Compare the molecular shape and hybrid orbitals of PF3 and PF5 molecules

Chemistry

2 answers:

MaRussiya [10]2 years ago

5 0

PF3 is phosphorustrifluoride with p in the center and three f bonding with it.PF5 is phosphoruspentafluoride with p in the middle and 5 f linking to bond

g100num [7]2 years ago

4 0

Explanation:

The phosphorus trifluoride ( $PF_3$ ) has a piramidal geometry as can be seen in the First figure with the three F bonded to the P.

The phosphorus pentafluoride ( $PF_5$ ) has a bipiramidal geometry as can be seen in the Second figure with all the F bonded to the P.

You might be interested in

Calculate the maximum KE and velocity of an electron from zinc by a 275nm photon

Dmitry [639]

Answer:

Kinetic energy: $6.024*10^{-20}$

Velocity: $3.64*10^{5}$

Explanation:

From the equations of the photo-electric effect,

We know:

$hv=hv_0+K.E$

Where,

1. $h$ is the Planck's constant which is $6.626*10^{-34}$

2. $v,v_0$ are the frequency of light emitted and threshold frequencies respectively.

3. $K.E$ is the kinetic energy of the electrons emitted.

By fact, we come to know that the threshold frequency of Zn is 300nm

And also $v=\frac{c}{d}$

Where ,

1. $c$ is the speed of light $=3*10^8$

2. $d$ is the wavelength.

Thus,

$\frac{hc}{d}=\frac{hc}{d_0}+K.E\\K.E=\frac{hc}{d}-\frac{hc}{d_0}\\K.E=hc*(\frac{1}{d}-\frac{1}{d_0})\\K.E=6.626*10^{-34}*3*10^8*(\frac{1}{275*10^{-9}}-\frac{1}{300*10^{-9}})\\K.E=6.024*10^{-20}kgm^2s^{-2}$