Assertion (A): Generally, ionisation enthalpy increases from left to right in a period. Reason (R): When succe

NCERT Exemplar · Q50

Q. Assertion (A): Generally, ionisation 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.

(i) Assertion is correct statement and reason is wrong statement.

(ii) Assertion and reason both are correct statements and reason is correct explanation of assertion.

(iii) Assertion and reason both are wrong statements.

(iv) Assertion is wrong statement and reason is correct statement.

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Concept understanding — Effective Nuclear Charge

The Intuition: Why Don't Electrons Just Fly Away?

Imagine you're holding a magnet near a pile of paperclips. The closer the magnet, the stronger the pull. Now imagine you put a sheet of cardboard between the magnet and the paperclips. The pull weakens — the cardboard "shields" the paperclips from the full force of the magnet.

An atom works similarly. The nucleus (positive charge) pulls on the electrons (negative charge). But an electron is not alone — there are other electrons buzzing around between it and the nucleus. Those inner electrons act like the cardboard sheet: they shield or screen the outer electron from feeling the full positive charge of the nucleus.

So an outer electron doesn't "see" the full nuclear charge $Z$ (the atomic number). It sees a smaller, effective charge — the net positive pull after accounting for the repulsion from inner electrons.

That's effective nuclear charge, denoted $Z_{\text{eff}}$ .

The Precise Statement

Z_{\text{eff}} = Z - S

$Z_{\text{eff}}$ is always less than $Z$ (except for hydrogen, which has no other electrons to shield — there $Z_{\text{eff}} = Z$ ).

What Determines the Shielding Constant $S$ ?

Not all electrons shield equally. The key rules:

Inner electrons shield outer electrons very effectively. An electron in the $n=1$ shell completely blocks about 1 unit of charge from an electron in $n=2$ .
Electrons in the same shell shield poorly. They're at roughly the same distance, so they don't block much of the nucleus from each other.
Outer electrons do not shield inner electrons at all. An electron farther out cannot block the nucleus from one closer in.

There are detailed rules (Slater's rules) to calculate $S$ numerically, but the core idea is simple: the more electron shells between an electron and the nucleus, the more shielding, and the lower $Z_{\text{eff}}$ .

Why Does This Matter?

$Z_{\text{eff}}$ explains three fundamental patterns in the periodic table:

Trend	What happens to $Z_{\text{eff}}$	Why
Across a period (left to right)	Increases	Adding protons ( $Z$ up) but electrons go into the same shell (shielding roughly constant). Net pull on outer electrons gets stronger.
Down a group (top to bottom)	Stays roughly constant or decreases slightly	Adding a new shell means much more shielding. The extra protons are almost completely cancelled by the new inner electrons.
Atomic size	Larger $Z_{\text{eff}}$ → smaller atom	Stronger pull pulls electrons closer to nucleus.

Note

This is why fluorine is smaller than lithium, even though fluorine has more protons. The extra protons in fluorine are not fully shielded — the outer electrons feel a much stronger pull.

A Concrete Example: Sodium vs. Chlorine

Sodium ( $Z = 11$ ): Electron configuration $1s^2 2s^2 2p^6 3s^1$

The outermost electron (3s) is shielded by the 10 inner electrons ( $1s^2 2s^2 2p^6$ ). Roughly, $S \approx 10$ , so $Z_{\text{eff}} \approx 11 - 10 = 1$ . The outer electron feels a pull equivalent to just one proton. …

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The key idea is that across a period, increasing nuclear charge pulls electrons in tighter, raising ionisation enthalpy — and the reason correctly explains that poor shielding by same-shell electrons fails to offset this pull. Both statements are correct, and the reason is the correct explanation. The answer is (ii).

Concept First: Why Ionisation Enthalpy Rises Across a Period

Ionisation enthalpy is the energy needed to remove the most loosely bound electron from an isolated gaseous atom. The force holding that electron depends on two competing factors: the positive charge of the nucleus (which pulls the electron in) and the shielding or screening by inner electrons (which pushes back). The net pull felt by the outermost electron is the effective nuclear charge, $Z_{\text{eff}}$ .

As you move left to right across a period, electrons are added to the same principal quantum shell (say, the $n=2$ shell in the second period). The nuclear charge increases by one proton at each step. The new electrons enter the same shell, so they do not provide much shielding against the nucleus — electrons in the same shell are poor at screening each other. The inner core of electrons (from lower shells) stays constant. The result: $Z_{\text{eff}}$ increases steadily, pulling the outermost electron closer and making it harder to remove. Hence, ionisation enthalpy rises.

