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

Chemistry

2 answers:

MaRussiya [10]3 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]3 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.

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Help please! A gas occupies 500 mL at 270 mm Hg and 55°C. If the pressure is changed to 1.8 atm and the temperature is

NNADVOKAT [17]

Answer:

V₂ = 112.14 mL

Explanation:

Given data:

Initial volume = 500 mL

Initial pressure = 270 mmHg (270/760 =0.355 atm)

Initial temperature = 55 °C (55 +273 = 328 K)

Final temperature = 100°C (100+273 = 373 K)

Final volume = ?

Final pressure = 1.8 atm

Formula:

P₁V₁/T₁ = P₂V₂/T₂

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

Solution:

V₂ = P₁V₁ T₂/ T₁ P₂

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6 0

3 years ago

In the reaction between bromine and sodium, a bromine atom gains an electron. What ion is formed? Is the bromine oxidized, or is

SVETLANKA909090 [29]

Answer: The ion formed after the reduction of bromine is $Br^-$

Explanation:

The electronic configuration of Sodium (Na) = $[Ne]3s^1$

The electronic configuration of Bromine (Br) = $[Ar]3d^{10}4s^24p^5$

From the above configurations, Sodium ion will loose 1 electron in order to gain stable electronic configuration and that electron is accepted by the Bromine atom because it is 1 electron short of the stable electronic configuration.

$Na\rightarrow Na^++e^-$ (oxidation reaction)