Q.Which of the following species will have the largest and the smallest size? $Mg,\ Mg^{2+},\ Al,\ Al^{3+}$.

Showing the 12 most recent of 23 on this concept.

1. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.Which of the following is not an example of covalent solid? (A) $\mathrm{SiC}$ (B) $\mathrm{SiO_2}$ (C) $\mathrm{AlN}$ (D) $\mathrm{ZnS}$

   ›Reveal solutionSolution

   Covalent solids are held together by a continuous network of covalent bonds. Among the options, ZnS is an ionic solid, not a covalent network solid. The correct answer is (D).

   The key to this question lies in understanding what defines a covalent solid (also called a network solid). These are substances where atoms are bonded to each other in a continuous, three-dimensional network entirely through covalent bonds. The entire crystal is essentially one giant molecule. Common examples include diamond (carbon), silicon carbide (SiC), and silicon dioxide (SiO₂).

   In contrast, many compounds of metals with non-metals form ionic solids, where the bonding is electrostatic between positive and negative ions. The distinction is not always sharp — some compounds have mixed bonding — but for the purpose of this classification, we look at the dominant bond type and the structure.

   Let’s examine each option:

   1. SiC (Silicon carbide) — Silicon and carbon are both non-metals with similar electronegativities. They form a tetrahedral network of covalent bonds, exactly like diamond but with alternating Si and C atoms. This is a classic covalent solid.

   2. SiO₂ (Silicon dioxide) — Each silicon is covalently bonded to four oxygen atoms, and each oxygen bridges two silicon atoms, creating a continuous covalent network. Quartz is a well-known example. Definitely a covalent solid.

   3. AlN (Aluminium nitride) — Aluminium and nitrogen have an electronegativity difference of about 1.5, which is borderline. However, AlN crystallizes in the wurtzite structure (like ZnS) but with significant covalent character due to the small size and high charge of the ions. In many exam contexts, AlN is considered a covalent network solid because it forms a continuous tetrahedral structure with strong directional bonds. It is often grouped with SiC and other III-V compounds as covalent.

   4. ZnS (Zinc sulfide) — Zinc is a metal, sulfur is a non-metal. The electronegativity difference is about 0.9, which is moderate. ZnS crystallizes in the zinc blende (sphalerite) structure, which is an ionic lattice with Zn²⁺ and S²⁻ ions held together by electrostatic forces. Although there is some covalent character, the bonding is predominantly ionic. In standard classification, ZnS is an ionic solid, not a covalent network solid.

   Watch out

   A common mistake is to think that any compound with a tetrahedral structure (like ZnS) must be covalent. But the zinc blende structure is also adopted by many ionic compounds. The nature of bonding — not just the geometry — decides the classification.

   Tip

   A quick rule of thumb: Covalent solids are typically formed between non-metals (e.g., C, Si, N, O) with small electronegativity differences. If a metal is involved, the compound is usually ionic unless the metal is very small and highly charged (like Al in AlN, which is a special case).

   ✓Final answer

   The compound that is not an example of a covalent solid is (D) ZnS.

2. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.The pair of elements which do not give super oxides with excess of oxygen is (A) K, Rb (B) Na, K (C) Li, Mg (D) K, Ba

   ›Reveal solutionSolution

   The key idea is that only large alkali metals (K, Rb, Cs) form stable superoxides; small alkali metals (Li, Na) and alkaline earth metals (Mg, Ba) form normal oxides or peroxides instead. The pair that does not give superoxides is Li and Mg — option (C).

   Concept & Intuition

   Superoxides ($MO_2$) contain the $O_2^-$ ion, which is large and requires a large cation to stabilize it via lattice energy. Small cations like Li⁺ and Mg²⁺ cannot stabilize the bulky $O_2^-$ ion; they instead form normal oxides ($M_2O$ or $MO$) or peroxides ($M_2O_2$). Only the largest alkali metals (K, Rb, Cs) readily form superoxides. Ba²⁺ is large enough to form a peroxide ($BaO_2$) but not a superoxide under normal conditions.

   Step-by-step reasoning

   1. Recall the trend in superoxide formation

      • Group 1 (alkali metals): Li forms only $Li_2O$; Na forms $Na_2O_2$ (peroxide); K, Rb, Cs form $MO_2$ (superoxides).
      • Group 2 (alkaline earth metals): Be, Mg form normal oxides ($MO$); Ca, Sr, Ba form peroxides ($MO_2$) but not superoxides (except under extreme conditions).

