Determine the increasing order of bond angles in the following compounds: (i) H2O (ii) ClO2 (iii) Cl2O (iv) OF2 (A) H2O < ClO2 < Cl2O < OF2 (B) ClO2 < Cl2O < H2O < OF2 (C) OF2 < Cl2O < ClO2 < H2O (D) OF2 < H2O < Cl2O < ClO2
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(ii) ClO2: Chlorine has one lone pair and two bond pairs, so it is sp2 hybridized. (iii) Cl2O: Chlorine has two bond pairs and no lone pairs, so it is sp hybridized. (iv) OF2: Oxygen has two bond pairs and two lone pairs, so it is sp3 hybridized. Now, let's Show more…
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The correct decreasing order of bond angles is : (a) $\mathrm{CIF}_{3}>\mathrm{PF}_{3}>\mathrm{NF}_{3}>\mathrm{BF}_{3}$ (b) $\mathrm{BF}_{3}>\mathrm{PF}_{3}>\mathrm{NF}_{3}>\mathrm{ClF}_{3}$ (c) $\mathrm{BF}_{3}>\mathrm{NF}_{3}>\mathrm{PF}_{3}>\mathrm{ClF}_{3}$ (d) $\mathrm{BF}_{3}>\mathrm{ClF}_{3}>\mathrm{PF}_{3}>\mathrm{NF}_{3}$ [Hint : Bond angles are : $\mathrm{BF}_{3}-120^{\circ}, \mathrm{NF}_{3}-106^{\circ}$, $\mathrm{PF}_{3}-101^{*}$, $\mathrm{ClF}_{3} \quad 90^{\circ} 1$
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