Q.CO is isoelectronic with (Note: more than one of the given options may be correct.)

Showing the 12 most recent of 13 on this concept.

1. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.Identify the correct pair of species having same number of valency electrons (A) $\mathrm{SF_4}$, $\mathrm{ClO_3^-}$ (B) $\mathrm{PO_4^{3-}}$, $\mathrm{SF_4}$ (C) $\mathrm{ClF_3}$, $\mathrm{SO_4^{2-}}$ (D) $\mathrm{CCl_4}$, $\mathrm{PO_4^{3-}}$

   ›Reveal solutionSolution

   Summing each atom's valence electrons (plus charge electrons for anions) for every species and comparing pairs shows only $\mathrm{CCl_4}$ and $\mathrm{PO_4^{3-}}$ share the same total (32).

   Concept and Intuition

   The total number of valence electrons in a species is found by adding up the valence-electron count (group number, essentially) of every atom present, then adding one electron per unit of negative charge (or subtracting one per unit of positive charge). This total valence-electron count is exactly what's used in VSEPR/Lewis-structure electron counting, and species with the same total often turn out to be isoelectronic in a valence sense (same shape family), which is why the question probes it directly.

   Step-by-Step Solution

   Compute total valence electrons for each species (group-valence electrons per atom: C=4, N=5, O=6, F=7, P=5, S=6, Cl=7):

   1. $\mathrm{SF_4}$: $6 + 4(7) = 6+28 = 34$
   2. $\mathrm{ClO_3^-}$: $7 + 3(6) + 1(\text{charge}) = 7+18+1 = 26$
   3. $\mathrm{PO_4^{3-}}$: $5 + 4(6) + 3(\text{charge}) = 5+24+3 = 32$
   4. $\mathrm{ClF_3}$: $7 + 3(7) = 7+21 = 28$
   5. $\mathrm{SO_4^{2-}}$: $6 + 4(6) + 2(\text{charge}) = 6+24+2 = 32$
   6. $\mathrm{CCl_4}$: $4 + 4(7) = 4+28 = 32$

   Now check each option's pair:

   • (A) $\mathrm{SF_4}$(34), $\mathrm{ClO_3^-}$(26) — unequal. …

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2. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.Which of the following sets of molecules / ions represent isoelectronic species? I. $NO^+, CN^-, CO, O_2^{2+}$ II. $NH_3, Al^{3+}, Ne, F^-$ III. $O_2^+, NO, N_2^-, H_2O$ The correct answer is (A) I, II only (B) I, II, III (C) II, III only (D) I, III only

   ›Reveal solutionSolution

   Isoelectronic species share the same total electron count; checking each set shows I and II qualify, but III fails because $H_2O$ doesn't match the electron count of the other three members.

   Concept and Intuition

   "Isoelectronic" means having the same number of electrons (and often, similar bonding/geometry). To check a set, add up (atomic number) for each atom and subtract the charge (add electrons for negative charge, remove for positive charge).

   Step-by-Step Solution

   1. Set I: $NO^+$: $7+8-1=14$. $CN^-$: $6+7+1=14$. $CO$: $6+8=14$. $O_2^{2+}$: $8+8-2=14$. All four = 14 electrons → isoelectronic. ✓
   2. Set II: $NH_3$: $7+3(1)=10$. $Al^{3+}$: $13-3=10$. $Ne$: $10$. $F^-$: $9+1=10$. All four = 10 electrons → isoelectronic. ✓
   3. Set III: $O_2^+$: $8+8-1=15$. $NO$: $7+8=15$. $N_2^-$: $7+7+1=15$. $H_2O$: $2(1)+8=10$. Three members have 15 electrons but $H_2O$ has only 10 → NOT isoelectronic. ✗ …

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3. AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
   Q.Identify the set containing isoelectronic species. (A) $\mathrm{N_2}$, $\mathrm{O_2^{2-}}$, $\mathrm{NO^+}$ (B) $\mathrm{N_2}$, $\mathrm{CO}$, $\mathrm{NO^+}$ (C) $\mathrm{F_2}$, $\mathrm{O_2^{2-}}$, $\mathrm{N_2}$ (D) $\mathrm{N_2}$, $\mathrm{O_2^{2+}}$, $\mathrm{C_2}$

   ›Reveal solutionSolution

   Isoelectronic species share the same electron count; checking each option, only $\mathrm{N_2}$, $\mathrm{CO}$, and $\mathrm{NO^+}$ all have exactly 14 electrons.