   2. Analyze each option

      • (A) K, Rb: Both are large alkali metals — they do form superoxides ($KO_2$, $RbO_2$).
      • (B) Na, K: Na forms peroxide, K forms superoxide — so K gives a superoxide, but Na does not. The pair includes one that does, so this is not the correct answer (the question asks for a pair where neither gives a superoxide).
      • (C) Li, Mg: Li forms only $Li_2O$; Mg forms only $MgO$ — neither forms a superoxide. This matches the requirement.
      • (D) K, Ba: K forms superoxide; Ba forms peroxide ($BaO_2$) but not superoxide — again, one does, so not the answer.

   3. Confirm the exception

      • Li⁺ is too small to stabilize $O_2^-$; Mg²⁺ is even smaller and has a +2 charge that would destabilize the large anion. Both are “hard” cations that prefer small oxide ions ($O^{2-}$).

   Watch out

   A common mistake is to think that all alkali metals form superoxides. Remember: only K, Rb, Cs do. Na gives peroxide, Li gives normal oxide.

   Tip

   A quick memory aid: “Superoxides are for super-sized cations” — the cation must be as large as K⁺ or bigger.

   ✓Final answer

   The correct option is (C).

   ANSWER: C

3. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.Identify the pair of metals which are refined by zone refining method? (A) Zr, Ti (B) Ga, Ti (C) Ga, In (D) Al, Cu

   ›Reveal solutionSolution

   Zone refining is used for metals that need ultra-high purity, typically those with low melting points like Ga and In; the correct pair is (C) Ga, In.

   Concept & Intuition

   Zone refining is a purification technique that exploits differences in solubility of impurities in the solid vs. liquid phase. A narrow molten zone is moved slowly along a rod of the metal; impurities concentrate in the liquid and are swept to one end. This method is most practical for metals that melt at moderate temperatures and are required in extremely pure form (e.g., for semiconductors). Common examples are gallium (Ga), indium (In), and sometimes silicon or germanium. Metals like zirconium (Zr), titanium (Ti), aluminum (Al), and copper (Cu) are typically refined by other methods (e.g., Kroll process, electrolytic refining) because of their high melting points or different impurity behavior.

   Step-by-step reasoning

   1. Recall the principle of zone refining – It works by repeatedly passing a heated zone along a solid rod. Impurities that lower the melting point stay in the liquid zone and are carried to the end. The process is slow and energy-intensive, so it’s reserved for materials where extreme purity is critical.

   2. Identify typical candidates – The metals most commonly refined by zone refining are those with relatively low melting points and high value in pure form: gallium (Ga, mp ~30°C), indium (In, mp ~157°C), and sometimes bismuth, antimony, or tellurium. Silicon and germanium are also zone-refined, but they are metalloids.

   3. Evaluate each option

      • (A) Zr, Ti – Both are refractory metals (high melting points: Zr ~1855°C, Ti ~1668°C). They are refined by chemical methods (e.g., Kroll process) or vacuum arc melting, not zone refining.
      • (B) Ga, Ti – Ga is zone-refined, but Ti is not (see above). Mixed pair, so incorrect.
      • (C) Ga, In – Both have low melting points and are classic examples of metals purified by zone refining. This matches the principle.
      • (D) Al, Cu – Al and Cu are refined electrolytically (Hall–Héroult for Al, electrorefining for Cu). Zone refining is not standard for them.

   4. Confirm the correct pair – Only option (C) lists two metals that are both routinely zone-refined for high-purity applications (e.g., in electronics and semiconductor doping).

   Watch out

   A common mistake is to think that any metal can be zone-refined. In reality, the method is impractical for high-melting-point metals because of equipment limitations and slow throughput.

   Tip

   Remember the mnemonic: “Gallium and Indium are zone buddies” – they melt in your hand or in hot water, making them ideal for this technique.

   ✓Final answer

   The correct option is (C).

   ANSWER: C

4. TG EAPCET 2026Set ap-2026-05-04-AN1 markMCQ
   Q.The metal oxides which cannot be reduced by CO are I. $\mathrm{Li_2O}$ II. $\mathrm{MgO}$ III. $\mathrm{Al_2O_3}$ IV. $\mathrm{ZnO}$ The correct answer is Options : (A) II, III, IV only (B) I, III, IV only (C) I, II, III only (D) II, III only

   ›Reveal solutionSolution

   Carbon monoxide (CO) can reduce metal oxides where the metal has a lower affinity for oxygen than carbon. This means CO cannot reduce oxides of highly reactive metals like lithium, magnesium, and aluminium. Therefore, $\mathrm{Li_2O}$, $\mathrm{MgO}$, and $\mathrm{Al_2O_3}$ cannot be reduced by CO. The correct option is (C).

