# How do you arrange NH_4^+, BF_3, H_2O, C_2H_2, in increasing order of bond angle?

Jun 5, 2017

Determine Geometry of each.
$N {H}_{4}^{+} \implies {\theta}_{b} = {109}^{o}$
$B {F}_{3} \implies {\theta}_{b} = {120}^{o}$
${H}_{2} O \implies {\theta}_{b} = {105}^{o}$
${C}_{2} {H}_{2} \implies H - C \equiv C - H \implies {\theta}_{b} = {180}^{o}$

${H}_{2} O < N {H}_{4}^{+} < B {F}_{3} < {C}_{2} {H}_{2}$ about both carbon centers.

#### Explanation:

Geometry of HOH => $A {X}_{2} {E}_{2}$ => Bent angular from an $A {X}_{4}$ Tetrahedral Parent. The 2 free pair repel the 2 bonded pair to compress the bond angle from ${109}^{o}$ in a tetrahedron to ${105}^{o}$ => Bent Angular Geometry.

Geometry of $N {H}_{4}^{+}$ is $A {X}_{4}$ tetrahedral => ${\theta}_{b} = {109}^{o}$

Geometry of $B {F}_{3}$ is $A {X}_{3}$ Trigonal Planar $A {X}_{3}$ => ${\theta}_{b} = {120}^{o}$

Geometry of ${C}_{2} {H}_{2}$ is linear through two carbon centers.
=> $H - C \equiv C - H \implies {\theta}_{b} = {180}^{o}$