   Concept and Intuition

   'Isoelectronic' means having the same number of electrons (and often a similar arrangement/bond order), even if the species are made of different elements or carry a charge. To check, sum the atomic numbers (electrons) of the neutral atoms and then add or subtract electrons for any ionic charge.

   Step-by-Step Solution

   1. $\mathrm{N_2}$: $Z_N=7$, total $=7+7=14$ electrons.
   2. Option (A): $\mathrm{N_2}(14)$, $\mathrm{O_2^{2-}}$: $Z_O=8$, $O_2=16$, plus 2 extra electrons (2− charge) $=18$; $\mathrm{NO^+}$: $7+8-1=14$. Not all equal (18 ≠ 14) — rejected.
   3. Option (B): $\mathrm{N_2}(14)$; $\mathrm{CO}$: $Z_C=6, Z_O=8$, total $=14$; $\mathrm{NO^+}$: $7+8-1=14$. All three $=14$ — matches.
   4. Option (C): $\mathrm{F_2}$: $9+9=18$; $\mathrm{O_2^{2-}} = 18$; $\mathrm{N_2}=14$. Not all equal — rejected. …

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4. AP EAPCET 2024Set eng-2024-05-20-AN1 markMCQ
   Q.Among the following species, correct set of isostructural pairs are $\mathrm{XeO_3, CO_3^{2-}, SO_3, H_3O^+, ClF_3}$ (A) $(\mathrm{XeO_3, CO_3^{2-}})$ & $(\mathrm{SO_3, H_3O^+})$ (B) $(\mathrm{XeO_3, SO_3})$ & $(\mathrm{CO_3^{2-}, H_3O^+})$ (C) $(\mathrm{XeO_3, H_3O^+})$ & $(\mathrm{SO_3, CO_3^{2-}})$ (D) $(\mathrm{SO_3, ClF_3})$ & $(\mathrm{XeO_3, CO_3^{2-}})$

   ›Reveal solutionSolution

   Grouping the five species by shape (via VSEPR) shows XeO3 and H3O+ are both pyramidal, while SO3 and CO3^2- are both trigonal planar — these are the isostructural pairs.

   Concept and Intuition

   "Isostructural" means the species share the same geometry (and typically the same hybridisation), which is dictated by the number of bond pairs and lone pairs on the central atom (VSEPR theory).

   Step-by-Step Solution

   1. $XeO_3$: Xe (group 18) contributes 8 valence electrons; with 3 Xe=O bonds using up 3 domains and 1 lone pair remaining → 4 domains total → sp3, pyramidal shape (like NH3).
   2. $CO_3^{2-}$: C has 4 valence electrons plus the charge; with 3 C–O bond domains and no lone pair on C (all electrons used in bonding/resonance) → sp2, trigonal planar.
   3. $SO_3$: S has 6 valence electrons; with 3 S=O bond domains and no lone pair on S → sp2, trigonal planar.
   4. $H_3O^+$: O has 6 valence electrons minus 1 (cation) = effectively 3 bond pairs + 1 lone pair → sp3, pyramidal (like NH3). …

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5. AP EAPCET 2024Set eng-2024-05-21-FN1 markMCQ
   Q.The number of d electrons in Fe is equal to which of the following? i. Total number of 's' electrons of Mg ii. Total number of 'p' electrons of Cl iii. Total number of 'p' electrons of Ne The correct option is (A) i, ii only (B) ii, iii only (C) i, iii only (D) i, ii, iii

   ›Reveal solutionSolution

   Fe has 6 d-electrons; comparing this against the total s-electrons of Mg (6), total p-electrons of Cl (11), and total p-electrons of Ne (6) shows it matches Mg and Ne but not Cl.