   Concept and Intuition

   Reduction, in the context of metal oxides, refers to the removal of oxygen from the metal oxide to obtain the free metal. A reducing agent is a substance that facilitates this process by taking away the oxygen. Carbon monoxide (CO) is a common reducing agent used in metallurgy.

   The ability of CO to reduce a metal oxide depends on the relative stability of the metal oxide compared to carbon dioxide ($\mathrm{CO_2}$). Essentially, for CO to reduce a metal oxide ($\mathrm{M_xO_y}$), the metal (M) must have a weaker affinity for oxygen than carbon does. If the metal has a stronger affinity for oxygen, its oxide will be more stable than $\mathrm{CO_2}$, and CO will not be able to reduce it.

   This concept is directly related to the position of the metal in the reactivity series. Metals high in the reactivity series (like alkali metals, alkaline earth metals, and aluminium) form very stable oxides because they have a strong affinity for oxygen. These oxides are difficult to reduce and typically require electrolytic reduction or reduction by very strong reducing agents (like active metals). Metals lower in the reactivity series (like zinc, iron, lead, copper) form less stable oxides, which can often be reduced by carbon or carbon monoxide at elevated temperatures.

   Important

   A general rule of thumb is that carbon (and carbon monoxide) can reduce oxides of metals that are less reactive than carbon itself. In the reactivity series, metals like K, Na, Ca, Mg, Al are more reactive than carbon, so their oxides cannot be reduced by carbon or CO. Metals like Zn, Fe, Pb, Cu are less reactive than carbon, and their oxides can be reduced by carbon or CO.

   Step-by-Step Analysis

   Let's examine each metal oxide based on the reactivity of the metal:

   1. $\mathrm{Li_2O}$ (Lithium oxide):

      Lithium is an alkali metal and is extremely high in the reactivity series. It has a very strong affinity for oxygen, forming a highly stable oxide. Therefore, $\mathrm{Li_2O}$ cannot be reduced by a moderately strong reducing agent like CO.

   2. $\mathrm{MgO}$ (Magnesium oxide):

      Magnesium is an alkaline earth metal, also very high in the reactivity series (just below calcium). It forms a very stable oxide due to its strong affinity for oxygen. Consequently, $\mathrm{MgO}$ cannot be reduced by CO.

   3. $\mathrm{Al_2O_3}$ (Aluminium oxide):

      Aluminium is a reactive metal, positioned high in the reactivity series. Its oxide, $\mathrm{Al_2O_3}$, is exceptionally stable and has a very high melting point. It is typically reduced by electrolysis (Hall-Héroult process) rather than by chemical reducing agents like carbon or CO. Thus, $\mathrm{Al_2O_3}$ cannot be reduced by CO.

   4. $\mathrm{ZnO}$ (Zinc oxide):

      Zinc is a moderately reactive metal, positioned below aluminium but above iron in the reactivity series. Its affinity for oxygen is less than that of lithium, magnesium, or aluminium. $\mathrm{ZnO}$ can be reduced by carbon monoxide at high temperatures (typically above $900 \ ^\circ\mathrm{C}$), as shown by the reaction:

   $\mathrm{ZnO(s) + CO(g) \xrightarrow{\text{high temp.}} Zn(s) + CO_2(g)}$

   Therefore, $\mathrm{ZnO}$ *can* be reduced by CO.
   

   Based on this analysis, the metal oxides which cannot be reduced by CO are $\mathrm{Li_2O}$, $\mathrm{MgO}$, and $\mathrm{Al_2O_3}$. These correspond to options I, II, and III.

   ✓Final answer

   The metal oxides which cannot be reduced by CO are $\mathrm{Li_2O}$, $\mathrm{MgO}$, and $\mathrm{Al_2O_3}$. The correct option is (C).

5. TG EAPCET 2026Set ap-2026-05-04-FN1 markMCQ
   Q.Dihydrogen reacted with elements of atomic number 15, 19 and 40 to form respective hydrides. The type of these hydrides respectively are (A) ionic, interstitial, electron rich (B) electron rich, ionic, interstitial (C) ionic, electron rich, interstitial (D) electron rich, interstitial, ionic

   ›Reveal solutionSolution

   The type of hydride depends on the element's electronegativity and position in the periodic table. The correct order is electron rich, ionic, interstitial, which corresponds to option (B).