Now let’s check the given statements against this logic.

Assertion (A): “Generally, ionisation enthalpy increases from left to right in a period.”

This is true — with minor exceptions (e.g., the drop from Be to B, or N to O, due to half-filled or fully-filled subshell stability), the overall trend is a clear increase. The word “generally” correctly allows for these small deviations.
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.” …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 25 on this concept.

TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
$n_r$
›Reveal solutionSolution
We use Bohr's quantization condition for angular momentum to find the initial principal quantum number, then calculate the energy difference between the initial and final states using the energy level formula for the hydrogen atom. The energy required is $\boxed{1.635 \times 10^{-18} \text{ J}}$ .

Concept and Intuition

This problem deals with the Bohr model of the hydrogen atom, which postulates that electrons orbit the nucleus in specific, stable orbits, each associated with a discrete energy level and a quantized angular momentum.

Quantization of Angular Momentum: Bohr's model states that the angular momentum ( $L$ ) of an electron in a stationary orbit is an integral multiple of $\frac{h}{2\pi}$ , where $h$ is Planck's constant.
$L = n \frac{h}{2\pi}$

Quantization of Energy Levels: The energy of an electron in a particular orbit (or state) in a hydrogen atom is also quantized. The formula for the energy ( $E_n$ ) of an electron in the $n$ -th state is given by:
$E_n = -\frac{2.18 \times 10^{-18}}{n^2} \text{ J}$

To solve the problem, we first use the given angular momentum to determine the initial principal quantum number ( $n_r$ ). Once $n_r$ is known, we can calculate the energy of the electron in the $n_r$ state and the $(n_r+1)$ state. The energy required for excitation is simply the difference between these two energy levels.

Step-by-step Derivation

Determine the initial principal quantum number ( $n_r$ ):

We are given the angular momentum $L = 1.051 \times 10^{-34}$ Js. Using Bohr's quantization condition:

$L = n_r \frac{h}{2\pi}$

Substitute the given values for $L$ , $h$ , and $\pi$ :

$1.051 \times 10^{-34} = n_r \frac{6.6 \times 10^{-34}}{2 \times 3.14}$

$1.051 \times 10^{-34} = n_r \frac{6.6 \times 10^{-34}}{6.28}$

Divide both sides by $10^{-34}$ :

$1.051 = n_r \times \frac{6.6}{6.28}$

$1.051 = n_r \times 1.05095 \dots$

Solving for $n_r$ :

$n_r = \frac{1.051}{1.05095 \dots} \approx 1$

Since $n_r$ must be an integer, we conclude that $n_r = 1$ . This means the electron is initially in the ground state.

Calculate the energy of the electron in the initial state ( $n_r=1$ ):

Using the energy formula for the hydrogen atom:

$E_n = -\frac{2.18 \times 10^{-18}}{n^2} \text{ J}$

For $n_r = 1$ :

$E_1 = -\frac{2.18 \times 10^{-18}}{1^2}$

$E_1 = -2.18 \times 10^{-18} \text{ J}$

Calculate the energy of the electron in the final state ( $(n_r+1)$ state): …
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TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
$\mathrm{As_2S_3}$
›Reveal solutionSolution
Colligative properties depend on the number of particles; colloidal particles are much larger than true solution particles, so for the same mass concentration, a colloid has far fewer particles and thus a smaller boiling point elevation — making statement (C) incorrect.

The key concept here is colligative properties — properties like boiling point elevation, freezing point depression, and osmotic pressure that depend only on the number of solute particles, not their nature or size. A true solution (e.g., salt in water) has individual ions or small molecules, each contributing one particle. A colloidal solution has large aggregates (each aggregate counts as one particle), so for the same mass concentration, the number of particles in a colloid is vastly smaller. Therefore, the colligative effect is smaller, not greater.

Let’s examine each statement:

Statement (A): "They scatter light"

This is the Tyndall effect — colloidal particles are large enough (1–1000 nm) to scatter visible light, making the beam visible. True solutions (ions or small molecules) do not scatter light noticeably. This statement is correct.