   Concept and Intuition

   This is a straightforward electron-counting exercise using the Aufbau principle. We need Fe's d-electron count as the reference number, then work out the s- or p-electron totals (summed across all shells, not just the outermost) for each of the three comparison elements, and check which ones equal that reference number.

   Step-by-Step Solution

   1. Fe (Z=26): configuration $1s^22s^22p^63s^23p^64s^23d^6$ → d-electrons $=6$.
   2. Mg (Z=12): $1s^22s^22p^63s^2$ → s-electrons $=2(1s)+2(2s)+2(3s)=6$. Matches Fe's d-count.
   3. Cl (Z=17): $1s^22s^22p^63s^23p^5$ → p-electrons $=6(2p)+5(3p)=11$. Does not match.
   4. Ne (Z=10): $1s^22s^22p^6$ → p-electrons $=6(2p)$. Matches Fe's d-count. …

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6. AP EAPCET 2023Set eng-2023-05-18-AN1 markMCQ
   Q.The ion with smallest radius among the following is (A) $\mathrm{Ca^{2+}}$ (B) $\mathrm{K^{+}}$ (C) $\mathrm{Ti^{4+}}$ (D) $\mathrm{Sc^{3+}}$

   ›Reveal solutionSolution

   Among isoelectronic ions, radius shrinks as nuclear charge increases; $Ti^{4+}$ (Z=22) has the smallest radius here.

   Concept and Intuition

   All four ions have exactly 18 electrons — the same electron configuration as argon. Since the electron cloud (number and arrangement of electrons) is identical, the deciding factor for size is how strongly the nucleus pulls on that cloud. A larger atomic number means more protons attracting the same 18 electrons, so the electron cloud contracts more — the ion becomes smaller.

   Step-by-Step Solution

   1. Confirm each ion has 18 electrons: $K^+$ ($19-1=18$), $Ca^{2+}$ ($20-2=18$), $Sc^{3+}$ ($21-3=18$), $Ti^{4+}$ ($22-4=18$).
   2. List nuclear charges: $K=19 < Ca=20 < Sc=21 < Ti=22$.
   3. For isoelectronic species, ionic radius decreases as $Z$ increases (greater effective nuclear charge per electron). …

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7. AP EAPCET 2022Set eng-2022-07-05-FN1 markMCQ
   Q.Identify the isoelectronic pair of ions from the following (A) $Pr^{3+}, Nd^{3+}$ (B) $Tb^{3+}, Dy^{2+}$ (C) $Eu^{2+}, Gd^{3+}$ (D) $Pr^{3+}, Ce^{4+}$

   ›Reveal solutionSolution

   Isoelectronic species have the same number of electrons; only the $\mathrm{Eu^{2+}}$/$\mathrm{Gd^{3+}}$ pair matches, both having 61 electrons and the extra-stable $4f^7$ configuration.

   Concept and Intuition

   Two species are isoelectronic if they have exactly the same number of electrons (and, ideally, the same electronic arrangement). For lanthanide ions, this is checked by taking the atomic number Z, subtracting the ionic charge, and comparing the resulting electron counts. A useful cross-check is the extra stability of a half-filled $4f^7$ subshell, which is why $\mathrm{Eu}$ (readily loses 2 electrons to give $\mathrm{Eu^{2+}}$, $4f^7$) and $\mathrm{Gd}$ (readily loses 3 electrons, including its one $5d$ electron, to give $\mathrm{Gd^{3+}}$, also $4f^7$) both favour that particular oxidation state.