   The question asks you to classify the hydrides formed when dihydrogen reacts with elements of atomic numbers 15, 19, and 40. The key is to identify each element first, then understand how its nature dictates the type of hydride it forms.

   Hydrides are broadly classified based on the bonding and structure:

   • Ionic (or saline) hydrides form with highly electropositive metals (typically group 1 and 2), where hydrogen acts as H⁻.
   • Covalent (or molecular) hydrides form with non-metals and metalloids. Within these, "electron-rich" hydrides have excess lone pairs (e.g., NH₃, H₂O, HF).
   • Interstitial (or metallic) hydrides form with transition metals, where hydrogen atoms occupy holes in the metal lattice without forming discrete molecules.

   Let's identify the elements and apply this logic.

   1. Atomic number 15: Phosphorus (P). Phosphorus is a non-metal in group 15. It forms a covalent hydride, PH₃ (phosphine). The phosphorus atom has a lone pair of electrons, making this an electron-rich hydride (since it has more than the required electrons for simple bonding). So the first hydride is electron-rich.

   2. Atomic number 19: Potassium (K). Potassium is an alkali metal in group 1. It is highly electropositive. When it reacts with hydrogen, it forms KH (potassium hydride), where hydrogen gains an electron to become H⁻. This is a classic ionic hydride. So the second hydride is ionic.

   3. Atomic number 40: Zirconium (Zr). Zirconium is a transition metal (group 4, period 5). Transition metals typically form interstitial hydrides (also called metallic hydrides). Hydrogen atoms occupy the interstitial sites in the metal lattice, and the compound retains metallic properties. So the third hydride is interstitial.

   Watch out

   A common mistake is to misidentify atomic number 40 as something else (like 40 is not Zr if you confuse it with atomic mass). Always use the periodic table: 15 = P, 19 = K, 40 = Zr. Also, don't confuse "electron-rich" with "electron-deficient" — electron-rich hydrides have lone pairs (like NH₃, PH₃, H₂O), while electron-deficient ones (like B₂H₆) have fewer electrons than needed for normal covalent bonds.

   Thus, the sequence is: electron-rich (for P), ionic (for K), interstitial (for Zr).

   ✓Final answer

   The correct option is (B).

6. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.Electron gain enthalpy values ($\Delta_{eg}H$) (in kJ mol$^{-1}$) of elements X, Y and Z are $-349$, $-200$ and $-295$ respectively. X, Y and Z are respectively (A) Cl, I, S (B) Cl, S, I (C) S, Se, Te (D) Na, K, Rb

   ›Reveal solutionSolution

   The key idea is that electron gain enthalpy becomes more negative (more exothermic) as we move up a group and across a period, so the most negative value corresponds to chlorine, the least negative to iodine, and the intermediate to sulfur — making the correct option (A).

   The concept here is electron gain enthalpy ($\Delta_{eg}H$): the energy change when an isolated gaseous atom gains an electron. A more negative value means the atom releases more energy, indicating a stronger tendency to accept an electron. For nonmetals, this becomes more negative as we go up a group (smaller atom, stronger nuclear pull) and across a period (higher effective nuclear charge). The given values are:

   • X: $-349$ kJ/mol
   • Y: $-200$ kJ/mol
   • Z: $-295$ kJ/mol

   We need to match these to the elements in the options.

   1. Identify the most negative value — X has $-349$ kJ/mol, the most exothermic. Among the given elements, chlorine (Cl) is known to have the most negative electron gain enthalpy in its group (Group 17), around $-349$ kJ/mol. So X is likely Cl.

   2. Identify the least negative value — Y has $-200$ kJ/mol, the least exothermic. In Group 17, iodine (I) has a less negative value than chlorine (about $-295$ kJ/mol for I, but actually iodine's value is around $-295$ kJ/mol — wait, that matches Z). Let's check: iodine's $\Delta_{eg}H$ is about $-295$ kJ/mol, so Z could be I. Then Y, with $-200$, must be something else. Among the options, sulfur (S) has $\Delta_{eg}H \approx -200$ kJ/mol, which fits Y perfectly.

   3. Confirm the intermediate — Z is $-295$ kJ/mol. Iodine's value is indeed around $-295$ kJ/mol, so Z = I. This gives the order: X = Cl, Y = S, Z = I.

   4. Check the options — (A) Cl, I, S — no, that order is Cl, I, S, but we have Cl, S, I. Wait, careful: The question says "X, Y and Z are respectively" meaning X is first, Y second, Z third. Our assignment: X = Cl, Y = S, Z = I. That matches option (B): Cl, S, I.