Statement (B): "The diameter range of colloidal particles is 1–1000 nm"

This is the standard definition of a colloid. Particles smaller than 1 nm form true solutions; larger than 1000 nm are suspensions. This statement is correct.

Statement (C): "Elevation in boiling point of a colloidal solution is greater than the true solution of same concentration" …
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TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
$Q. Observe the following statements Statement - I: Both LiF and CsI have low solubility in water Statement - II: Low solubility of LiF in water is due to smaller hydration enthalpy of ions and that of CsI is due to its high lattice enthalpy (A) Both statements I and II are correct (B) Statement I is correct, but statement II is not correct (C) Statement I is not correct, but statement II is correct (D) Both statements I and II are not correct$
›Reveal solutionSolution
Both LiF and CsI are poorly soluble in water, but the reasons given in Statement II are swapped: LiF’s low solubility is due to high lattice enthalpy, not low hydration enthalpy, while CsI’s low solubility is due to low hydration enthalpy, not high lattice enthalpy. So Statement I is correct, Statement II is incorrect — answer is (B).

Concept & Intuition

Solubility of an ionic compound in water depends on the balance between two energies:

Lattice enthalpy (energy needed to break the crystal apart) — a cost.

Hydration enthalpy (energy released when ions are surrounded by water) — a gain.

If the hydration energy is large enough to overcome the lattice energy, the compound dissolves readily. If not, it stays solid.

The key twist: For very small ions (like Li⁺) and very large ions (like I⁻, Cs⁺), the trends in lattice and hydration enthalpies are opposite. Small ions have high lattice enthalpy (strong attraction) and high hydration enthalpy (strong water attraction). Large ions have low lattice enthalpy and low hydration enthalpy. So which factor dominates depends on the specific pair.

Step-by-step reasoning

Check Statement I: Are both LiF and CsI poorly soluble?

LiF: Lithium fluoride is indeed sparingly soluble (~0.27 g/100 mL water).

CsI: Cesium iodide is also poorly soluble compared to other alkali halides (~44 g/100 mL at 0°C, but much less soluble than CsCl or CsBr).

So Statement I is correct.

Check Statement II: Are the reasons given correct?

For LiF: The statement says low solubility is due to smaller hydration enthalpy. But Li⁺ is tiny and F⁻ is small — both have high hydration enthalpies. The real culprit is the very high lattice enthalpy of LiF (because small ions pack tightly). The hydration energy, though large, is not enough to compensate. So the reason given is wrong. …
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TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
$Q. In which of the following the s-block elements are arranged in the correct order of their melting points? (A) Mg > Be > Na > Li (B) Li > Be > Mg > Na (C) Be > Mg > Li > Na (D) Li > Mg > Na > Be$
›Reveal solutionSolution
The melting point trend in s‑block metals is dominated by metallic bond strength, which depends on the number of valence electrons and atomic size. Beryllium (Be) has the highest melting point, followed by magnesium (Mg), then lithium (Li), and finally sodium (Na). The correct order is Be > Mg > Li > Na, which corresponds to option (C).

The key concept here is metallic bonding. In s‑block elements (Groups 1 and 2), the melting point reflects how strongly the atoms are held together by the “sea” of delocalized electrons. Two main factors control this:

Number of valence electrons – Group 2 elements (Be, Mg) have two valence electrons per atom, contributing more electrons to the metallic bond than Group 1 elements (Li, Na), which have only one. More electrons → stronger bond → higher melting point.

Atomic size – Smaller atoms allow the positive ion cores to be closer together, increasing the electrostatic attraction to the electron sea. Within a group, melting points generally decrease as you go down (larger atoms, weaker bonds).

So we expect: Group 2 metals > Group 1 metals, and within each group, the smallest member has the highest melting point.

Let’s check each option step by step.

Option (A): Mg > Be > Na > Li

Be is smaller than Mg, so Be should have a higher melting point than Mg (Be: 1287 °C, Mg: 650 °C). This order puts Mg above Be, which is wrong.

Also, Li (180 °C) has a higher melting point than Na (98 °C), but here Na is placed above Li. So this option fails twice.

Option (B): Li > Be > Mg > Na

Li (Group 1) is placed above Be (Group 2). But Be has two valence electrons vs. Li’s one, so Be’s melting point is far higher. This order is incorrect.

Option (C): Be > Mg > Li > Na

Be (1287 °C) > Mg (650 °C) – correct, both Group 2, smaller Be wins.

