The decreasing values of bond angles from NH3 (106°) to SbH3 (101°) down the group 15 of the periodic table is due to

Question

The decreasing values of bond angles from NH3 (106°) to SbH3 (101°) down the group 15 of the periodic table is due to (A) decreasing Ip-bp repulsion (B) decreasing electronegativity (C) increasing Ip-bp repulsion (D) increasing p-orbital character in sp3

Tardigrade Admin · Accepted Answer

With the increase in electronegativity, bond pairs move farther and farther away from the central atom. As a result bp - bp repulsion decreases and the angle decreases accordingl

Q. The decreasing values of bond angles from $N H_{3}$ (106 $^{\circ}$ ) to $S b H_{3}$ (101 $^{\circ}$ ) down the group 15 of the periodic table is due to

decreasing Ip-bp repulsion

decreasing electronegativity

increasing Ip-bp repulsion

increasing p-orbital character in $s p^{3}$

With the increase in electronegativity, bond pairs move farther and farther away from the central atom. As a result bp - bp repulsion decreases and the angle decreases accordingl

Q. The decreasing values of bond angles from NH3​ (106∘) to SbH3​ (101∘) down the group 15 of the periodic table is due to

decreasing Ip-bp repulsion

decreasing electronegativity

increasing Ip-bp repulsion

increasing p-orbital character in sp3

With the increase in electronegativity, bond pairs move farther and farther away from the central atom. As a result bp - bp repulsion decreases and the angle decreases accordingl

Q. The decreasing values of bond angles from $N H_{3}$ (106 $^{\circ}$ ) to $S b H_{3}$ (101 $^{\circ}$ ) down the group 15 of the periodic table is due to

increasing p-orbital character in $s p^{3}$