   5. Verify other options — (C) S, Se, Te: these have values around $-200$, $-195$, $-190$ — none as negative as $-349$. (D) Na, K, Rb: these are metals with positive or slightly negative values (Na: $-53$, K: $-48$, Rb: $-47$), nowhere near $-349$. So only (B) fits.

   Watch out

   A common mistake is to assume that the most negative value always belongs to the smallest element in a group, but remember that fluorine is an exception — its electron gain enthalpy is less negative than chlorine due to electron-electron repulsion in its compact 2p subshell. Here, chlorine is correctly the most negative.

   Tip

   You can memorize the approximate $\Delta_{eg}H$ values for Group 17: Cl ($-349$), F ($-328$), Br ($-325$), I ($-295$). For Group 16: O ($-141$), S ($-200$), Se ($-195$), Te ($-190$). This makes matching quick.

   ✓Final answer

   The correct option is (B).

   ANSWER: B

7. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
   Q.Consider the following pairs of elements and identify the pairs of elements which have nearly same atomic radius. I. Y, La II. Zr, Hf III. Mo, W IV. Cr, Mo (A) II & III (B) I & II (C) I & III (D) III & IV

   ›Reveal solutionSolution

   The key idea is that lanthanoid contraction causes elements in the same group of the second and third transition series (like Zr–Hf and Mo–W) to have nearly identical atomic radii. Pairs I (Y, La) and IV (Cr, Mo) are from different periods and do not show this effect. The correct answer is (A).

   ---

   Concept & Intuition

   Why do Zr and Hf have almost the same size, but Y and La do not? The answer lies in a subtle but powerful phenomenon called lanthanoid contraction. As we move across the lanthanide series (elements 58–71), the 4f orbitals are being filled. These f-electrons are poor at shielding the nuclear charge, so the effective nuclear pull on the outer electrons increases, causing the atomic radius to shrink gradually across the series. This contraction is so significant that it almost exactly cancels the expected size increase from moving down one full period. Consequently, elements directly below each other in the second and third transition series (e.g., Zr below Hf, Mo below W) end up with nearly identical radii. This is not true for the first transition series (like Cr) compared to the second (Mo), nor for elements from different groups (like Y and La).

   ---

   Step-by-Step Reasoning

   1. Understand the pairs in context of the periodic table.

      • Pair I: Y (Group 3, Period 5) and La (Group 3, Period 6) — same group, different periods.
      • Pair II: Zr (Group 4, Period 5) and Hf (Group 4, Period 6) — same group, different periods.
      • Pair III: Mo (Group 6, Period 5) and W (Group 6, Period 6) — same group, different periods.
      • Pair IV: Cr (Group 6, Period 4) and Mo (Group 6, Period 5) — same group, different periods.

   2. Apply the lanthanoid contraction effect.

      The lanthanoid contraction occurs between Periods 5 and 6. Elements in Period 6 (like Hf, W) come after the 4f series, so their radii are unexpectedly small. For Zr and Hf, the expected increase in radius from Period 5 to Period 6 is almost completely offset by the contraction, making their radii nearly equal (~160 pm). The same holds for Mo and W (~139 pm).

   3. Check Pair I: Y and La.

      Y is in Period 5, La is in Period 6 but before the 4f series begins (La is actually the first element of the lanthanides). La does not experience the lanthanoid contraction relative to Y — in fact, La is significantly larger than Y (atomic radii: Y ~180 pm, La ~187 pm). So they do not have nearly the same radius.

   4. Check Pair IV: Cr and Mo.

      Cr is in Period 4 (first transition series), Mo in Period 5 (second transition series). There is no lanthanoid contraction between Periods 4 and 5 — only between 5 and 6. So Cr (~128 pm) and Mo (~139 pm) have a noticeable size difference. They are not nearly the same.

   5. Identify the correct pairs.

      Only pairs II (Zr, Hf) and III (Mo, W) satisfy the condition of nearly identical atomic radii due to lanthanoid contraction.

   Tip

   A quick memory aid: The "twin pairs" from lanthanoid contraction are always the second and third transition series in the same group — Zr–Hf, Nb–Ta, Mo–W, Tc–Re, Ru–Os, Rh–Ir, Pd–Pt. The first transition series (Period 4) is never involved.

   Watch out

   A common mistake is to think that Y and La are similar because they are both in Group 3. But La is the first lanthanide and is noticeably larger than Y. The contraction only kicks in after La, so the "twin" of Y is actually Lu (Lutetium), not La.

   ---

   ✓Final answer

   The correct option is (A).

