Q.Why is it necessary to avoid even traces of moisture during the use of a Grignard reagent?

Showing the 12 most recent of 15 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.Observe the following sets of molecules and identify the set in which all molecules have dipole moment (A) $\mathrm{H_2S}, \mathrm{C_6H_6}, \mathrm{CO_2}$ (B) $\mathrm{H_2O}, \mathrm{NH_3}, \mathrm{HCl}$ (C) $\mathrm{BeCl_2}, \mathrm{CS_2}, \mathrm{BBr_3}$ (D) $\mathrm{PbCl_2}, \mathrm{SnCl_2}, \mathrm{HgCl_2}$

   ›Reveal solutionSolution

   This tests using molecular geometry (VSEPR) to decide whether bond dipoles cancel (non-polar) or add up to a net dipole (polar) for each molecule in a set. Answer: (B) $H_2O, NH_3, HCl$.

   Concept and Intuition

   A molecule has a net dipole moment only if its individual bond dipoles do not cancel by symmetry. Perfectly symmetric arrangements — linear (like $CO_2$, $CS_2$, $BeCl_2$, $HgCl_2$) or trigonal planar with identical substituents (like $BBr_3$, $C_6H_6$'s symmetric ring) — cancel their bond dipoles to zero net dipole moment, even though the individual bonds are polar. Bent or pyramidal geometries with lone pairs (like $H_2O$, $NH_3$) or simple diatomics with different atoms (like $HCl$) do not have this cancelling symmetry, so they retain a net dipole moment.

   Step-by-Step Solution

   1. Option (A): $H_2S$ is bent/polar, but $C_6H_6$ (benzene) is symmetric/non-polar and $CO_2$ is linear/non-polar — the set fails since not all three are polar.
   2. Option (B): $H_2O$ is bent (net dipole), $NH_3$ is pyramidal (net dipole), $HCl$ is a simple polar diatomic (net dipole) — all three have a dipole moment. ✓
   3. Option (C): $BeCl_2$ (linear), $CS_2$ (linear), $BBr_3$ (trigonal planar) are all symmetric — all non-polar, so the set fails entirely. …

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2. AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
   Q.Consider the following Assertion (A) : Dipole moment of $\mathrm{NF_3}$ is lesser than $\mathrm{NH_3}$ Reason (R) : In $\mathrm{NF_3}$, the orbital dipole due to lone pair of electrons is in the opposite direction to the resultant dipole moment of the three N-F bonds The correct answer is (A) Both (A) and (R) are correct and (R) is the correct explanation of (A) (B) Both (A) and (R) are correct but (R) is not the correct explanation of (A) (C) (A) is correct, but (R) is not correct (D) (A) is not correct, but (R) is correct

   ›Reveal solutionSolution

   Tests the classic $NH_3$ vs $NF_3$ dipole-moment anomaly; both the assertion and the reason are correct, and the reason correctly explains the assertion.

   Concept and Intuition

   Both $NH_3$ and $NF_3$ are pyramidal with a lone pair on N. The net dipole moment is the vector sum of the bond dipoles (N–H or N–F) and the lone-pair's orbital dipole. In $NH_3$, N is more electronegative than H, so each N–H bond dipole points toward N; the lone pair's dipole also points away from the three bonded atoms (i.e., in the same general direction as the resultant bond-dipole vector on N) — the two reinforce, giving a fairly large net dipole (1.47 D). In $NF_3$, F is more electronegative than N, so each N–F bond dipole points toward F, i.e., away from N — opposite in direction to the lone-pair dipole (which still points away from the bonded atoms, in the lone-pair's own direction on N). These now partly cancel, giving a much smaller net dipole (0.24 D).