   Step-by-Step Solution

   1. $\mathrm{Pr}$ ($Z=59$): $\mathrm{Pr^{3+}}$ has $59-3=56$ electrons. $\mathrm{Nd}$ ($Z=60$): $\mathrm{Nd^{3+}}$ has $60-3=57$ electrons. Not equal → option (A) fails.
   2. $\mathrm{Tb}$ ($Z=65$): $\mathrm{Tb^{3+}}$ has $65-3=62$ electrons. $\mathrm{Dy}$ ($Z=66$): $\mathrm{Dy^{2+}}$ has $66-2=64$ electrons. Not equal → option (B) fails. …

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8. AP EAPCET 2021Set ap-2021-09-06-FN1 markMCQ
   Q.Which of the following species is iso-electronic with $\mathrm{Cr}$? (A) $\mathrm{Fe^{3+}}$ (B) $\mathrm{Co}$ (C) $\mathrm{Mn^{+}}$ (D) $\mathrm{Mn^{+2}}$

   ›Reveal solutionSolution

   Iso-electronic species have the same number of electrons. Neutral Cr has 24 electrons, and $Mn^+$ (25−1=24) is the only species among the options with exactly 24.

   Concept and Intuition

   "Iso-electronic" simply means having the same total electron count — it does not require the same element or the same electron configuration details, just the same number of electrons. To check each option, take the atomic number of the base element and subtract (or add) electrons for the stated charge.

   Step-by-Step Solution

   1. Cr, $Z=24$, neutral $\Rightarrow$ 24 electrons — this is the target count.
   2. Fe³⁺: Fe has $Z=26$; losing 3 electrons gives $26-3=23$. Not a match.
   3. Co: $Z=27$, neutral, so 27 electrons. Not a match.
   4. Mn⁺: Mn has $Z=25$; losing 1 electron gives $25-1=24$. Matches Cr's 24. …

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9. AP EAPCET 2021Set eng-2021-08-19-FN1 markMCQ
   Q.The spectrum of Helium is expected to be similar to that of __________ (A) $Li^+$ (B) $H$ (C) $Na$ (D) $He^+$

   ›Reveal solutionSolution

   Neutral Helium (2 electrons) is isoelectronic with $Li^+$ (also 2 electrons), so their spectra are expected to resemble each other.

   Concept and Intuition

   Atomic/ionic spectra depend heavily on the number and arrangement of electrons (electron configuration), since spectral lines arise from transitions between energy levels of the electrons present. Species with the same number of electrons and the same ground-state configuration (isoelectronic species) show qualitatively similar spectral patterns, even though the actual energies scale with nuclear charge.

   Step-by-Step Solution

   1. Neutral He has $Z=2$, 2 electrons, configuration $1s^2$.
   2. Among the options: $H$ has 1 electron; $He^+$ has 1 electron (hydrogen-like, single-electron ion); $Na$ has 11 electrons; $Li^+$ has $Z=3$ minus 1 electron $=2$ electrons, configuration $1s^2$.
   3. $Li^+$ is the only species with the same electron count/configuration as neutral He — they are isoelectronic. …

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10. AP EAPCET 2021Set eng-2021-08-20-FN1 markMCQ
    Q.Match the following species with the correct number of electrons present in them: Species:

    (i) $Be^{2+}$

    (ii) $H^+$

    (iii) $Na^+$

    (iv) $Mg^+$; Number of Electrons:

    (a) 0

    (b) 10

    (c) 2

    (d) 11

    (e) 4 (A) (i - d), (ii - c), (iii - b), (iv - a) (B) (i - a), (ii - b), (iii - c), (iv - d) (C) (i - e), (ii - d), (iii - a), (iv - c) (D) (i - c), (ii - a), (iii - b), (iv - d)

    ›Reveal solutionSolution

    This tests computing electron counts for ions by subtracting/adding charge from the atomic number: $Be^{2+}{=}2$, $H^+{=}0$, $Na^+{=}10$, $Mg^+{=}11$ electrons — answer (D).