   ANSWER: A

8. TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
   Q.The order of negative standard potential values of Li, Na, K is (A) Li $>$ Na $>$ K (B) K $>$ Na $>$ Li (C) Na $>$ K $>$ Li (D) Li $>$ K $>$ Na

   ›Reveal solutionSolution

   Standard reduction potentials generally become more negative going down Group 1, but lithium is an exception because of its very high hydration enthalpy. The actual order of negative standard potential values is Li > K > Na, so option (D) is correct.

   The question asks for the order of negative standard reduction potential values ($E^\circ$) for Li, Na, and K.

   1. Expected trend from ionization energy alone. Ionization energy decreases down Group 1 (Li > Na > K), so a simple extrapolation would suggest reducing power — and hence the negativity of $E^\circ$ — should increase down the group, giving K as the most negative and Li as the least.

   2. The role of hydration energy. The overall process $M(s) \rightarrow M^+(aq) + e^-$ depends on sublimation energy, ionization energy, and hydration energy. Lithium's very small ionic size gives it an exceptionally large hydration enthalpy, which more than compensates for its high ionization energy. This makes Li's overall oxidation the most thermodynamically favourable of the three, giving it the most negative $E^\circ$.

   3. Actual standard reduction potentials:

      • Li⁺/Li: −3.04 V
      • K⁺/K: −2.93 V
      • Na⁺/Na: −2.71 V

   4. Order of negativity (largest negative magnitude first):

   $\text{Li} (-3.04\text{ V}) > \text{K} (-2.93\text{ V}) > \text{Na} (-2.71\text{ V})$

   This matches option (D): Li > K > Na.

   Watch out

   A common mistake is to assume the trend follows ionization energy alone (which would predict K as most negative). Lithium's anomalously high hydration enthalpy overrides this and makes it the most negative of the three.

   Tip

   Whenever alkali-metal reduction potentials are compared, remember that Li is the exception to the simple "more negative down the group" rule because of its small size and huge hydration enthalpy.

   ✓Final answer

   The correct option is (D).

   ANSWER: D

9. TG EAPCET 2025Set ap-2025-04-30-FN1 markMCQ
   Q.Among the following the isoelectronic species are I. $\mathrm{Mg^{2+},\ Al^{3+},\ F^-}$ II. $\mathrm{O^{2-},\ F^-,\ N^{3-}}$ III. $\mathrm{K^+,\ Ca^{2+},\ Sc^{3+}}$ IV. $\mathrm{Mn^{2+},\ Fe^{3+},\ V^{3+}}$ (A) II, III only (B) I, II, III only (C) II, III, IV only (D) I, III, IV only

   ›Reveal solutionSolution

   The key idea is that isoelectronic species have the same number of electrons. Counting electrons for each ion shows that sets II and III are isoelectronic, while I and IV are not. The correct answer is (A).

   The concept here is straightforward: "isoelectronic" means "same electronic configuration" — which boils down to having the same total number of electrons. For ions, you start with the atomic number (number of protons in the neutral atom) and then add or subtract electrons based on the charge. A positive charge means you've lost that many electrons; a negative charge means you've gained that many.

   Let's check each set one by one.

   1. Set I: $\mathrm{Mg^{2+},\ Al^{3+},\ F^-}$

      • $\mathrm{Mg}$ has atomic number 12. $\mathrm{Mg^{2+}}$ has lost 2 electrons: $12 - 2 = 10$ electrons.
      • $\mathrm{Al}$ has atomic number 13. $\mathrm{Al^{3+}}$ has lost 3 electrons: $13 - 3 = 10$ electrons.
      • $\mathrm{F}$ has atomic number 9. $\mathrm{F^-}$ has gained 1 electron: $9 + 1 = 10$ electrons. So all three have 10 electrons — they are isoelectronic. But wait, the question asks "among the following the isoelectronic species are" and lists four sets. We need to see which sets are entirely isoelectronic within themselves. Set I is isoelectronic, but we must check the others to see which options match.

   2. Set II: $\mathrm{O^{2-},\ F^-,\ N^{3-}}$

      • $\mathrm{O}$ has atomic number 8. $\mathrm{O^{2-}}$: $8 + 2 = 10$ electrons.
      • $\mathrm{F}$ has atomic number 9. $\mathrm{F^-}$: $9 + 1 = 10$ electrons.
      • $\mathrm{N}$ has atomic number 7. $\mathrm{N^{3-}}$: $7 + 3 = 10$ electrons. All have 10 electrons — isoelectronic.

