The angle between two hybrid orbitals is 105^ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The relationship between the bond angle $(	heta)$ and the $s-$character $(s)$ in hybrid orbitals is given by the formula: $\cos 	heta = \frac{s}

Vedclass · Accepted Answer

$22 - 23 \%$

Vedclass · Answer

(C) The relationship between the bond angle $(	heta)$ and the $s-$character $(s)$ in hybrid orbitals is given by the formula: $\cos 	heta = \frac{s}{s-1}$.Given $	heta = 105^o$,we have $\cos(105^o) = -0.2588$.Substituting this into the equation: $-0.2588 = \frac{s}{s-1}$.$-0.2588(s-1) = s$.$-0.2588s + 0.2588 = s$.$1.2588s = 0.2588$.$s = \frac{0.2588}{1.2588} \approx 0.2056$.Converting to percentage,$s \approx 20.56 \%$.Among the given options,the value closest to this result is $22 - 23 \%$.

The angle between two hybrid orbitals is $105^o$. The $s-$character of the hybrid orbitals will be ..............

Similar Questions

Identify the correct set of molecules with different geometries and central atoms with different hybridisations.

What type of bonds can the given electronic configuration of the excited state of $C$ form?

The compounds with $sp^2$ hybridized central atom among the following are:
$(A)$ $H_2CO_3$
$(B)$ $SiF_4$
$(C)$ $BF_3$
$(D)$ $HClO_2$

If the dipole moment of an $MX_3$ molecule is zero,then the orbitals used by $M$ $(atomic\ number < 21)$ in sigma bonding are ....

In which one of the following pairs do the central atoms exhibit $sp^2$ hybridization?

Vedclass Test Series

Exam Paper Generator

Online Exam Module