    Concept and Intuition

    The number of electrons in an ion equals the parent atom's atomic number (number of protons, which never changes in ionization) minus the ion's positive charge (or plus the magnitude for a negative ion). This is just electron bookkeeping: removing one electron per unit of positive charge from the neutral atom's electron count.

    Step-by-Step Solution

    1. $Be$: $Z=4$. $Be^{2+}$ has lost 2 electrons: $4-2=2$ electrons. → matches (c) 2.
    2. $H$: $Z=1$. $H^+$ has lost 1 electron: $1-1=0$ electrons. → matches (a) 0.
    3. $Na$: $Z=11$. $Na^+$ has lost 1 electron: $11-1=10$ electrons. → matches (b) 10.
    4. $Mg$: $Z=12$. $Mg^+$ has lost 1 electron: $12-1=11$ electrons. → matches (d) 11. …

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11. AP EAPCET 2021Set eng-2021-08-23-FN1 markMCQ
    Q.The ions $S^{2-}, Cl^{-}, K^{+}, Ca^{2+}$ are iso-electronic. Their ionic radii show ________ (A) a decrease from $S^{2-}$ to $Cl^{-}$ and then increase from $K^{+}$ to $Ca^{2+}$ (B) an increase from $S^{2-}$ to $Cl^{-}$ and then decrease from $K^{+}$ to $Ca^{2+}$ (C) a significant decrease from $S^{2-}$ to $Ca^{2+}$ (D) a significant increase from $S^{2-}$ to $Ca^{2+}$

    ›Reveal solutionSolution

    Tests isoelectronic-series radius trends: with electron count fixed, radius shrinks as nuclear charge increases. Answer: a steady, significant decrease across the whole series.

    Concept and Intuition

    All four species — $S^{2-}$, $Cl^-$, $K^+$, $Ca^{2+}$ — have exactly 18 electrons (the argon electron configuration), since $S$ (Z=16) gains 2 electrons, $Cl$ (Z=17) gains 1, $K$ (Z=19) loses 1, and $Ca$ (Z=20) loses 2, all landing at 18 electrons. In such an isoelectronic series, the electron cloud is essentially the same size and shape, but the nuclear charge pulling on it steadily increases from $Z=16$ to $Z=20$. More protons pull the same number of electrons in tighter, so the ionic radius steadily shrinks.

    Step-by-Step Solution

    1. Confirm isoelectronic status: $S^{2-}$ (16+2=18e⁻), $Cl^-$ (17+1=18e⁻), $K^+$ (19−1=18e⁻), $Ca^{2+}$ (20−2=18e⁻) — all 18 electrons.
    2. Nuclear charge increases monotonically: $Z=16\to17\to19\to20$. …

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12. AP EAPCET 2021Set eng-2021-08-23-FN1 markMCQ
    Q.Which among the following iso-electronic species has the smallest size? $O^{2-}, F^{-}, Ne, Na^{+}, Mg^{2+}, Al^{3+}, Si^{4+}$ (A) $F^{-}$ (B) $Ne$ (C) $Si^{4+}$ (D) $Na^{+}$

    ›Reveal solutionSolution

    Tests isoelectronic radius trends again: same 10-electron count, so the ion with the highest nuclear charge is the smallest. Answer: $Si^{4+}$.

    Concept and Intuition

    $O^{2-}, F^-, Ne, Na^+, Mg^{2+}, Al^{3+}, Si^{4+}$ are all isoelectronic with 10 electrons (neon's configuration). In an isoelectronic series, radius decreases as nuclear charge increases, because more protons pull the same electron cloud in more tightly.

    Step-by-Step Solution

    1. Verify electron counts: $O^{2-}$(8+2), $F^-$(9+1), $Ne$(10), $Na^+$(11−1), $Mg^{2+}$(12−2), $Al^{3+}$(13−3), $Si^{4+}$(14−4) — all equal 10 electrons.
    2. List nuclear charges: $Z = 8, 9, 10, 11, 12, 13, 14$ respectively. …

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