   3. Set III: $\mathrm{K^+,\ Ca^{2+},\ Sc^{3+}}$

      • $\mathrm{K}$ has atomic number 19. $\mathrm{K^+}$: $19 - 1 = 18$ electrons.
      • $\mathrm{Ca}$ has atomic number 20. $\mathrm{Ca^{2+}}$: $20 - 2 = 18$ electrons.
      • $\mathrm{Sc}$ has atomic number 21. $\mathrm{Sc^{3+}}$: $21 - 3 = 18$ electrons. All have 18 electrons — isoelectronic.

   4. Set IV: $\mathrm{Mn^{2+},\ Fe^{3+},\ V^{3+}}$

      • $\mathrm{Mn}$ has atomic number 25. $\mathrm{Mn^{2+}}$: $25 - 2 = 23$ electrons.
      • $\mathrm{Fe}$ has atomic number 26. $\mathrm{Fe^{3+}}$: $26 - 3 = 23$ electrons.
      • $\mathrm{V}$ has atomic number 23. $\mathrm{V^{3+}}$: $23 - 3 = 20$ electrons. Here, $\mathrm{Mn^{2+}}$ and $\mathrm{Fe^{3+}}$ both have 23 electrons, but $\mathrm{V^{3+}}$ has only 20. So they are not all isoelectronic.

   Watch out

   A common mistake is to forget that $\mathrm{V}$ has atomic number 23, not 24 or 25. Always double-check atomic numbers for transition metals — they're easy to mix up.

   So the sets that are isoelectronic are I, II, and III. But look at the options:

   • (A) II, III only
   • (B) I, II, III only
   • (C) II, III, IV only
   • (D) I, III, IV only

   Set I is isoelectronic, so (A) is wrong because it excludes I. Option (B) includes I, II, and III — that matches our finding. Option (C) includes IV which is not isoelectronic. Option (D) includes IV and excludes II.

   Tip

   You can quickly check isoelectronic sets by noticing patterns: ions from consecutive elements in the same period with appropriate charges often give the same electron count. For example, $\mathrm{O^{2-}, F^-, Na^+, Mg^{2+}, Al^{3+}}$ all have 10 electrons.

   ✓Final answer

   The correct option is (B).

10. TG EAPCET 2024Set ap-2024-05-08-FN1 markMCQ
    Q.The products formed in the reaction of BeCl$_2$ with LiAlH$_4$ are (A) Be, Li[AlCl$_4$], H$_2$ (B) BeH$_2$, LiCl, AlCl$_3$ (C) Be, AlH$_3$, LiCl, HCl (D) BeH$_2$, Li[AlCl$_4$]

    ›Reveal solutionSolution

    $2\,\text{BeCl}_2 + \text{LiAlH}_4 \rightarrow 2\,\text{BeH}_2 + \text{LiCl} + \text{AlCl}_3$, so the products are BeH$_2$, LiCl and AlCl$_3$ — option (B).

    Concept: Beryllium hydride (BeH$_2$) is prepared by treating beryllium chloride with lithium aluminium hydride, which acts as a hydride (H$^-$) donor. The chloride is transferred to lithium and aluminium.

    Balanced reaction:

    $2\,\text{BeCl}_2 + \text{LiAlH}_4 \longrightarrow 2\,\text{BeH}_2 + \text{LiCl} + \text{AlCl}_3$

    Thus the products are BeH$_2$, LiCl and AlCl$_3$.

    ✓Final answer

    The products are BeH$_2$, LiCl and AlCl$_3$, option (B).

    ANSWER: B

11. TG EAPCET 2024Set eng-2024-05-09-AN1 markMCQ
    Q.In which of the following species, the ratio of s - electrons to p - electrons is same? (A) $K^+, Cr^{3+}$ (B) $Zn, Fe^{2+}$ (C) $Zn, Cr^{3+}$ (D) $Na^+, K^+$

    ›Reveal solutionSolution

    Write the electron configuration of each species and count s- vs p-electrons. $K^+$ and $Cr^{3+}$ both give an s : p ratio of $6:12 = 1:2$ — option (A).

    Step-by-step solution

    Count electrons occupying $s$ orbitals against those in $p$ orbitals.

    $K^+$ ($Z=19$, $18$ electrons): $\text{1s}^2\,\text{2s}^2\,\text{2p}^6\,\text{3s}^2\,\text{3p}^6$.

    • $s$-electrons: $2+2+2 = 6$; $\ p$-electrons: $6+6 = 12$. Ratio $= 6:12 = 1:2$.