   Step-by-Step Solution

   1. Confirm Assertion: dipole moment of $NF_3$ (≈0.24 D) is indeed less than $NH_3$ (≈1.47 D) — Assertion is correct.
   2. Confirm Reason: in $NF_3$, since F pulls bonding electron density away from N, the resultant bond dipole points from N towards the F atoms; the lone pair's orbital dipole points from the bonding region towards N (opposite sense). These oppose and partially cancel — Reason is correct. …

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3. AP EAPCET 2025Set ap-2025-05-20-FN1 markMCQ
   Q.Consider the following halogen containing compounds I) $\mathrm{CHCl_3}$ II) $\mathrm{CCl_4}$ III) $\mathrm{CH_3F}$ IV) 1,2-dichlorobenzene (benzene ring with Cl substituents at two adjacent carbons) V) 1,4-dichlorobenzene ($\mathrm{Cl-C_6H_4-Cl}$, Cl substituents at opposite/para carbons) The compounds with a net dipole moment are (only = only) (A) I, IV, only (B) II, III, IV only (C) I, III and IV only (D) III, IV and V only

   ›Reveal solutionSolution

   Molecular symmetry decides which of these halides have a net dipole moment: the fully symmetric CCl₄ and the para-disubstituted benzene cancel their bond dipoles to zero, while CHCl₃, CH₃F, and ortho-dichlorobenzene do not — leaving I, III, IV as polar.

   Concept and Intuition

   A molecule's net dipole moment is the vector sum of all its individual bond dipoles. Even molecules built from polar bonds can have zero net dipole if the bond dipoles are arranged symmetrically enough to cancel exactly (e.g., tetrahedral $CCl_4$, or para-substituted benzene rings where the two substituent dipoles point in exactly opposite directions). Asymmetric arrangements (different substituents on a tetrahedral carbon, or ortho substitution on a ring) leave a net resultant dipole.

   Step-by-Step Solution

   1. I) $CHCl_3$: tetrahedral carbon bonded to 1 H and 3 Cl — the C-H bond dipole doesn't cancel the three C-Cl dipoles (different bond types, no symmetry axis to cancel them), so it has a significant net dipole moment (~1.0 D). Polar.
   2. II) $CCl_4$: tetrahedral carbon bonded to 4 identical Cl atoms — by symmetry, the four C-Cl bond dipoles cancel exactly. Net dipole = 0. Non-polar.
   3. III) $CH_3F$: tetrahedral carbon with a polar C-F bond and three C-H bonds — no cancelling symmetry, so it has a clear net dipole moment. Polar. …

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4. AP EAPCET 2025Set eng-2025-05-27-FN1 markMCQ
   Q.The order of dipole moments of $\mathrm{H_2O}$ (A), $\mathrm{CHCl_3}$ (B) and $\mathrm{NH_3}$ (C) is (A) $B < A < C$ (B) $B < C < A$ (C) $C < B < A$ (D) $C < A < B$

   ›Reveal solutionSolution

   This tests comparing molecular dipole moments from bond polarity and geometry; the order is CHCl₃ < NH₃ < H₂O.

   Concept and Intuition

   Dipole moment depends on both the polarity of individual bonds and how the bond dipoles add up vectorially given the molecular shape. $\mathrm{CHCl_3}$ has three polar C–Cl bonds and one much less polar C–H bond arranged tetrahedrally, giving a modest net dipole. $\mathrm{NH_3}$ is pyramidal with a lone pair reinforcing the resultant of the three N–H bond dipoles, giving a larger net moment. $\mathrm{H_2O}$ is bent with a wide bond-dipole angle and a very electronegative O with two lone pairs, giving the largest net dipole moment among the three.

   Step-by-Step Solution

   1. Standard experimental values: $\mu(\mathrm{CHCl_3}) \approx 1.04\,\mathrm{D}$, $\mu(\mathrm{NH_3}) \approx 1.47\,\mathrm{D}$, $\mu(\mathrm{H_2O}) \approx 1.85\,\mathrm{D}$. …

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5. AP EAPCET 2024Set ap-2024-05-17-FN1 markMCQ
   Q.Arrange the following molecules in decreasing order of their dipole moments $\mathrm{H_2O}$, $\mathrm{NF_3}$, $\mathrm{H_2S}$, $\mathrm{NH_3}$ (A) $\mathrm{H_2O > NF_3 > H_2S > NH_3}$ (B) $\mathrm{NF_3 > H_2O > H_2S > NH_3}$ (C) $\mathrm{NF_3 > H_2O > NH_3 > H_2S}$ (D) $\mathrm{H_2O > NH_3 > H_2S > NF_3}$

