 pm Ordering them in increasing radius: C (70) < B (85) < Si (110) < Al (125). The correct order is C < B < Si < Al.
Atomic radius decreases across a period and increases down a group. The increase in radius down a group due to adding a new electron shell is usually more significant than the decrease across a period due to increasing nuclear charge.
Exceptions and nuances exist in atomic radius trends, especially for transition metals and noble gases. The type of radius (covalent, ionic, van der Waals) also matters depending on the bonding situation.

40. A species that is not isoelectronic with $^{35}_{17}$Cl$^-$ is

A species that is not isoelectronic with $^{35}_{17}$Cl$^-$ is

$^{31}_{15}$P$^{3-}$
$^{32}_{16}$S$^{2-}$
$^{40}_{18}$Ar$^{2+}$
$^{39}_{19}$K$^+$
This question was previously asked in
UPSC Geoscientist – 2022
Isoelectronic species are atoms or ions that have the same number of electrons. First, determine the number of electrons in $^{35}_{17}$Cl$^-$:
– Atomic number (Z) = 17 (number of protons)
– Charge = -1
– Number of electrons = Protons – Charge = 17 – (-1) = 18 electrons.
Now check the electron count for each option:
– A) $^{31}_{15}$P$^{3-}$: Z=15. Electrons = 15 – (-3) = 18 electrons. (Isoelectronic)
– B) $^{32}_{16}$S$^{2-}$: Z=16. Electrons = 16 – (-2) = 18 electrons. (Isoelectronic)
– C) $^{40}_{18}$Ar$^{2+}$: Z=18. Electrons = 18 – (+2) = 16 electrons. (Not isoelectronic)
– D) $^{39}_{19}$K$^+$: Z=19. Electrons = 19 – (+1) = 18 electrons. (Isoelectronic)
The species that is not isoelectronic with Cl$^-$ is $^{40}_{18}$Ar$^{2+}$ with 16 electrons.
– Isoelectronic species have the same electron configuration.
– The number of electrons in an ion is calculated by subtracting the charge from the number of protons (atomic number).
Noble gases, like Argon (Ar), are often found as isoelectronic species with neighboring elements when they form ions, as they achieve a stable electron configuration. For instance, Cl$^-$ and K$^+$ are isoelectronic with neutral Argon (Ar), which has 18 electrons (Z=18). The mass number (superscript, e.g., 35) indicates the total number of protons and neutrons and is not relevant for determining the number of electrons.