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Q.Assertion (A): Formaldehyde molecule has a planer geometry.
Reason (R): Carbon present in it is sp³ hybridised.

(a)
(i) Both A and R are correct and R is the correct explanation of A.
(b)
(ii) Both A and R are correct but R is not the correct explanation of A.
(c)
(iii) A is correct but R is incorrect.
(d)
(iv) Both A and R are incorrect.
Uttarakhand UbseUttarakhand Board Intermediate (Class 12) 2026MCQ· 1mImportance★★★★★
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Formaldehyde is planar (A true), but the carbon is sp2sp^{2} hybridised, not sp3sp^{3} (R false). Correct option: (iii).

Concept. In methanal, H2C=O\text{H}_2\text{C}=\text{O}, the carbonyl carbon forms three σ\sigma-bonds (two C–H and one C–O) and one π\pi-bond. Three σ\sigma-bonds + one lone-region-free geometry means three electron domains ⇒\Rightarrow sp2sp^{2} hybridisation, trigonal planar, bond angles ≈120∘\approx 120^\circ.

Analysis of the statements.

  • Assertion: "Formaldehyde has planar geometry" — TRUE. All four atoms lie in one plane. …

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