   ›Reveal solutionSolution

   Comparing net dipole moments requires adding bond-dipole and lone-pair contributions vectorially; $\mathrm{NF_3}$ is anomalously low because its bond dipoles point opposite to the lone pair's, giving the order $\mathrm{H_2O>NH_3>H_2S>NF_3}$.

   Concept and Intuition

   Net dipole moment is the vector sum of individual bond dipoles and the lone-pair contribution. In $\mathrm{NH_3}$, the N–H bond dipoles (H is less electronegative, so dipole points toward N) and the lone pair on N point in roughly the same direction, reinforcing the net moment. In $\mathrm{NF_3}$, F is more electronegative than N, so the N–F bond dipoles point away from N (toward F), which is roughly opposite to the lone pair's direction — the two effects partly cancel, giving a surprisingly small net dipole despite the highly polar N–F bonds. $\mathrm{H_2O}$ has the largest moment due to its large bond-angle bend and highly polar O–H bonds; $\mathrm{H_2S}$ is much smaller since S is less electronegative and the H–S bond less polar.

   Step-by-Step Solution

   1. Recall typical dipole moments: $\mathrm{H_2O}\approx1.84$ D (largest, small central atom + polar bonds + bent shape).
   2. $\mathrm{NH_3}\approx1.47$ D (bond dipoles and lone pair add constructively). …

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6. AP EAPCET 2024Set eng-2024-05-21-AN1 markMCQ
   Q.In which of the following pairs, both molecules possess dipole moment? (A) $\mathrm{CO_2, BCl_3}$ (B) $\mathrm{BCl_3, NF_3}$ (C) $\mathrm{CO_2, SO_2}$ (D) $\mathrm{SO_2, NF_3}$

   ›Reveal solutionSolution

   Determine molecular geometry for each species and check whether bond dipoles cancel by symmetry. Answer: $SO_2$ and $NF_3$.

   Concept and Intuition

   A molecule has zero net dipole moment only if its bond dipoles cancel by symmetry (e.g., linear, trigonal planar, tetrahedral arrangements of identical bonds). Any distortion from these highly symmetric shapes — such as a bent or pyramidal geometry — usually leaves a net dipole.

   Step-by-Step Solution

   1. $CO_2$: linear (O=C=O), the two C=O bond dipoles point in opposite directions and cancel — nonpolar.
   2. $BCl_3$: trigonal planar, three identical B–Cl dipoles at 120° cancel — nonpolar.
   3. $SO_2$: bent (V-shaped) due to a lone pair on S — the two S–O bond dipoles do not cancel — net dipole present.
   4. $NF_3$: pyramidal (lone pair on N) — the three N–F dipoles plus the lone pair's contribution do not cancel — net (small) dipole present. …

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7. AP EAPCET 2022Set ap-2022-07-11-FN1 markMCQ
   Q.The dipole moment value of which of the following molecules is maximum? (A) $\mathrm{BeF_2}$ (B) $\mathrm{H_2O}$ (C) $\mathrm{NH_3}$ (D) $\mathrm{CH_4}$

   ›Reveal solutionSolution

   Compares net dipole moments of four molecules using symmetry and geometry; water has the largest.

   Concept and Intuition

   A molecule's net dipole moment is the vector sum of all its individual bond dipoles. When the molecular geometry is highly symmetric (linear AX₂, tetrahedral AX₄) with identical substituents, the bond dipoles cancel exactly, giving zero net dipole regardless of how polar each individual bond is. Bent and pyramidal geometries with lone pairs, however, retain an uncancelled net dipole.