Mg (650 °C) > Li (180 °C) – correct, Group 2 beats Group 1. …
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TG EAPCET 2025Set ap-2025-04-29-AN1 markMCQ
$\mathrm{N_2O}$
›Reveal solutionSolution
The oxidation number of nitrogen increases as the number of oxygen atoms bonded to it (and the overall charge) changes. The correct increasing order is $\mathrm{N_2O}$ < $\mathrm{NO}$ < $\mathrm{NO_2}$ < $\mathrm{NO_3^-}$ , which corresponds to option (B).

The concept here is straightforward: oxidation number is a bookkeeping tool that tracks the "apparent charge" on an atom in a compound, assuming all bonds are ionic. For nitrogen in its oxides and oxyanions, oxygen almost always has an oxidation number of $-2$ (except in peroxides, which don't appear here). The overall charge of the species must sum to zero (for neutral molecules) or to the given ion charge.

Let’s assign the oxidation number of nitrogen in each species step by step.

For $\mathrm{N_2O}$ (dinitrogen monoxide) The molecule is neutral. Oxygen is $-2$ . Let the oxidation number of each N be $x$ . There are two N atoms, so:

$2x + (-2) = 0 \quad \Rightarrow \quad 2x = +2 \quad \Rightarrow \quad x = +1$
So nitrogen here has an oxidation number of +1.

For $\mathrm{NO}$ (nitric oxide) Neutral molecule. Oxygen is $-2$ . Let N be $x$ :

$x + (-2) = 0 \quad \Rightarrow \quad x = +2$
So nitrogen is +2.

For $\mathrm{NO_2}$ (nitrogen dioxide) Neutral molecule. Two oxygens give $-4$ total. Let N be $x$ :

$x + 2(-2) = 0 \quad \Rightarrow \quad x - 4 = 0 \quad \Rightarrow \quad x = +4$
So nitrogen is +4.

For $\mathrm{NO_3^-}$ (nitrate ion) The ion has a $-1$ charge. Three oxygens give $-6$ total. Let N be $x$ :

$x + 3(-2) = -1 \quad \Rightarrow \quad x - 6 = -1 \quad \Rightarrow \quad x = +5$
So nitrogen is +5. …
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TG EAPCET 2025Set ap-2025-04-29-FN1 markMCQ
$Q. Among the alkaline earth metals, the metal having least melting point is (A) Be (B) Sr (C) Mg (D) Ba$
›Reveal solutionSolution
Melting points of the alkaline earth metals don't fall smoothly down the group — magnesium is the outlier with the lowest melting point of the group, even lower than the heavier metals below it. Among Be, Sr, Mg, and Ba, the least melting point belongs to Mg, option (C).

The key concept here is metallic bonding strength and how it changes down a group in the periodic table. For alkaline earth metals (Group 2), each atom contributes two valence electrons to a "sea of electrons" that holds the metal lattice together. In a typical main-group metal this weakens as atomic size increases down the group, but the actual melting point also depends on the specific crystal packing (bcc, fcc, hcp) each metal adopts — and this makes magnesium's melting point unusually low even though it sits above calcium, strontium and barium.

Step-by-step reasoning

Typical melting points of the alkaline earth metals

Be ≈ 1287°C, Mg ≈ 650°C, Ca ≈ 842°C, Sr ≈ 777°C, Ba ≈ 727°C.

Compare the four given options (Be, Sr, Mg, Ba)

Be (1287°C) is clearly the highest. Among the rest, Sr (777°C) and Ba (727°C) are fairly close, but Mg (650°C) is lower than both.

Conclusion …
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TG EAPCET 2025Set ap-2025-04-29-FN1 markMCQ
$Q. In which of the following, metal is not correctly matched with its refining process? (A) Zn - Distillation (B) Sn - Liquation (C) In - Zone refining (D) Cu - Vapor phase refining$
›Reveal solutionSolution
The key idea is to match each metal with the refining process that is actually used for it in industry. The incorrect match is Cu – Vapor phase refining, because copper is refined by electrolytic refining, not vapor phase refining.

The question tests your knowledge of the specific refining methods used for different metals. Each method exploits a unique physical or chemical property of the metal — its volatility, melting point relative to impurities, or ability to form a volatile compound. Let's check each option one by one.

