Match list I with list II and select the correct answer List -I (Molecule / Species) List -II (P) NO_2 (Unpaired electron in) (1) Vacant p- orbital involved in hybridization (Q) B_2H_6 (2) sp^2- orbital (R) ClO_3 (Unpaired electron in) (3) sp^3-orbital (S) CH_3^+ (4) 120^o Bond angle

Match list I with list II and select the correct answer List -I (…

MCQ

Match list $I$ with list $II$ and select the correct answer

List $-I$ (Molecule / Species)	List $-II$
$(P)$ $NO_2$ (Unpaired electron in)	$(1)$ Vacant $p-$ orbital involved in hybridization
$(Q)$ $B_2H_6$	$(2)$ $sp^2-$ orbital
$(R)$ $ClO_3$ (Unpaired electron in)	$(3)$ $sp^3-$orbital
$(S)$ $CH_3^+$	$(4)$ $120^o$ Bond angle

✓

Correct option: B.

$P-2, Q-1, R-3, S-4$

b
$\dot{\mathrm{NO}}_{2}=$ in $sp ^2$ hybrid orbital

$\dot{\mathrm{C}} 1 \mathrm{O}_{3}=$ in $\mathrm{sp}^{3}$ hybrid orbital

$\mathrm{B}_{2} \mathrm{H}_{6}=$ vacant orbital involved in bonding

(Banana bonding)

$ \mathrm{CH}_{3}^{+} =\mathrm{sp}^{2} $

$=120^{o} \text { angle exact }$

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LIST $I$ (Precipitating reagent and conditions)	LIST $II$ (Cation)
$A$ $\mathrm{NH}_4 \mathrm{Cl}+\mathrm{NH}_4 \mathrm{OH}$	$I$ $\mathrm{Mn}^{2+}$
$B$ $\mathrm{NH}_4 \mathrm{OH}+\mathrm{Na}_2 \mathrm{CO}_3$	$II$ $\mathrm{Pb}^{2+}$
$C$ $\mathrm{NH}_4 \mathrm{OH}+\mathrm{NH}_4 \mathrm{Cl}+\mathrm{H}_2 \mathrm{~S}$ gas	$III$ $\mathrm{Al}^{3+}$
$D$ dilute $\mathrm{HCl}$	$IV$ $\mathrm{Sr}^{2+}$