    $Cr^{3+}$ ($Z=24$, $21$ electrons): $\text{1s}^2\,\text{2s}^2\,\text{2p}^6\,\text{3s}^2\,\text{3p}^6\,\text{3d}^3$.

    • $s$-electrons: $2+2+2 = 6$; $\ p$-electrons: $6+6 = 12$. Ratio $= 6:12 = 1:2$.

    Both species share the ratio $1:2$.

    Checking the others rules them out — e.g. $Zn$ ($Z=30$) has $8$ $s$-electrons and $12$ $p$-electrons (ratio $2:3$), while $Fe^{2+}$ has $6:12 = 1:2$, so option (B) is unequal.

    ✓Final answer

    $K^+$ and $Cr^{3+}$ have the same s : p electron ratio ($1:2$) — option (A).

12. TG EAPCET 2024Set eng-2024-05-09-AN1 markMCQ
    Q.Identify the pair in which difference in bond order value is maximum (A) $O_2^-, O_2^+$ (B) $O_2^{2-}, O_2^{2+}$ (C) $O_2, O_2^{2+}$ (D) $O_2^+, O_2^{2+}$

    ›Reveal solutionSolution

    The bond order difference is largest for the pair $O_2^{2-}$ and $O_2^{2+}$ because their electron counts differ by 4, giving bond orders of 1 and 3 respectively — a difference of 2.

    Concept & Intuition

    Bond order tells us the net number of bonding pairs between two atoms. For diatomic molecules like oxygen, we use molecular orbital (MO) theory. The key is that adding electrons to $O_2$ fills antibonding orbitals (which decrease bond order), while removing electrons removes antibonding electrons (which increase bond order). So the farther apart two species are in electron count, the larger the bond order difference — but only if the electrons are added/removed from the right orbitals.

    ---

    Step-by-step reasoning

    1. Recall the MO configuration for neutral $O_2$ $O_2$ has 16 electrons total. Its MO configuration is:

    $(\sigma_{1s})^2(\sigma_{1s}^*)^2(\sigma_{2s})^2(\sigma_{2s}^*)^2(\sigma_{2p_z})^2(\pi_{2p_x})^2(\pi_{2p_y})^2(\pi_{2p_x}^*)^1(\pi_{2p_y}^*)^1$

    Bond order = $\frac{\text{bonding} - \text{antibonding}}{2} = \frac{10 - 6}{2} = 2$.

    2. Understand how charge changes the electron count

       Each positive charge means one fewer electron; each negative charge means one extra electron. For $O_2$ species:

       • $O_2^+$: 15 electrons
       • $O_2^{2+}$: 14 electrons
       • $O_2^-$: 17 electrons
       • $O_2^{2-}$: 18 electrons

    3. Determine bond orders for each species

       The last electrons added/removed come from the $\pi^*$ orbitals (antibonding). So:

       • $O_2^+$ (15 e⁻): remove one $\pi^*$ electron → bonding = 10, antibonding = 5 → bond order = $(10-5)/2 = 2.5$
       • $O_2^{2+}$ (14 e⁻): remove both $\pi^*$ electrons → bonding = 10, antibonding = 4 → bond order = $(10-4)/2 = 3$
       • $O_2^-$ (17 e⁻): add one $\pi^*$ electron → bonding = 10, antibonding = 7 → bond order = $(10-7)/2 = 1.5$
       • $O_2^{2-}$ (18 e⁻): add two $\pi^*$ electrons → bonding = 10, antibonding = 8 → bond order = $(10-8)/2 = 1$

    4. Compute the bond order differences for each pair

       • (A) $O_2^-$ (1.5) and $O_2^+$ (2.5): difference = $|1.5 - 2.5| = 1$
       • (B) $O_2^{2-}$ (1) and $O_2^{2+}$ (3): difference = $|1 - 3| = 2$
       • (C) $O_2$ (2) and $O_2^{2+}$ (3): difference = $|2 - 3| = 1$
       • (D) $O_2^+$ (2.5) and $O_2^{2+}$ (3): difference = $|2.5 - 3| = 0.5$

    5. Identify the maximum

       The largest difference is 2, which occurs for pair (B).

    Watch out

    A common mistake is to think $O_2^{2-}$ has bond order 0 (like $Ne_2$), but $O_2^{2-}$ still has 10 bonding and 8 antibonding electrons — bond order = 1, not 0.

    Tip

    The bond order difference is simply half the difference in the number of electrons in antibonding orbitals between the two species, because bonding electrons are the same for all these $O_2$ variants.

    ✓Final answer

    The correct option is (B).

    ANSWER: B