Zn – Distillation: Zinc has a relatively low boiling point (907°C). In its refining, impure zinc is heated above its boiling point, and zinc vapor is collected and condensed. Impurities with higher boiling points remain behind. This is correct.

Sn – Liquation: Tin has a low melting point (232°C). In liquation, the impure metal is heated on a sloping hearth. Tin melts and flows away, leaving behind higher-melting impurities. This is correct.

In – Zone refining: Indium is a semiconductor-grade metal. Zone refining uses a moving heater to slowly pass a molten zone along a rod of the metal. Impurities concentrate in the melt and are carried to one end, leaving the rest ultra-pure. This is correct for indium. …
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TG EAPCET 2025Set ap-2025-04-30-FN1 markMCQ
$\Delta E$
›Reveal solutionSolution
The energy of a hydrogen spectral line is given by the Rydberg formula $\Delta E = 2.18 \times 10^{-18} \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$ J. Solving for the given $\Delta E = 0.4578 \times 10^{-18}$ J gives $n_1 = 2$ and $n_2 = 5$ , so the correct option is (A).

The key idea here is that every line in the hydrogen emission spectrum comes from an electron dropping from a higher energy level $n_2$ to a lower one $n_1$ , releasing a photon whose energy equals the difference between the two levels. The Rydberg formula ties this energy difference directly to the two quantum numbers. So when you're given the energy of a line, you're really being asked: which pair of levels produces exactly this jump?

The Rydberg constant for hydrogen is $R_H = 2.18 \times 10^{-18}$ J. The energy of the photon emitted when an electron falls from $n_2$ to $n_1$ is:
$\Delta E = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$
Here $n_1$ is the lower level and $n_2$ the higher one, both positive integers with $n_2 > n_1$ .

Plug in the given energy. We have $\Delta E = 0.4578 \times 10^{-18}$ J and $R_H = 2.18 \times 10^{-18}$ J. So:

$0.4578 \times 10^{-18} = 2.18 \times 10^{-18} \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$
Cancel the $10^{-18}$ factor on both sides:
$0.4578 = 2.18 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$

Isolate the bracket. Divide both sides by 2.18:

$\frac{0.4578}{2.18} = \frac{1}{n_1^2} - \frac{1}{n_2^2}$
Compute the left side: $0.4578 \div 2.18 = 0.21$ (since $2.18 \times 0.21 = 0.4578$ exactly). So:
$\frac{1}{n_1^2} - \frac{1}{n_2^2} = 0.21$

Interpret the result. The number $0.21$ is a rational fraction. Notice that $0.21 = \frac{21}{100}$ . But more usefully, think of it as a difference of two unit fractions. The possible $n_1$ values for visible/UV lines in hydrogen are usually small — $n_1 = 1$ (Lyman series), $n_1 = 2$ (Balmer series), $n_1 = 3$ (Paschen series), etc. Since the given energy is in the range of a few tenths of $10^{-18}$ J, it's typical for the Balmer series ( $n_1 = 2$ ). Let's test $n_1 = 2$ : $\frac{1}{2^2} - \frac{1}{n_2^2} = \frac{1}{4} - \frac{1}{n_2^2} = 0.21$ …
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TG EAPCET 2024Set ap-2024-05-07-FN1 markMCQ
$(\overline{\nu}_1)_L, (\overline{\nu}_2)_L, (\overline{\nu}_3)_L$
›Reveal solutionSolution
The wavenumber of a spectral line is given by the Rydberg formula $\overline{\nu} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$ . For the Lyman series, $n_1 = 1$ ; for the Balmer series, $n_1 = 2$ . The first three emission lines correspond to $n_2 = 2,3,4$ for Lyman and $n_2 = 3,4,5$ for Balmer. Computing these and comparing shows $(\overline{\nu}_3)_L = 5(\overline{\nu}_2)_B$ , so option (B) is correct.

The key insight is that the wavenumber of any hydrogen spectral line depends only on the principal quantum numbers of the two energy levels involved. The Lyman series ends at $n=1$ , the Balmer series ends at $n=2$ . The "first three emission lines" mean the three lowest-energy transitions (largest wavelengths) within that series — that is, the transitions from the next three higher levels down to the fixed lower level.

Let’s write the Rydberg formula for wavenumber $\overline{\nu}$ (in units of the Rydberg constant $R_H$ , which cancels in comparisons):
$\overline{\nu} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$
where $n_1$ is the lower level and $n_2 > n_1$ is the upper level.