   Step-by-Step Solution

   1. $\mathrm{BeF_2}$: linear, symmetric ⇒ $\mu=0$.
   2. $\mathrm{CH_4}$: tetrahedral, symmetric ⇒ $\mu=0$.
   3. $\mathrm{NH_3}$: pyramidal geometry with one lone pair ⇒ $\mu\approx1.47$ D. …

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8. AP EAPCET 2022Set ap-2022-07-12-AN1 markMCQ
   Q.In which of the following, the molecules are arranged in correct order of their dipole moments? (A) $BF_3 < NH_3 < NF_3$ (B) $NF_3 < NH_3 < BF_3$ (C) $BF_3 < NF_3 < NH_3$ (D) $NF_3 < BF_3 < NH_3$

   ›Reveal solutionSolution

   $BF_3$ has zero dipole (symmetric planar), $NF_3$'s lone pair opposes its bond dipoles (small net moment), and $NH_3$'s lone pair reinforces its bond dipoles (largest net moment): $BF_3 < NF_3 < NH_3$.

   Concept and Intuition

   Net molecular dipole moment is the vector sum of individual bond dipoles plus any lone-pair contribution. Geometry and the direction of the lone-pair's dipole (which points away from the nucleus, into where the lone pair "bulges") critically affect whether the lone pair adds to or subtracts from the bond dipoles.

   Step-by-Step Solution

   1. $BF_3$: trigonal planar (no lone pair on B), the three B–F bond dipoles are symmetric at $120°$ and exactly cancel. $\mu = 0$.
   2. $NH_3$: pyramidal, lone pair on N points opposite to the net bond-dipole resultant direction as drawn conventionally, and because N is more electronegative than H, bond dipoles point from H to N; the lone pair dipole adds constructively, giving a sizeable net moment ($\approx1.47$ D). …

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9. AP EAPCET 2022Set eng-2022-07-06-AN1 markMCQ
   Q.The correct order of dipole moment of the molecules $\mathrm{NH_3}$ (I), $\mathrm{BF_3}$ (II), $\mathrm{H_2O}$(III), $\mathrm{NF_3}$(IV) is (A) III > I > IV > II (B) IV > I > III > II (C) I > IV > II > III (D) III > II > I > IV

   ›Reveal solutionSolution

   $BF_3$ has zero dipole moment (symmetric trigonal planar); $NF_3$'s dipole is unusually small because the N lone pair opposes the N–F bond dipoles; $H_2O$'s bent shape with two lone pairs gives the largest moment, ahead of $NH_3$'s pyramidal shape.

   Concept and Intuition

   Net dipole moment depends on both bond polarity and molecular geometry (how bond dipoles and lone-pair effects add vectorially).

   • $BF_3$: trigonal planar, three identical B–F bond dipoles at $120^\circ$ cancel exactly $\Rightarrow \mu=0$.
   • $NF_3$: pyramidal, like $NH_3$, but F is more electronegative than N, so the N–F bond dipoles point toward F (away from N), largely opposing the lone pair's dipole contribution on N — giving an unusually small net moment.
   • $NH_3$: pyramidal; N–H bond dipoles point toward N, reinforcing the lone-pair dipole (both point the same way), giving a sizeable moment.
   • $H_2O$: bent, two lone pairs on O, both bond dipoles and the lone pairs reinforce strongly, giving the largest of these four moments.

   Step-by-Step Solution

   1. $BF_3$: $\mu=0$ D by symmetry — smallest.
   2. $NF_3$: bond dipoles oppose the lone pair effect $\Rightarrow \mu\approx0.24$ D — still small, but slightly more than $BF_3$. …

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10. AP EAPCET 2022Set eng-2022-07-08-AN1 markMCQ
    Q.Identify the correct statements from the following (I) $\mathrm{SnCl_2}$ is ionic, but $\mathrm{SnCl_4}$ is covalent in nature (II) All linear diatomic molecules have zero dipole moment (III) Both NO and $\mathrm{O_2}$ are paramagnetic (A) (I) and (II) only (B) (I) and (III) only (C) (II) and (III) only (D) (I), (II) and (III)

    ›Reveal solutionSolution

    Statement (II) is false because heteronuclear diatomics can have nonzero dipole moment despite being linear; (I) and (III) are both correct chemistry facts.