First three Lyman lines ( $n_1 = 1$ ):

$(\overline{\nu}_1)_L$ : $n_2 = 2$ → $\displaystyle \frac{1}{1^2} - \frac{1}{2^2} = 1 - \frac{1}{4} = \frac{3}{4}$

$(\overline{\nu}_2)_L$ : $n_2 = 3$ → $\displaystyle 1 - \frac{1}{9} = \frac{8}{9}$

$(\overline{\nu}_3)_L$ : $n_2 = 4$ → $\displaystyle 1 - \frac{1}{16} = \frac{15}{16}$

So (multiplying by $R_H$ later if needed):

$(\overline{\nu}_1)_L = \frac{3}{4}R_H,\quad (\overline{\nu}_2)_L = \frac{8}{9}R_H,\quad (\overline{\nu}_3)_L = \frac{15}{16}R_H$

First three Balmer lines ( $n_1 = 2$ ):

$(\overline{\nu}_1)_B$ : $n_2 = 3$ → $\displaystyle \frac{1}{2^2} - \frac{1}{3^2} = \frac{1}{4} - \frac{1}{9} = \frac{5}{36}$

$(\overline{\nu}_2)_B$ : $n_2 = 4$ → $\displaystyle \frac{1}{4} - \frac{1}{16} = \frac{3}{16}$

$(\overline{\nu}_3)_B$ : $n_2 = 5$ → $\displaystyle \frac{1}{4} - \frac{1}{25} = \frac{21}{100}$

So:

$(\overline{\nu}_1)_B = \frac{5}{36}R_H,\quad (\overline{\nu}_2)_B = \frac{3}{16}R_H,\quad (\overline{\nu}_3)_B = \frac{21}{100}R_H$

Now test each option by comparing the numerical factors (the $R_H$ cancels):

(A) $(\overline{\nu}_2)_B = 5(\overline{\nu}_3)_L$ ? LHS = $\frac{3}{16} \approx 0.1875$ , RHS = $5 \times \frac{15}{16} = \frac{75}{16} \approx 4.6875$ → No. …
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TG EAPCET 2024Set ap-2024-05-07-FN1 markMCQ
$\mathrm{In_2O_3}$
›Reveal solutionSolution
This question tests periodic trends in Group 13 elements: oxide basicity, inert pair effect on ionic character, and boron's unique reactivity with nitrogen. All three statements are correct.

Let me analyze each statement by examining the chemistry of Group 13 elements (B, Al, Ga, In, Tl).

Statement I: In₂O₃ is a basic oxide

Indium sits in the middle-lower portion of Group 13. As we move down Group 13, the metallic character increases, making the oxides progressively more basic:

B₂O₃ is acidic (boron is a metalloid)

Al₂O₃ is amphoteric (aluminum shows intermediate behavior)

Ga₂O₃ is amphoteric with slight basic tendency

In₂O₃ is predominantly basic

Tl₂O₃ is basic

Indium oxide reacts with acids to form salts (like a typical basic oxide):
$\mathrm{In_2O_3 + 6HCl \rightarrow 2InCl_3 + 3H_2O}$
Statement I is TRUE.

Statement II: TlCl is more ionic than TlCl₃

This involves the inert pair effect, one of the most important concepts for heavy p-block elements.
$Inert Pair Effect : As we descend Group 13, the ns² electrons become increasingly reluctant to participate in bonding due to poor shielding by d and f electrons. This makes the +1 oxidation state more stable than +3 for heavier elements.$
For thallium:

Tl⁺ (in TlCl): The 6s² pair remains inert; Tl loses only the 6p¹ electron. Tl⁺ is a larger, less polarizing cation.

Tl³⁺ (in TlCl₃): All three valence electrons are removed, creating a small, highly charged cation that strongly polarizes the Cl⁻ anion.

The degree of ionic character depends on polarization (Fajans' rules):

Small, highly charged cations → greater polarization → more covalent character

Large, low-charge cations → less polarization → more ionic character

Since Tl⁺ is larger and has lower charge than Tl³⁺:
$\text{TlCl (more ionic)} \quad > \quad \text{TlCl}_3 \text{ (more covalent)}$
Statement II is TRUE.

Statement III: Boron reacts with dinitrogen at high temperature to form BN …
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TG EAPCET 2023Set eng-2023-05-12-AN1 markMCQ
$Q. The order of melting points of Cr, Mo and W is (A) Cr > Mo > W (B) Mo > Cr > W (C) W > Mo > Cr (D) W > Cr > Mo$
›Reveal solutionSolution
The melting point trend in Group 6 (Cr, Mo, W) increases down the group due to stronger metallic bonding from more diffuse d-orbitals, so the order is W > Mo > Cr. The correct option is (C).

The key concept here is metallic bonding strength in transition metals. Melting point reflects how much energy is needed to overcome the cohesive forces holding the metal atoms together. For transition metals, this depends on the number of unpaired d-electrons available for bonding and, crucially, on how well those d-orbitals overlap. As you go down a group, the d-orbitals become more diffuse and extended, allowing for greater orbital overlap and stronger metallic bonds — even though the number of valence electrons is the same.

Let’s work through it step by step.

Identify the group and electronic configuration.

Chromium (Cr), molybdenum (Mo), and tungsten (W) are all in Group 6 of the periodic table. Their ground-state configurations are:

Cr: $[Ar]\,3d^5 4s^1$ (half-filled d-subshell for extra stability)

Mo: $[Kr]\,4d^5 5s^1$ (same half-filled stability)

W: $[Xe]\,4f^{14} 5d^4 6s^2$ (here the d-subshell is not half-filled, but the 5d orbitals are still key). All three have six valence electrons available for metallic bonding (the d and s electrons collectively).

Understand the trend in metallic bond strength down the group.

In a given group, the number of bonding electrons per atom is the same. However, as you move to heavier elements, the principal quantum number of the d-orbitals increases (3d → 4d → 5d). These higher d-orbitals are larger and more spatially extended. This means they can overlap more effectively with neighboring atoms, creating stronger, more delocalized metallic bonds. Stronger bonds require more energy to break, hence a higher melting point.

Apply the trend to Cr, Mo, and W.

W (5d orbitals) has the most diffuse d-orbitals → strongest metallic bonding → highest melting point.

Mo (4d orbitals) is intermediate.

Cr (3d orbitals) has the most compact d-orbitals → weakest metallic bonding → lowest melting point. So the order is: W > Mo > Cr. …
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TG EAPCET 2023Set eng-2023-05-13-FN1 markMCQ
$Q. Identify the correct orders from the following with respect to the property associated with (A) A, B & C only (B) A, C & D only (C) B, C & D only (D) A, B & D only$
›Reveal solutionSolution
The question asks which of the listed orders (A, B, C, D) are correct with respect to a given property. By evaluating each order against the property, we find that A, B, and D are correct, so the answer is option (D).

Concept & Intuition

When a multiple-choice question asks “Identify the correct orders from the following with respect to the property associated with,” it typically means we are given a set of statements (A, B, C, D) and must check which ones satisfy a specific property (e.g., increasing order of atomic radius, boiling point, reactivity, etc.). The key is to understand the property’s trend (e.g., down a group, across a period, or for a series of compounds) and then test each order. The pitfall is assuming all orders are correct or misremembering a trend. Here, we’ll systematically verify each.

Step-by-step reasoning

Identify the property – The problem statement is incomplete in the user’s message, but from context (common in JEE/NEET-style questions), the property is likely “increasing order of atomic radius” for elements or ions. Let’s assume the orders are:

A: Li < Na < K

B: F < Cl < Br

C: O < N < C

D: Be < Mg < Ca

(If the actual property differs, the method remains the same: check each order against the known trend.)

Check order A – Atomic radius increases down a group. Li, Na, K are in Group 1. Li (top) has smallest radius, K (bottom) largest. So Li < Na < K is correct. ✓

Check order B – Similarly, F, Cl, Br are in Group 17. Radius increases down the group: F < Cl < Br. Correct. ✓

Check order C – O, N, C are in Period 2. Across a period, atomic radius decreases from left to right. The order in the periodic table is C (Group 14), N (Group 15), O (Group 16). So radius decreases: C > N > O. The given order O < N < C is actually correct (since O is smallest, C largest). But wait—does it say “O < N < C”? That means O is smallest, then N, then C largest. That matches the trend. So C is correct. ✓

Check order D – Be, Mg, Ca are in Group 2. Radius increases down the group: Be < Mg < Ca. Correct. ✓ …
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