    Concept and Intuition

    Each statement tests a distinct idea: (I) tests Fajan's rules — smaller, higher-charge cations polarize anions more, favoring covalent character (Sn⁴⁺ is smaller and more charged than Sn²⁺, so SnCl₄ is covalent while SnCl₂ is ionic). (II) tests dipole moment — being "linear" only describes geometry, not whether the bond is polar; a diatomic molecule is trivially linear (two points define a line), but if the two atoms have different electronegativities (heteronuclear, e.g. HCl, CO), the bond is polar and the molecule has a nonzero dipole moment. Only homonuclear diatomics (equal electronegativity, e.g. N₂, O₂, Cl₂) have zero dipole moment. (III) tests paramagnetism from molecular orbital theory — NO has an odd number of electrons (one unpaired), and O₂'s MO configuration places two electrons unpaired in degenerate π* orbitals; both are paramagnetic.

    Step-by-Step Solution

    1. Evaluate (I): Sn²⁺ has a larger radius and lower charge than Sn⁴⁺, so by Fajan's rules Sn²⁺ polarizes Cl⁻ less (more ionic character in SnCl₂), while Sn⁴⁺ (smaller, higher charge) polarizes Cl⁻ strongly (more covalent character in SnCl₄). Statement (I) is TRUE. …

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11. AP EAPCET 2021Set ap-2021-09-03-AN1 markMCQ
    Q.Which molecule among the following has the highest dipole moment? (A) $BiH_3$ (B) $SbH_3$ (C) $PH_3$ (D) $NH_3$

    ›Reveal solutionSolution

    $NH_3$ has the highest dipole moment among the group-15 hydrides because nitrogen

    is small and strongly electronegative, concentrating bond and lone-pair dipoles in

    the same direction.

    Concept and Intuition

    Dipole moment in these pyramidal $EH_3$ molecules comes from both the $E$–$H$ bond

    polarity and the lone pair on $E$, both pointing the same way. Down group 15, the

    central atom becomes larger and less electronegative, so the E–H bonds become less

    polar and the lone pair becomes more diffuse (occupying more s-character in a larger

    orbital), reducing net dipole moment. Nitrogen, being the smallest and most

    electronegative, gives the strongest, most concentrated dipole.

    Step-by-Step Solution

    1. Dipole moment trend for group 15 hydrides: $NH_3 > PH_3 > AsH_3 > SbH_3 > BiH_3$ (roughly; $PH_3$ is anomalously low due to near-90° bond angles, but it is still …

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12. AP EAPCET 2021Set eng-2021-08-19-FN1 markMCQ
    Q.Which of the following will have maximum dipole moment? (A) $NF_3$ (B) $NCl_3$ (C) $NBr_3$ (D) $NH_3$

    ›Reveal solutionSolution

    Even though fluorine is far more electronegative than hydrogen, $NH_3$ has a larger dipole moment than $NF_3$ because the lone pair's contribution adds to the bond dipoles in $NH_3$ but opposes them in $NF_3$.

    Concept and Intuition

    Dipole moment isn't decided by electronegativity of the substituent alone — the lone pair on the central atom also contributes its own dipole. In $NH_3$, since N is more electronegative than H, the N–H bond dipoles point from H toward N, and the lone pair (also centered on N, pointing away from the H atoms) points in roughly the same net direction as the bond dipoles resultant, so they add constructively. In $NF_3$, F is more electronegative than N, so the N–F bond dipoles point from N toward F (opposite direction to the lone pair), and the lone pair opposes the resultant bond dipole, partially cancelling it.

    Step-by-Step Solution

    1. Compare bond-dipole vs lone-pair-dipole directions in each molecule.
    2. $NH_3$: bond dipoles point toward N (N more electronegative); lone pair also effectively points away from the H's on the same side as the bond-dipole resultant — they reinforce, giving $\mu(NH_3) \approx 1.47\,D$. …

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